📖Topic Explanations

🌐 Overview
Hello students! Welcome to Relative Lowering of Vapour Pressure!

Get ready to unravel one of the fundamental mysteries of solutions that governs many everyday phenomena, laying a crucial foundation for your JEE and board exam success!

Have you ever wondered why adding salt to water changes its boiling point? Or why an open container of petrol evaporates much faster than a bowl of sugar solution? These seemingly simple observations are governed by profound chemical principles, and at the heart of many such phenomena lies the concept of Vapour Pressure.

Imagine a liquid in a closed container. Some of its molecules constantly escape from the surface into the gaseous phase, creating pressure above the liquid – this is its vapour pressure. Now, what happens if you dissolve something in that liquid, especially something that doesn't easily evaporate itself, like sugar or common salt?

Intuitively, you might guess it changes things. And you'd be absolutely right! When a non-volatile solute is added to a pure solvent, it occupies some space on the liquid surface. This reduces the number of solvent molecules that can escape into the vapor phase, thereby causing a lowering of the vapour pressure of the solution compared to the pure solvent.

But the story doesn't end there. The *extent* to which the vapour pressure is lowered is not just a random value; it's proportional to the amount of solute present, and remarkably, independent of the nature of the solute! This makes it a colligative property – a property that depends only on the number of solute particles, not their identity.

More specifically, we will delve into the concept of Relative Lowering of Vapour Pressure (RLVP) – the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent. This specific ratio becomes an incredibly powerful tool because it is directly related to the mole fraction of the solute.

Why is this so important for your studies? RLVP is not just an abstract concept; it's a foundational pillar for understanding all other colligative properties – boiling point elevation, freezing point depression, and osmotic pressure. Mastering RLVP will give you a strong conceptual base to tackle related topics with confidence and score well in your exams. In the real world, this principle is crucial in various applications, from determining the molar masses of unknown substances to understanding the physical chemistry of biological systems and industrial processes like desalination.

In this section, we will embark on a fascinating journey to:

  • Understand the precise definition of vapour pressure and its lowering.

  • Explore Raoult's Law, the quantitative relationship that governs RLVP.

  • Learn how to calculate the molar mass of non-volatile solutes using this property.

  • Appreciate its significance in both theoretical chemistry and practical applications.


So, prepare to dive deep into the world of solutions, where a simple addition can lead to profound changes. Let's unlock the secrets of Relative Lowering of Vapour Pressure together and build an unshakeable understanding!
📚 Fundamentals
Hello, future JEE champions! Welcome to our foundational journey into the fascinating world of Solutions. Today, we're diving into a very special property that solutions exhibit, especially when we mix something "lazy" into something "energetic." Our topic for today is Relative Lowering of Vapour Pressure. Sounds fancy, right? But trust me, by the end of this session, you'll find it as intuitive as making your morning tea!

Let's start from the very beginning, shall we?

### Chapter 1: The Invisible Force - Vapour Pressure

Imagine you have a glass of water, and you cover it with a lid. What happens?
You've probably noticed tiny droplets forming on the underside of the lid after some time. Where do these droplets come from? They are water molecules that have escaped from the liquid surface, turned into vapor, and then condensed back into liquid.

This whole process is key to understanding Vapour Pressure.

1. Evaporation: At any given temperature, the molecules in a liquid are constantly moving. Some molecules near the surface have enough kinetic energy to break free from the attractive forces of their neighbors and escape into the space above the liquid as a gas (vapor). This is evaporation.
2. Condensation: In a closed container, these escaped vapor molecules are trapped. They move randomly and eventually collide with the liquid surface, losing energy and returning to the liquid state. This is condensation.
3. Equilibrium: Initially, evaporation is faster. But as more and more vapor molecules accumulate, the rate of condensation increases. Eventually, the rate of evaporation becomes equal to the rate of condensation. At this point, we say the system has reached dynamic equilibrium.

The pressure exerted by the vapor molecules in equilibrium with the liquid at a given temperature is called its Vapour Pressure.

Analogy Time!
Think of a classroom with an open door. Students (water molecules) are constantly trying to leave (evaporate). If you close the door (lid), the students who left are now stuck in the hallway (vapor phase). Eventually, some might decide to go back into the classroom (condense). When the number of students leaving the classroom per minute equals the number of students entering per minute, that's equilibrium. The "pressure" they exert in the hallway is like vapour pressure.

Key Takeaways:
* Vapour pressure is a property of a liquid at a specific temperature.
* It increases with temperature (more energy, more evaporation).
* It depends on the nature of the liquid (liquids with weaker intermolecular forces have higher vapour pressure because molecules escape more easily).

### Chapter 2: The Intruder - Non-Volatile Solute

Now, let's introduce an "intruder" into our pure solvent. This intruder is called a solute. But not just any solute – we're talking about a non-volatile solute.

What does "non-volatile" mean?
It simply means that the solute does not readily evaporate or turn into vapor at the given temperature. Think of sugar, salt, or urea. If you leave a bowl of sugar out, it doesn't just disappear into the air like water does, right? That's because it's non-volatile.

So, we have:
* Volatile Solvent: Our active, energetic liquid (like water) that loves to evaporate.
* Non-Volatile Solute: Our lazy, non-evaporating substance (like sugar) that we're dissolving into the solvent.

When we mix these two, we form a solution.

### Chapter 3: The Big Reveal - Lowering of Vapour Pressure

Here's where the magic (or rather, the chemistry!) happens. When you add a non-volatile solute (like sugar) to a volatile solvent (like water), the vapour pressure of the solution becomes lower than the vapour pressure of the pure solvent.

Why does this happen? Let's use another analogy!

Imagine our classroom again, but this time, the "students" are the solvent molecules, and they want to escape through the door (surface) to the hallway (vapor phase).

1. Pure Solvent: The entire door is available for students to exit. There are many students at the door, trying to get out.
2. Solution (with non-volatile solute): Now, imagine we've added some "furniture" (the non-volatile solute molecules) into the classroom, specifically near the door. This furniture doesn't move or try to leave. It just occupies space at the doorway.
* Less Surface Area: With the furniture blocking parts of the door, fewer "students" (solvent molecules) can reach the exit (surface) at any given moment to escape into the hallway.
* Intermolecular Attractions: Also, the solvent molecules are now attracted not only to other solvent molecules but also to the solute molecules. These new attractions can slightly "hold back" some solvent molecules, making it a bit harder for them to escape.

Because fewer solvent molecules can escape into the vapor phase, the number of vapor molecules in the hallway (and thus the pressure they exert) will be less. Hence, the vapour pressure of the solution is lowered.

In simple terms: The non-volatile solute molecules occupy some of the surface area of the liquid, effectively reducing the number of solvent molecules exposed at the surface that can evaporate. Less evaporation means less vapor, which means lower vapour pressure.

### Chapter 4: Quantifying the Change - Relative Lowering of Vapour Pressure

So, we understand *why* the vapour pressure is lowered. Now, let's talk about *how much* it's lowered.

Let's denote:
* $mathbf{P^0}$ = Vapour pressure of the pure solvent
* $mathbf{P_s}$ = Vapour pressure of the solution

The lowering of vapour pressure is simply the difference: $mathbf{Delta P = P^0 - P_s}$

But here's a crucial insight, thanks to a brilliant scientist named Raoult. Raoult's Law (which we'll explore in more detail later) essentially states that for a solution containing a non-volatile solute, the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.

This means the *extent* of lowering isn't just an absolute value ($Delta P$), but rather its proportion relative to the original vapour pressure. This leads us to Relative Lowering of Vapour Pressure.

Relative Lowering of Vapour Pressure (RLVP) is defined as the ratio of the lowering of vapour pressure ($Delta P$) to the vapour pressure of the pure solvent ($mathbf{P^0}$).

Mathematically, it's expressed as:
$mathbf{ ext{Relative Lowering of Vapour Pressure} = frac{P^0 - P_s}{P^0}}$

This ratio, $frac{P^0 - P_s}{P^0}$, is a very important quantity because it's directly related to the concentration of the solute.

### Chapter 5: The "Number Game" - Colligative Property

Why do we care so much about this "Relative Lowering of Vapour Pressure"? Because it's one of the four famous Colligative Properties!

What are colligative properties?
These are properties of solutions that depend only on the number of solute particles (molecules or ions) present in a given amount of solvent, and not on the nature or identity of the solute particles.

Think about our "furniture blocking the door" analogy. It doesn't matter if the furniture is a chair, a table, or a cupboard – what matters is *how much space* it occupies at the door, i.e., *how many* pieces of furniture are there. Similarly, for relative lowering of vapour pressure, it doesn't matter if you add sugar or urea (both non-volatile), as long as you add the same *number* of particles, the lowering will be the same.

Key Characteristics of RLVP as a Colligative Property:
* It depends on the mole fraction of the solute (which is a measure of the number of solute particles relative to total particles).
* It does NOT depend on whether the solute is sugar, salt, or any other non-volatile substance; only on *how many* particles are present.

This property is incredibly useful because by measuring the relative lowering of vapour pressure, we can actually determine the unknown molar mass of a non-volatile solute – a powerful application in chemistry!

### Wrapping Up the Fundamentals!

So, to summarize our foundational understanding:


  1. Vapour Pressure is the pressure exerted by vapor in equilibrium with its liquid at a given temperature.

  2. Adding a non-volatile solute to a volatile solvent reduces the solvent's ability to evaporate from the surface.

  3. This reduction leads to a Lowering of Vapour Pressure of the solution compared to the pure solvent.

  4. The Relative Lowering of Vapour Pressure ($frac{P^0 - P_s}{P^0}$) quantifies this change and is a crucial measure.

  5. This property is a Colligative Property, meaning it depends on the number of solute particles, not their identity.



You've now laid a strong foundation for understanding this critical concept. In the next sections, we'll delve deeper into its mathematical expression (Raoult's Law) and explore its applications, especially from a JEE perspective. Keep these fundamentals clear in your mind, and the advanced concepts will be a breeze!
🔬 Deep Dive
Alright, my bright young chemists! Let's embark on a deep dive into one of the most fundamental colligative properties: Relative Lowering of Vapour Pressure. This concept is crucial, not just for your exams, but for understanding how solutions behave at a molecular level. So, grab your notebooks and let's get started!

### Understanding Vapour Pressure: A Quick Recap

Before we talk about *lowering* vapour pressure, we must firmly grasp what vapour pressure itself is. Imagine you have a pure liquid, say water, in a closed container at a constant temperature. What happens?
* Some liquid molecules at the surface gain enough kinetic energy to escape into the gaseous phase (evaporation).
* At the same time, some gas molecules lose energy and return to the liquid phase (condensation).

Eventually, a dynamic equilibrium is established where the rate of evaporation equals the rate of condensation. The pressure exerted by the vapour molecules in this equilibrium state is called the vapour pressure of the liquid.

Key factors affecting vapour pressure:
1. Temperature: Higher temperature means more molecules have enough energy to escape, thus higher vapour pressure.
2. Nature of the liquid: Liquids with weaker intermolecular forces (e.g., ether) evaporate more easily and thus have higher vapour pressure than liquids with strong intermolecular forces (e.g., water, which has hydrogen bonding).

### The Phenomenon: Lowering of Vapour Pressure

Now, let's introduce a twist. What happens when we add a non-volatile solute to our pure solvent? A non-volatile solute is one that does not easily evaporate (e.g., sugar, urea, common salt).

When a non-volatile solute is dissolved in a solvent, the vapour pressure of the resulting solution is *lower* than the vapour pressure of the pure solvent at the same temperature. This phenomenon is called lowering of vapour pressure.

Why does this happen? Let's visualize it:
* Surface Area Blockage: In a pure solvent, the entire surface is occupied by solvent molecules, all capable of evaporating. When you add a non-volatile solute, some of the solvent molecules at the surface are replaced by solute particles. These solute particles *do not* evaporate. This effectively reduces the fraction of the surface area available for the solvent molecules to escape into the vapour phase.
* Reduced Escaping Tendency: The attractive forces between solvent and solute molecules also play a role. The presence of solute molecules hinders the solvent molecules from escaping the liquid phase, thereby reducing their "escaping tendency."

Fewer solvent molecules escaping means fewer molecules in the vapour phase at equilibrium, which translates to a lower vapour pressure for the solution compared to the pure solvent.

### Raoult's Law: Quantifying Vapour Pressure Lowering

This empirical observation was quantified by French chemist François-Marie Raoult in 1887, leading to Raoult's Law.

Raoult's Law for a non-volatile solute states that the partial vapour pressure of each volatile component in a solution is directly proportional to its mole fraction in the solution.

For a solution containing a non-volatile solute (let's call it component B) and a volatile solvent (component A):
The vapour phase above the solution will consist *only* of solvent molecules (since the solute is non-volatile).
According to Raoult's Law, the vapour pressure of the solvent in the solution, $P_A$, is given by:

$mathbf{P_A = x_A cdot P_A^circ}$

Where:
* $mathbf{P_A}$ = Vapour pressure of the solvent above the solution.
* $mathbf{x_A}$ = Mole fraction of the solvent in the solution.
* $mathbf{P_A^circ}$ = Vapour pressure of the *pure* solvent at the same temperature.

From this equation, we can immediately see that since $x_A$ (mole fraction) is always less than 1 (as long as there is some solute), $P_A$ will always be less than $P_A^circ$. This mathematically confirms the lowering of vapour pressure!

### Derivation of Relative Lowering of Vapour Pressure (RLVP)

Now, let's get to the star of our show: Relative Lowering of Vapour Pressure. This is a direct consequence of Raoult's Law and the definition of colligative properties.

1. Lowering of Vapour Pressure ($Delta P$):
The *lowering* of vapour pressure is simply the difference between the vapour pressure of the pure solvent and the vapour pressure of the solution:
$Delta P = P_A^circ - P_A$

2. Substituting Raoult's Law:
We know $P_A = x_A cdot P_A^circ$. Let's substitute this into the equation for $Delta P$:
$Delta P = P_A^circ - (x_A cdot P_A^circ)$
$Delta P = P_A^circ (1 - x_A)$

3. Introducing Mole Fraction of Solute ($x_B$):
For a binary solution (solvent A and solute B), the sum of mole fractions is always 1:
$x_A + x_B = 1$
Therefore, $1 - x_A = x_B$.

4. Connecting $Delta P$ to $x_B$:
Substitute $(1 - x_A)$ with $x_B$ in the $Delta P$ equation:
$mathbf{Delta P = x_B cdot P_A^circ}$
This equation tells us that the *lowering* of vapour pressure is directly proportional to the mole fraction of the solute and the vapour pressure of the pure solvent.

5. Defining Relative Lowering of Vapour Pressure (RLVP):
The term "relative lowering" implies we are comparing the lowering of vapour pressure to the vapour pressure of the pure solvent. So, RLVP is defined as:
$mathbf{ ext{RLVP} = frac{Delta P}{P_A^circ}}$

6. The Fundamental Equation for RLVP:
Now, substitute $Delta P = x_B cdot P_A^circ$ into the RLVP definition:
$ ext{RLVP} = frac{x_B cdot P_A^circ}{P_A^circ}$
$mathbf{ ext{RLVP} = x_B}$

This is a profoundly important result! It states that the relative lowering of vapour pressure is equal to the mole fraction of the solute in the solution.

### Why is RLVP a Colligative Property?

A colligative property is one that depends only on the *number of solute particles* in the solution, and not on their chemical nature or identity.

Look at the final equation: $mathbf{frac{Delta P}{P_A^circ} = x_B}$.
The mole fraction of the solute, $x_B$, is defined as:
$x_B = frac{ ext{moles of solute}}{ ext{total moles (moles of solute + moles of solvent)}}$
As you can see, $x_B$ clearly depends on the *number of moles* (and thus the number of particles) of the solute and solvent. It does *not* depend on whether the solute is sugar, urea, or any other non-volatile substance, as long as it doesn't dissociate.

Therefore, Relative Lowering of Vapour Pressure is a colligative property.

CBSE vs. JEE Focus:
* For CBSE, understanding the definition, Raoult's Law, and the final formula $frac{Delta P}{P_A^circ} = x_B$ is key. You'll solve problems primarily using this formula.
* For JEE Main & Advanced, you need to not only know the formula but also understand its derivation, its assumptions (ideal solution, non-volatile solute, dilute solutions), and its application in determining the molar mass of unknown solutes, sometimes in more complex scenarios involving multiple solutes or non-ideal behavior (though non-ideal behavior itself is a separate topic, recognizing when Raoult's law might deviate is important).

### Applications: Determining Molar Mass of an Unknown Solute

One of the most significant applications of RLVP is the determination of the molar mass of an unknown non-volatile solute.
Let's expand the mole fraction term ($x_B$):
$x_B = frac{n_B}{n_A + n_B}$
Where:
* $n_A$ = moles of solvent = $frac{w_A}{M_A}$ (mass of solvent / molar mass of solvent)
* $n_B$ = moles of solute = $frac{w_B}{M_B}$ (mass of solute / molar mass of solute)

So, $frac{P_A^circ - P_A}{P_A^circ} = frac{n_B}{n_A + n_B}$

For very dilute solutions, $n_B$ is much smaller than $n_A$ ($n_B ll n_A$). In such cases, we can approximate $n_A + n_B approx n_A$.
Therefore, for dilute solutions:
$frac{P_A^circ - P_A}{P_A^circ} approx frac{n_B}{n_A} = frac{w_B/M_B}{w_A/M_A}$
$mathbf{frac{P_A^circ - P_A}{P_A^circ} = frac{w_B cdot M_A}{M_B cdot w_A}}$

This simplified expression is extremely useful for calculating $M_B$ (molar mass of solute) when all other quantities are known. Remember, this approximation is valid only for dilute solutions. For concentrated solutions, the exact formula using $(n_A + n_B)$ must be used.

### Solved Examples

Let's put this knowledge to the test with some examples.

Example 1: Basic Calculation of Vapour Pressure

The vapour pressure of pure benzene ($C_6H_6$) at a certain temperature is 640 mmHg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mmHg. Calculate the molar mass of the solid substance. (Molar mass of benzene = 78 g/mol).

Solution:
Given:
* $P_A^circ$ (vapour pressure of pure benzene) = 640 mmHg
* $P_A$ (vapour pressure of solution) = 600 mmHg
* $w_B$ (mass of solute) = 2.175 g
* $w_A$ (mass of solvent, benzene) = 39.0 g
* $M_A$ (molar mass of benzene) = 78 g/mol

We need to find $M_B$ (molar mass of solute).

Using the formula for Relative Lowering of Vapour Pressure:
$frac{P_A^circ - P_A}{P_A^circ} = frac{n_B}{n_A + n_B}$

First, calculate moles of solvent ($n_A$):
$n_A = frac{w_A}{M_A} = frac{39.0 ext{ g}}{78 ext{ g/mol}} = 0.5 ext{ mol}$

Now, substitute the known values into the RLVP equation:
$frac{640 ext{ mmHg} - 600 ext{ mmHg}}{640 ext{ mmHg}} = frac{n_B}{0.5 ext{ mol} + n_B}$
$frac{40}{640} = frac{n_B}{0.5 + n_B}$
$frac{1}{16} = frac{n_B}{0.5 + n_B}$

Cross-multiply:
$0.5 + n_B = 16 n_B$
$0.5 = 15 n_B$
$n_B = frac{0.5}{15} = frac{1}{30} ext{ mol}$

Now, calculate the molar mass of the solute ($M_B$):
$M_B = frac{w_B}{n_B} = frac{2.175 ext{ g}}{1/30 ext{ mol}} = 2.175 imes 30 ext{ g/mol}$
$M_B = 65.25 ext{ g/mol}$

The molar mass of the solid substance is 65.25 g/mol.

Note: In this problem, we did *not* use the dilute solution approximation, as the problem didn't specify it, and the difference between $n_A + n_B$ and $n_A$ might be significant enough. Always use the exact formula unless specifically stated or if the numbers clearly indicate extreme dilution.

Example 2: Vapour Pressure of a Solution with Urea

The vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of a solution containing 6 g of urea ($NH_2CONH_2$) in 100 g of water at the same temperature. (Molar mass of water = 18 g/mol, Molar mass of urea = 60 g/mol).

Solution:
Given:
* $P_A^circ$ (vapour pressure of pure water) = 17.535 mmHg
* $w_B$ (mass of urea) = 6 g
* $w_A$ (mass of water) = 100 g
* $M_A$ (molar mass of water) = 18 g/mol
* $M_B$ (molar mass of urea) = 60 g/mol

We need to find $P_A$ (vapour pressure of the solution).

Calculate moles of solute ($n_B$):
$n_B = frac{w_B}{M_B} = frac{6 ext{ g}}{60 ext{ g/mol}} = 0.1 ext{ mol}$

Calculate moles of solvent ($n_A$):
$n_A = frac{w_A}{M_A} = frac{100 ext{ g}}{18 ext{ g/mol}} approx 5.555 ext{ mol}$

Now, calculate the mole fraction of solute ($x_B$):
$x_B = frac{n_B}{n_A + n_B} = frac{0.1}{5.555 + 0.1} = frac{0.1}{5.655} approx 0.01768$

Using the RLVP formula:
$frac{P_A^circ - P_A}{P_A^circ} = x_B$
$frac{17.535 - P_A}{17.535} = 0.01768$

Now, solve for $P_A$:
$17.535 - P_A = 0.01768 imes 17.535$
$17.535 - P_A approx 0.3097$
$P_A = 17.535 - 0.3097$
$P_A approx 17.2253 ext{ mmHg}$

The vapour pressure of the solution is approximately 17.225 mm Hg.

Example 3: Comparing RLVP for different solutes (JEE advanced concept)

At a particular temperature, the vapour pressure of pure water is 50 mm Hg. 18 g of glucose (Molar mass = 180 g/mol) is added to 180 g of water. Calculate the relative lowering of vapour pressure. If 18 g of urea (Molar mass = 60 g/mol) was used instead of glucose, what would be the relative lowering of vapour pressure?

Solution:
Given:
* $P_A^circ$ (vapour pressure of pure water) = 50 mmHg
* $w_A$ (mass of water) = 180 g
* $M_A$ (molar mass of water) = 18 g/mol

Calculate moles of solvent ($n_A$):
$n_A = frac{180 ext{ g}}{18 ext{ g/mol}} = 10 ext{ mol}$

Case 1: Glucose as solute
* $w_B$ (mass of glucose) = 18 g
* $M_B$ (molar mass of glucose) = 180 g/mol

Calculate moles of glucose ($n_B$):
$n_B = frac{18 ext{ g}}{180 ext{ g/mol}} = 0.1 ext{ mol}$

Relative Lowering of Vapour Pressure ($RLVP_1$) = $x_B = frac{n_B}{n_A + n_B}$
$RLVP_1 = frac{0.1}{10 + 0.1} = frac{0.1}{10.1} approx 0.0099$

Case 2: Urea as solute
* $w_B'$ (mass of urea) = 18 g
* $M_B'$ (molar mass of urea) = 60 g/mol

Calculate moles of urea ($n_B'$):
$n_B' = frac{18 ext{ g}}{60 ext{ g/mol}} = 0.3 ext{ mol}$

Relative Lowering of Vapour Pressure ($RLVP_2$) = $x_B' = frac{n_B'}{n_A + n_B'}$
$RLVP_2 = frac{0.3}{10 + 0.3} = frac{0.3}{10.3} approx 0.0291$

Results:
* For glucose solution, RLVP is approximately 0.0099.
* For urea solution, RLVP is approximately 0.0291.

This example clearly demonstrates that for the same mass of solute, the relative lowering of vapour pressure is different because the *number of moles* (and thus the mole fraction) of the solute is different, highlighting its nature as a colligative property. Urea, having a smaller molar mass, provides more moles for the same mass, leading to a greater relative lowering of vapour pressure.

### Conclusion

Relative lowering of vapour pressure is a powerful concept that showcases how the mere presence of solute particles affects the physical properties of a solvent. Remember its core definition, Raoult's Law, and the critical relationship $mathbf{frac{Delta P}{P_A^circ} = x_B}$. Master this, and you've got a strong foundation for understanding other colligative properties! Keep practicing these derivations and numerical problems to solidify your understanding.
🎯 Shortcuts

Mnemonics & Shortcuts for Relative Lowering of Vapour Pressure (RLVP)


Relative Lowering of Vapour Pressure (RLVP) is a fundamental colligative property. Mastering its formula and application is crucial for both JEE Main and CBSE Board exams. Here are some effective mnemonics and shortcuts to help you remember the key aspects.



1. Mnemonic for the RLVP Formula


The core formula for Relative Lowering of Vapour Pressure, as given by Raoult's Law, is:



(P° - P) / P° = Xsolute



  • P°: Vapour pressure of the pure solvent

  • P: Vapour pressure of the solution

  • Xsolute: Mole fraction of the solute


To remember this formula, use the following mnemonic:



  • Pure Pressure Minus Solution Pressure Over Pure Pressure Equals eX-Solute


Breaking it down:



  • "Pure Pressure" stands for P°

  • "Minus Solution Pressure" stands for - P

  • "Over Pure Pressure" stands for / P°

  • "Equals eX-Solute" stands for = Xsolute



2. Mnemonic for Identifying RLVP as a Colligative Property


Colligative properties depend only on the number of solute particles, not their nature.



  • Really Lovely Vapour Properties Depend On Particles.


This helps you recall that RLVP is a colligative property because its effect ("Depend On Particles") is about the count of solute particles.



3. Shortcut for Dilute Solutions (JEE Main & CBSE)


For very dilute solutions, the mole fraction of solute (Xsolute) can be approximated:



Xsolute = nsolute / (nsolute + nsolvent) ≈ nsolute / nsolvent


Therefore, for dilute solutions, the RLVP formula becomes:



(P° - P) / P° ≈ nsolute / nsolvent


To remember this approximation:



  • Dilute Solutions Always Lazy – Ignore Solute In Denominator.


This means when the solution is dilute, we can effectively ignore the number of moles of solute (nsolute) in the denominator (nsolute + nsolvent), simplifying it to just nsolvent.



4. Quick Tip for Problem Solving (JEE Specific)




  • Always use the exact formula: Unless explicitly stated that the solution is *very dilute* or the approximation is specified, always use the exact mole fraction formula for Xsolute = nsolute / (nsolute + nsolvent). While the dilute approximation often yields similar results, using the exact formula is safer, especially for JEE questions where precision matters or for concentrated solutions.




  • Watch out for Non-Volatile Solute: The RLVP applies to non-volatile solutes dissolved in a volatile solvent. If the solute is also volatile, Raoult's Law for ideal solutions with two volatile components needs to be applied, which is a different scenario.






CBSE vs JEE Callout: Both CBSE and JEE require understanding the RLVP formula and its application. For CBSE, the dilute approximation is frequently used. For JEE, be prepared to use the exact formula, especially if the solution is not explicitly dilute, and understand its derivation from Raoult's Law.


💡 Quick Tips

Quick Tips for Relative Lowering of Vapour Pressure (RLVP)



Relative Lowering of Vapour Pressure is a fundamental colligative property that frequently appears in JEE and Board examinations. Mastering its core concepts and applications is crucial for scoring well.

1. Understand the Core Principle




  • Definition: RLVP is defined as the ratio of the lowering of vapour pressure of the solvent (∆P) to the vapour pressure of the pure solvent (P°).


    ∆P / P° = (P° - P₃) / P°, where P₃ is the vapour pressure of the solution.


  • Raoult's Law (for non-volatile solutes): According to Raoult's law, the relative lowering of vapour pressure is equal to the mole fraction of the solute (X₂).


    (P° - P₃) / P° = X₂


  • Colligative Nature: RLVP depends solely on the number of solute particles, not on their nature. This is key for understanding its application in determining molar masses.



2. Key Formulae and Their Application




  • Basic Formula:


    (P° - P₃) / P° = n₂ / (n₁ + n₂)


    Where n₁ = moles of solvent, n₂ = moles of solute.


  • For Dilute Solutions (Approximation): For very dilute solutions, n₂ ≪ n₁, so (n₁ + n₂) ≈ n₁.


    This simplifies to: (P° - P₃) / P° ≈ n₂ / n₁


    This approximation is particularly useful when calculating the molar mass of a non-volatile solute (M₂):


    (P° - P₃) / P° = (W₂ / M₂) / (W₁ / M₁)


    where W is mass and M is molar mass. Always check if the "dilute solution" approximation is valid or explicitly asked for.



3. Handling Electrolytes (JEE Specific)




  • Van't Hoff Factor (i): For electrolytes (solutes that dissociate or associate), the number of particles changes. The RLVP formula must be modified using the van't Hoff factor.


    (P° - P₃) / P° = i × X₂


    or, more generally, (P° - P₃) / P° = i × n₂ / (n₁ + i × n₂).


    For dilute solutions: (P° - P₃) / P° ≈ i × n₂ / n₁


  • Calculating 'i': Remember i = 1 + (n-1)α for dissociation (n = number of ions produced, α = degree of dissociation) and i = 1 + (1/n - 1)α for association (n = number of molecules associating, α = degree of association).



4. Common Pitfalls and Tips for Problem Solving




  • Units: Ensure consistency in pressure units (e.g., all in mmHg, Pa, or bars). Molar masses should be in g/mol.


  • Solvent vs. Solute: Clearly identify which components are solvent (1) and solute (2).


  • Non-Volatile Solute Assumption: RLVP theory strictly applies to non-volatile solutes. If the solute is volatile, Raoult's law for ideal solutions with two volatile components must be used, which is a different concept.


  • Concentration Terms: RLVP directly relates to mole fraction of solute. If concentrations are given in molality or molarity, convert them to mole fractions or use the dilute solution approximation carefully.


  • Direct Calculation: Sometimes, instead of molar mass, you might be asked to find the vapour pressure of the solution (P₃). Rearrange the formula: P₃ = P° (1 - X₂).



5. CBSE vs. JEE Main Focus



























Aspect CBSE Board JEE Main
Conceptual Understanding Definition, Raoult's Law statement, dilute solution approximation. In-depth understanding of deviations from ideality, role of intermolecular forces.
Problem Complexity Direct application of formulae for non-electrolytes; molar mass calculation. Problems involving van't Hoff factor (electrolytes), multiple steps, and less obvious data.
Approximation Often explicitly stated or implied for dilute solutions. Requires judgment on when to use approximation vs. exact formula.


Practice a variety of numerical problems, especially those involving electrolytes, to solidify your understanding and improve problem-solving speed.
🧠 Intuitive Understanding

Understanding the concept of Relative Lowering of Vapour Pressure (RLVP) is crucial for both board exams and JEE. Let's build an intuitive understanding of why adding a non-volatile solute lowers the vapour pressure of a solvent.



What is Vapour Pressure? (A Quick Recap)


Imagine a pure liquid in a closed container. Some molecules at the surface have enough kinetic energy to escape into the space above the liquid, forming vapour. These vapour molecules, in turn, collide with the liquid surface and return to the liquid phase. Eventually, an equilibrium is established where the rate of evaporation equals the rate of condensation. The pressure exerted by the vapour at this equilibrium is called the vapour pressure.



  • Essentially, it's a measure of the "tendency" of liquid molecules to escape into the gaseous phase.



The Intuition Behind Lowering of Vapour Pressure


Now, let's add a non-volatile solute to our pure solvent. A non-volatile solute is one that does not readily evaporate (e.g., sugar, urea). Here's what happens:



  1. Surface Occupation: When you add a non-volatile solute, its particles occupy some of the space at the liquid's surface. Think of it like a crowded street – some spots previously occupied by pure solvent molecules are now taken by solute molecules.

  2. Reduced Surface Area for Solvent: Because solute particles are at the surface and don't evaporate, fewer solvent molecules are available at the surface per unit area to escape into the vapour phase. The "active" surface area for solvent evaporation is reduced.

  3. Fewer Escaping Solvent Molecules: Since fewer solvent molecules are at the surface ready to evaporate, the rate at which solvent molecules escape into the vapour phase decreases.

  4. Lower Vapour Pressure: A reduced rate of evaporation means fewer vapour molecules in the space above the liquid at equilibrium. Consequently, the pressure exerted by these vapour molecules (i.e., the vapour pressure) will be lower than that of the pure solvent.


Key Intuitive Point: The non-volatile solute particles are like "obstructions" or "blockers" at the surface. They don't contribute to the vapour pressure themselves, but they hinder the solvent molecules from escaping, thus reducing the overall vapour pressure.



What makes it "Relative"?


The term "relative lowering" means we are comparing the *decrease* in vapour pressure to the *original* vapour pressure of the pure solvent.
It's not just the absolute decrease (P° - Ps), but rather the ratio of this decrease to the pure solvent's vapour pressure (P°):


Relative Lowering = (P° - Ps) / P°



Where:

  • P° = Vapour pressure of pure solvent

  • Ps = Vapour pressure of solution



Raoult's Law beautifully quantifies this intuitive understanding:


Raoult's Law (for non-volatile solute): The relative lowering of vapour pressure is equal to the mole fraction of the solute.


(P° - Ps) / P° = xsolute


This means the extent of vapour pressure lowering depends directly on the number of solute particles (represented by mole fraction), not on their identity. This is why it's a colligative property.



CBSE vs. JEE Focus:



  • CBSE: Focuses on understanding the qualitative reason for lowering and applying Raoult's law for calculations.

  • JEE: Requires a deeper understanding of the underlying principles, deviations from Raoult's law, and more complex problem-solving involving combinations of colligative properties. The intuitive understanding is the foundation for advanced problems.


This intuitive grasp helps solidify the concept before diving into the mathematical derivations and problem-solving, ensuring you remember *why* RLVP occurs.

🌍 Real World Applications

The concept of Relative Lowering of Vapour Pressure (RLVP), a colligative property, extends beyond theoretical understanding into numerous practical and industrial applications. It states that the addition of a non-volatile solute to a pure solvent lowers the solvent's vapour pressure in proportion to the mole fraction of the solute. This principle is fundamental to various processes we encounter in daily life and scientific research.



Real World Applications of Relative Lowering of Vapour Pressure




  • 1. Determination of Molar Mass of Non-Volatile Solutes:

    One of the primary applications of RLVP in chemistry is the accurate determination of the molar mass of unknown non-volatile solutes. By precisely measuring the lowering of vapour pressure of a solvent upon adding a known mass of the solute, the mole fraction of the solute can be calculated using Raoult's law:


    (P° - P) / P° = X_solute


    Where P° is the vapour pressure of the pure solvent, P is the vapour pressure of the solution, and X_solute is the mole fraction of the solute. From the mole fraction and the known masses, the molar mass of the solute can be derived. This method is particularly useful for substances that are unstable at higher temperatures or have high molar masses, making other methods challenging.


    JEE/CBSE Significance: This forms the basis for numerical problems, testing your ability to apply Raoult's law and molar mass calculations.




  • 2. Food Preservation (Salting/Sugaring):

    The age-old techniques of preserving food by salting (e.g., curing meat, pickling vegetables) or sugaring (e.g., making jams, candies) are direct applications of the RLVP principle. Adding a high concentration of salt or sugar (non-volatile solutes) to food products significantly lowers the "water activity" within the food. Lower water activity implies a lower vapour pressure of water in the food.


    Microorganisms, such as bacteria and fungi, require a certain level of water activity to grow and multiply. By reducing the water activity below this critical threshold, their growth is inhibited, thus preserving the food from spoilage. This method relies on creating an environment where water effectively moves out of microbial cells due to osmosis (driven by vapour pressure difference), leading to their dehydration.




  • 3. Humidity Control Using Desiccants:

    Desiccants are substances used to absorb moisture from the air, thereby controlling humidity in various environments (e.g., laboratories, electronics packaging, storage areas). Many effective desiccants, such as anhydrous calcium chloride (CaCl₂) or silica gel, work on the principle of RLVP.


    When these substances absorb water, they form a solution (or a hydrated compound) whose vapour pressure is significantly lower than the partial pressure of water vapour in the surrounding humid air. This difference in vapour pressure drives the movement of water molecules from the air into the desiccant until equilibrium is reached, effectively drying the air. The lower the vapour pressure of the desiccant solution, the more efficiently it can absorb moisture.




Understanding these real-world applications helps in appreciating the practical significance of colligative properties beyond textbook definitions.

🔄 Common Analogies

Common Analogies for Relative Lowering of Vapour Pressure


Understanding complex concepts like Colligative Properties can often be simplified through relatable analogies. For Relative Lowering of Vapour Pressure (RLVP), the core idea is how the presence of a non-volatile solute hinders the solvent's ability to escape into the gaseous phase. Here’s a powerful analogy to help solidify your understanding:



The Dance Floor Analogy


Imagine a bustling dance floor (representing the surface of the pure solvent). The dancers on this floor are energetic and constantly trying to jump off the floor onto a higher stage (representing the vapour phase).



  • Pure Solvent (Only Dancers): When the floor is full of only dancers, they have maximum space and opportunity to reach the edge and jump off to the higher stage. This corresponds to the vapour pressure of the pure solvent, where solvent molecules have maximum freedom to escape.

  • Adding Non-Volatile Solute (Non-Dancers): Now, imagine you introduce some people who are not dancing; they just stand around, occupying space on the dance floor (these are the non-volatile solute molecules). These non-dancers do not jump to the higher stage themselves, nor do they help the dancers. They simply take up valuable space.

  • Lowered Vapour Pressure: With non-dancers occupying parts of the floor, fewer dancers can reach the edges to jump off. Even though the dancers are still as energetic, their access to the "exit points" (the surface area from which molecules escape) is reduced. This reduced escape rate directly correlates to the lowering of the vapour pressure. The solvent molecules find it harder to escape to the vapour phase.

  • Relative Lowering: The extent to which the vapour pressure is lowered depends directly on the proportion of the dance floor occupied by the non-dancers. If 10% of the floor is taken by non-dancers, the reduction in escape opportunity is roughly 10%. This precisely mirrors Raoult's Law, where the relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. The more non-dancers (solute particles) you have, the greater the fraction of the surface they occupy, and thus the greater the relative lowering.




JEE & CBSE Relevance: This analogy provides a strong intuitive foundation for Raoult's Law. For competitive exams like JEE, a clear conceptual understanding is crucial for solving problems, even if direct analogy questions aren't asked. For CBSE, it aids in grasping the core principle behind the formula.


By visualizing the dance floor, you can easily grasp why a non-volatile solute reduces the vapour pressure and how this reduction is *relative* to the amount of solute present.


📋 Prerequisites

To effectively grasp the concept of Relative Lowering of Vapour Pressure (RLVP), a strong foundation in certain fundamental topics is essential. These prerequisites ensure that you understand the underlying principles and can apply them correctly in problem-solving.



Prerequisites for Relative Lowering of Vapour Pressure




  • Vapour Pressure of Pure Liquids:

    • Understand what vapour pressure is: The pressure exerted by the vapour in equilibrium with its liquid phase at a given temperature.

    • Factors affecting vapour pressure: Temperature (increases with temperature) and nature of the liquid (depends on intermolecular forces – weaker forces lead to higher vapour pressure).

    • This concept is fundamental as RLVP deals with the *change* in this pressure.




  • Solutions and Their Components:

    • Basic definitions of a solution, solvent, and solute.

    • Focus on binary solutions (one solvent and one solute), as RLVP typically discusses a non-volatile solute dissolved in a volatile solvent.




  • Concentration Terms (Specifically Mole Fraction):

    • Mole fraction ($X$) is the most crucial concentration term for understanding RLVP. It is defined as the ratio of the number of moles of one component to the total number of moles of all components in the solution.

    • JEE Focus: Be proficient in calculating mole fractions for both solvent and solute. Errors in these calculations are common mistakes.

    • Understand that the sum of mole fractions of all components in a solution is always 1 ($X_{ ext{solvent}} + X_{ ext{solute}} = 1$).




  • Concept of Volatile and Non-Volatile Substances:

    • Volatile substance: A substance that readily vaporizes at a given temperature and contributes to the total vapour pressure.

    • Non-volatile substance: A substance that does not readily vaporize and thus does not contribute to the vapour pressure of the solution. RLVP specifically deals with the effect of adding such a solute.




  • Raoult's Law (for solutions containing non-volatile solutes):

    • This is the core law upon which RLVP is based. Raoult's Law states that for a solution of a non-volatile solute in a volatile solvent, the vapour pressure of the solution is directly proportional to the mole fraction of the solvent.

    • Mathematically: $P_{ ext{solution}} = X_{ ext{solvent}} imes P^0_{ ext{solvent}}$, where $P^0_{ ext{solvent}}$ is the vapour pressure of the pure solvent.

    • CBSE & JEE: A thorough understanding of Raoult's Law is critical for both board and competitive exams, including its implications and derivations.




Mastering these foundational concepts will provide a clear pathway to understanding the derivation, applications, and problem-solving aspects of Relative Lowering of Vapour Pressure.

🔗 Related Topics

Related Topics to Relative Lowering of Vapour Pressure


Understanding the concept of Relative Lowering of Vapour Pressure (RLVP) is enhanced by its connections to several fundamental and advanced topics in physical chemistry. These related concepts provide the necessary groundwork and expand upon the principles governing solutions.





  • Vapour Pressure of Pure Liquids:

    Before delving into RLVP, a solid understanding of a pure liquid's vapour pressure is crucial. This includes its definition (pressure exerted by the vapour in equilibrium with its liquid phase at a given temperature) and factors affecting it (temperature, intermolecular forces, and nature of the liquid).




  • Raoult's Law:

    RLVP is essentially a specific application of Raoult's Law. Raoult's Law states that for a solution of volatile components, the partial vapour pressure of each component in the solution is directly proportional to its mole fraction. For a non-volatile solute, Raoult's Law is simplified to: Psolution = Xsolvent * Posolvent. The lowering of vapour pressure (ΔP = Posolvent - Psolution) directly follows from this, and the relative lowering (ΔP/Posolvent) is equal to the mole fraction of the solute (Xsolute).




  • Ideal and Non-Ideal Solutions:

    The derivations and equations for RLVP are based on the assumption of ideal dilute solutions. An ideal solution obeys Raoult's Law over the entire range of concentrations and temperatures. Non-ideal solutions, which exhibit positive or negative deviations from Raoult's Law, will not strictly follow the standard RLVP equation. Understanding these deviations helps predict real-world behaviour of solutions.




  • Mole Fraction (X):

    Mole fraction is the primary concentration term used in the mathematical expression of RLVP (RLVP = Xsolute). Its accurate calculation is fundamental to solving problems related to colligative properties. Students must be proficient in converting other concentration units (like molality or molarity) to mole fraction for these calculations, especially in JEE Main problems.




  • Other Colligative Properties:

    RLVP is one of the four colligative properties, all of which depend on the number of solute particles, not their identity, in a given amount of solvent. The other three are:



    • Elevation in Boiling Point (ΔTb): Related to the lowering of vapour pressure, as a lower vapour pressure requires a higher temperature to reach the atmospheric pressure.

    • Depression in Freezing Point (ΔTf): Also linked to vapour pressure, as a lower vapour pressure leads to a lower freezing point.

    • Osmotic Pressure (Π): This is the pressure difference required to prevent the flow of solvent into the solution through a semipermeable membrane. Like RLVP, it directly depends on the number of solute particles.


    Understanding the interconnectedness of these properties is vital for JEE Main and CBSE Board exams, as problems often involve relating one colligative property to another.




  • Van't Hoff Factor (i):

    For electrolyte solutions (ionic compounds that dissociate in solution) or solutes that associate in solution, the number of particles effectively changes. The Van't Hoff factor 'i' accounts for this. It is incorporated into all colligative property equations, including RLVP, to correct for the actual number of particles. This factor is particularly important for JEE Main questions involving ionic compounds.




Mastering these related concepts ensures a comprehensive understanding of RLVP and its applications in various chemical scenarios.

⚠️ Common Exam Traps

Understanding "Relative Lowering of Vapour Pressure" (RLVP) is fundamental, but exams often set up traps to test conceptual clarity and attention to detail. Identifying these common pitfalls can significantly improve your score.



Here are the common exam traps related to Relative Lowering of Vapour Pressure:





  • Confusion between (P° - P)/P° and (P° - P)/P:

    • Mistake: Students often confuse the definition of relative lowering. The correct definition for RLVP is (P° - P) / P°, where P° is the vapor pressure of the pure solvent and P is the vapor pressure of the solution. Sometimes, candidates mistakenly use (P° - P) / P, which is incorrect.

    • Correction: Always remember the exact formula derived from Raoult's Law: (P° - P) / P° = Xsolute (mole fraction of non-volatile solute). The denominator must be the vapor pressure of the pure solvent (P°).




  • Ignoring the van't Hoff factor (i) for Electrolytes:

    • Mistake: This is a major trap in JEE Main and advanced level questions. If the solute is an electrolyte (like NaCl, CaCl2, glucose, etc.), it dissociates in solution, increasing the effective number of particles. Many students forget to incorporate the van't Hoff factor (i).

    • Correction: For electrolytic or associating solutes, the modified formula for all colligative properties must be used: (P° - P) / P° = i × Xsolute. Calculate 'i' based on the degree of dissociation/association or assume complete dissociation/association if not specified.




  • Incorrectly Calculating Mole Fraction (Xsolute):

    • Mistake: Errors often arise in determining the moles of solute or solvent, especially when masses are given. Forgetting to convert mass to moles (mass/molar mass) is common. Sometimes, students might accidentally use molarity or molality values directly without converting them to mole fractions.

    • Correction: Meticulously calculate: Xsolute = nsolute / (nsolute + nsolvent). Always convert given masses into moles first.




  • Approximation for Dilute Solutions used Incorrectly:

    • Mistake: For very dilute solutions, Xsolute ≈ nsolute / nsolvent. This approximation simplifies calculations but is only valid when nsolute << nsolvent. Applying this approximation when the solution is not dilute enough leads to inaccurate answers.

    • Correction: While useful for quick estimates, it's generally safer to use the exact formula for mole fraction unless the question explicitly suggests a very dilute solution or for specific contexts where the approximation is highly accurate.




  • Misinterpreting the Question's Requirement:

    • Mistake: A question might ask for the lowering of vapour pressure (P° - P) but students provide the relative lowering (P° - P)/P°, or vice-versa.

    • Correction: Read the question carefully! Pay attention to keywords like "lowering," "relative lowering," or "vapour pressure of solution."




  • Units and Molar Mass Errors:

    • Mistake: Not being careful with units of vapor pressure (mmHg, torr, kPa, bar) or making calculation mistakes with molar masses of solute or solvent.

    • Correction: Ensure all vapor pressure units are consistent throughout the calculation. Always double-check the molar masses of the compounds involved.




By being aware of these common traps and practicing meticulously, you can avoid losing marks on RLVP questions.

⭐ Key Takeaways

Key Takeaways: Relative Lowering of Vapour Pressure (RLVP)



Understanding Relative Lowering of Vapour Pressure (RLVP) is fundamental to the study of Colligative Properties. This section summarizes the essential concepts, formulas, and implications for both CBSE and JEE Main examinations.



  • Definition: When a non-volatile solute is added to a pure solvent, the vapour pressure of the resulting solution decreases. The relative lowering of vapour pressure (RLVP) is defined as the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent.


    Mathematically: RLVP = (P° - Ps) / P°


    Where P° is the vapour pressure of the pure solvent and Ps is the vapour pressure of the solution.


  • Raoult's Law (for Solutions with Non-Volatile Solute): According to Raoult's Law, for a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute.


    (P° - Ps) / P° = Xsolute


    Where Xsolute is the mole fraction of the solute in the solution.


  • Colligative Property: RLVP is one of the four colligative properties. This means it depends only on the number of solute particles present in a given amount of solvent, and not on their chemical nature or identity. This is a crucial concept for both CBSE and JEE.


  • Expanded Formula for Molar Mass Determination:
    Since Xsolute = nsolute / (nsolute + nsolvent), where n is the number of moles.


    For dilute solutions (nsolute << nsolvent), the formula simplifies to:


    (P° - Ps) / P° ≈ nsolute / nsolvent = (wsolute / Msolute) / (wsolvent / Msolvent)


    This simplified form is very useful for determining the molar mass (Msolute) of an unknown non-volatile solute from experimental RLVP data.


  • Conditions of Applicability:

    • The solute must be non-volatile (does not contribute to the vapour pressure of the solution).

    • The solute must be a non-electrolyte (does not dissociate or associate in the solvent). If dissociation or association occurs, the van't Hoff factor (i) must be introduced into the formula.

    • The solution should ideally be dilute for the approximation to hold true.




  • JEE Main vs. CBSE Focus:

    • CBSE: Emphasis is on understanding the definition, Raoult's law statement, the formula, and basic numerical problems involving molar mass calculation. Direct application of the formula is common.

    • JEE Main: Questions can be more complex, often involving the calculation of van't Hoff factor (i) for electrolytic solutes or combining RLVP with other colligative properties. Knowledge of ideal vs. non-ideal solutions and deviations from Raoult's law might be tested in a broader context.





Mastering RLVP lays the groundwork for understanding other colligative properties. Always ensure you identify the nature of the solute (non-volatile, non-electrolyte) before applying the standard formulas.

🧩 Problem Solving Approach

Problem Solving Approach for Relative Lowering of Vapour Pressure (RLVP)



Understanding how to systematically approach problems related to Relative Lowering of Vapour Pressure (RLVP) is crucial for both JEE and CBSE exams. This section provides a step-by-step methodology to tackle such questions effectively.

1. Grasp the Core Concept: Raoult's Law


The fundamental principle governing RLVP is Raoult's Law for non-volatile solutes. It states that the relative lowering of vapour pressure of a solution containing a non-volatile solute is equal to the mole fraction of the solute.

* Formula: $frac{P_A^0 - P_A}{P_A^0} = X_B = frac{n_B}{n_A + n_B}$
* $P_A^0$: Vapour pressure of the pure solvent
* $P_A$: Vapour pressure of the solution
* $X_B$: Mole fraction of the solute
* $n_A$: Moles of solvent
* $n_B$: Moles of solute

For dilute solutions, where $n_B ll n_A$, the denominator can be approximated as $n_A$.
* Approximate Formula (for dilute solutions): $frac{P_A^0 - P_A}{P_A^0} approx frac{n_B}{n_A}$
* Caution: Use the exact formula unless explicitly stated that the solution is very dilute or if the options are widely spaced.

2. Step-by-Step Problem-Solving Strategy


Follow these steps to solve RLVP problems:

1. Read the Question Carefully:
* Identify all given quantities: initial vapour pressure of pure solvent ($P_A^0$), mass of solute, mass of solvent, molar mass of solvent.
* Identify what needs to be calculated: molar mass of solute, vapour pressure of solution ($P_A$), or RLVP itself.
* Confirm the nature of the solute (must be non-volatile and non-electrolyte for direct application of the formula).

2. Calculate Moles of Solvent ($n_A$):
* $n_A = frac{ ext{Mass of solvent}}{ ext{Molar mass of solvent}}$

3. Calculate Moles of Solute ($n_B$):
* If molar mass of solute is given, $n_B = frac{ ext{Mass of solute}}{ ext{Molar mass of solute}}$.
* If molar mass of solute is unknown, express $n_B$ in terms of its unknown molar mass.

4. Set up the Equation:
* Substitute the values of $P_A^0$, $P_A$ (if known), $n_A$, and $n_B$ into the exact Raoult's Law formula: $frac{P_A^0 - P_A}{P_A^0} = frac{n_B}{n_A + n_B}$.
* Ensure all vapour pressure units are consistent (e.g., all in mmHg, or all in bar).

5. Solve for the Unknown:
* Rearrange the equation algebraically to solve for the required quantity (e.g., $P_A$, molar mass of solute, etc.).

6. Check Units and Significant Figures:
* Ensure the final answer has appropriate units and significant figures.

3. Important Considerations & JEE Specifics



* Units Consistency: Always use consistent units for vapour pressure throughout the calculation.
* Nature of Solute: The formula is strictly for non-volatile, non-dissociating (non-electrolyte) solutes. For electrolytes, the van't Hoff factor ($i$) must be included: $frac{P_A^0 - P_A}{P_A^0} = i cdot X_B$. (JEE Focus)
* Dilute vs. Concentrated: While the approximate formula is handy, stick to the exact formula for higher accuracy, especially in JEE problems where small differences can matter.
* Temperature: Vapour pressure is highly temperature-dependent. Ensure $P_A^0$ and $P_A$ are considered at the same temperature.
* Interconversion of Concentration Terms: JEE problems may require converting molality or molarity to mole fraction before applying the RLVP formula.
* Example: If molality ($m$) is given, assume 1 kg of solvent, calculate moles of solute ($n_B = m imes 1 ext{ kg}$ solvent), and moles of solvent ($n_A = 1000 ext{ g} / ext{Molar mass of solvent}$).

Example Problem Walkthrough:


Question: The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm Hg. Calculate the molar mass of the solid substance. (Molar mass of benzene = 78 g/mol).

Solution:
1. Given:
* $P_A^0$ (pure benzene) = 640 mm Hg
* $P_A$ (solution) = 600 mm Hg
* Mass of solute ($W_B$) = 2.175 g
* Mass of solvent (benzene, $W_A$) = 39.0 g
* Molar mass of solvent ($M_A$) = 78 g/mol
* Molar mass of solute ($M_B$) = ?

2. Calculate moles of solvent ($n_A$):
* $n_A = frac{W_A}{M_A} = frac{39.0 ext{ g}}{78 ext{ g/mol}} = 0.5 ext{ mol}$

3. Express moles of solute ($n_B$):
* $n_B = frac{W_B}{M_B} = frac{2.175 ext{ g}}{M_B}$

4. Apply Raoult's Law:
* $frac{P_A^0 - P_A}{P_A^0} = frac{n_B}{n_A + n_B}$
* $frac{640 - 600}{640} = frac{2.175/M_B}{0.5 + 2.175/M_B}$
* $frac{40}{640} = frac{1}{16} = frac{2.175/M_B}{0.5 + 2.175/M_B}$

5. Solve for $M_B$:
* $0.5 + frac{2.175}{M_B} = 16 imes frac{2.175}{M_B}$
* $0.5 = frac{16 imes 2.175}{M_B} - frac{2.175}{M_B}$
* $0.5 = frac{(16-1) imes 2.175}{M_B}$
* $0.5 = frac{15 imes 2.175}{M_B}$
* $M_B = frac{15 imes 2.175}{0.5} = 15 imes 4.35 = 65.25 ext{ g/mol}$

By following these steps, you can confidently approach and solve problems involving Relative Lowering of Vapour Pressure. Keep practicing, and success will follow!
📝 CBSE Focus Areas

CBSE Focus Areas: Relative Lowering of Vapour Pressure (RLVP)



For CBSE board examinations, the topic of Relative Lowering of Vapour Pressure (RLVP) is fundamental and frequently tested. Students must have a clear understanding of its definition, underlying law, and practical applications, especially in determining the molar mass of a non-volatile solute.



Key Concepts for CBSE



  • Definition of RLVP: Understand that when a non-volatile solute is added to a pure solvent, its vapour pressure decreases. The ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent is known as Relative Lowering of Vapour Pressure.

  • Raoult's Law (for non-volatile solute): This is the cornerstone for RLVP.

    • Statement: For a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute.

    • Mathematical Form:


      ℁ $frac{P^circ - P_s}{P^circ} = X_{solute}$


      Where:



      • $P^circ$ = Vapour pressure of pure solvent

      • $P_s$ = Vapour pressure of solution

      • $(P^circ - P_s)$ = Lowering of vapour pressure

      • $X_{solute}$ = Mole fraction of solute






  • Determination of Molar Mass of Solute: A very common numerical application in CBSE.

    • The mole fraction of solute ($X_{solute}$) can be expressed as $n_{solute} / (n_{solute} + n_{solvent})$.

    • For dilute solutions (a common assumption in CBSE problems), $n_{solute} ll n_{solvent}$, so $n_{solute} + n_{solvent} approx n_{solvent}$.

    • This simplifies the formula to:


      $frac{P^circ - P_s}{P^circ} = frac{n_{solute}}{n_{solvent}} = frac{w_B/M_B}{w_A/M_A}$


      Where:



      • $w_B$, $M_B$ = Mass and molar mass of solute

      • $w_A$, $M_A$ = Mass and molar mass of solvent




    • CBSE questions often provide experimental data (vapour pressures, masses) and ask you to calculate the molar mass of the unknown non-volatile solute.





Important Considerations for CBSE



  • Numerical Proficiency: Be prepared to solve numerical problems directly applying the RLVP formula to find $M_B$, $P_s$, or $X_{solute}$. Ensure correct units and significant figures.

  • Assumptions: Remember that Raoult's law for RLVP is strictly applicable to ideal solutions containing non-volatile and non-associating/non-dissociating solutes. For numerical problems, the "dilute solution" approximation is often used implicitly or explicitly.

  • Comparison with JEE Main: While CBSE focuses on the direct application of Raoult's Law and numerical calculations for molar mass, JEE Main might extend to more complex scenarios involving non-ideal solutions, solutions with volatile solutes (where Raoult's law is applied to each component), or integration with other colligative properties in multi-step problems. For CBSE, the emphasis remains on the foundational aspect of RLVP.



Mastering these concepts will ensure you confidently tackle RLVP questions in your CBSE board exams!


🎓 JEE Focus Areas

🎯 JEE Focus Areas: Relative Lowering of Vapour Pressure


The concept of Relative Lowering of Vapour Pressure (RLVP) is a fundamental colligative property and a high-yield topic for JEE Main and Advanced. Mastery of its definition, formula, and applications, especially with respect to non-volatile solutes and electrolytes, is crucial.



1. Understanding Relative Lowering of Vapour Pressure (RLVP)



  • Definition: When a non-volatile solute is added to a pure solvent, the vapour pressure of the solvent decreases. This decrease in vapour pressure is known as the lowering of vapour pressure ($Delta P = P^0 - P_s$). The relative lowering of vapour pressure is defined as the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent ($frac{P^0 - P_s}{P^0}$).

  • Raoult's Law (for non-volatile solute): It states that for a solution containing a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute ($X_B$) in the solution.



2. Key Formula and Applications


The most important formula for RLVP, directly derived from Raoult's Law, is:



$frac{P^0 - P_s}{P^0} = X_B$


Where:



  • $P^0$ = Vapour pressure of pure solvent

  • $P_s$ = Vapour pressure of solution

  • $X_B$ = Mole fraction of solute ($n_B / (n_A + n_B)$, where $n_A$ and $n_B$ are moles of solvent and solute, respectively).


For dilute solutions, where $n_B ll n_A$, the formula can be approximated as:



$frac{P^0 - P_s}{P^0} = frac{n_B}{n_A}$ (This approximation is less common for RLVP calculations in JEE; the direct mole fraction is preferred)


Primary Application: This property is extensively used to determine the molar mass of an unknown non-volatile, non-electrolyte solute.



  • JEE Tip: Always use the exact mole fraction formula for accuracy, unless specifically asked for an approximation.



3. Important Considerations for JEE



  • Colligative Nature: RLVP depends only on the number of solute particles, not on their nature. This is why it's a colligative property.

  • Non-Volatile Solute: The formula is strictly applicable only when the solute is non-volatile. If both components are volatile, Raoult's law for volatile components applies.

  • Electrolytes and van't Hoff Factor (i): For solutions containing electrolytes (which dissociate or associate), the effective number of particles changes. Hence, the van't Hoff factor 'i' must be incorporated:


    $frac{P^0 - P_s}{P^0} = i cdot X_B$


    Where $X_B$ is the mole fraction calculated assuming no dissociation/association.

  • Units: Ensure consistency in units for vapour pressure (e.g., mmHg, atm, bar). Mole fraction is unitless.



4. Common Mistakes & JEE Traps



  • Confusion between $Delta P$ and $frac{Delta P}{P^0}$ : Remember, RLVP is $frac{P^0 - P_s}{P^0}$, not just $P^0 - P_s$. The latter is merely "lowering of vapour pressure" and is not directly proportional to the mole fraction of solute.

  • Ignoring 'i' for Electrolytes: A very common mistake. Always check if the solute is an electrolyte (e.g., NaCl, $MgCl_2$, $CH_3COOH$) and calculate 'i' appropriately.

  • Using Incorrect Moles: Ensure you correctly calculate moles of solute ($n_B$) and solvent ($n_A$) from given masses and molar masses. For water, the molar mass is 18 g/mol.

  • Applying to Volatile Solutes: Do not use this formula if the solute is also volatile.



💡 Example Application (JEE Style)


A solution containing 30 g of a non-volatile solute in 90 g of water has a vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is added to the solution, and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate the molar mass of the solute ($M_B$). (Given: Vapour pressure of pure water at 298 K = 3.0 kPa).


This type of problem tests your ability to set up two equations based on RLVP for two different solution compositions and solve for the unknown molar mass and pure solvent vapour pressure (if not given).


Mastering RLVP ensures you confidently tackle numerical problems related to molar mass determination and understanding the effect of solutes on solvent properties. Practice with problems involving both non-electrolytes and electrolytes.


📊 AI Generation Metadata

AI Model: Gemini Full Parallel API Client
Status: AI Generated
Version: 1
Generated: Oct 22, 2025
🌐 Overview
Relative lowering of vapour pressure (RLVP) in an ideal solution: (P^* − P)/P^* = x_solute for a non-volatile solute (Raoult’s law). It depends only on the number of solute particles (colligative), not their nature—enabling molar mass determination.
📚 Fundamentals
• RLVP: (P^* − P)/P^* = x_solute (ideal solution).
• x_solute = n_solute / (n_solute + n_solvent).
• For dilute solutions: x_solute ≈ n_solute / n_solvent.
• Electrolytes: effective particles increase by i (van ’t Hoff).
🔬 Deep Dive
• Derivation from Raoult’s law and chemical potential.
• Non-ideal corrections via activity coefficients (qualitative).
🎯 Shortcuts
“Lowering tracks the fraction of solute: RLVP = x_solute.”
💡 Quick Tips
• Confirm solute is non-volatile.
• For dilute solutions, n_solute ≪ n_solvent simplifies algebra.
• Carefully compute i from dissociation/association.
🧠 Intuitive Understanding
Solute particles “occupy” surface sites and reduce solvent molecules escaping to vapour—thus lowering vapour pressure. The effect scales with how many solute particles are present (mole fraction).
🌍 Real World Applications
• Determination of unknown molar mass.
• Industrial solution design where vapour pressure control matters.
• Foundations for other colligative properties (EB, DF, osmotic pressure).
🔄 Common Analogies
• Fewer exits open: adding “inert blockers” at the surface means fewer solvent molecules can leave.
• Crowded doorway effect reduces escape rate.
📋 Prerequisites
Raoult’s law, ideal solutions, mole fraction, vapour pressure concepts, non-volatile solute assumption.
🔗 Related Topics
Elevation of boiling point; depression of freezing point; osmotic pressure; van ’t Hoff factor for electrolytes.
⚠️ Common Exam Traps
• Using mass fraction instead of mole fraction.
• Ignoring electrolyte dissociation.
• Applying to volatile solutes without modification.
⭐ Key Takeaways
• Colligative property—depends on number of particles.
• RLVP enables molar mass measurement.
• For electrolytes, account for dissociation/association via i.
🧩 Problem Solving Approach
1) Write RLVP relation.
2) Express mole fraction via given masses and molar masses.
3) For dilute solutions, use approximation if valid.
4) Include i when solute is electrolyte.
5) Solve for the unknown and check units.
📝 CBSE Focus Areas
Statement, formula, simple molar mass problems, and qualitative understanding.
🎓 JEE Focus Areas
Numericals including electrolytes; use of approximations and error checks; conceptual edge cases.

📊 AI Generation Metadata

Difficulty: Intermediate
Estimated Time: 120 mins
AI Model: ChatGPT 5 (Copilot Agent)
Status: AI Generated
Version: 1
Generated: Oct 23, 2025
🌐 Overview
Relative Lowering of Vapour Pressure

When a non-volatile solute is dissolved in a volatile solvent, the vapour pressure of the solution decreases. This phenomenon is called lowering of vapour pressure, and the ratio of this decrease to the original vapour pressure is the relative lowering of vapour pressure.

Quick Example 1: Raoult's Law Application
Pure water at 25°C has vapour pressure P₀ = 23.8 mm Hg.
After dissolving 18 g glucose (C₆H₁₂O₆, M=180) in 90 g water:
- Moles of glucose = 18/180 = 0.1 mol
- Moles of water = 90/18 = 5 mol
- Mole fraction of glucose χ_solute = 0.1/5.1 ≈ 0.0196
- Relative lowering = (P₀ - P_s)/P₀ = χ_solute = 0.0196
- New vapour pressure P_s = P₀(1 - 0.0196) = 23.3 mm Hg

Quick Example 2: Finding Molar Mass
If relative lowering of vapour pressure = 0.02 when 6 g of a substance is dissolved in 100 g water, find molar mass.
χ_solute = 0.02 = n_solute/(n_solute + n_solvent)
For dilute solution: χ_solute ≈ n_solute/n_solvent
0.02 = (6/M)/(100/18) where M is molar mass
M = 6 × 18/(0.02 × 100) = 54 g/mol
📚 Fundamentals
Fundamental Concepts

1. Vapour Pressure:
Pressure exerted by vapour in equilibrium with its liquid at a given temperature.
- Pure solvent: has vapour pressure P₀
- After adding non-volatile solute: vapour pressure decreases to P_s

2. Lowering of Vapour Pressure:
ΔP = P₀ - P_s

Where:
- P₀ = vapour pressure of pure solvent
- P_s = vapour pressure of solution
- ΔP = lowering of vapour pressure (absolute)

3. Relative Lowering of Vapour Pressure:
(P₀ - P_s)/P₀ = ΔP/P₀

4. Raoult's Law (for non-volatile solute):
P_s = P₀ × χ_solvent

Where χ_solvent = mole fraction of solvent = n_solvent/(n_solvent + n_solute)

Alternatively:
(P₀ - P_s)/P₀ = χ_solute

Where χ_solute = mole fraction of solute

5. For Dilute Solutions:
When n_solute << n_solvent:
χ_solute ≈ n_solute/n_solvent

Therefore:
ΔP/P₀ = n_solute/n_solvent = (w₂/M₂)/(w₁/M₁)

Where:
- w₁, M₁ = mass and molar mass of solvent
- w₂, M₂ = mass and molar mass of solute

6. Important Conditions:
- Solute must be non-volatile (doesn't evaporate)
- Solute must be non-electrolyte (or use van't Hoff factor for electrolytes)
- Solution should be dilute for approximations
- Temperature constant throughout
🔬 Deep Dive
Advanced Theory and Derivations

1. Derivation of Raoult's Law from Thermodynamics

At equilibrium, chemical potential of solvent in liquid phase = chemical potential in vapour phase:
μ_liquid = μ_vapour

For ideal solution:
μ_liquid = μ*_liquid + RT ln(χ_solvent)

For pure solvent vapour:
μ_vapour = μ*_vapour = μ*_liquid + RT ln(P_s/P₀)

Equating:
μ*_liquid + RT ln(χ_solvent) = μ*_liquid + RT ln(P_s/P₀)

RT ln(χ_solvent) = RT ln(P_s/P₀)

χ_solvent = P_s/P₀

Therefore: P_s = P₀ × χ_solvent

This is Raoult's Law.

2. Derivation of Relative Lowering Formula:

P_s = P₀ × χ_solvent
P_s = P₀ × [n_solvent/(n_solvent + n_solute)]

P₀ - P_s = P₀[1 - n_solvent/(n_solvent + n_solute)]
= P₀[(n_solvent + n_solute - n_solvent)/(n_solvent + n_solute)]
= P₀[n_solute/(n_solvent + n_solute)]
= P₀ × χ_solute

Therefore: (P₀ - P_s)/P₀ = χ_solute

3. Molecular Interpretation:

Why does vapour pressure decrease?

- In pure solvent: all surface molecules are solvent → maximum evaporation rate
- In solution: some surface positions occupied by non-volatile solute → reduced number of solvent molecules at surface → lower evaporation rate
- At equilibrium: lower vapour pressure

Fraction of surface covered by solvent ≈ χ_solvent
Vapour pressure proportional to surface coverage
Hence: P_s = P₀ × χ_solvent

4. Non-Ideal Solutions:

For real solutions showing deviation from Raoult's Law:

Positive Deviation: P_s > P₀ × χ_solvent
- Solvent-solute attraction < Solvent-solvent attraction
- Examples: Acetone + CS₂, Ethanol + Water

Negative Deviation: P_s < P₀ × χ_solvent
- Solvent-solute attraction > Solvent-solvent attraction
- Examples: Chloroform + Acetone, HNO₃ + Water

5. Van't Hoff Factor for Electrolytes:

For electrolytes that dissociate:
χ_solute(effective) = i × χ_solute

Where i = van't Hoff factor (number of particles per formula unit)

Example: NaCl → Na⁺ + Cl⁻ (i ≈ 2)
Relative lowering = i × (moles of NaCl)/(total moles)

6. Relation to Other Colligative Properties:

All colligative properties depend on number of solute particles:
- Relative lowering of VP: (P₀-P_s)/P₀ = χ_solute
- Elevation of BP: ΔT_b = K_b × m
- Depression of FP: ΔT_f = K_f × m
- Osmotic pressure: π = CRT

All are interrelated through χ_solute or molality m.
🎯 Shortcuts
Mnemonics and Memory Aids

1. "RLVP = Chi-Solute" (Core Formula):
Relative Lowering of Vapour Pressure = χ_solute (mole fraction)
Simplest way to remember Raoult's Law for this property.

2. "Lowering is Relative to Original":
Relative lowering = (Lowering) / (Original)
= (P₀ - P_s) / P₀
Always divide by the original (pure solvent) value.

3. "Non-Volatile Solute Only":
Think NV = Non-Volatile
If solute evaporates, Raoult's Law (this form) doesn't apply.

4. "Dilute: Drop the Denominator Addition":
For dilute solutions:
χ_solute = n_solute/(n_solute + n_solvent) ≈ n_solute/n_solvent
Drop the n_solute from denominator (it's negligible).

5. "Pressure Down, Solute Around":
When solute is around, vapour pressure goes down.
Simple rhyme to remember the effect.

6. "Molar Mass from RLVP: W2M1 over M2W1":
For finding M₂ (molar mass of solute):
(P₀-P_s)/P₀ = (w₂/M₂)/(w₁/M₁)
Cross-multiply to get: M₂ = (w₂×M₁) / [(P₀-P_s)/P₀ × w₁]

7. "Electrolytes: i-times the Effect":
For electrolytes, multiply χ_solute by i (van't Hoff factor)
NaCl: i≈2, CaCl₂: i≈3, C₆H₁₂O₆: i=1
💡 Quick Tips
Quick Tips

- Tip 1: Relative lowering is always a fraction between 0 and 1 (or percentage 0-100%). If you get value > 1, check calculation

- Tip 2: For water as solvent, M₁ = 18 g/mol is commonly used. Keep this memorized for quick calculations

- Tip 3: In dilute solutions (χ_solute < 0.1), use approximation χ_solute ≈ n_solute/n_solvent to save time

- Tip 4: Solute molar mass problems: isolate M₂ first algebraically, then substitute numbers. Less error-prone

- Tip 5: If given relative lowering as percentage, convert to decimal: 5% = 0.05

- Tip 6: Remember P_s = P₀(1 - χ_solute) gives solution VP directly. Faster than calculating lowering first

- Tip 7: For comparing solutions: same χ_solute gives same relative lowering regardless of P₀ (colligative property!)

- Tip 8: When solute dissociates (electrolyte), effective mole fraction = i × χ_solute. Don't forget i factor!

- Tip 9: Units: Vapour pressure can be in mm Hg, atm, Pa, kPa. Relative lowering is unitless, so pressure units cancel

- Tip 10: Double-check: P_s must be LESS than P₀. If you get P_s > P₀, you made an error

- Tip 11: For mixed solutes, calculate total moles of all solutes and use total χ_solute

- Tip 12: Quick estimate: For every 1 mole solute per 10 moles solvent, relative lowering ≈ 0.1 (10%)
🧠 Intuitive Understanding
Building Intuition

The Crowded Swimming Pool Analogy:

Imagine a swimming pool (the solvent) with people (solvent molecules) trying to jump out (evaporate):
- Pure solvent: All positions at the edge are accessible for jumping out → maximum rate
- With solute: Some edge positions are blocked by non-swimmers (solute molecules that don't jump) → fewer can jump out → lower "jumping rate" (vapour pressure)

The relative lowering = fraction of edge positions blocked = mole fraction of non-jumpers (solute).

Surface Area Perspective:

Think of the liquid surface as a checkerboard:
- Each square can be occupied by solvent (can evaporate) or solute (cannot)
- Probability a square has solvent = χ_solvent
- Evaporation rate ∝ number of solvent squares
- Vapour pressure ∝ evaporation rate
- Hence: P_s/P₀ = χ_solvent

Why Only Mole Fraction Matters (Not Mass):

Vapour pressure depends on:
- How many solute particles block surface positions
- NOT how heavy they are

A large molecule (high molar mass) and small molecule (low molar mass) have the SAME effect if present in same number (moles). This is why colligative properties depend on particle count, not mass.

The Equilibrium Shift:

Liquid ⇌ Vapour

Adding non-volatile solute:
- Doesn't affect vapour side (solute doesn't evaporate)
- Reduces effective concentration of solvent in liquid phase
- Equilibrium shifts left (less vapour)
- Result: lower vapour pressure

Dilute Solution Approximation:

For dilute solutions:
χ_solute = n_solute/(n_solute + n_solvent) ≈ n_solute/n_solvent

Why? Because n_solute << n_solvent, so denominator ≈ n_solvent.

This simplification makes calculations easier and is valid when solute concentration is low.
🌍 Real World Applications
Real-World Applications

1. Food Preservation:
- Adding salt or sugar to jams, pickles lowers water vapour pressure
- Reduces water activity, prevents microbial growth
- Principle behind dehydration preservation

2. Humidification and Dehumidification:
- Humidity control in industrial processes
- Air conditioning systems monitor vapour pressure
- Preventing condensation in storage facilities

3. Chemical Industry:
- Determining purity of solvents (pure solvent has maximum VP)
- Quality control in solution preparation
- Distillation process design

4. Pharmaceutical Industry:
- Drug formulation stability depends on vapour pressure
- Moisture content control in tablets and capsules
- Shelf-life determination based on vapour pressure changes

5. Meteorology:
- Understanding dew point and humidity
- Predicting fog and cloud formation
- Sea water (dissolved salts) has lower VP than fresh water

6. Automotive Industry:
- Antifreeze solutions (coolants) have modified vapour pressures
- Prevents radiator fluid evaporation
- Related to elevation of boiling point

7. Paint and Coating Industry:
- Solvent evaporation rates in paints
- Drying time calculations
- Formulation of quick-dry vs slow-dry paints

8. Desalination:
- Multi-stage flash distillation uses vapour pressure differences
- Reverse osmosis related to osmotic pressure (another colligative property)
- Energy-efficient water purification
🔄 Common Analogies
Common Analogies

1. Parking Lot Analogy:
Liquid surface is like a parking lot. Solvent molecules (cars that can leave) and solute molecules (permanent fixtures). Fewer cars can leave if more spaces are occupied by fixtures. Exit rate (vapour pressure) decreases proportionally to fixtures' fraction.
Limitation: Doesn't explain why ratio equals mole fraction specifically.

2. Concert Exit Analogy:
People (molecules) leaving a concert venue (liquid phase). If some doors are blocked (solute present), fewer people exit per minute. Exit rate ∝ fraction of open doors ∝ χ_solvent.
Limitation: Doors are discrete; actual molecular distribution is continuous.

3. Lottery Ticket Analogy:
Each surface molecule has a "ticket" to evaporate. In pure solvent, all tickets are valid. With solute, some tickets are invalid (solute can't evaporate). Probability of evaporation = fraction of valid tickets = χ_solvent.
Limitation: Evaporation isn't truly random; depends on kinetic energy.

4. Dilution Reduces Potency:
Like diluting concentrated juice reduces its strength, adding solute "dilutes" the solvent's tendency to evaporate. The more you dilute, the weaker the vapour pressure.
Limitation: Doesn't capture the quantitative relationship.

5. Blocking Sunlight:
Solute molecules are like clouds blocking sunlight. The more clouds (higher χ_solute), the less sunlight (lower vapour pressure) reaches ground.
Limitation: Clouds absorb/reflect; solute molecules don't absorb vapour pressure but prevent evaporation.

6. Team Selection:
If a team of 11 must be selected from solvent and solute molecules, probability of selecting solvent = χ_solvent. Vapour phase "team" composition reflects this probability.
Limitation: Vapour doesn't contain solute at all in ideal case.
📋 Prerequisites
Prerequisites

1. Mole Concept:
Fluent in calculating moles from mass: n = w/M. Understanding of Avogadro's number and molar mass.

2. Mole Fraction:
Comfortable calculating mole fractions: χ_A = n_A/(n_A + n_B + ...). Know that Σχ = 1.

3. Solutions Basics:
Understand solvent, solute, and solution. Know concentration terms: molarity, molality, mass percent.

4. States of Matter:
Familiarity with liquid-vapour equilibrium. Basic understanding of evaporation and condensation.

5. Vapor Pressure Concept:
Know that liquids have characteristic vapour pressures at given temperatures. Understand equilibrium between liquid and vapour.

6. Ideal vs Non-Ideal Behavior:
Basic awareness that real solutions may deviate from ideal behavior.

7. Basic Algebra:
Solving equations with ratios and fractions. Rearranging formulas to isolate unknowns.

8. Colligative Properties Overview:
General understanding that some properties depend only on particle count, not identity.
🔗 Related Topics
Related Topics

Previous Topics to Review:
- Mole concept and molar mass (Topic 46)
- Concentration terms: molarity, molality (Topic 52)
- Ideal and non-ideal solutions (Topic 129)
- Raoult's Law basics

Current Topic Connections:
- Colligative properties (umbrella topic)
- Mole fraction and partial pressures
- Chemical equilibrium concepts

Next Topics to Study:
- Elevation of boiling point (Topic 137)
- Depression of freezing point (Topic 137)
- Osmosis and osmotic pressure (Topic 138)
- van't Hoff factor for electrolytes

Advanced Extensions:
- Thermodynamics of solutions
- Activity and activity coefficients
- Gibbs-Duhem equation
- Phase diagrams and vapor-liquid equilibria
- Fractional distillation principles
⚠️ Common Exam Traps
Common Exam Traps

1. Confusing Absolute and Relative Lowering:
Trap: Writing ΔP = χ_solute (wrong!)
Correct: ΔP/P₀ = χ_solute (relative lowering = χ_solute)

2. Wrong Mole Fraction Formula:
Trap: Writing χ_solute = n_solute/n_solvent
Correct: χ_solute = n_solute/(n_solute + n_solvent) [Denominator has both!]

3. Approximation Misuse:
Trap: Using χ_solute ≈ n_solute/n_solvent for concentrated solutions
Correct: Approximation only valid when n_solute << n_solvent (dilute solution)

4. Forgetting van't Hoff Factor:
Trap: Treating NaCl (electrolyte) same as glucose (non-electrolyte)
Correct: For NaCl: effective χ = i × χ_NaCl where i ≈ 2

5. Units of Vapour Pressure:
Trap: Mixing units: P₀ in mm Hg, P_s in atm
Correct: Use same units for P₀ and P_s (units cancel in relative lowering anyway)

6. Molar Mass Calculation Error:
Trap: Writing M₂ = (w₂/w₁) × M₁ × (P₀-P_s)/P₀
Correct: M₂ = (w₂ × M₁) / [(P₀-P_s)/P₀ × w₁] [reciprocal relationship]

7. Sign Convention:
Trap: Writing P_s - P₀ (positive number?)
Correct: P₀ - P_s (lowering is positive; P₀ > P_s always)

8. Volatile Solute Assumption:
Trap: Using formula for solutions where solute also evaporates
Correct: Raoult's Law (this form) only for non-volatile solutes

9. Percentage vs Decimal:
Trap: If relative lowering = 5%, using 5 instead of 0.05 in calculations
Correct: Convert percentage to decimal: 5% = 0.05

10. Mole Fraction > 1:
Trap: Calculation giving χ_solute > 1 and not noticing error
Correct: χ must be between 0 and 1. If χ > 1, check your calculation

11. Solvent vs Solute Confusion:
Trap: Using χ_solvent in formula (P₀-P_s)/P₀ = χ_solvent
Correct: (P₀-P_s)/P₀ = χ_solute (not solvent!)

12. Multiple Solutes:
Trap: Calculating χ for each solute separately and adding
Correct: Calculate total solute moles first: n_total_solute = n₁ + n₂ + ..., then χ_total = n_total/(n_total + n_solvent)
⭐ Key Takeaways
Key Takeaways

- Vapour pressure of solution P_s < vapour pressure of pure solvent P₀ when non-volatile solute is added
- Relative lowering of vapour pressure = (P₀ - P_s)/P₀ = ΔP/P₀
- Raoult's Law: (P₀ - P_s)/P₀ = χ_solute (mole fraction of solute)
- Alternatively: P_s = P₀ × χ_solvent
- Relative lowering depends only on mole fraction, not on nature of solute (colligative property)
- For dilute solutions: χ_solute ≈ n_solute/n_solvent (denominator simplification)
- Formula: (P₀ - P_s)/P₀ = (w₂/M₂)/(w₁/M₁) for dilute solutions
- Solute must be non-volatile (doesn't contribute to vapour pressure itself)
- For electrolytes: multiply χ_solute by van't Hoff factor i
- Relative lowering is dimensionless (ratio of two pressures)
- Used to determine molar mass of unknown non-volatile solutes
- Foundation for understanding other colligative properties (BP elevation, FP depression, osmotic pressure)
🧩 Problem Solving Approach
Problem-Solving Approach

Algorithm:

Step 1: Read and Identify
- What's given: masses, molar masses, vapour pressures?
- What to find: relative lowering, molar mass, vapour pressure, mole fraction?
- Is solute an electrolyte (need i factor)?

Step 2: Calculate Moles
- n_solvent = w_solvent / M_solvent
- n_solute = w_solute / M_solute (if M known)
- Total moles = n_solvent + n_solute

Step 3: Calculate Mole Fractions
- χ_solute = n_solute / (n_solvent + n_solute)
- χ_solvent = n_solvent / (n_solvent + n_solute) = 1 - χ_solute
- For dilute solutions: χ_solute ≈ n_solute / n_solvent

Step 4: Apply Raoult's Law
- (P₀ - P_s)/P₀ = χ_solute
- Or: P_s = P₀(1 - χ_solute) = P₀χ_solvent

Step 5: Solve for Unknown
- Rearrange equation as needed
- Substitute known values
- Calculate final answer

Step 6: Check Units and Reasonableness
- Relative lowering is dimensionless (0 < value < 1)
- P_s should be less than P₀
- Molar mass should be positive and reasonable

Worked Example:

Problem: 12 g of urea (NH₂CONH₂) is dissolved in 90 g of water at 25°C. Calculate:
(a) Relative lowering of vapour pressure
(b) Vapour pressure of solution if P₀(water) = 23.8 mm Hg

Solution:

Given:
- Mass of urea (w₂) = 12 g
- Molar mass of urea (M₂) = 60 g/mol
- Mass of water (w₁) = 90 g
- Molar mass of water (M₁) = 18 g/mol
- P₀ = 23.8 mm Hg

Step 1: Calculate moles
n_urea = 12/60 = 0.2 mol
n_water = 90/18 = 5 mol
Total moles = 0.2 + 5 = 5.2 mol

Step 2: Calculate mole fraction
χ_urea = 0.2/5.2 = 0.0385

Step 3: Apply Raoult's Law

(a) Relative lowering:
(P₀ - P_s)/P₀ = χ_urea = 0.0385 or 3.85%

(b) Vapour pressure of solution:
P_s = P₀(1 - χ_urea)
= 23.8(1 - 0.0385)
= 23.8 × 0.9615
= 22.88 mm Hg

Verification:
P_s < P₀ ✓
Lowering = 23.8 - 22.88 = 0.92 mm Hg
Relative lowering = 0.92/23.8 = 0.0387 ≈ 0.0385 ✓
📝 CBSE Focus Areas
CBSE Focus Areas

1. Raoult's Law Statement (2 marks):
- State Raoult's law for solutions containing non-volatile solute
- Write mathematical expression: (P₀-P_s)/P₀ = χ_solute
- Command words: "State", "Write the expression for"

2. Numerical Problems - Standard Type (3 marks):
- Calculate relative lowering given masses and molar masses
- Find vapour pressure of solution
- Calculate mole fractions
- Command words: "Calculate", "Find", "Determine"

3. Molar Mass Determination (3-4 marks):
- Given relative lowering, find molar mass of unknown solute
- Very common board exam question
- Command words: "Calculate the molar mass", "Determine molecular mass"

4. Conceptual Questions (2-3 marks):
- Why does vapour pressure decrease on adding non-volatile solute?
- Why is it called a colligative property?
- Define relative lowering of vapour pressure
- Command words: "Explain", "Why", "Define"

5. Comparison Questions (2 marks):
- Compare lowering for different solutions
- Which has higher vapour pressure?
- Command words: "Compare", "Which solution has"

6. Typical Problem Structures:
- "When w g of solute (M = M₂) is dissolved in W g of solvent (M = M₁), relative lowering is found to be x. Calculate..."
- "Vapour pressure of pure water is P₀. When y g of glucose is added to z g water, find vapour pressure of solution."

7. Important Formulas to Write:
- (P₀ - P_s)/P₀ = χ_solute = n₂/(n₁ + n₂)
- For dilute: ≈ n₂/n₁ = (w₂/M₂)/(w₁/M₁)

8. Presentation Tips:
- Always write "Given" and "To find" clearly
- Show mole calculations step by step
- Write formula before substitution
- Box final answer with correct units (or mention "dimensionless")
🎓 JEE Focus Areas
JEE Focus Areas

1. Advanced Numerical Problems:
- Multi-step calculations involving conversions
- Problems with multiple solutes (additive effect)
- Reverse calculations: finding required mass for target lowering
- Combination with other colligative properties

2. Conceptual Depth:
- Molecular-level explanation of vapour pressure lowering
- Thermodynamic derivation of Raoult's Law
- Ideal vs non-ideal solutions: positive and negative deviations
- Why colligative properties depend only on particle number

3. Electrolytes and van't Hoff Factor:
- Calculating effective mole fraction for dissociating solutes
- Degree of dissociation and ionization
- Comparing i values for different electrolytes
- Weak electrolytes: partial dissociation

4. Mixture Problems:
- Two or more non-volatile solutes in same solvent
- Additive nature of mole fractions
- Finding composition of mixture from colligative data

5. Graph-Based Questions:
- P_s vs χ_solute plot (straight line through origin with slope P₀)
- Deviation plots for non-ideal solutions
- Interpreting slopes and intercepts

6. Integration with Other Topics:
- Connection to elevation of BP and depression of FP
- Osmotic pressure problems using same mole fraction concepts
- Phase diagrams and vapor-liquid equilibrium
- Chemical equilibrium: partial pressures

7. Limiting Cases:
- Very dilute solutions: χ_solute ≈ n_solute/n_solvent
- Concentrated solutions: no approximation possible
- Binary solutions with two volatile components (Raoult's Law for both)

8. Tricky Scenarios:
- Solute that partially evaporates (volatile)
- Solute that associates or dissociates
- Temperature dependence of vapour pressure
- Solutions showing maximum or minimum VP

9. Problem-Solving Strategies:
- Identifying when to use exact formula vs approximation
- Dimensional analysis to verify formulas
- Using ratios to simplify calculations
- Recognizing patterns in colligative property problems

10. Typical JEE Question Types:
- Single Correct MCQ: Numerical calculation with specific answer
- Multiple Correct MCQ: Comparing different solutions or effects
- Integer Type: Ratio of quantities (requires exact value)
- Assertion-Reason: Conceptual understanding
- Numerical (JEE Main): Direct application with 2-3 steps
- Subjective (JEE Advanced): Multi-concept integration, derivations

11. Common Tricky Elements:
- Given vapour pressure as lowering (ΔP) instead of P_s
- Multiple solvents or multiple solutes
- Mass given in different units (g, kg, mg)
- Molar mass to be calculated from given relative lowering

12. Time Management:
- Standard calculation: 2-3 minutes
- Molar mass problem: 3-4 minutes
- Conceptual MCQ: 1-2 minutes
- Multi-step integration: 5-7 minutes

📊 AI Generation Metadata

Difficulty: Intermediate
Estimated Time: 220 mins
AI Model: Claude Sonnet 4.5 (Copilot Agent)
Status: AI Generated
Version: 1
Generated: Oct 20, 2025

📝CBSE 12th Board Problems (13)

Problem 255
Easy 3 Marks
The vapor pressure of pure benzene at a certain temperature is 0.850 bar. A non-volatile, non-electrolyte solid weighing 0.5 g when added to 39.0 g of benzene (Molar mass of benzene = 78 g/mol). The vapor pressure of the solution is 0.845 bar. What is the molar mass of the solid substance?
Show Solution
1. Calculate moles of solvent: n_solvent = w_solvent / M_solvent. 2. Calculate the relative lowering of vapor pressure: (P° - P_solution) / P°. 3. Apply Raoult's Law for relative lowering of vapor pressure: (P° - P) / P° = X_solute = n_solute / (n_solute + n_solvent). 4. For dilute solutions, approximate: (P° - P) / P° ≈ n_solute / n_solvent. 5. Substitute n_solute = w_solute / M_solute and solve for M_solute.
Final Answer: 170 g/mol
Problem 255
Easy 3 Marks
At 298 K, the vapor pressure of water is 23.75 mm Hg. Calculate the vapor pressure of a solution containing 5 g of urea (NH₂CONH₂) dissolved in 100 g of water. Molar mass of urea = 60 g/mol, Molar mass of water = 18 g/mol.
Show Solution
1. Calculate moles of urea (solute): n_solute = w_solute / M_solute. 2. Calculate moles of water (solvent): n_solvent = w_solvent / M_solvent. 3. Calculate mole fraction of solute (X_solute) using X_solute = n_solute / (n_solute + n_solvent). 4. Apply Raoult's Law for relative lowering of vapor pressure: (P° - P) / P° = X_solute. 5. Solve for P_solution.
Final Answer: 23.40 mm Hg
Problem 255
Easy 2 Marks
The vapor pressure of pure water at 25°C is 30.0 mm Hg. The vapor pressure of a solution containing a non-volatile non-electrolyte solute in 108 g of water is 29.0 mm Hg. Calculate the number of moles of solute present in the solution. (Molar mass of water = 18 g/mol)
Show Solution
1. Calculate moles of water (solvent): n_solvent = w_solvent / M_solvent. 2. Calculate the relative lowering of vapor pressure: (P° - P_solution) / P°. 3. Apply Raoult's Law: (P° - P) / P° = n_solute / (n_solute + n_solvent). 4. Substitute known values and solve for n_solute.
Final Answer: 0.207 moles
Problem 255
Easy 2 Marks
The vapor pressure of pure liquid 'A' is 0.70 bar. A non-volatile solute 'B' is added to 'A' such that its mole fraction is 0.2. Calculate the vapor pressure of the solution.
Show Solution
1. Determine the mole fraction of the solvent (X_A) using X_A + X_B = 1. 2. Apply Raoult's Law for the vapor pressure of a solution containing a non-volatile solute: P_solution = P°_A * X_A.
Final Answer: 0.56 bar
Problem 255
Easy 3 Marks
Calculate the mass of a non-volatile solute (molar mass 40 g/mol) which should be dissolved in 114 g of octane to reduce its vapor pressure to 80% of its pure state.
Show Solution
1. Calculate moles of solvent (octane): n_solvent = w_solvent / M_solvent. 2. Determine the relative lowering of vapor pressure from the given percentage reduction. 3. Apply Raoult's Law: (P° - P) / P° = n_solute / (n_solute + n_solvent). 4. Substitute n_solute = w_solute / M_solute and solve for w_solute.
Final Answer: 10 g
Problem 255
Easy 2 Marks
An aqueous solution of a non-volatile solute has a vapor pressure of 750 mm Hg at 373 K. If the vapor pressure of pure water at 373 K is 760 mm Hg, calculate the mole fraction of the solute.
Show Solution
1. Identify the given vapor pressure of the solution (P_solution) and pure solvent (P°). 2. Apply Raoult's Law for relative lowering of vapor pressure directly: (P° - P) / P° = X_solute.
Final Answer: 0.0132
Problem 255
Hard 4 Marks
The vapour pressure of pure benzene at a certain temperature is 640 mmHg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 630 mmHg. Calculate the molar mass of the solid substance.
Show Solution
Calculate the number of moles of solvent (benzene): n₁ = w₁ / M₁ = 39.0 g / 78 g/mol = 0.5 mol. Apply Raoult's Law for relative lowering of vapour pressure: (P° - P_s) / P° = n₂ / (n₁ + n₂). Substitute the given values: (640 - 630) / 640 = (w₂/M₂) / (0.5 + w₂/M₂). Simplify the left side: 10 / 640 = 1 / 64. Let n₂ = w₂/M₂. So, 1/64 = n₂ / (0.5 + n₂). Cross-multiply: 0.5 + n₂ = 64 n₂. Rearrange to solve for n₂: 0.5 = 63 n₂. Calculate n₂: n₂ = 0.5 / 63 ≈ 0.007936 mol. Calculate M₂: M₂ = w₂ / n₂ = 2.175 g / 0.007936 mol. Final calculation: M₂ ≈ 274.06 g/mol.
Final Answer: The molar mass of the solid substance is approximately 274.06 g/mol.
Problem 255
Hard 3 Marks
An aqueous solution contains 10% by mass of urea (NH₂CONH₂). The vapour pressure of pure water at a certain temperature is 100 mmHg. Calculate the vapour pressure of the solution at the same temperature. (Molar mass of urea = 60 g/mol, Molar mass of water = 18 g/mol)
Show Solution
Assume 100 g of solution. Mass of urea (w₂) = 10 g; Mass of water (w₁) = 100 - 10 = 90 g. Calculate moles of urea (n₂): n₂ = w₂ / M₂ = 10 g / 60 g/mol = 0.1667 mol. Calculate moles of water (n₁): n₁ = w₁ / M₁ = 90 g / 18 g/mol = 5.0 mol. Calculate mole fraction of solute (X₂): X₂ = n₂ / (n₁ + n₂) = 0.1667 / (5.0 + 0.1667) = 0.1667 / 5.1667 ≈ 0.03226. Apply Raoult's Law for relative lowering of vapour pressure: (P° - P_s) / P° = X₂. Substitute values: (100 - P_s) / 100 = 0.03226. Solve for P_s: 100 - P_s = 100 * 0.03226 = 3.226. P_s = 100 - 3.226 = 96.774 mmHg.
Final Answer: The vapour pressure of the solution is approximately 96.77 mmHg.
Problem 255
Hard 3 Marks
What mass of a non-volatile solute (molar mass 60 g/mol) should be dissolved in 114 g of octane (C₈H₁₈) to reduce its vapour pressure to 80% of its pure value? (Molar mass of octane = 114 g/mol)
Show Solution
From the given information, P_s = 0.8 P°. This means (P° - P_s) / P° = (P° - 0.8 P°) / P° = 0.2. Calculate moles of solvent (octane): n₁ = w₁ / M₁ = 114 g / 114 g/mol = 1 mol. Apply Raoult's Law: (P° - P_s) / P° = n₂ / (n₁ + n₂). Substitute known values: 0.2 = (w₂/60) / (1 + w₂/60). Let x = w₂/60. So, 0.2 = x / (1 + x). Solve for x: 0.2 (1 + x) = x. 0.2 + 0.2x = x. 0.2 = x - 0.2x = 0.8x. x = 0.2 / 0.8 = 0.25 mol. Calculate mass of solute (w₂): w₂ = x * M₂ = 0.25 mol * 60 g/mol = 15 g.
Final Answer: The mass of the non-volatile solute required is 15 g.
Problem 255
Hard 4 Marks
The vapour pressure of pure water at 298 K is 23.75 mmHg. 10 g of a non-volatile organic substance is dissolved in 100 g of water. The vapour pressure of the solution is 23.37 mmHg. Calculate the molar mass of the organic substance.
Show Solution
Calculate the number of moles of solvent (water): n₁ = w₁ / M₁ = 100 g / 18 g/mol ≈ 5.5556 mol. Calculate the relative lowering of vapour pressure: (P° - P_s) / P° = (23.75 - 23.37) / 23.75 = 0.38 / 23.75 ≈ 0.016. Apply Raoult's Law for dilute solutions: (P° - P_s) / P° = n₂ / n₁ (approximation for dilute solutions). Substitute values: 0.016 = n₂ / 5.5556. Solve for n₂: n₂ = 0.016 * 5.5556 ≈ 0.08889 mol. Calculate molar mass of solute (M₂): M₂ = w₂ / n₂ = 10 g / 0.08889 mol ≈ 112.5 g/mol. For a more precise calculation using (P° - P_s) / P° = n₂ / (n₁ + n₂): 0.016 = (10/M₂) / (5.5556 + 10/M₂). 0.016 * (5.5556 + 10/M₂) = 10/M₂. 0.08889 + 0.16/M₂ = 10/M₂. 0.08889 = (10 - 0.16) / M₂ = 9.84 / M₂. M₂ = 9.84 / 0.08889 ≈ 110.69 g/mol. (Using the exact formula is better for 'hard' difficulty). Let's stick to the exact one for hard problems as it demonstrates thorough understanding.
Final Answer: The molar mass of the organic substance is approximately 110.69 g/mol.
Problem 255
Hard 3 Marks
At 300 K, 30 g of glucose (C₆H₁₂O₆) in 1000 g of water causes a lowering of vapour pressure by 0.018 mmHg. Calculate the vapour pressure of pure water at 300 K. (Molar mass of glucose = 180 g/mol, Molar mass of water = 18 g/mol)
Show Solution
Calculate moles of glucose (n₂): n₂ = w₂ / M₂ = 30 g / 180 g/mol = 0.1667 mol. Calculate moles of water (n₁): n₁ = w₁ / M₁ = 1000 g / 18 g/mol ≈ 55.5556 mol. Apply Raoult's Law for relative lowering of vapour pressure: (P° - P_s) / P° = n₂ / (n₁ + n₂). We are given (P° - P_s) = 0.018 mmHg. So, 0.018 / P° = n₂ / (n₁ + n₂). Calculate the mole fraction of solute: n₂ / (n₁ + n₂) = 0.1667 / (55.5556 + 0.1667) = 0.1667 / 55.7223 ≈ 0.002991. Substitute this into the equation: 0.018 / P° = 0.002991. Solve for P°: P° = 0.018 / 0.002991 ≈ 6.018 mmHg.
Final Answer: The vapour pressure of pure water at 300 K is approximately 6.02 mmHg.
Problem 255
Hard 4 Marks
A 5% (w/w) aqueous solution of a non-volatile solute has a vapour pressure of 23.37 mmHg at 298 K. The vapour pressure of pure water at 298 K is 23.75 mmHg. Calculate the molar mass of the solute.
Show Solution
Assume 100 g of solution. Mass of solute (w₂) = 5 g; Mass of water (w₁) = 100 - 5 = 95 g. Calculate moles of water (n₁): n₁ = w₁ / M₁ = 95 g / 18 g/mol ≈ 5.2778 mol. Calculate the relative lowering of vapour pressure: (P° - P_s) / P° = (23.75 - 23.37) / 23.75 = 0.38 / 23.75 ≈ 0.016. Apply Raoult's Law: (P° - P_s) / P° = n₂ / (n₁ + n₂). Substitute values: 0.016 = (5/M₂) / (5.2778 + 5/M₂). Let x = 5/M₂. So, 0.016 = x / (5.2778 + x). Solve for x: 0.016 * (5.2778 + x) = x. 0.0844448 + 0.016x = x. 0.0844448 = x - 0.016x = 0.984x. x = 0.0844448 / 0.984 ≈ 0.085818 mol. Calculate M₂: M₂ = w₂ / x = 5 g / 0.085818 mol ≈ 58.26 g/mol.
Final Answer: The molar mass of the solute is approximately 58.26 g/mol.
Problem 255
Hard 5 Marks
An aqueous solution of a non-volatile electrolyte (which dissociates completely) contains 1.17 g of NaCl in 100 g of water. If the vapour pressure of pure water at the given temperature is 20 mmHg, calculate the vapour pressure of the solution. (Molar mass of NaCl = 58.5 g/mol, Molar mass of water = 18 g/mol)
Show Solution
Calculate moles of NaCl (n₂): n₂ = w₂ / M₂ = 1.17 g / 58.5 g/mol = 0.02 mol. Calculate moles of water (n₁): n₁ = w₁ / M₁ = 100 g / 18 g/mol ≈ 5.5556 mol. Determine the van't Hoff factor (i) for NaCl: NaCl → Na⁺ + Cl⁻. Since it dissociates completely, i = 2. Apply Raoult's Law for electrolytes: (P° - P_s) / P° = i * n₂ / (n₁ + i * n₂). Substitute values: (20 - P_s) / 20 = (2 * 0.02) / (5.5556 + 2 * 0.02). Simplify the right side: (2 * 0.02) / (5.5556 + 0.04) = 0.04 / 5.5956 ≈ 0.007149. So, (20 - P_s) / 20 = 0.007149. Solve for P_s: 20 - P_s = 20 * 0.007149 = 0.14298. P_s = 20 - 0.14298 = 19.85702 mmHg.
Final Answer: The vapour pressure of the solution is approximately 19.86 mmHg.

🎯IIT-JEE Main Problems (6)

Problem 255
Easy 4 Marks
The vapour pressure of pure benzene at a certain temperature is 640 mm of Hg. A non-volatile, non-electrolyte solid weighing 2.175 g is added to 39.0 g of benzene. The vapour pressure of the solution is 600 mm of Hg. What is the molar mass of the solid substance? (Molar mass of benzene = 78 g/mol)
Show Solution
1. Calculate moles of solvent (benzene). 2. Apply Raoult's Law for Relative Lowering of Vapour Pressure (RLVP): (P⁰ - Pₛ) / P⁰ = X₂ = n₂ / (n₁ + n₂). 3. Substitute n₂ = w₂ / M₂ and solve for M₂.
Final Answer: 65.25 g/mol
Problem 255
Easy 4 Marks
The vapour pressure of pure water at 25°C is 23.76 mm Hg. Calculate the vapour pressure of a solution containing 5 g of urea (NH₂CONH₂) dissolved in 90 g of water. (Molar mass of urea = 60 g/mol, Molar mass of water = 18 g/mol)
Show Solution
1. Calculate moles of solute (urea) and solvent (water). 2. Calculate the mole fraction of solute (X₂). 3. Apply Raoult's Law: (P⁰ - Pₛ) / P⁰ = X₂. 4. Solve for Pₛ.
Final Answer: 23.37 mm Hg
Problem 255
Easy 4 Marks
At 300 K, the vapour pressure of pure liquid A is 0.50 atm. A non-volatile solute B is added to A. The vapour pressure of the solution becomes 0.40 atm. What is the mole fraction of solute B in the solution?
Show Solution
1. Apply Raoult's Law for Relative Lowering of Vapour Pressure (RLVP): (P⁰ - Pₛ) / P⁰ = X_B. 2. Substitute the given values and solve for X_B.
Final Answer: 0.20
Problem 255
Easy 4 Marks
The vapour pressure of pure water at a given temperature is 50 mm Hg. What mass of glucose (C₆H₁₂O₆) should be added to 180 g of water to lower its vapour pressure by 1 mm Hg? (Molar mass of glucose = 180 g/mol, Molar mass of water = 18 g/mol)
Show Solution
1. Calculate the relative lowering of vapour pressure. 2. This equals the mole fraction of solute (X₂). 3. Calculate moles of solvent (n₁). 4. Use X₂ = n₂ / (n₁ + n₂) to find moles of solute (n₂). 5. Calculate mass of solute (w₂) from n₂.
Final Answer: 36.72 g
Problem 255
Easy 4 Marks
The vapour pressure of pure liquid X is 100 mm Hg at a certain temperature. If the mole fraction of a non-volatile solute in a solution prepared with liquid X is 0.1, what is the vapour pressure of the solution?
Show Solution
1. Apply Raoult's Law for Relative Lowering of Vapour Pressure (RLVP): (P⁰ - Pₛ) / P⁰ = X_solute. 2. Substitute the given values and solve for Pₛ.
Final Answer: 90 mm Hg
Problem 255
Easy 4 Marks
A solution is prepared by dissolving 1.5 moles of a non-volatile solute in 8.5 moles of a solvent. Calculate the relative lowering of vapour pressure for this solution.
Show Solution
1. Calculate the total moles in the solution. 2. Calculate the mole fraction of the solute (X_solute). 3. The relative lowering of vapour pressure is equal to X_solute as per Raoult's Law.
Final Answer: 0.15

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📐Important Formulas (6)

Raoult's Law for Solution Vapour Pressure (Non-Volatile Solute)
P_{solution} = X_{solvent} imes P_{solvent}^0
Text: P_solution = X_solvent * P_solvent^0
This fundamental law states that the vapour pressure of a solution (P_solution) containing a <strong>non-volatile solute</strong> is directly proportional to the mole fraction of the solvent (X_solvent). <span style='color: #1E90FF;'>P_solvent^0 is the vapour pressure of the pure solvent.</span>
Variables: Use this formula to calculate the vapour pressure of a solution when the vapour pressure of the pure solvent and the mole fraction of the solvent in the solution are known.
Lowering of Vapour Pressure (ΔP)
Delta P = P_{solvent}^0 - P_{solution}
Text: ΔP = P_solvent^0 - P_solution
This formula calculates the <strong>absolute decrease</strong> in vapour pressure observed when a non-volatile solute is added to a pure solvent. <span style='color: #FF0000;'>It's not a colligative property on its own, but its relative form is.</span>
Variables: To quantify the extent of reduction in vapour pressure of a solvent after adding a non-volatile solute.
Relative Lowering of Vapour Pressure (RLVP) - Definition
frac{Delta P}{P_{solvent}^0} = frac{P_{solvent}^0 - P_{solution}}{P_{solvent}^0}
Text: (P_solvent^0 - P_solution) / P_solvent^0
This is the definition of <strong>Relative Lowering of Vapour Pressure</strong>, which is one of the four colligative properties. It is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent. Being a ratio of same units, it is <strong>dimensionless</strong>.
Variables: Use this to define or calculate the relative lowering given the pure solvent's vapour pressure and the solution's vapour pressure.
Raoult's Law - RLVP in terms of Solute Mole Fraction
frac{P_{solvent}^0 - P_{solution}}{P_{solvent}^0} = X_{solute}
Text: (P_solvent^0 - P_solution) / P_solvent^0 = X_solute
This is the most important form of Raoult's Law for non-volatile solutes, stating that the <strong>relative lowering of vapour pressure is equal to the mole fraction of the solute (X_solute)</strong> in an ideal solution. Here, X_solute = n_solute / (n_solute + n_solvent).
Variables: This is the primary formula for solving problems involving RLVP, especially when direct calculation of the solute's mole fraction or molar mass is required.
RLVP for Dilute Solutions (Molar Mass Determination)
frac{P_{solvent}^0 - P_{solution}}{P_{solvent}^0} approx frac{n_{solute}}{n_{solvent}} = frac{w_{solute}/M_{solute}}{w_{solvent}/M_{solvent}} = frac{w_{solute} imes M_{solvent}}{M_{solute} imes w_{solvent}}
Text: (P_solvent^0 - P_solution) / P_solvent^0 ≈ (w_solute * M_solvent) / (M_solute * w_solvent)
For <strong>very dilute solutions</strong>, the number of moles of solute (n_solute) is negligible compared to the number of moles of solvent (n_solvent). Thus, X_solute ≈ n_solute/n_solvent. This approximated form is commonly used to determine the <strong>molar mass (M_solute) of an unknown non-volatile solute</strong>. <span style='color: #FF0000;'>Caution: For concentrated solutions, use the exact formula: X_solute = n_solute / (n_solute + n_solvent).</span>
Variables: Primarily used in JEE Main/Advanced and CBSE for calculating the molar mass of an unknown solute, especially when masses of solute and solvent are given.
RLVP with Van't Hoff Factor (for Electrolytes/Association)
frac{P_{solvent}^0 - P_{solution}}{P_{solvent}^0} = i imes X_{solute}
Text: (P_solvent^0 - P_solution) / P_solvent^0 = i * X_solute
For solutions of <strong>electrolytes</strong> (which dissociate into ions) or solutes that undergo <strong>association</strong>, the colligative properties are affected. The <strong>Van't Hoff factor (i)</strong> accounts for the actual number of particles in solution. 'i' > 1 for dissociation, 'i' < 1 for association, and 'i' = 1 for non-electrolytes.
Variables: This modified formula is essential when dealing with ionic compounds (e.g., NaCl, CaCl2) or organic solutes that show association (e.g., carboxylic acids in non-polar solvents).

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⚠️Common Mistakes to Avoid (56)

Minor Other

❌ Misinterpreting the 'Relative' in Relative Lowering of Vapour Pressure

Students often incorrectly assume that 'relative lowering of vapour pressure' refers to the lowering of vapour pressure (ΔP) divided by the final vapour pressure of the solution (P_s), instead of the vapour pressure of the pure solvent (P₀). This leads to fundamental errors in applying Raoult's Law.
💭 Why This Happens:
This mistake stems from a lack of clarity on the reference state for the term 'relative'. Students might oversimplify or memorize the formula (ΔP/P₀) without fully understanding that P₀ specifically denotes the vapour pressure of the pure solvent. Confusion between 'lowering of vapour pressure' (ΔP) and 'relative lowering of vapour pressure' (ΔP/P₀) is a common root cause.
✅ Correct Approach:
The term 'relative lowering' specifically means the lowering of vapour pressure (ΔP) divided by the vapour pressure of the pure solvent (P₀). According to Raoult's Law for ideal solutions with a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute (xsolute): (P₀ - Ps) / P₀ = xsolute. Here, P₀ is the vapour pressure of the pure solvent and Ps is the vapour pressure of the solution. This ratio is a colligative property, dependent only on the number of solute particles.
📝 Examples:
❌ Wrong:
If the vapour pressure of pure water (P₀) is 100 mmHg and that of an aqueous solution (Ps) is 90 mmHg, the lowering of vapour pressure (ΔP) is 10 mmHg. A common mistake would be to calculate the 'relative lowering' as ΔP / Ps = 10 / 90 ≈ 0.111.
✅ Correct:
Using the same values (P₀ = 100 mmHg, Ps = 90 mmHg), the correct relative lowering of vapour pressure is calculated as ΔP / P₀ = (P₀ - Ps) / P₀ = (100 - 90) / 100 = 10 / 100 = 0.1. This value (0.1) is then equal to the mole fraction of the solute in the solution.
💡 Prevention Tips:
  • Tip 1: Understand Definitions: Clearly differentiate between 'lowering of vapour pressure' (ΔP) and 'relative lowering of vapour pressure' (ΔP/P₀). The denominator is crucial.
  • Tip 2: Master Raoult's Law: Always recall the exact form of Raoult's Law for non-volatile solutes: (P₀ - Ps) / P₀ = xsolute. Pay attention to the P₀ in the denominator.
  • Tip 3: Conceptual Clarity: Remember that RLVP is a colligative property because it depends on the mole fraction of the solute, which is a measure of the number of solute particles.
JEE_Advanced
Minor Conceptual

❌ Confusing Lowering of Vapour Pressure with Relative Lowering

Students often incorrectly interchange the terms 'lowering of vapour pressure' (ΔP) and 'relative lowering of vapour pressure' (RLVP = ΔP/P₀). This leads to fundamental errors in applying Raoult's Law for non-volatile solutes and understanding colligative properties.

  • Many assume that ΔP itself is a colligative property, which is incorrect.
  • They might directly equate ΔP with the mole fraction of the solute (Xsolute), misapplying the core formula.

JEE Main Relevance: This confusion can lead to incorrect calculations, especially when questions test the definitions and dependencies of colligative properties.

💭 Why This Happens:

This mistake primarily stems from a lack of precise conceptual understanding and not distinguishing between the absolute change and the relative change. Students often:

  • Overlook Definitions: Fail to internalize the exact definitions of ΔP (P₀ - Ps) and RLVP ((P₀ - Ps)/P₀).

  • Formula Memorization without Understanding: Simply memorize Raoult's Law (RLVP = Xsolute) without grasping what 'relative' signifies, leading to an incorrect substitution of ΔP for RLVP.

  • Misconception of Colligative Properties: Incorrectly assume that any lowering of vapour pressure (ΔP) is a colligative property, while only the relative lowering (RLVP) is truly independent of the solvent's specific nature (P₀) for a given mole fraction of solute, making it a colligative property.

✅ Correct Approach:

The key is to understand the distinct nature of both terms:

  • Lowering of Vapour Pressure (ΔP): This is the absolute decrease in vapour pressure, calculated as ΔP = P₀ - Ps (where P₀ is the vapour pressure of pure solvent and Ps is the vapour pressure of the solution). Important: ΔP is NOT a colligative property as its value depends on the initial vapour pressure of the pure solvent (P₀), which varies with the solvent.

  • Relative Lowering of Vapour Pressure (RLVP): This is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent. RLVP = ΔP/P₀ = (P₀ - Ps)/P₀. According to Raoult's Law, for dilute solutions of non-volatile solutes, RLVP is equal to the mole fraction of the solute (Xsolute). Important: RLVP IS a colligative property because its value depends only on the number of solute particles (represented by Xsolute) and is independent of the nature of the solvent.

📝 Examples:
❌ Wrong:
A student incorrectly states:

"If 1 mole of a non-volatile solute is added to 10 moles of water, the lowering of vapour pressure (ΔP) will be equal to the mole fraction of the solute."

This is wrong because ΔP is not equal to Xsolute; it's the relative lowering (ΔP/P₀) that equals Xsolute.

✅ Correct:
A student correctly states:

"If 1 mole of a non-volatile solute is added to 10 moles of water, the relative lowering of vapour pressure (ΔP/P₀) will be equal to the mole fraction of the solute, which is 1/(1+10) = 1/11."

Therefore, ΔP = P₀ * Xsolute, meaning the actual lowering depends on the pure solvent's vapour pressure.

💡 Prevention Tips:
  • Distinguish Definitions: Always define ΔP and RLVP separately in your mind. ΔP is an absolute value, RLVP is a ratio.
  • Colligative Property Test: Remind yourself that a property is colligative only if it depends on the number of solute particles, not the nature of the solvent. Only RLVP passes this test for vapour pressure.
  • Formula Application Check: When using Raoult's Law, ensure you're applying ΔP/P₀ = Xsolute and not ΔP = Xsolute.
  • Practice with Varied Solvents: Solve problems where different solvents are used. This will highlight that ΔP changes with P₀, but RLVP remains the same for the same mole fraction of solute.
JEE_Main
Minor Calculation

❌ Incorrectly Using Mass Instead of Moles, or Confusing Solute/Solvent in Mole Fraction Calculation

Students frequently make calculation errors by directly substituting the masses of solute and solvent into the mole fraction formula ($X_{solute} = frac{n_{solute}}{n_{solute} + n_{solvent}}$) instead of first converting them into moles. Another common slip is interchanging the roles of solute and solvent, especially in the denominator where the sum of moles is required.
💭 Why This Happens:
This mistake primarily stems from exam pressure, leading to a hurried approach, or a fundamental misunderstanding of the definition of mole fraction. Sometimes, students might remember the overall formula for relative lowering of vapour pressure but overlook the crucial first step of converting mass to moles for each component.
✅ Correct Approach:
Always convert the given masses of both the non-volatile solute and the volatile solvent into their respective moles using their molar masses ($n = frac{ ext{mass}}{ ext{molar mass}}$). Only then should these mole values be used to calculate the mole fraction of the solute, which is equivalent to the relative lowering of vapour pressure for dilute solutions.
For JEE Main, accuracy in these initial conversion steps is paramount for correct final answers.
📝 Examples:
❌ Wrong:
Consider 18 g of glucose (Molar Mass = 180 g/mol) dissolved in 90 g of water (Molar Mass = 18 g/mol).
Incorrect Calculation:
Relative Lowering = $X_{glucose} = frac{ ext{mass}_{glucose}}{ ext{mass}_{glucose} + ext{mass}_{water}} = frac{18 ext{ g}}{18 ext{ g} + 90 ext{ g}} = frac{18}{108} = mathbf{0.1667}$
✅ Correct:
Using the same scenario: 18 g of glucose in 90 g of water.
Correct Calculation:
1. Calculate moles of glucose ($n_{glucose}$): $n_{glucose} = frac{18 ext{ g}}{180 ext{ g/mol}} = 0.1 ext{ mol}$
2. Calculate moles of water ($n_{water}$): $n_{water} = frac{90 ext{ g}}{18 ext{ g/mol}} = 5 ext{ mol}$
3. Relative Lowering = $X_{glucose} = frac{n_{glucose}}{n_{glucose} + n_{water}} = frac{0.1}{0.1 + 5} = frac{0.1}{5.1} = mathbf{0.0196}$
💡 Prevention Tips:
  • Double-Check Units: Always ensure all quantities are in consistent units and converted to moles before substitution.
  • Formula Awareness: Clearly recall the definition of mole fraction as 'moles of component / total moles'.
  • Labeling: Explicitly label $n_{solute}$ and $n_{solvent}$ during intermediate steps to avoid confusion.
  • Practice: Solve a variety of problems focusing on initial mass-to-mole conversions.
JEE_Main
Minor Formula

❌ Incorrect Denominator and Mole Fraction Usage in Raoult's Law for Relative Lowering of Vapour Pressure

Students often make two common errors: (1) using the vapour pressure of the solution (Ps) as the denominator instead of the vapour pressure of the pure solvent (Po) when calculating relative lowering, and (2) confusing the mole fraction of the solute (χsolute) with the mole fraction of the solvent (χsolvent) in the formula.
💭 Why This Happens:
This mistake stems from a fundamental misunderstanding of the term 'relative lowering'. Relative lowering is always with respect to the pure solvent's vapour pressure. Students might mistakenly think it's relative to the solution's vapour pressure. The confusion with mole fraction often arises from not remembering that Raoult's Law for a non-volatile solute directly links relative lowering to the mole fraction of the solute.
✅ Correct Approach:
The correct formula for relative lowering of vapour pressure (RLVP) for a non-volatile solute is:

(Po - Ps) / Po = χsolute

Where:
  • Po = Vapour pressure of the pure solvent.
  • Ps = Vapour pressure of the solution.
  • (Po - Ps) = Lowering of vapour pressure.
  • χsolute = Mole fraction of the solute (moles of solute / total moles).

For electrolytes, remember to include the van't Hoff factor (i):
(Po - Ps) / Po = i ⋅ χsolute
📝 Examples:
❌ Wrong:
A student might write: (Po - Ps) / Ps = χsolute
Or: (Po - Ps) / Po = χsolvent
✅ Correct:
Given Po (pure water) = 20 mmHg, Ps (solution) = 18 mmHg.
The correct calculation for relative lowering is:
(20 - 18) / 20 = 2 / 20 = 0.1
This 0.1 is equal to χsolute (or i ⋅ χsolute if it's an electrolyte).
💡 Prevention Tips:
  • Memorize the formula correctly: Focus on the pure solvent's vapour pressure (Po) in the denominator.
  • Understand the 'relative' term: It always implies comparison with the original or pure state.
  • Practice derivation: Knowing how Raoult's law is derived helps in understanding why it's mole fraction of solute.
  • Check for electrolytes: Always remember to incorporate the van't Hoff factor (i) for ionic solutes, especially in JEE problems.
JEE_Main
Minor Unit Conversion

❌ <strong>Inconsistent Units in Mole Fraction Calculation</strong>

Students frequently overlook unit consistency when calculating the number of moles of solute and solvent, which are crucial for determining the mole fraction in relative lowering of vapour pressure problems. This often occurs when masses, volumes, or densities are provided in non-standard or mixed units.
💭 Why This Happens:
This mistake typically arises from:
  • Haste: Rushing through problems without a careful unit check.
  • Assumption: Assuming all given values are in 'standard' units (e.g., grams for mass) without verification.
  • Lack of attention: Not realizing that molar masses are almost universally expressed in g/mol, requiring other mass units (like kg) to be converted.
✅ Correct Approach:
Always convert all quantities involved in mole calculations (mass, volume, density) to a consistent system of units before proceeding. For instance, if molar mass is in g/mol, ensure all masses are in grams. If density is in g/mL, ensure volume is in mL to get mass in grams.
📝 Examples:
❌ Wrong:

Problem: A solution contains 5 kg of a non-volatile solute (Molar mass = 100 g/mol) in 1000 g of solvent.
Incorrect Calculation: Moles of solute = 5 kg / 100 g/mol = 0.05 mol (Incorrect, units of mass and molar mass are inconsistent).

✅ Correct:

Problem: A solution contains 5 kg of a non-volatile solute (Molar mass = 100 g/mol) in 1000 g of solvent.
Correct Calculation: First, convert mass of solute to grams: 5 kg = 5000 g.
Moles of solute = 5000 g / 100 g/mol = 50 mol (Correct, units are consistent).

💡 Prevention Tips:
  • Initial Unit Check: Before any calculation, write down all given quantities along with their units.
  • Standardize Units: Decide on a consistent set of units (e.g., all masses in grams, volumes in mL) and convert all input data to these units.
  • Molar Mass Awareness: Remember that molar mass is almost always in g/mol. Any mass given in kg must be converted to grams.
  • Dimensional Analysis: Practice tracking units throughout your calculations. If units don't cancel out or combine correctly, it's a red flag.
JEE_Main
Minor Sign Error

❌ Sign Errors in Relative Lowering of Vapour Pressure Calculations

Students frequently make sign errors when dealing with the 'lowering' aspect of vapour pressure. This typically involves incorrectly calculating the change in vapour pressure (ΔP) or misinterpreting the direction of change, leading to incorrect values or signs for the relative lowering.
💭 Why This Happens:
This error stems from a fundamental misunderstanding of the term 'lowering'. While it implies a decrease, some students might subtract the pure solvent's vapour pressure from the solution's vapour pressure (Psolution - Ppure) instead of the correct order (Ppure - Psolution). This results in a negative value for a quantity that should inherently be positive (the *amount* of lowering). Another common mistake is confusing the denominator for relative lowering – it must always be the vapour pressure of the pure solvent (P₀), not the solution.
✅ Correct Approach:
Always remember that the addition of a non-volatile solute decreases the vapour pressure of the solvent. Therefore:
  • The lowering of vapour pressure (ΔP) is always calculated as: ΔP = P₀ - Psolution, where P₀ is the vapour pressure of the pure solvent and Psolution is the vapour pressure of the solution. ΔP must always be a positive value.
  • The relative lowering of vapour pressure (RLVP) is the ratio of the lowering of vapour pressure to the vapour pressure of the pure solvent: RLVP = (P₀ - Psolution) / P₀. This is equal to the mole fraction of the solute (Xsolute) in an ideal dilute solution (Raoult's Law).
📝 Examples:
❌ Wrong:

If pure water has P₀ = 100 mmHg and a solution has Psolution = 95 mmHg:

  • Incorrect Calculation of Lowering: ΔP = Psolution - P₀ = 95 - 100 = -5 mmHg. This is wrong; lowering cannot be negative.
  • Incorrect Calculation of Relative Lowering: RLVP = (P₀ - Psolution) / Psolution = (100 - 95) / 95 = 5/95 ≈ 0.0526. This is wrong; the denominator must be P₀.
✅ Correct:

Using the same values (P₀ = 100 mmHg, Psolution = 95 mmHg):

  • Correct Calculation of Lowering: ΔP = P₀ - Psolution = 100 - 95 = 5 mmHg. This is the correct positive value for the lowering.
  • Correct Calculation of Relative Lowering: RLVP = (P₀ - Psolution) / P₀ = (100 - 95) / 100 = 5/100 = 0.05. This correctly represents the relative lowering, equal to Xsolute.
💡 Prevention Tips:
  • Conceptual Clarity: Always visualize that the vapour pressure decreases upon adding a non-volatile solute.
  • Formula Check: Memorize and consistently use the correct formulas: ΔP = P₀ - Psolution and RLVP = (P₀ - Psolution) / P₀.
  • Positive Values: The 'lowering' and 'relative lowering' are scalar quantities and must always be positive. If your calculation yields a negative value, you've made a sign error.
  • JEE Main Relevance: These conceptual errors are easy to avoid with careful attention, ensuring you don't lose marks on straightforward questions.
JEE_Main
Minor Approximation

❌ Misapplying Dilute Solution Approximation in RLVN

Students frequently misuse or fail to identify the dilute solution approximation in Relative Lowering of Vapour Pressure (RLVN) problems. This involves incorrectly assuming nsolute + nsolvent ≈ nsolvent for solutions that are not sufficiently dilute, or conversely, not applying it when appropriate, leading to errors or inefficient calculations.
💭 Why This Happens:
  • Lack of Criteria Understanding: Not knowing the conditions (nsolute << nsolvent) for valid approximation.
  • Over-simplification: Tendency to always use the simpler form without verification.
  • Rushing: Failing to analyze the magnitude of moles before calculation.
✅ Correct Approach:

The exact RLVN formula is (P° - P) / P° = Xsolute = nsolute / (nsolute + nsolvent).

  • Exact for Concentrated Solutions: Always use the full denominator (nsolute + nsolvent) unless the solution is explicitly very dilute.
  • JEE Tip - Dilute Approximation: For very dilute solutions (where nsolvent ≥ 100 × nsolute), the denominator simplifies to nsolvent. Then, (P° - P) / P° ≈ nsolute / nsolvent. This is often expected for faster problem solving in JEE.
📝 Examples:
❌ Wrong:

For a solution with 1 mol solute and 10 mol solvent, approximating mole fraction of solute as:

Xsolute ≈ 1 / 10 = 0.1

This approximation is incorrect as 1 is not much smaller than 10 (nsolvent is only 10 times nsolute).

✅ Correct:

For the same solution (1 mol solute, 10 mol solvent), the correct mole fraction is:

Xsolute = 1 / (1 + 10) = 1 / 11 ≈ 0.0909

However, if it were 0.01 mol solute and 10 mol solvent, then the approximation 0.01/10 = 0.001 would be valid as nsolvent (10) is 1000 times nsolute (0.01). The exact value 0.01/10.01 ≈ 0.000999, is very close.

💡 Prevention Tips:
  • Quantify Dilution: Before approximating, compare nsolute and nsolvent. If nsolvent is at least 100 times nsolute, proceed with approximation.
  • Read Problem Context: Look for explicit mentions of 'dilute' or 'very dilute' in the question.
  • Practice: Work through diverse problems to build discernment for approximation scenarios.
JEE_Main
Minor Other

❌ Misinterpreting the Solute's Role in Relative Lowering of Vapour Pressure

Students often incorrectly assume that the relative lowering of vapour pressure (RLVP) depends on specific properties of the non-volatile solute, such as its molecular size, polarity, or the strength of its intermolecular forces. They might believe that a 'heavier' or 'more polar' solute will cause a greater RLVP for the same mole fraction.
💭 Why This Happens:
This mistake stems from an incomplete understanding of colligative properties. While the nature of the solvent dictates its pure vapour pressure (P°), the relative lowering of this pressure is solely dependent on the number of solute particles in the solution, not their chemical identity or physical attributes (as long as the solute is non-volatile and a non-electrolyte). Students sometimes over-extend the concept of intermolecular forces from pure liquids to solutions.
✅ Correct Approach:
Understand that RLVP is a colligative property. For a given solvent at a specific temperature, the relative lowering of vapour pressure depends only on the mole fraction of the non-volatile solute, irrespective of the solute's specific chemical nature (e.g., glucose vs. sucrose, as long as both are non-volatile and non-electrolytes). The crucial factor is the concentration of solute particles, not their individual characteristics.
📝 Examples:
❌ Wrong:
If 1 mole of glucose and 1 mole of urea are dissolved separately in 1 kg of water, the solution with urea will have a higher relative lowering of vapour pressure because urea is smaller and forms stronger hydrogen bonds. (Incorrect reasoning)
✅ Correct:
If 1 mole of glucose (180 g/mol) and 1 mole of urea (60 g/mol) are dissolved separately in 1 kg of water, both solutions will exhibit the same relative lowering of vapour pressure. This is because both are non-volatile, non-electrolytes, and their mole fractions in their respective solutions (with 1 kg water) will be identical for 1 mole of solute. RLVP depends only on the mole fraction of the solute. (Correct)
💡 Prevention Tips:
  • Reinforce Colligative Definition: Always remember that colligative properties depend only on the number of solute particles, not their identity.
  • Focus on Mole Fraction: When calculating or comparing RLVP, concentrate solely on the mole fraction (χsolute), as it's the direct measure of solute particle concentration.
  • CBSE vs. JEE: Both CBSE and JEE emphasize this fundamental aspect. For JEE, this understanding is critical for solving comparison-based problems and avoiding distractors.
CBSE_12th
Minor Approximation

❌ Incorrect Application of Approximation for Solute Mole Fraction in Dilute Solutions

Students often make the minor mistake of incorrectly assuming that for all solutions, the mole fraction of solute (Xsolute) can be approximated as the ratio of moles of solute to moles of solvent (nsolute / nsolvent). While this approximation is valid and simplifies calculations for very dilute solutions, applying it to moderately concentrated solutions will lead to inaccurate final results.
💭 Why This Happens:
This error stems from a lack of clear understanding regarding the conditions under which the approximation nsolute << nsolvent holds true. Students might over-simplify concepts, apply a learned shortcut without considering its general applicability, or not recognize when a solution is sufficiently dilute to warrant the approximation.
✅ Correct Approach:
Always use the exact formula for mole fraction: Xsolute = nsolute / (nsolute + nsolvent). Only employ the approximation Xsolute ≈ nsolute / nsolvent when explicitly dealing with very dilute solutions, where the moles of solute are negligibly small compared to the moles of solvent (typically when nsolute < 0.01 * nsolvent). For CBSE, problems generally state 'dilute solution' if the approximation is intended.
📝 Examples:
❌ Wrong:
Consider a solution with 1 mole of urea (solute) and 10 moles of water (solvent).
Incorrect approximation: Xurea ≈ 1 / 10 = 0.1.
✅ Correct:
For the same solution (1 mole urea, 10 moles water):
Correct Calculation: Xurea = 1 / (1 + 10) = 1 / 11 ≈ 0.0909.
The approximate value (0.1) differs significantly from the exact value (0.0909).

However, for a very dilute solution, e.g., 0.01 mole urea and 10 moles water:
Exact Xurea = 0.01 / (0.01 + 10) = 0.01 / 10.01 ≈ 0.000999.
Approximate Xurea ≈ 0.01 / 10 = 0.001.
In this dilute case, the approximation is very close and acceptable, demonstrating its utility.
💡 Prevention Tips:
  • Understand the Derivation: Remember that Relative Lowering of Vapour Pressure is exactly equal to the mole fraction of solute. The approximation occurs only in simplifying the mole fraction calculation.
  • Check for 'Dilute': In CBSE problems, if the term 'dilute solution' is explicitly used, the approximation is generally safe. Otherwise, use the exact formula.
  • Compare Magnitudes: Before applying the approximation (nsolute + nsolvent ≈ nsolvent), quickly compare the magnitudes of nsolute and nsolvent. If nsolute is less than 1% of nsolvent, the approximation is usually acceptable and will simplify calculations without significant loss of accuracy.
CBSE_12th
Minor Sign Error

❌ Sign Error in Calculating 'Lowering of Vapour Pressure'

Students frequently make a sign error by treating the 'lowering of vapour pressure' (ΔP) as a negative quantity (i.e., P - P°) instead of correctly defining it as the positive difference (P° - P), which represents the magnitude of the decrease.
💭 Why This Happens:
This error often arises from interpreting 'lowering' literally as a negative change, similar to 'decrease' in other contexts. Consequently, they subtract the initial (pure solvent) vapour pressure from the final (solution) vapour pressure, leading to a negative value. This propagates into the calculation of relative lowering, sometimes yielding illogical results like negative molar masses.
✅ Correct Approach:
Always define the lowering of vapour pressure (ΔP) as P° - P, where P° is the vapour pressure of the pure solvent and P is the vapour pressure of the solution. This ensures that ΔP is always a positive value, as the vapour pressure of a solution (P) is inherently lower than that of the pure solvent (P°). The formula for Relative Lowering of Vapour Pressure (RLVP) is thus correctly written as (P° - P) / P° = X_solute.
📝 Examples:
❌ Wrong:
If the vapour pressure of pure water (P°) is 50 mmHg and that of a solution (P) is 45 mmHg, a student might incorrectly calculate the lowering as P - P° = 45 - 50 = -5 mmHg. This negative value is then used in the numerator for relative lowering.
✅ Correct:
Given P° = 50 mmHg and P = 45 mmHg, the correct lowering of vapour pressure (ΔP) is P° - P = 50 - 45 = 5 mmHg. This positive value is correctly used in the formula for relative lowering: (P° - P) / P° = 5 / 50 = 0.1.
💡 Prevention Tips:
Always remember that 'lowering' refers to the magnitude of decrease, hence it must be a positive value.
Consistently use the formula ΔP = P° - P to ensure a positive value for the lowering.
For CBSE and JEE, understand that mole fraction (X_solute) and molar mass are always positive quantities, so calculations derived from them must also result in positive values.
Double-check your subtractions: Ensure the larger value (P°) is always the minuend when calculating the lowering.
CBSE_12th
Minor Unit Conversion

❌ Inconsistent Pressure Units in RLVP Calculations

Students often overlook or incorrectly convert pressure units (e.g., kPa, mmHg, atm, bar) when given different units for the vapour pressure of the pure solvent and the solution. The formula for relative lowering of vapour pressure, (P° - P) / P°, requires both P° (vapour pressure of pure solvent) and P (vapour pressure of solution) to be in the same units for the ratio to be dimensionless and correct. Failing to do so leads to incorrect numerical values for RLVP and subsequently, for molar mass calculations.
💭 Why This Happens:
This mistake primarily occurs due to:
  • Carelessness: Not carefully reading the units specified for each pressure value.
  • Lack of Unit Awareness: Assuming all given pressure values are already in consistent units.
  • Incorrect Conversion Factors: Using wrong conversion factors between different pressure units (e.g., confusing kPa and bar, or mmHg and atm).
  • Rushing: Hastily plugging values into the formula without a preliminary unit check.
✅ Correct Approach:
Always ensure that all pressure values used in the relative lowering of vapour pressure formula are expressed in the same unit before performing any calculations. If different units are provided, convert one or more values so that they are all consistent. Common conversions include:
  • 1 atm = 760 mmHg
  • 1 atm = 101.325 kPa
  • 1 bar = 100 kPa
  • 1 bar = 750.062 mmHg

Choose a convenient unit and convert all other values to it.
📝 Examples:
❌ Wrong:
Problem: Vapour pressure of pure water at 298 K is 23.8 mmHg. When a non-volatile solute is added, the vapour pressure of the solution becomes 3.1 kPa. Calculate the relative lowering of vapour pressure.
Wrong Calculation:
P° = 23.8 mmHg
P = 3.1 kPa
RLVP = (23.8 - 3.1) / 23.8 = 20.7 / 23.8 ≈ 0.87
This is incorrect as the units are mixed.
✅ Correct:
Problem: Vapour pressure of pure water at 298 K is 23.8 mmHg. When a non-volatile solute is added, the vapour pressure of the solution becomes 3.1 kPa. Calculate the relative lowering of vapour pressure.
Correct Calculation:
Convert P from kPa to mmHg:
1 atm = 101.325 kPa = 760 mmHg
So, 1 kPa = 760 / 101.325 mmHg ≈ 7.50 mmHg
P = 3.1 kPa * (7.50 mmHg / 1 kPa) = 23.25 mmHg

Now, both are in mmHg:
P° = 23.8 mmHg
P = 23.25 mmHg
RLVP = (P° - P) / P° = (23.8 - 23.25) / 23.8 = 0.55 / 23.8 ≈ 0.0231
This is the correct approach.
💡 Prevention Tips:
  • Double-Check Units: Before starting any calculation, always verify that all given pressure values have consistent units.
  • List Conversion Factors: Keep common pressure unit conversion factors handy and memorize them for the exam.
  • Highlight Units: In the exam, underline or circle the units of each given value to ensure they are addressed.
  • Unit Cancellation: Mentally or physically write out the units during conversion steps to ensure they cancel out correctly.
CBSE_12th
Minor Formula

❌ Confusing Lowering of Vapour Pressure with Relative Lowering of Vapour Pressure Formula

Students often incorrectly equate the lowering of vapour pressure (ΔP = P° - P) directly to the mole fraction of the solute, instead of using the relative lowering of vapour pressure (ΔP / P° = (P° - P) / P°).
💭 Why This Happens:
This error typically arises from an incomplete understanding of Raoult's Law for non-volatile solutes. Students might recall that the vapour pressure change is related to the mole fraction of the solute but forget the precise form of the relationship, specifically the division by P° (vapour pressure of pure solvent) to make it 'relative'. The terms 'lowering' and 'relative lowering' are often used interchangeably in their minds, leading to formula misuse.
✅ Correct Approach:
According to Raoult's Law, it is the relative lowering of vapour pressure that is equal to the mole fraction of the solute for dilute solutions of non-volatile solutes.

The correct formula is:
(P° - P) / P° = Xsolute = nsolute / (nsolute + nsolvent)
Where:
  • P° = Vapour pressure of the pure solvent
  • P = Vapour pressure of the solution
  • Xsolute = Mole fraction of the solute
  • nsolute = Moles of solute
  • nsolvent = Moles of solvent
Always remember that ΔP = (P° - P) is the lowering, while ΔP/P° is the relative lowering.
📝 Examples:
❌ Wrong:
A student is given: Lowering of Vapour Pressure (ΔP) = 0.5 atm, Vapour Pressure of pure solvent (P°) = 2.0 atm.
Student's calculation: Xsolute = ΔP = 0.5.
This is incorrect because ΔP alone is not equal to Xsolute.
✅ Correct:
Given: Lowering of Vapour Pressure (ΔP) = 0.5 atm, Vapour Pressure of pure solvent (P°) = 2.0 atm.
Correct calculation: Xsolute = (P° - P) / P° = ΔP / P° = 0.5 / 2.0 = 0.25.
This correctly applies the relative lowering of vapour pressure formula.
💡 Prevention Tips:
  • Mnemonic for Raoult's Law: Think 'Relative Lowering R-elates to X-solute'.
  • Write Formulas Explicitly: Always write out the full formula, including the denominator P°, when solving problems. For CBSE, clearly define each term.
  • Unit Analysis: Notice that Xsolute is dimensionless, so (P° - P) / P° must also be dimensionless, which only happens when you divide by P°.
  • Practice with Definitions: When practicing, consciously state whether you are calculating 'lowering' or 'relative lowering' to reinforce the distinction.
CBSE_12th
Minor Calculation

❌ Incorrect Calculation of Solute Mole Fraction (X_solute)

Students frequently err in calculating the mole fraction of the solute (X_solute), which is the cornerstone of Raoult's Law for relative lowering of vapour pressure. This minor error often arises from either misapplying the dilute solution approximation or inaccuracies in determining the number of moles for solute or solvent.
💭 Why This Happens:
  • Misapplication of Dilute Solution Approximation: Students sometimes recall the simplified formula (P₀ - Ps) / P₀ ≈ nsolute / nsolvent without recognizing that this is strictly valid only for very dilute solutions where nsolute ≪ nsolvent.
  • Carelessness in Mole Calculation: Errors in using the correct mass (e.g., using total solution mass instead of solvent mass) or molar mass for solute or solvent when calculating their respective moles (n = mass / molar_mass).
  • Conceptual Confusion: Lack of a clear understanding of the definition of mole fraction as moles of a component divided by the total moles of all components in the solution.
✅ Correct Approach:
The relative lowering of vapour pressure is precisely given by Raoult's Law as:
(P₀ - Ps) / P₀ = Xsolute
Where P₀ is the vapour pressure of the pure solvent and Ps is the vapour pressure of the solution. The mole fraction of solute (Xsolute) is always calculated as:
Xsolute = nsolute / (nsolute + nsolvent)
Here, nsolute = masssolute / molar_masssolute and nsolvent = masssolvent / molar_masssolvent. CBSE vs JEE: Both exams expect accurate mole fraction calculations. While JEE might use the dilute approximation for specific problem types, understanding its limits is key. For CBSE, consistently using the full formula is recommended unless extreme dilution is explicitly stated.
📝 Examples:
❌ Wrong:
Problem: Calculate Xsolute for a solution containing 18g glucose (M.M = 180 g/mol) in 90g water (M.M = 18 g/mol).
Incorrect Calculation:
nsolute = 18/180 = 0.1 mol
nsolvent = 90/18 = 5 mol
Xsolute ≈ nsolute / nsolvent = 0.1 / 5 = 0.02 (This applies the dilute approximation inappropriately as the solution is not extremely dilute).
✅ Correct:
Problem: Calculate Xsolute for the same solution (18g glucose in 90g water).
Correct Calculation:
nsolute = 18/180 = 0.1 mol
nsolvent = 90/18 = 5 mol
Xsolute = nsolute / (nsolute + nsolvent) = 0.1 / (0.1 + 5) = 0.1 / 5.1 ≈ 0.0196
Note the subtle but important difference in the result, which can affect the final answer significantly.
💡 Prevention Tips:
  • Always use the full mole fraction formula: Stick to Xsolute = nsolute / (nsolute + nsolvent) unless the problem explicitly states or clearly implies a very dilute solution where the approximation is negligible.
  • Double-check masses and molar masses: Carefully verify that you are using the correct mass and molar mass for both the solute and the solvent.
  • Write down intermediate steps: Explicitly calculate nsolute and nsolvent separately before combining them in the mole fraction formula.
  • Understand the 'dilute' condition: Recognize that the approximation nsolute ≈ nsolute / nsolvent is generally valid only when nsolute is less than about 5% of nsolvent.
CBSE_12th
Minor Conceptual

❌ Confusing 'Lowering of Vapour Pressure' with 'Relative Lowering of Vapour Pressure'

Students often interchange the terms 'Lowering of Vapour Pressure' (ΔP) and 'Relative Lowering of Vapour Pressure' (ΔP/P° or (P° - P)/P°). This leads to incorrect application of Raoult's Law and errors in calculating properties related to colligative properties, especially in numerical problems. The fundamental conceptual difference lies in the reference point: one is an absolute change, the other is a fractional change relative to the pure solvent's vapour pressure.
💭 Why This Happens:
This confusion typically arises from an incomplete understanding of Raoult's Law for non-volatile solutes. Students might correctly identify the formula for ΔP as P° - P (where P° is vapour pressure of pure solvent and P is vapour pressure of solution) but then incorrectly use this value directly in calculations where relative lowering (a dimensionless quantity) is required. Sometimes, they might mistakenly use the solution's vapour pressure (P) as the denominator instead of the pure solvent's vapour pressure (P°).
✅ Correct Approach:
Understand that Lowering of Vapour Pressure (ΔP) = P° - P. This is an absolute decrease. Relative Lowering of Vapour Pressure (RLVP) = (P° - P) / P°. This is a ratio and according to Raoult's Law, for dilute solutions, it is equal to the mole fraction of the non-volatile solute (χsolute). Always ensure you use P° (vapour pressure of pure solvent) in the denominator for relative lowering. For JEE Advanced, precision in these definitions is critical for applying Raoult's Law accurately.
📝 Examples:
❌ Wrong:
A student might calculate the lowering of vapour pressure (ΔP = 10 mmHg) and then, to find the mole fraction of solute, wrongly write χsolute = ΔP / P, where P is the vapour pressure of the solution. This is incorrect because the denominator should be P° (pure solvent's vapour pressure).
✅ Correct:
If the vapour pressure of pure water (P°) is 760 mmHg and the vapour pressure of a solution (P) is 750 mmHg, then:
Lowering of Vapour Pressure (ΔP) = P° - P = 760 - 750 = 10 mmHg.
Relative Lowering of Vapour Pressure (RLVP) = (P° - P) / P° = 10 / 760.
According to Raoult's Law, this RLVP (10/760) is equal to the mole fraction of the solute (χsolute).
💡 Prevention Tips:
  • Clear Definitions: Memorize and clearly distinguish between the definitions of ΔP and RLVP.
  • Formula Application: Always write down the full formula for RLVP: (P° - P) / P° = χsolute. Do not omit the denominator.
  • Unit Analysis: Remember ΔP has units (e.g., mmHg, kPa), while RLVP is a dimensionless ratio. This can help you identify if you're using the correct quantity.
  • Practice: Solve various numerical problems, paying close attention to what the question asks for (lowering vs. relative lowering).
JEE_Advanced
Minor Calculation

❌ Arithmetic Errors in Calculating Moles of Solvent or Solute

Students frequently make minor arithmetic errors while determining the number of moles (n) for either the solute or, more commonly, the solvent. This can involve using an incorrect molar mass, performing division incorrectly, or misplacing decimals, leading to an inaccurate mole fraction (Xsolute) and consequently, an incorrect relative lowering of vapour pressure value. This mistake is particularly common when dealing with large quantities of solvent or complex molar masses.
💭 Why This Happens:
These errors primarily stem from haste, lack of thorough double-checking, or simple arithmetic oversights under exam pressure. Sometimes, students might confuse the molar masses of common substances or rely too heavily on quick mental calculations that are prone to error, especially without a calculator in JEE Advanced.
✅ Correct Approach:
Always explicitly write down the formula for moles (n = mass / molar mass) and substitute values carefully. For common solvents like water, specifically state its molar mass (18 g/mol) to avoid errors. Perform divisions meticulously, breaking down complex numbers if necessary, and always double-check your result before proceeding to the next step of the problem.
📝 Examples:
❌ Wrong:
A student is given 360g of water and a non-volatile solute. They need to find nsolvent.
Wrong Calculation: nwater = 360g / 180 g/mol (mistakenly using 180 as water's molar mass or a large integer) = 2 mol, OR nwater = 360g / 16 g/mol (molar mass of Oxygen only) = 22.5 mol.
✅ Correct:
Given 360g of water. The molar mass of water (H₂O) = 2(1) + 16 = 18 g/mol.
Correct Calculation: nwater = 360g / 18 g/mol = 20 mol. This correct value must then be accurately used in the denominator for mole fraction calculation: Xsolute = nsolute / (nsolute + 20).
💡 Prevention Tips:
Explicitly State Molar Masses: Before starting calculations, list the molar masses of all relevant compounds. For JEE, learn common molar masses by heart (e.g., H₂O = 18, CO₂ = 44, C₆H₁₂O₆ = 180).
Double-Check Basic Arithmetic: Even simple divisions and additions should be quickly reviewed. Practice mental arithmetic or concise rough work.
Box Intermediate Results: Clearly mark calculated moles of solute and solvent before using them in the final formula for relative lowering of vapour pressure. This makes it easier to track and verify values.
JEE_Advanced
Minor Formula

❌ Incorrect Denominator or Mole Fraction in RLVVP Formula

Students commonly confuse the denominator in the Relative Lowering of Vapour Pressure (RLVVP) formula, using $P_s$ (solution's vapour pressure) instead of $P_0$ (pure solvent's vapour pressure). This leads to $(P_0 - P_s)/P_s$ instead of the correct $(P_0 - P_s)/P_0$. Another error is in calculating the solute's mole fraction ($X_{solute}$), sometimes using the solvent's mole fraction or an incorrect approximation, especially for dilute solutions (where approximations for other colligative properties might be misapplied).
💭 Why This Happens:

  • Conceptual Gap: Not clearly distinguishing 'lowering' of vapour pressure ($P_0 - P_s$) from 'relative lowering' of vapour pressure ($(P_0 - P_s)/P_0$).

  • Rote Memorization: Incomplete understanding of each term's precise definition ($P_0$, $P_s$, $X_{solute}$) and their roles in the formula.

  • Cross-Topic Confusion: Misapplying approximations or formula structures from other colligative properties (e.g., elevation in boiling point for dilute solutions) to RLVVP where the exact mole fraction is always required.

✅ Correct Approach:
The denominator for Relative Lowering of Vapour Pressure must always be the vapour pressure of the pure solvent ($P_0$).
The mole fraction of the solute ($X_{solute}$) is precisely calculated using the exact formula:


$$ ext{X}_{solute} = frac{ ext{moles of solute (}n_{solute} ext{)}}{ ext{moles of solute (}n_{solute} ext{)} + ext{moles of solvent (}n_{solvent} ext{)}}$$


Thus, the correct RLVVP formula is:


$$frac{P_0 - P_s}{P_0} = frac{n_{solute}}{n_{solute} + n_{solvent}}$$

(JEE Advanced Specific): Always ensure you're using the exact mole fraction. Approximations are generally not used for the RLVVP formula itself, unlike some derivations involving other colligative properties.
📝 Examples:
❌ Wrong:
If the pure solvent's vapour pressure ($P_0$) is $100 ext{ mmHg}$ and the solution's vapour pressure ($P_s$) is $90 ext{ mmHg}$:
Incorrect RLVVP calculation: $frac{100 - 90}{90} = frac{10}{90} approx 0.111$. (The denominator should be $P_0$).
✅ Correct:
Using the same values ($P_0 = 100 ext{ mmHg}$, $P_s = 90 ext{ mmHg}$):


Correct RLVVP = $frac{P_0 - P_s}{P_0} = frac{100 - 90}{100} = frac{10}{100} = mathbf{0.1}$.


Also, for a solution with $n_{solute}=1$ mole and $n_{solvent}=9$ moles:


Correct $X_{solute}$ = $frac{n_{solute}}{n_{solute} + n_{solvent}} = frac{1}{1+9} = frac{1}{10} = mathbf{0.1}$.
💡 Prevention Tips:

  • Understand Definitions Thoroughly: Master the precise meanings of $P_0$ (pure solvent vapour pressure), $P_s$ (solution vapour pressure), and $X_{solute}$ (mole fraction of solute).

  • Denominator Rule: Always remember that $P_0$ (pure solvent's VP) goes in the denominator for RLVVP.

  • Exact Mole Fraction: For RLVVP, always use the exact mole fraction formula for the solute: $n_{solute} / (n_{solute} + n_{solvent})$.

  • Practice Consistently: Solve a variety of problems from both CBSE and JEE Advanced to solidify correct formula application and avoid conceptual blurs.

JEE_Advanced
Minor Unit Conversion

❌ Inconsistent Pressure Units in Relative Lowering of Vapour Pressure Calculations

Students frequently make mistakes by using inconsistent units for vapour pressures (P° and Ps) when calculating the relative lowering of vapour pressure, (P° - Ps)/P°. For instance, one pressure might be given in mmHg and another in atm or kPa, and they are used directly in the formula without proper conversion.
💭 Why This Happens:
This error often stems from a lack of attention to detail, especially under exam pressure, or an incomplete understanding that for ratio-based calculations, all components of the ratio must be expressed in the same units. Sometimes, students might confuse the absolute value of pressure with its numerical value without considering the unit context. JEE Advanced problems often provide data in varied units to test this very understanding.
✅ Correct Approach:
Always convert all pressure values to a single, consistent unit before performing any calculations. Standard units like Pascals (Pa), atmospheres (atm), or millimeters of mercury (mmHg) are commonly used. Choose one unit and convert all other given pressure values into that chosen unit. Remember the conversion factors: 1 atm = 760 mmHg = 1.01325 x 105 Pa = 1.01325 bar.
📝 Examples:
❌ Wrong:
A non-volatile solute is added to water. The vapour pressure of pure water (P°) is 760 mmHg. The vapour pressure of the solution (Ps) is 0.9 atm.
Relative Lowering = (760 - 0.9) / 760 (Incorrect, units are mixed)
✅ Correct:
A non-volatile solute is added to water. The vapour pressure of pure water (P°) is 760 mmHg. The vapour pressure of the solution (Ps) is 0.9 atm.
Step 1: Convert Ps to mmHg: Ps = 0.9 atm * 760 mmHg/atm = 684 mmHg.
Step 2: Calculate Relative Lowering = (P° - Ps) / P° = (760 mmHg - 684 mmHg) / 760 mmHg = 76 / 760 = 0.1.
Alternatively, convert P° to atm: P° = 760 mmHg / 760 mmHg/atm = 1 atm.
Relative Lowering = (1 atm - 0.9 atm) / 1 atm = 0.1 / 1 = 0.1.
In both cases, the result is consistent as units are uniform.
💡 Prevention Tips:
  • Always write units: When writing down values, always include their units. This visual reminder helps in identifying inconsistencies.
  • Box/Highlight units: Before starting a calculation, quickly scan all given values and mentally (or physically) highlight their units.
  • Standardize early: Make it a habit to convert all values to a common, preferred unit at the very beginning of the problem-solving process.
  • Practice with varied units: Solve problems where different pressure units are intentionally mixed to reinforce the conversion habit.
JEE_Advanced
Minor Sign Error

❌ Incorrect Sign Convention in Relative Lowering of Vapour Pressure (RLVP) Calculations

Students frequently make a sign error by incorrectly calculating the 'lowering' of vapor pressure or the relative lowering. Instead of (P° - Ps), they might use (Ps - P°), leading to a negative value for a quantity that must always be positive. This can occur when expressing ΔP or ΔP/P°.
💭 Why This Happens:
This error primarily stems from:
  • A misunderstanding of the term 'lowering', which implies a positive decrease.
  • Carelessness in substituting values into Raoult's Law, particularly when distinguishing between pure solvent vapor pressure (P°) and solution vapor pressure (Ps).
  • Forgetting that the mole fraction of solute (Xsolute) is always positive, and since RLVP = Xsolute, RLVP must also be positive.
✅ Correct Approach:
Always define the lowering of vapor pressure (ΔP) as the difference between the vapor pressure of the pure solvent (P°) and the vapor pressure of the solution (Ps): ΔP = P° - Ps. Since the addition of a non-volatile solute always decreases vapor pressure, P° > Ps, making ΔP a positive quantity. Consequently, the Relative Lowering of Vapor Pressure (RLVP) given by ΔP / P° = (P° - Ps) / P° must also be positive. This is consistent with Raoult's Law: (P° - Ps) / P° = Xsolute, where Xsolute is always positive.
📝 Examples:
❌ Wrong:
If pure solvent has P° = 120 mmHg and the solution has Ps = 108 mmHg:
An incorrect calculation of relative lowering would be: (108 - 120) / 120 = -12 / 120 = -0.1. This negative value is erroneous for RLVP.
✅ Correct:
Using the same values: P° = 120 mmHg and Ps = 108 mmHg:
The correct relative lowering is: (120 - 108) / 120 = 12 / 120 = 0.1. This positive value is consistent with the definition and Raoult's Law.
💡 Prevention Tips:
  • Remember: Both ΔP (lowering of vapour pressure) and RLVP (relative lowering of vapour pressure) are always positive quantities.
  • Strictly adhere to the formula: RLVP = (P° - Ps) / P°. Always subtract the solution's vapor pressure from the pure solvent's vapor pressure in the numerator.
  • Cross-check with Mole Fraction: If you've calculated RLVP, ensure it matches the positive mole fraction of the solute.
  • JEE Advanced context: While a minor error, a negative sign can lead to incorrect options in MCQ or a deduction of marks in numerical answer types. Always perform a quick check of the sign.
JEE_Advanced
Minor Approximation

❌ Incorrect Application of Approximation for Relative Lowering of Vapour Pressure

Students frequently approximate the mole fraction of the solute (X_solute) as n_solute / n_solvent (moles of solute per moles of solvent) directly when dealing with Relative Lowering of Vapour Pressure (RLVP). While this approximation is valid for extremely dilute solutions, its misapplication in moderately dilute or concentrated solutions leads to errors. The exact expression for RLVP, according to Raoult's Law, is (P° - P)/P° = X_solute, where X_solute = n_solute / (n_solute + n_solvent).
💭 Why This Happens:
This mistake stems from a lack of complete understanding of the derivation of Raoult's Law and its approximate forms. Students often memorize the simplified formula without fully grasping the conditions under which the approximation (n_solute <<< n_solvent, allowing n_solute + n_solvent ≈ n_solvent) is valid. Time pressure in JEE Advanced can also lead to quick, but often incorrect, simplifications.
✅ Correct Approach:
Always start with the fundamental definition from Raoult's Law: (P° - P)/P° = X_solute. Substitute the exact expression for mole fraction: X_solute = n_solute / (n_solute + n_solvent). Only apply the approximation n_solute + n_solvent ≈ n_solvent if the solution is explicitly stated to be 'very dilute' or if the mole fraction of solute is negligible (typically <1-2%). For JEE Advanced, unless the concentration is extremely low, it's safer to use the exact mole fraction expression to ensure accuracy.
📝 Examples:
❌ Wrong:
Consider a solution containing 1 mole of a non-volatile solute in 9 moles of solvent.
Wrong Approach (Approximation):
RLVP = n_solute / n_solvent = 1 / 9 ≈ 0.111
✅ Correct:
Using the same solution containing 1 mole of non-volatile solute in 9 moles of solvent.
Correct Approach (Exact Mole Fraction):
RLVP = X_solute = n_solute / (n_solute + n_solvent) = 1 / (1 + 9) = 1 / 10 = 0.100
The difference (0.111 vs 0.100) is significant and can lead to incorrect options in multi-choice questions.
💡 Prevention Tips:
  • Understand the Derivation: Know why and when the approximation n_solute + n_solvent ≈ n_solvent is made.
  • Prioritize Exact Formula: Always write down the exact Raoult's Law expression first: (P° - P)/P° = n_solute / (n_solute + n_solvent).
  • Critical Assessment: Before approximating, check the concentration. If the mole fraction of solute is not extremely small, use the full expression. For JEE Advanced, precision is often key.
  • Practice with Varied Concentrations: Solve problems involving both very dilute and moderately dilute solutions to build intuition.
JEE_Advanced
Important Calculation

❌ Misapplication of Van't Hoff Factor 'i' and Incorrect Mole Fraction Usage in Relative Lowering of Vapour Pressure Calculations

Students frequently make two critical errors in Relative Lowering of Vapour Pressure (RLVP) calculations for JEE Advanced:

  • Failing to incorporate the Van't Hoff factor 'i' for electrolytic solutes, treating them as non-electrolytes.

  • Confusing the mole fraction of the solute (Xsolute) with the mole fraction of the solvent (Xsolvent) in the relative lowering formula.


These mistakes lead to significantly incorrect answers, as colligative properties depend on the effective number of particles.
💭 Why This Happens:

  • Conceptual Gap: Lack of deep understanding that colligative properties depend on the number of particles in solution, not just the moles of the initial substance.

  • Careless Reading: Overlooking keywords like "electrolyte," "ionic compound," or information indicating dissociation/association.

  • Formula Misapplication: Rote memorization of (P₀ - P)/P₀ = Xsolute without recalling the essential 'i' for non-ideal solutions or understanding that it's Xsolute, not Xsolvent, that directly relates to relative lowering.

✅ Correct Approach:

Always remember that Relative Lowering of Vapour Pressure (RLVP) is given by:


(P₀ - P) / P₀ = i * Xsolute



  • Identify Solute Type: Determine if the solute is an electrolyte (e.g., NaCl, BaCl₂, acids, bases) or a non-electrolyte (e.g., urea, glucose).

  • Calculate 'i':

    • For non-electrolytes, i = 1.

    • For electrolytes, calculate 'i' based on dissociation: If 100% dissociation, 'i' equals the number of ions produced (e.g., for NaCl, i=2; for BaCl₂, i=3). For partial dissociation (common in JEE Advanced), i = 1 + (n-1)α, where 'n' is the number of ions and 'α' is the degree of dissociation.



  • Use Solute's Mole Fraction: Ensure you are using Xsolute = moles of solute / (moles of solute + moles of solvent).

📝 Examples:
❌ Wrong:

Calculating the relative lowering of vapor pressure for a 0.2 molal aqueous solution of CaCl₂ (an electrolyte) by incorrectly using Xsolute directly (assuming i=1) or by using Xsolvent in the formula.


Incorrect Calculation: (P₀ - P) / P₀ = Xsolute (ignoring 'i')


Incorrect Calculation: (P₀ - P) / P₀ = Xsolvent (fundamental error in formula application)

✅ Correct:

For a 0.2 molal aqueous solution of CaCl₂ (a strong electrolyte, assuming 100% dissociation, thus i=3):


  • Moles of CaCl₂ = 0.2 moles (in 1 kg of solvent)

  • Moles of water (in 1 kg solvent) = 1000g / 18 g/mol = 55.55 moles

  • Xsolute = 0.2 / (0.2 + 55.55) ≈ 0.00359

  • Correct Relative Lowering = i * Xsolute = 3 * (0.2 / (0.2 + 55.55)) ≈ 3 * 0.00359 ≈ 0.01077


Contrast this with the incorrect approach which would yield a value roughly one-third of the correct answer (if 'i' is ignored) or a completely different value if Xsolvent is used.
💡 Prevention Tips:

  • Systematic Approach: Before any calculation, always identify the solute type, then determine 'i' (if applicable), and finally apply the correct mole fraction of the solute.

  • Formula Precision: Always explicitly write down the complete formula (P₀ - P) / P₀ = i * Xsolute before substituting values to act as a checklist.

  • Practice Electrolyte Problems: Specifically solve a variety of problems involving strong and weak electrolytes to build confidence in calculating and using 'i'.

  • Double Check: Always verify if the calculated 'i' value makes sense for the given solute and whether you've used Xsolute correctly.

JEE_Advanced
Important Formula

❌ Ignoring Van't Hoff Factor 'i' for Electrolytes and Misapplying Exact vs. Approximate RLVP Formulas

Students frequently make two critical errors concerning the formula for Relative Lowering of Vapour Pressure (RLVP):
  1. Forgetting the Van't Hoff factor (i): For electrolytic solutes (e.g., NaCl, CaCl2), the RLVP depends on the effective number of particles in solution. Many students directly use the moles of solute added without multiplying by 'i'.
  2. Incorrectly applying the approximate formula: For non-volatile solutes, Raoult's Law states (P° - Psolution) / P° = i * χsolute, where χsolute = nsolute / (nsolute + nsolvent). A common approximation used for very dilute solutions is (P° - Psolution) / P° ≈ i * (nsolute / nsolvent). Students often use this approximation even when the solution isn't dilute enough, leading to incorrect results in JEE Advanced where precision matters.
💭 Why This Happens:
  • Lack of clear understanding that colligative properties depend on the number of solute particles, not just the moles added. Electrolytes dissociate, increasing the particle count.
  • Confusing the exact definition of mole fraction with the simplified ratio of moles (solute/solvent) often seen in introductory problems or for extremely dilute cases.
  • Over-reliance on the approximate formula without checking its applicability conditions, especially for moderately concentrated solutions that might appear in JEE Advanced.
  • Failure to identify whether the given solute is an electrolyte or non-electrolyte.
✅ Correct Approach:
  • Always identify if the solute is an electrolyte (dissociates/associates) or a non-electrolyte. If it's an electrolyte, determine the Van't Hoff factor 'i'. For strong electrolytes, 'i' is often equal to the number of ions formed (e.g., NaCl, i=2; CaCl2, i=3). For weak electrolytes or association, 'i' must be calculated using the degree of dissociation/association.
  • The universally applicable and exact formula for RLVP is: (P° - Psolution) / P° = i * (nsolute / (nsolute + nsolvent)).
  • Use the approximation (P° - Psolution) / P° ≈ i * (nsolute / nsolvent) only if explicitly stated that the solution is extremely dilute, or if the molarity is very low (typically < 0.1 M). In JEE Advanced, it's safer to use the exact formula unless the approximation clearly simplifies calculations without significant error.
📝 Examples:
❌ Wrong:
A student calculates the RLVP for a 0.5 molal aqueous NaCl solution at 25°C using the formula (P° - Psolution) / P° = (nNaCl / nwater) and 'i' = 1 (ignoring dissociation) or using the exact mole fraction formula but still with 'i'=1.
✅ Correct:
For the same 0.5 molal aqueous NaCl solution at 25°C, the correct approach is:
  • Identify NaCl as a strong electrolyte, so i ≈ 2 (assuming complete dissociation).
  • Calculate moles: 0.5 mol NaCl per 1000g water (nwater = 1000/18 = 55.56 mol).
  • Use the exact formula: (P° - Psolution) / P° = i * (nNaCl / (nNaCl + nwater))
  • = 2 * (0.5 / (0.5 + 55.56)) = 2 * (0.5 / 56.06) ≈ 0.0178.
  • If using the approximate formula (only for very dilute solutions): ≈ 2 * (0.5 / 55.56) ≈ 0.0180. Notice the slight difference even in this case, which can be significant in multi-step problems or where options are close.
💡 Prevention Tips:
  • Tip 1 (JEE Advanced Focus): Always assume the exact formula unless the approximation leads to negligible difference or is specifically warranted.
  • Tip 2: Practice identifying electrolytes vs. non-electrolytes and calculating 'i' for various compounds, including those with incomplete dissociation/association.
  • Tip 3: Clearly write down the formula you are using before plugging in values, making sure to include 'i' if applicable and the correct denominator for mole fraction.
  • Tip 4: When in doubt, always use the full mole fraction formula (nsolute / (nsolute + nsolvent)) for χsolute.
JEE_Advanced
Important Sign Error

❌ Incorrect Interpretation of 'Lowering of Vapour Pressure' vs. 'Vapour Pressure of Solution'

Students frequently confuse the term 'lowering of vapour pressure' (ΔP = P° - P) with the 'vapour pressure of the solution' (P) itself. This leads to incorrect substitution into the Relative Lowering of Vapour Pressure (RLVAP) formula, especially in JEE Advanced problems where nuanced phrasing is common. A common sign error also involves mistakenly calculating (P - P°) instead of (P° - P), resulting in a negative 'lowering' value.
💭 Why This Happens:
  • Lack of Conceptual Clarity: Not distinguishing clearly between the 'change' (lowering) and the 'final state' (solution's VP).
  • Misreading Problem Statements: Rushing leads to misinterpreting keywords and directly using a value provided for ΔP as P or vice-versa.
  • Algebraic Errors: Incorrectly rearranging the RLVAP formula or solving for P after determining ΔP.
  • Fundamental Misconception: For non-volatile solutes, the solution's vapour pressure (P) *must* always be less than the pure solvent's (P°). Failing to check this can hide a sign error.
✅ Correct Approach:
Always remember that for a non-volatile solute, the vapour pressure of the solution (P) is always less than the vapour pressure of the pure solvent (P°). The lowering of vapour pressure (ΔP) is P° - P and must always be a positive value. The RLVAP formula is (P° - P) / P° = Xsolute. Carefully identify what each given value represents before substitution.
📝 Examples:
❌ Wrong:

If the vapour pressure of pure water (P°) is 50 mmHg and the problem states 'the lowering of vapour pressure is 5 mmHg'.

Incorrect Substitution: A student might mistakenly substitute 5 mmHg as P (vapour pressure of solution) into the RLVAP formula:

(50 - 5) / 50 = Xsolute (This implies P = 5 mmHg, which is wrong, as 5 mmHg is the lowering, ΔP)

✅ Correct:

Given P° = 50 mmHg and 'lowering of vapour pressure' (ΔP = P° - P) = 5 mmHg.

Correct Approach:

  1. First, calculate the actual vapour pressure of the solution (P): P = P° - ΔP = 50 - 5 = 45 mmHg.
  2. Then, correctly substitute P into the RLVAP formula: (P° - P) / P° = (50 - 45) / 50 = 5 / 50 = Xsolute.
💡 Prevention Tips:
  • Deconstruct the Problem: Before calculations, clearly list P°, P, and ΔP, identifying which values are given and which need to be found.
  • Check for Sanity: Always verify that your calculated P (solution VP) is less than P° (pure solvent VP). If P > P°, a fundamental error has occurred.
  • Practice Algebraic Manipulations: Be comfortable rearranging the RLVAP formula to solve for any variable (P, P°, Xsolute).
  • JEE Advanced Note: Questions can be tricky; a phrase like 'the relative lowering is 0.1' means (P° - P) / P° = 0.1, not P = 0.1.
JEE_Advanced
Important Approximation

❌ Incorrect Approximation of Solute Mole Fraction in Dilute Solutions

Students frequently misuse or misapply the approximation for the mole fraction of solute (Xsolute) in the relative lowering of vapour pressure (RLVP) formula. This often happens by directly equating Xsolute to nsolute/nsolvent without considering the actual diluteness of the solution.
💭 Why This Happens:
This mistake stems from a misunderstanding of the conditions under which the approximation is valid. While the exact formula for mole fraction is Xsolute = nsolute / (nsolute + nsolvent), for very dilute solutions, the nsolute term in the denominator becomes negligible compared to nsolvent, allowing the approximation Xsolute ≈ nsolute / nsolvent. Students often forget this critical condition and apply the approximation universally or when the solution is not sufficiently dilute.
✅ Correct Approach:
Always start with the fundamental definition of mole fraction:
Xsolute = nsolute / (nsolute + nsolvent)
The relative lowering of vapour pressure (ΔP/P°) is precisely equal to Xsolute.
The approximation Xsolute ≈ nsolute / nsolvent is valid only when nsolute << nsolvent, meaning the number of moles of solute is significantly smaller than the number of moles of solvent. For JEE Advanced, it is generally safer to use the exact formula unless the problem explicitly states 'very dilute' or the options provided imply such an approximation.
📝 Examples:
❌ Wrong:
A student calculates the mole fraction of a solution containing 1 mole of solute in 10 moles of solvent as Xsolute = 1/10 = 0.1, assuming nsolute is negligible in the denominator. This is an incorrect approximation as 1 is not negligible compared to 10.
✅ Correct:
For a solution containing 1 mole of solute in 10 moles of solvent:
Xsolute = 1 / (1 + 10) = 1/11 ≈ 0.0909
Using the approximation 1/10 = 0.1 introduces a significant error of about 10%.
Caution: In JEE Advanced, such a difference can lead to choosing the wrong option. Only approximate if nsolute/nsolvent is extremely small, typically less than 0.01.
💡 Prevention Tips:
  • Always write down the full mole fraction formula: Xsolute = nsolute / (nsolute + nsolvent).
  • Evaluate the magnitude of nsolute relative to nsolvent before deciding to apply any approximation.
  • JEE Advanced Strategy: When in doubt, or if the ratio of moles is not extremely small (e.g., nsolute/nsolvent > 0.01), stick to the exact mole fraction calculation. Approximation should be a last resort or clearly indicated by the problem context.
  • Practice problems where both exact and approximate values are calculated to understand the error introduced by approximation.
JEE_Advanced
Important Other

❌ Confusing Mole Fraction of Solute with Solvent in Relative Lowering of Vapour Pressure

A common and significant error made by students is incorrectly equating the relative lowering of vapour pressure (RLVP) to the mole fraction of the solvent instead of the solute. While Raoult's Law connects the vapour pressure of a solution to the mole fraction of the solvent, the definition of relative lowering specifically involves the solute's concentration.
💭 Why This Happens:
This mistake primarily stems from a partial understanding or misapplication of Raoult's Law. Students often recall that the vapour pressure of the solution (Ps) is given by Ps = Xsolvent * P°solvent (for a non-volatile solute), and then mistakenly assume that the 'lowering' aspect also directly relates to the solvent's mole fraction in the final RLVP expression. The subtle distinction between the vapour pressure of the solution and the relative lowering of vapour pressure often gets overlooked. For JEE Advanced, precision in definition is key.
✅ Correct Approach:
According to Raoult's Law for a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute. The correct formula is:

(P° - Ps) / P° = Xsolute

Where:
  • P° = Vapour pressure of the pure solvent
  • Ps = Vapour pressure of the solution
  • Xsolute = Mole fraction of the solute (nsolute / (nsolute + nsolvent))
This formula is a colligative property and is independent of the nature of the solute, depending only on its concentration (mole fraction).
📝 Examples:
❌ Wrong:
A student calculates the relative lowering of vapour pressure using the formula: (P° - Ps) / P° = Xsolvent. This would lead to an incorrect value, as Xsolvent and Xsolute are related by Xsolvent + Xsolute = 1, hence Xsolvent is generally much larger than Xsolute for dilute solutions.
✅ Correct:
Given the vapour pressure of pure water (P°) is 50 mmHg and a solution containing a non-volatile solute has a vapour pressure (Ps) of 48 mmHg. The relative lowering of vapour pressure is (50 - 48) / 50 = 2/50 = 0.04. According to Raoult's Law, this value (0.04) is the mole fraction of the solute (Xsolute) present in the solution.
💡 Prevention Tips:
  • Memorize the Definition: Always recall that 'relative lowering of vapour pressure equals the mole fraction of the solute'.
  • Understand the Derivation: Revisit the derivation of Raoult's Law for non-volatile solutes to understand why the solute's mole fraction appears in the final RLVP expression.
  • Formula Checklist: Before solving, explicitly write down the formula: (P° - Ps) / P° = Xsolute.
  • Practice JEE Advanced Problems: Work through problems that test this specific understanding to solidify the concept.
JEE_Advanced
Important Conceptual

❌ <span style='color: red;'>Confusing Lowering of Vapour Pressure (ΔP) with Relative Lowering of Vapour Pressure (RLVP) and Incorrect Mole Fraction Usage</span>

Students frequently confuse the absolute lowering of vapour pressure (ΔP = P° - P_solution) with the relative lowering of vapour pressure (RLVP = ΔP/P°). A critical error arises when they incorrectly substitute the mole fraction of the solvent (X_solvent) where the mole fraction of the solute (X_solute) is required for RLVP, or vice-versa. This fundamental conceptual misunderstanding often leads to incorrect application of Raoult's Law in problems.
💭 Why This Happens:
  • Lack of clear conceptual distinction between P° (vapour pressure of pure solvent), P_solution (vapour pressure of solution), ΔP, and RLVP.
  • Misinterpretation of the exact statement of Raoult's Law for non-volatile solutes, which directly equates RLVP to the mole fraction of the solute.
  • Forgetting that P_solution = P° * X_solvent, and therefore ΔP = P°(1 - X_solvent) = P° * X_solute. The 'relative' aspect normalizes this lowering by P°.
  • Over-reliance on approximated forms (e.g., for very dilute solutions where X_solute ≈ n_solute/n_solvent) without understanding the core relationship.
✅ Correct Approach:
  • Understand that Raoult's Law for a non-volatile solute states: P_solution = P° ⋅ X_solvent, where P° is the vapour pressure of the pure solvent and X_solvent is its mole fraction in the solution.
  • The lowering of vapour pressure (ΔP) is the difference: ΔP = P° - P_solution. Substituting P_solution, we get ΔP = P° - (P° ⋅ X_solvent) = P°(1 - X_solvent). Since X_solute + X_solvent = 1, it follows that ΔP = P° ⋅ X_solute.
  • The relative lowering of vapour pressure (RLVP) is defined as the ratio ΔP/P°. Substituting ΔP, we get:
    RLVP = (P° ⋅ X_solute) / P° = X_solute.
  • JEE Advanced Focus: This relationship, (P° - P_solution)/P° = X_solute, is a colligative property and is exact for ideal solutions with non-volatile solutes. Always use this precise form.
📝 Examples:
❌ Wrong:
A student calculates the relative lowering of vapour pressure for a solution by using the formula (P° - P_solution) / P_solution = X_solute, or by simply stating that ΔP = P° ⋅ X_solvent is the relative lowering.
✅ Correct:
Given pure solvent vapour pressure P° = 100 mmHg and solution vapour pressure P_solution = 95 mmHg. The correct calculation for the mole fraction of the non-volatile solute, X_solute, would be:
(P° - P_solution) / P° = X_solute
(100 mmHg - 95 mmHg) / 100 mmHg = X_solute
5 / 100 = 0.05 = X_solute
💡 Prevention Tips:
  • Precise Definitions: Clearly define and differentiate between P°, P_solution, ΔP, and RLVP in your notes.
  • Revisit Derivation: Understand the step-by-step derivation of Raoult's Law for non-volatile solutes. This clarifies why RLVP equals X_solute.
  • Formula Association: Always associate 'Relative Lowering of Vapour Pressure' directly with 'mole fraction of solute' (X_solute).
  • Practice Problems: Work through various problems, explicitly writing down the correct formula before substituting values.
  • JEE vs. CBSE: While CBSE might tolerate some approximations for very dilute solutions, JEE Advanced expects a rigorous application of the exact formulas.
JEE_Advanced
Important Unit Conversion

❌ Inconsistent Mass Units for Mole Calculation in RLV Problems

A frequent error observed in JEE Main for 'Relative Lowering of Vapour Pressure' (RLVP) problems is the inconsistent use of mass units when calculating the number of moles (n = mass / molar mass). Students often mix units like grams (g) and kilograms (kg) without proper conversion, especially when molar masses are typically given in g/mol. This directly impacts the mole fraction calculation, leading to an incorrect RLVP value.
💭 Why This Happens:
  • Rushing and Oversight: Under exam pressure, students often overlook the units provided for masses (e.g., 1 kg of solvent) and use them directly with molar masses given in g/mol.
  • Assumption of Consistency: Incorrectly assuming all given quantities are already in compatible units.
  • Lack of Unit Tracking: Not writing down units at each step of the calculation makes it difficult to catch inconsistencies.
✅ Correct Approach:
Always convert all given masses to a single, consistent unit (preferably grams, as molar masses are typically in g/mol) before calculating the number of moles for both solute and solvent. The mole fraction, Xsolute = nsolute / (nsolute + nsolvent), is crucial for RLVP, and its accuracy depends entirely on correct mole calculations. Remember, RLVP is dimensionless, but intermediate mole calculations require careful unit management.
📝 Examples:
❌ Wrong:

Problem: Calculate the relative lowering of vapour pressure for a solution containing 10 g of a non-volatile solute (Molar mass = 100 g/mol) in 1 kg of water (Molar mass = 18 g/mol).

Incorrect Calculation:
nsolute = 10 g / 100 g/mol = 0.1 mol
nsolvent = 1 kg / 18 g/mol = 0.0556 mol (Incorrect: 1 kg directly used with g/mol)
Xsolute = 0.1 / (0.1 + 0.0556) = 0.1 / 0.1556 ≈ 0.64

✅ Correct:

Problem: Calculate the relative lowering of vapour pressure for a solution containing 10 g of a non-volatile solute (Molar mass = 100 g/mol) in 1 kg of water (Molar mass = 18 g/mol).

Correct Calculation:
nsolute = 10 g / 100 g/mol = 0.1 mol
Convert solvent mass: 1 kg = 1000 g
nsolvent = 1000 g / 18 g/mol = 55.56 mol
Xsolute = 0.1 / (0.1 + 55.56) = 0.1 / 55.66 ≈ 0.0018

Observation: The correct relative lowering of vapour pressure (mole fraction) is significantly different due to proper unit conversion.

💡 Prevention Tips:
  • Check Units First: Before starting any calculation, explicitly write down the units of all given quantities.
  • Standardize Units: Convert all masses to grams (or kilograms) consistently at the very beginning of the problem. For JEE, assume molar mass is in g/mol unless specified.
  • Unit Tracking: Carry units through your calculations. This helps in identifying when a unit mismatch occurs (e.g., g/g/mol results in mol, but kg/g/mol does not).
  • JEE Specific: While RLVP itself is a ratio, ensuring correct moles is paramount. Always double-check mass units when molar masses are provided.
JEE_Main
Important Other

❌ <strong>Incorrect Mole Fraction Usage in Relative Lowering of Vapour Pressure</strong>

Students frequently make the error of incorrectly using the mole fraction of the solvent (Xsolvent) instead of the mole fraction of the solute (Xsolute) when applying the formula for relative lowering of vapour pressure. This leads to fundamental errors in calculating colligative properties.
💭 Why This Happens:
  • Memorization vs. Understanding: Many students memorize the final formula for Raoult's Law without understanding its derivation. Raoult's Law itself states that the vapour pressure of the solution is proportional to the mole fraction of the solvent (Ps = P0 Xsolvent). However, the *relative lowering* is then derived from this as (P0 - Ps) / P0 = 1 - Xsolvent = Xsolute.
  • Symbol Confusion: Different texts may use '1' for solvent and '2' for solute, or vice-versa, causing interchangeability issues.
  • Ignoring Non-Volatile Solute's Role: The presence of a non-volatile solute *lowers* the solvent's vapor pressure, and the extent of this lowering is directly related to how many solute particles are present relative to total particles.
✅ Correct Approach:
Always remember that for a non-volatile solute, the relative lowering of vapour pressure is equal to the mole fraction of the solute. The precise formula is:
(P0 - Ps) / P0 = Xsolute = nsolute / (nsolute + nsolvent)
Where:
  • P0 = Vapour pressure of pure solvent
  • Ps = Vapour pressure of solution
  • Xsolute = Mole fraction of solute
  • nsolute = Moles of solute
  • nsolvent = Moles of solvent
📝 Examples:
❌ Wrong:
If a solution contains 1 mole of solute and 9 moles of solvent, a student might incorrectly calculate the relative lowering of vapor pressure as Xsolvent = 9 / (1 + 9) = 0.9. This would yield an incorrect result for the relative lowering.
✅ Correct:
For the same solution with 1 mole of solute and 9 moles of solvent, the mole fraction of solute Xsolute = 1 / (1 + 9) = 0.1. Therefore, the relative lowering of vapour pressure is 0.1. This correctly reflects the definition.
💡 Prevention Tips:
  • Understand the Derivation (CBSE & JEE): Grasp how (P0 - Ps) / P0 = Xsolute is derived from Raoult's Law (Ps = P0 Xsolvent) and the fact that Xsolute + Xsolvent = 1.
  • Clearly Label Variables: Always identify and label nsolute and nsolvent to avoid confusion.
  • JEE Specific - Dilute Solutions: While the exact formula uses nsolute / (nsolute + nsolvent), for very dilute solutions (nsolute <<< nsolvent), the approximation (P0 - Ps) / P0 ≈ nsolute / nsolvent is often used. Know when this approximation is valid.
  • Practice Application: Solve problems where you need to find the molar mass of the solute or the vapour pressure of the solution to reinforce correct formula usage.
JEE_Main
Important Sign Error

❌ Incorrect Sign/Order in Relative Lowering of Vapour Pressure Formula

Students frequently make sign errors or incorrect ordering of terms when calculating the relative lowering of vapour pressure (RLVP). This typically involves confusing the vapour pressure of the pure solvent (P°) with that of the solution (P_s), or incorrectly formulating the 'lowering' part of the equation, often resulting in a negative value for a quantity that must be positive.
💭 Why This Happens:
This error stems from a lack of conceptual clarity regarding the definition of RLVP and Raoult's Law. Students often:
  • Misinterpret 'lowering' as simply the difference `(P_s - P°)` instead of `(P° - P_s)`.
  • Forget that for a non-volatile solute, the vapour pressure of the solution (`P_s`) is always less than the vapour pressure of the pure solvent (`P°`).
  • Rush through formula application without internalizing the physical meaning of each term.
✅ Correct Approach:
Always remember that relative lowering of vapour pressure is defined as the ratio of the lowering in vapour pressure (`ΔP`) to the vapour pressure of the pure solvent (`P°`).
The lowering in vapour pressure (`ΔP`) is given by `P° - P_s`. Since `P° > P_s`, `ΔP` must always be a positive quantity. The formula for RLVP is:
RLVP = `(P° - P_s) / P° = X_solute`
where `X_solute` is the mole fraction of the solute. Ensure `P°` is in the denominator for relative lowering.
📝 Examples:
❌ Wrong:
If P° = 100 mmHg and P_s = 90 mmHg, a common mistake is to calculate RLVP as:
`RLVP = (P_s - P°) / P° = (90 - 100) / 100 = -10 / 100 = -0.1`
This yields a negative value, which is physically incorrect for 'relative lowering'.
✅ Correct:
Given P° = 100 mmHg and P_s = 90 mmHg, the correct calculation for RLVP is:
`ΔP = P° - P_s = 100 - 90 = 10 mmHg`
`RLVP = ΔP / P° = 10 / 100 = 0.1`
This positive value correctly represents the relative lowering.
💡 Prevention Tips:
  • Understand the Basics: Clearly define `P°` (pure solvent) and `P_s` (solution) and their relationship (`P° > P_s`).
  • Memorize Correct Formula: Drill the formula `(P° - P_s) / P°` and its equivalence to `X_solute`.
  • Check Sign: After calculation, always verify that the 'lowering' term (`P° - P_s`) and the overall RLVP are positive. If not, re-check your terms.
  • Conceptual Reinforcement (JEE Specific): For JEE, this understanding is critical as sign errors can drastically alter multiple-choice answers, often with options including the inverted or negative values.
JEE_Main
Important Formula

❌ Incorrect Denominator and Neglecting van't Hoff Factor in Relative Lowering of Vapour Pressure (RLVP) Formula

A frequent error in JEE Main is using the vapour pressure of the solution (Ps) instead of the pure solvent (Po) in the denominator of the RLVP formula. Additionally, students often fail to account for the van't Hoff factor (i) for electrolytic solutes, treating them as non-electrolytes.
💭 Why This Happens:
This mistake stems from a misunderstanding of the term 'relative' – it specifically refers to the pure solvent's vapour pressure. The neglect of 'i' arises from overlooking the dissociation or association characteristics of the solute in solution, which directly impacts the number of particles.
✅ Correct Approach:
The exact formula for Relative Lowering of Vapour Pressure, especially for JEE, is given by Raoult's Law for non-volatile solutes:
(Po - Ps) / Po = i * Xsolute
Where:
  • Po = Vapour pressure of the pure solvent
  • Ps = Vapour pressure of the solution
  • i = van't Hoff factor (1 for non-electrolytes; >1 for electrolytes that dissociate; <1 for solutes that associate)
  • Xsolute = Mole fraction of the solute, calculated as nsolute / (nsolute + nsolvent)

Important JEE Note: Only for extremely dilute solutions can the approximation Xsolute ≈ nsolute / nsolvent be considered, but it's always safer to use the exact mole fraction formula.
📝 Examples:
❌ Wrong:
Calculating RLVP using (Po - Ps) / Ps or for a 0.1 M NaCl solution, using XNaCl instead of i * XNaCl (where i ≈ 2).
✅ Correct:
To find the vapour pressure of a 0.1 M NaCl solution (i ≈ 2), given Po, one must use (Po - Ps) / Po = 2 * XNaCl , and solve for Ps. The mole fraction XNaCl should be calculated using nNaCl / (nNaCl + nsolvent) .
💡 Prevention Tips:
  • Memorize the Formula Precisely: Always remember the denominator is Po.
  • Identify Solute Type: Check if the solute is an electrolyte (ionic compound, strong acid/base) and incorporate the correct van't Hoff factor 'i'.
  • Exact Mole Fraction: Unless explicitly stated for extreme dilution, always use the exact mole fraction formula, nsolute / (nsolute + nsolvent).
  • Practice Numerical Problems: Solve diverse problems covering both non-electrolytes and electrolytes to solidify your understanding.
JEE_Main
Important Calculation

❌ Incorrect application of mole fraction in Relative Lowering of Vapour Pressure (RLVPP) calculations

Students frequently make errors in using the correct mole fraction term when calculating RLVPP. A common mistake is using the mole fraction of the solute, x_solute = n_solute / (n_solute + n_solvent), directly in the denominator for all scenarios, even when the dilute solution approximation (n_solute / n_solvent) is more appropriate and simplifies calculations, especially for molar mass determination in JEE Main. Another error is confusing RLVPP ( (P° - P) / P° ) with the simple lowering of vapour pressure ( P° - P ).
💭 Why This Happens:
This often stems from a lack of clarity on the exact derivation of the RLVPP formula from Raoult's Law and not fully grasping the conditions under which approximations are valid. Students might memorize the formula (P° - P) / P° = x_solute but then either incorrectly substitute x_solute as n_solute / n_solvent when the solution is not dilute enough, or more commonly, fail to utilize the simpler n_solute / n_solvent approximation for dilute solutions, which is crucial for quick problem-solving in competitive exams like JEE.
✅ Correct Approach:
The RLVPP is fundamentally defined as (P° - P) / P° = x_solute, where P° is the vapor pressure of the pure solvent, P is the vapor pressure of the solution, and x_solute is the mole fraction of the solute (n_solute / (n_solute + n_solvent)).

For dilute solutions, where the number of moles of solute (n_solute) is significantly less than the number of moles of solvent (n_solvent), the mole fraction of solute can be accurately approximated as:
x_solute = n_solute / (n_solute + n_solvent) ≈ n_solute / n_solvent.
Therefore, for dilute solutions, the RLVPP formula simplifies to: (P° - P) / P° ≈ n_solute / n_solvent. This approximation is widely used in JEE problems for calculating molar masses of non-volatile solutes.
📝 Examples:
❌ Wrong:
A 10g non-volatile solute is dissolved in 100g water (M = 18 g/mol). The RLVPP is 0.01. To find the molar mass (M_solute) of the solute, a student might incorrectly set up the equation without using the dilute approximation, leading to complex algebra:
0.01 = (10/M_solute) / ( (10/M_solute) + (100/18) )
This approach, though technically exact, is time-consuming and prone to calculation errors in an exam setting when the solution is dilute.
✅ Correct:
For the same problem, given RLVPP is 0.01 and assuming a dilute solution (as is common for such problems in JEE):
Number of moles of solvent (n_solvent) = 100g / 18 g/mol ≈ 5.55 moles
Number of moles of solute (n_solute) = 10g / M_solute

Using the dilute solution approximation for RLVPP:
0.01 = n_solute / n_solvent
0.01 = (10/M_solute) / (100/18)
0.01 = (10/M_solute) / 5.55
M_solute = 10 / (0.01 * 5.55) ≈ 180.18 g/mol.
This streamlined approach is expected in JEE Main unless the problem specifically indicates a concentrated solution requiring the exact formula.
💡 Prevention Tips:
  • Understand the derivation: Clearly understand how RLVPP relates to Raoult's Law and the definition of mole fraction, along with the conditions for approximations.
  • Identify dilute solutions: For most JEE problems involving RLVPP and molar mass calculations, assume the solution is dilute unless explicitly stated otherwise. This allows the use of the simpler approximation (P° - P) / P° ≈ n_solute / n_solvent.
  • Practice with both forms: Solve problems using both the exact formula and the dilute approximation to understand when each is appropriate.
  • Be mindful of Molar Mass calculations: RLVPP is frequently used to determine molar mass of non-volatile solutes; the correct application of the dilute approximation is key here.
JEE_Main
Important Conceptual

❌ <h3><span style='color: #FF0000;'>Confusing Lowering of Vapour Pressure (ΔP) with Relative Lowering of Vapour Pressure (RLVP)</span></h3>

Students frequently confuse lowering of vapour pressure (ΔP), which is the absolute decrease (P⁰ - Pₛ), with relative lowering of vapour pressure (RLVP), defined as (P⁰ - Pₛ) / P⁰. They might incorrectly equate ΔP directly to the mole fraction of the solute or use Pₛ in the denominator instead of P⁰ when calculating RLVP.

💭 Why This Happens:

This conceptual error often arises from an incomplete understanding of the term 'relative'. While it's clear that the vapour pressure decreases, students fail to internalize that 'relative' specifies the reference point—the vapour pressure of the pure solvent (P⁰). They might also mistakenly believe that any change in a colligative property is directly proportional to solute mole fraction, overlooking the specific definition of RLVP.

✅ Correct Approach:

Understand that Relative Lowering of Vapour Pressure (RLVP) is a unique colligative property defined as (P⁰ - Pₛ) / P⁰. This dimensionless quantity is equal to the mole fraction of the non-volatile solute (χ_solute) for dilute solutions. Lowering of vapour pressure (ΔP) itself (P⁰ - Pₛ), is simply the difference and carries units of pressure, not a dimensionless ratio.

📝 Examples:
❌ Wrong:

A common incorrect assumption is stating or applying P⁰ - Pₛ = χ_solute. This is dimensionally incorrect (pressure unit on LHS, dimensionless on RHS) and conceptually wrong for defining RLVP.

✅ Correct:

The correct expression, fundamental to Raoult's Law for non-volatile solutes, is (P⁰ - Pₛ) / P⁰ = χ_solute, where P⁰ is the vapour pressure of the pure solvent, Pₛ is the vapour pressure of the solution, and χ_solute is the mole fraction of the non-volatile solute.

💡 Prevention Tips:

  • Always remember the precise definition: 'Relative' means divided by the pure solvent's vapour pressure (P⁰). This is crucial for both CBSE and JEE Main.

  • Pay close attention to the units; RLVP is dimensionless, while ΔP has units of pressure.

  • For JEE Main problems, ensure you correctly identify whether the question asks for the absolute lowering (ΔP) or the relative lowering (ΔP/P⁰) before applying formulas or calculating unknown quantities like molar mass.

  • Practice deriving the relationship from Raoult's Law to solidify the understanding that Pₛ = P⁰ * χ_solvent, which leads to (P⁰ - Pₛ) / P⁰ = χ_solute.

JEE_Main
Important Conceptual

❌ Confusing Lowering of Vapour Pressure with Relative Lowering and Misunderstanding its Dependence

Students frequently interchange 'lowering of vapour pressure' (ΔP) with 'relative lowering of vapour pressure' (ΔP/P°). Additionally, they sometimes incorrectly assume that Relative Lowering of Vapour Pressure (RLVP) depends on the specific chemical nature of the non-volatile solute or the solvent, rather than just the number of solute particles.
💭 Why This Happens:
This confusion stems from an inadequate understanding of the definitions and the core principles of colligative properties. Students might not fully grasp that RLVP is a dimensionless ratio and a colligative property, meaning it depends only on the mole fraction of the solute, not its specific identity (as long as it's non-volatile and non-electrolyte).
✅ Correct Approach:
  • Understand that Lowering of Vapour Pressure (ΔP = P° - P_s) is the absolute decrease in vapour pressure, expressed in pressure units (e.g., mmHg, kPa).
  • Recognize that Relative Lowering of Vapour Pressure (ΔP/P° = (P° - P_s)/P°) is the ratio of the lowering to the vapour pressure of the pure solvent. This is a dimensionless quantity, directly equal to the mole fraction of the solute (χ_solute) for ideal dilute solutions according to Raoult's Law.
  • Remember that RLVP is a colligative property, depending only on the number of solute particles (mole fraction), not their specific identity, for a given solvent and temperature.
📝 Examples:
❌ Wrong:
A student calculates the vapour pressure of a solution to be 40 mmHg, while the pure solvent's vapour pressure is 50 mmHg. The student then states, 'The relative lowering of vapour pressure is 10 mmHg.'
✅ Correct:
Using the same values:
  • Lowering of Vapour Pressure (ΔP) = P° - P_s = 50 mmHg - 40 mmHg = 10 mmHg.
  • Relative Lowering of Vapour Pressure (ΔP/P°) = (P° - P_s)/P° = (50 - 40)/50 = 10/50 = 0.2.
The relative lowering is 0.2 (or 20%), a dimensionless quantity, not an absolute pressure value.
💡 Prevention Tips:
  • Distinguish Clearly: Explicitly write down the definitions and formulas for ΔP and ΔP/P° separately.
  • Focus on Raoult's Law: Internalize that for a non-volatile solute, ΔP/P° = χ_solute. This emphasizes its dependence solely on solute mole fraction.
  • Unit Awareness: Always check units. Lowering of vapour pressure has pressure units; relative lowering is dimensionless.
  • Practice Problems: Solve numericals where you are asked to calculate both, ensuring you use the correct formula and units for each.
CBSE_12th
Important Calculation

❌ Confusing 'Lowering of Vapour Pressure' with 'Relative Lowering of Vapour Pressure'

Students frequently interchange the terms 'lowering of vapour pressure' (ΔP) and 'relative lowering of vapour pressure' (ΔP/P°). This leads to incorrect application of formulas and calculation errors, especially when asked to determine molar mass of a solute or its mole fraction using Raoult's Law.
💭 Why This Happens:
This mistake stems from a lack of precise understanding of the definitions. 'Lowering' refers to the absolute decrease in vapour pressure (P° - P), while 'relative lowering' is the fractional decrease relative to the pure solvent's vapour pressure, i.e., (P° - P)/P°. Students often miss the denominator (P°) when calculating the relative lowering.
✅ Correct Approach:
Always carefully read the question to ascertain whether 'lowering of vapour pressure' or 'relative lowering of vapour pressure' is required.
  • Lowering of Vapour Pressure (ΔP) is simply the difference between the vapour pressure of the pure solvent (P°) and the vapour pressure of the solution (P): ΔP = P° - P.
  • Relative Lowering of Vapour Pressure (ΔP/P°) is the lowering of vapour pressure divided by the vapour pressure of the pure solvent: (ΔP/P°) = (P° - P) / P°. According to Raoult's Law for dilute solutions, this is equal to the mole fraction of the solute (χsolute): (P° - P) / P° = χsolute.
📝 Examples:
❌ Wrong:
If the vapour pressure of pure water (P°) is 100 mmHg and a solution has a vapour pressure (P) of 95 mmHg. A student might incorrectly state that the 'relative lowering of vapour pressure' is 5 mmHg (i.e., P° - P) when asked for relative lowering.
✅ Correct:
Using the same data: P° = 100 mmHg, P = 95 mmHg. The 'lowering of vapour pressure' (ΔP) is 100 - 95 = 5 mmHg. The 'relative lowering of vapour pressure' is (100 - 95) / 100 = 5 / 100 = 0.05 .
💡 Prevention Tips:
  • Distinguish Keywords: Pay close attention to the exact phrasing in the question. 'Lowering' vs. 'Relative Lowering' are critical.
  • Formula Recall: Clearly recall the two distinct formulas: ΔP = P° - P and (P° - P)/P° = χsolute.
  • Unit Check: Relative lowering is a dimensionless quantity (a ratio), while lowering of vapour pressure has units of pressure (e.g., mmHg, kPa, bar). This can be a quick check.
  • Practice Problems: Solve numerical problems specifically designed to test this distinction.
CBSE_12th
Important Formula

❌ Confusing the Denominator in Relative Lowering of Vapour Pressure (RLVP) Formula

Students frequently use the vapour pressure of the solution (P_s) in the denominator instead of the vapour pressure of the pure solvent (P°) when calculating the Relative Lowering of Vapour Pressure. The correct definition of RLVP is the lowering of vapour pressure relative to the pure solvent's vapour pressure, not the solution's.
💭 Why This Happens:
This mistake often arises from a superficial understanding of the term 'relative'. Students might assume 'relative to the final state' (solution) rather than 'relative to the initial reference state' (pure solvent). Hurried application of the formula without fully grasping its derivation or the meaning of each term also contributes to this error.
✅ Correct Approach:
The Relative Lowering of Vapour Pressure (RLVP) is defined as the ratio of the lowering of vapour pressure (P° - P_s) to the vapour pressure of the pure solvent (P°). According to Raoult's Law for non-volatile solutes, this is equal to the mole fraction of the solute (x_solute).

The correct formula is: RLVP = (P° - P_s) / P° = x_solute

CBSE & JEE Tip: Always remember that the denominator is the vapour pressure of the pure solvent. Using P_s in the denominator leads to an incorrect value, which can be critical for subsequent calculations like molar mass determination.
📝 Examples:
❌ Wrong:
A common incorrect application would be: RLVP = (P° - P_s) / P_s. This formula is incorrect and will yield a wrong numerical value for the relative lowering.
✅ Correct:
If the vapour pressure of pure water (P°) is 100 mmHg and the vapour pressure of an aqueous solution (P_s) is 90 mmHg, then:
Lowering of vapour pressure = P° - P_s = 100 - 90 = 10 mmHg.
Relative Lowering of Vapour Pressure = (P° - P_s) / P° = 10 / 100 = 0.1.
This correctly represents the relative lowering with respect to the pure solvent.
💡 Prevention Tips:
  • Understand the Definition: Clearly define RLVP as 'lowering relative to the pure solvent'.
  • Derivation Recall: Briefly recall the derivation of Raoult's Law which establishes the P° in the denominator.
  • Formula Memorization: Explicitly memorize the correct formula: (P° - P_s) / P°.
  • Unit Consistency: Ensure all pressure units are consistent before calculation.
  • Practice Numerical Problems: Solve various problems focusing on correct formula application to reinforce understanding.
CBSE_12th
Important Unit Conversion

❌ Inconsistent Pressure Units in Relative Lowering of Vapour Pressure Calculations

A common mistake in problems involving the Relative Lowering of Vapour Pressure (RLVP) is failing to use consistent units for vapour pressure values. Students often use the vapour pressure of the pure solvent (P°) and the vapour pressure of the solution (P_s) in different units (e.g., mmHg for one and bar for another) directly in the calculation, leading to incorrect results.
💭 Why This Happens:
This error primarily stems from a lack of attention to the units provided in the problem statement. Students might overlook the unit specified with each numerical value, or forget the necessary conversion factors between different pressure units (e.g., atm, mmHg, bar, kPa, Pa). Sometimes, there's an assumption that all given pressure values are implicitly in the same unit.
✅ Correct Approach:
Always ensure that all pressure values used in the RLVP formula, (P° - P_s) / P° = X_solute, are expressed in a single, consistent unit before performing any calculations. This might involve converting one or more given pressure values to match the unit of another, or converting all to a common standard unit like Pascals (Pa) or atmospheres (atm).
📝 Examples:
❌ Wrong:
If P° = 760 mmHg and P_s = 0.9 bar, an incorrect approach would be directly calculating (760 - 0.9) / 760. This is dimensionally incorrect and will yield a wrong answer.
✅ Correct:
Given P° = 760 mmHg and P_s = 0.9 bar.
Step 1: Convert P_s to mmHg. We know 1 bar ≈ 750.06 mmHg.
P_s = 0.9 bar × (750.06 mmHg / 1 bar) ≈ 675.05 mmHg.
Step 2: Apply the formula with consistent units.
RLVP = (760 mmHg - 675.05 mmHg) / 760 mmHg = 84.95 / 760 ≈ 0.1118
💡 Prevention Tips:
  • List Units Explicitly: When noting down given data, always write the numerical value along with its unit (e.g., P° = 760 mmHg).
  • Identify Target Unit: Decide on a common unit for all pressure values (e.g., convert everything to mmHg or Pa).
  • Memorize Conversions: Familiarize yourself with standard pressure conversion factors (e.g., 1 atm = 760 mmHg = 1.01325 bar = 101325 Pa).
  • Perform Conversions First: Complete all necessary unit conversions before substituting values into the RLVP formula.
  • JEE Specific: In competitive exams, multi-step problems might require conversions at different stages. Be extra vigilant.
CBSE_12th
Important Sign Error

❌ Confusing Terms and Incorrect Sign in Relative Lowering of Vapour Pressure

Students frequently make errors by incorrectly identifying the vapour pressure of the pure solvent (P°) and the vapour pressure of the solution (P). This often leads to an incorrect sign for the lowering of vapour pressure (ΔP). They might inadvertently use P - P° instead of P° - P, or misinterpret 'lowering' as a negative quantity directly, which is a significant conceptual flaw for CBSE 12th exams.
💭 Why This Happens:
  • Lack of Conceptual Clarity: Not fully grasping that the vapour pressure of a pure solvent is always higher than that of its solution containing a non-volatile solute.
  • Misinterpretation of 'Lowering': Students sometimes associate 'lowering' with a negative sign, leading them to subtract the larger value from the smaller one.
  • Careless Substitution: Rushing through problems and substituting values into the formula ΔP / P° = X_solute without confirming that ΔP = P° - P is positive.
✅ Correct Approach:
Always remember that the vapour pressure of a pure solvent (P°) is fundamentally higher than the vapour pressure of its solution (P). Therefore, the lowering of vapour pressure (ΔP) must always be calculated as P° - P, ensuring a positive value. Consequently, the relative lowering of vapour pressure, (P° - P) / P°, will also always be a positive fraction (or a dimensionless positive number). For JEE, this understanding is crucial for correctly interpreting multi-step problems.
📝 Examples:
❌ Wrong:
Consider a problem where the vapour pressure of pure benzene (P°) is 100 mmHg and the vapour pressure of its solution (P) is 90 mmHg. A common mistake would be to calculate the lowering of vapour pressure as:
ΔP = P - P° = 90 - 100 = -10 mmHg
This negative value for 'lowering' is incorrect and indicates a misunderstanding of the concept.
✅ Correct:
Using the same values: P° = 100 mmHg and P = 90 mmHg.
The correct calculation for the lowering of vapour pressure is:
ΔP = P° - P = 100 - 90 = 10 mmHg
The relative lowering of vapour pressure would then be 10 / 100 = 0.1, which is a positive and correct dimensionless value. This approach is essential for both CBSE and JEE numerical problems.
💡 Prevention Tips:
  • Clearly Label P° and P: Always assign P° to the pure solvent and P to the solution at the beginning of any problem.
  • Formula Reinforcement: Commit ΔP = P° - P to memory and apply it consistently.
  • Check for Positivity: After calculation, verify that ΔP and the relative lowering of vapour pressure are positive. A negative result immediately signals a sign error.
  • Conceptual Understanding: Revisit Raoult's Law and understand why a non-volatile solute always decreases the solvent's vapour pressure.
CBSE_12th
Important Other

❌ Confusing Lowering of Vapour Pressure with Relative Lowering and Incorrect Formula Application

Students frequently confuse lowering of vapour pressure (ΔP) with relative lowering of vapour pressure (RLVP). They often use ΔP (P° - P) where RLVP, (P° - P)/P°, is required. Another common error is using the incorrect denominator, e.g., (P° - P)/P instead of (P° - P)/P°. Furthermore, the crucial condition that Raoult's Law for RLVP applies to non-volatile solutes is sometimes overlooked, or the Van't Hoff factor (i) for electrolytes is ignored.
💭 Why This Happens:
This mistake stems from a lack of precise understanding of the definitions and the conditions under which Raoult's Law for RLVP is applied. Hasty formula substitution without analyzing the nature of the solute (volatile vs. non-volatile, electrolyte vs. non-electrolyte) leads to errors. The visual similarity between P° and P in the denominator can also cause confusion.
✅ Correct Approach:
Always distinguish between lowering of vapour pressure, ΔP = P° - P, and relative lowering of vapour pressure, RLVP = (P° - P)/P°. Raoult's Law for relative lowering of vapour pressure states:
(P° - P)/P° = χ_solute (for non-volatile, non-electrolyte solutes)
For solutions with a non-volatile electrolyte solute, the equation becomes:
(P° - P)/P° = i × χ_solute
where P° is the vapour pressure of the pure solvent, P is the vapour pressure of the solution, and χ_solute is the mole fraction of the solute. Remember that P° is always in the denominator for relative lowering.
📝 Examples:
❌ Wrong:
A student calculates the lowering of vapour pressure ΔP = (P° - P) and equates this directly to the mole fraction of the solute, or uses the formula as (P° - P)/P = χ_solute, or fails to multiply by 'i' for an electrolyte like NaCl.
✅ Correct:
Consider a solution of NaCl (an electrolyte) in water. To find the relative lowering of vapour pressure, the correct formula to use is (P° - P)/P° = i × χ_NaCl. Here, 'i' would be approximately 2 for complete dissociation of NaCl, and χ_NaCl is the mole fraction of NaCl.
💡 Prevention Tips:
  • Define Terms: Clearly understand the difference between 'lowering' and 'relative lowering' of vapour pressure.
  • Formula Precision: Memorize and apply the correct form of Raoult's Law for RLVP, ensuring P° is in the denominator.
  • Solute Analysis: Always identify if the solute is non-volatile or volatile. For CBSE, RLVP questions usually involve non-volatile solutes.
  • Electrolyte Check: For JEE and CBSE, always check if the solute is an electrolyte (dissociates) or a non-electrolyte. If it's an electrolyte, incorporate the Van't Hoff factor 'i'.
  • Practice: Solve numerical problems consistently, paying attention to these details.
CBSE_12th
Critical Calculation

❌ Incorrect Application of Mole Fraction in Relative Lowering of Vapour Pressure

Students often make critical calculation errors by incorrectly defining or applying the mole fraction of the solute (χsolute) when using Raoult's law for relative lowering of vapour pressure. This usually involves two main scenarios:
1. Using an approximate formula like nsolute / nsolvent instead of the precise nsolute / (nsolute + nsolvent).
2. Confusing and using the mole fraction of the solvent instead of the solute in the equation (P° - P) / P° = χsolute.
💭 Why This Happens:
This mistake stems from a fundamental misunderstanding of the definition of mole fraction and its direct application in Raoult's Law. Sometimes, students recall approximations valid only for very dilute solutions (where nsolute << nsolvent, so nsolute + nsolvent ≈ nsolvent) and apply them universally, leading to significant errors in less dilute cases. Furthermore, a lack of clarity on which component's mole fraction drives the colligative property contributes to the confusion.
✅ Correct Approach:
Always remember that the relative lowering of vapour pressure is equal to the mole fraction of the non-volatile solute. The correct and exact definition of mole fraction for the solute (χsolute) is:
χsolute = nsolute / (nsolute + nsolvent)
where nsolute is the number of moles of solute and nsolvent is the number of moles of solvent. Always use this precise formula for calculations unless explicitly stated that an approximation is acceptable (which is rare in CBSE/JEE problems requiring exact values).
📝 Examples:
❌ Wrong:
Problem: Calculate the relative lowering of vapour pressure when 18 g of glucose (Molar Mass = 180 g/mol) is dissolved in 180 g of water (Molar Mass = 18 g/mol).
Wrong Calculation:
nglucose = 18/180 = 0.1 mol
nwater = 180/18 = 10 mol
χglucose = nglucose / nwater = 0.1 / 10 = 0.01
Therefore, RLVP = 0.01
✅ Correct:
Problem: Calculate the relative lowering of vapour pressure when 18 g of glucose (Molar Mass = 180 g/mol) is dissolved in 180 g of water (Molar Mass = 18 g/mol).
Correct Calculation:
nglucose = 18 g / 180 g/mol = 0.1 mol
nwater = 180 g / 18 g/mol = 10 mol
Total moles = nglucose + nwater = 0.1 + 10 = 10.1 mol
χglucose = nglucose / (nglucose + nwater) = 0.1 / 10.1 ≈ 0.0099
Therefore, Relative Lowering of Vapour Pressure ≈ 0.0099.
Notice the difference from the wrong calculation (0.01 vs 0.0099), which can be significant in multi-step problems or for higher precision.
💡 Prevention Tips:
  • Master Definitions: Ensure a crystal-clear understanding of the definition of mole fraction and Raoult's law for relative lowering of vapour pressure.
  • Identify Components: Clearly label 'solute' and 'solvent' and their respective moles (nsolute, nsolvent) before starting calculations.
  • Formula Application: Always write down the full formula (P° - P) / P° = χsolute = nsolute / (nsolute + nsolvent) before substituting values.
  • Avoid Approximations: Unless explicitly justified or the problem context demands it (e.g., extremely dilute solutions for certain derivations), stick to the exact mole fraction formula for accuracy in CBSE/JEE exams.
  • Practice Diligently: Solve a variety of problems to solidify your understanding and avoid common pitfalls.
CBSE_12th
Critical Conceptual

❌ Confusing 'Lowering of Vapour Pressure' with 'Relative Lowering of Vapour Pressure' and Incorrectly Relating it to Solvent's Mole Fraction

A critical conceptual error students make is failing to differentiate between the lowering of vapour pressure (ΔP) and the relative lowering of vapour pressure (ΔP/P°solvent). Often, they incorrectly state that the relative lowering of vapour pressure is equal to the mole fraction of the solvent (Xsolvent), or they use Psolution in the denominator instead of P°solvent, which fundamentally misinterprets Raoult's Law for non-volatile solutes.
💭 Why This Happens:
This mistake primarily stems from:
  • Lack of clear distinction: Students often memorize formulas without understanding the precise definitions of ΔP and ΔP/P°solvent.
  • Confusion with Raoult's Law: While Raoult's Law for a non-volatile solute states Psolution = Xsolvent * P°solvent, students mistakenly extend the 'Xsolvent' part directly to relative lowering.
  • Misinterpretation of 'relative': The term 'relative' implies a ratio with respect to the initial or pure state, which is P°solvent, not Psolution.
✅ Correct Approach:
For a non-volatile solute, the correct understanding is:
1. Lowering of Vapour Pressure (ΔP): ΔP = P°solvent - Psolution = Xsolute * P°solvent
2. Relative Lowering of Vapour Pressure: ΔP / P°solvent
By substituting ΔP, we get: (Xsolute * P°solvent) / P°solvent = Xsolute
Therefore, the relative lowering of vapour pressure is directly equal to the mole fraction of the solute (Xsolute). This is a fundamental aspect of colligative properties.
📝 Examples:
❌ Wrong:
A student states, 'If the mole fraction of the solvent is 0.9, then the relative lowering of vapour pressure is 0.9.' (This is incorrect because it relates relative lowering to Xsolvent).
✅ Correct:
A student states, 'If the mole fraction of the solute is 0.1, then the relative lowering of vapour pressure is 0.1.' (This is correct, as relative lowering = Xsolute).
💡 Prevention Tips:
  • Understand Definitions: Clearly define and differentiate between 'lowering' and 'relative lowering' of vapour pressure.
  • Derive the Formula: Practice the derivation of Raoult's Law for non-volatile solutes to understand why relative lowering equals Xsolute.
  • Focus on 'Relative' Aspect: Always remember that 'relative lowering' is a ratio with respect to the vapour pressure of the pure solvent (P°solvent).
  • JEE & CBSE Focus: Both exams test this fundamental conceptual understanding, so a clear grasp is essential.
CBSE_12th
Critical Formula

❌ Incorrect Application of Denominator and Mole Fraction in Relative Lowering of Vapour Pressure Formula

Students frequently misinterpret the 'relative' aspect of Raoult's Law, often using the vapor pressure of the solution (Ps) instead of the pure solvent (P°) in the denominator of the relative lowering formula. Additionally, there's a common error of equating relative lowering to the mole fraction of the solvent (Xsolvent) instead of the mole fraction of the solute (Xsolute).
💭 Why This Happens:
  • Conceptual Confusion: Lack of clear distinction between 'lowering of vapor pressure' (P° - Ps) and 'relative lowering of vapor pressure' ((P° - Ps) / P°).
  • Misunderstanding Raoult's Law: For a non-volatile solute, Raoult's Law states that the vapor pressure of the solution is proportional to the mole fraction of the solvent (Ps = XsolventP°). Students sometimes incorrectly extend this to (P° - Ps) / P° = Xsolvent.
  • Formula Memorization without Derivation: Rote learning the formula without understanding its derivation leads to incorrect substitutions.
✅ Correct Approach:
The correct formula for relative lowering of vapor pressure (RLVP) for a non-volatile solute is:

$frac{ ext{P}^circ - ext{P}_{ ext{s}}}{ ext{P}^circ} = ext{X}_{ ext{solute}}$

Where:
  • P° is the vapor pressure of the pure solvent.
  • Ps is the vapor pressure of the solution.
  • Xsolute is the mole fraction of the solute.
This formula directly relates a colligative property (RLVP) to the number of solute particles (represented by Xsolute).
📝 Examples:
❌ Wrong:
A student attempts to calculate the relative lowering by using:

$frac{ ext{P}^circ - ext{P}_{ ext{s}}}{ ext{P}_{ ext{s}}} = ext{X}_{ ext{solute}}$

OR
$frac{ ext{P}^circ - ext{P}_{ ext{s}}}{ ext{P}^circ} = ext{X}_{ ext{solvent}}$

Both are fundamentally incorrect. The first changes the definition of 'relative', and the second confuses solute with solvent.
✅ Correct:
Given P° = 760 mm Hg, Ps = 700 mm Hg.
The correct relative lowering of vapor pressure is:

$frac{760 - 700}{760} = frac{60}{760} approx 0.0789$

This value (0.0789) is equal to the mole fraction of the solute (Xsolute).

$ ext{X}_{ ext{solute}} = 0.0789$
💡 Prevention Tips:
  • Understand the Derivation: Derive Raoult's Law for non-volatile solutes to clearly see why P° is in the denominator and why it equals Xsolute.
  • Define Terms Carefully: Always explicitly identify P°, Ps, Xsolute, and Xsolvent before substitution.
  • Practice with Both Types: Solve problems involving 'lowering of vapour pressure' (P° - Ps) and 'relative lowering of vapour pressure' ((P° - Ps) / P° = Xsolute) to distinguish them.
CBSE_12th
Critical Unit Conversion

❌ Inconsistent Mass Units in Mole Calculations (Critical for CBSE & JEE)

A very common and critical error in problems involving Relative Lowering of Vapour Pressure (RLVP) is the inconsistent use of mass units when calculating the number of moles. Students frequently use mass values directly as given in kilograms (kg) or milligrams (mg) without converting them to grams (g), while molar mass is always expressed in grams per mole (g/mol). This leads to incorrect mole values for both solute and solvent, thereby propagating significant errors throughout the entire RLVP calculation, ultimately yielding an incorrect final answer for the molar mass of the solute or the vapour pressure of the solution.
💭 Why This Happens:
  • Lack of Unit Awareness: Students often overlook the units provided in the problem statement, assuming all given masses are in grams.
  • Rushing Calculations: In an attempt to save time, unit conversions are sometimes skipped or performed incorrectly.
  • Conceptual Blurry: A weak understanding that molar mass (g/mol) necessitates mass to be in grams for mole calculations.
✅ Correct Approach:

Always ensure that the mass of the substance (solute or solvent) is consistently in grams (g) when calculating the number of moles using the formula: Moles (n) = Mass (W in g) / Molar Mass (M in g/mol).

  • If mass is given in kilograms (kg), multiply by 1000 to convert to grams (1 kg = 1000 g).
  • If mass is given in milligrams (mg), divide by 1000 to convert to grams (1 mg = 0.001 g).

This ensures that the units cancel out correctly to yield moles.

📝 Examples:
❌ Wrong:

❌ Incorrect Example:

Problem: Calculate the moles of solvent if 0.5 kg of water (Molar Mass = 18 g/mol) is used.

Incorrect Calculation:
nsolvent = Mass / Molar Mass = 0.5 kg / 18 g/mol = 0.0278 mol (This is wrong because 0.5 is not in grams).

✅ Correct:

✅ Correct Example:

Problem: Calculate the moles of solvent if 0.5 kg of water (Molar Mass = 18 g/mol) is used.

Correct Calculation:
Step 1: Convert mass to grams.
Mass of water = 0.5 kg × 1000 g/kg = 500 g

Step 2: Calculate moles.
nsolvent = Mass / Molar Mass = 500 g / 18 g/mol = 27.78 mol (Correct answer).

Using the incorrect mole value will significantly alter the calculated relative lowering of vapour pressure or the molar mass of the solute, leading to a completely wrong final answer in the board or JEE exam.

💡 Prevention Tips:
  • Read Carefully: Always read the problem statement thoroughly, paying close attention to the units of all given quantities.
  • Unit Conversion Table: Before starting calculations, list all given values and convert them into consistent base units (e.g., all masses to grams, all volumes to liters, all pressures to a single unit like bar or atm).
  • Write Units at Each Step: Include units with every numerical value during your calculation steps. This helps in visual verification that units are cancelling correctly.
  • JEE & CBSE Alert: This mistake is especially penalized in both CBSE board exams (for final answer accuracy) and JEE (where precision is key), often leading to a loss of significant marks.
CBSE_12th
Critical Sign Error

❌ Incorrect Sign in Relative Lowering of Vapour Pressure Calculations

Students frequently make a critical sign error by incorrectly calculating the 'lowering of vapour pressure' (ΔP) or 'relative lowering of vapour pressure'. This often manifests as reporting a negative value for these quantities, which are inherently positive.
💭 Why This Happens:
This mistake stems from a fundamental misunderstanding that the addition of a non-volatile solute always lowers the vapour pressure of the solvent. Students might mistakenly subtract the solvent's vapour pressure from the solution's (P - P⁰) instead of the correct (P⁰ - P), or simply fail to interpret the 'lowering' aspect correctly.
✅ Correct Approach:
Always remember that the vapour pressure of the pure solvent (P⁰) is always greater than the vapour pressure of the solution (P) containing a non-volatile solute. Therefore, the 'lowering of vapour pressure' (ΔP = P⁰ - P) and consequently the 'relative lowering of vapour pressure' ((P⁰ - P) / P⁰) must always be positive quantities.
📝 Examples:
❌ Wrong:
If the vapour pressure of pure water (P⁰) is 760 mmHg and the vapour pressure of an aqueous solution (P) is 740 mmHg, a common mistake is to calculate the relative lowering as:
(740 - 760) / 760 = -20 / 760 ≈ -0.026
The negative sign is incorrect.
✅ Correct:
Using the same values: P⁰ = 760 mmHg, P = 740 mmHg.
The correct calculation for relative lowering of vapour pressure is:
(P⁰ - P) / P⁰ = (760 - 740) / 760 = 20 / 760 ≈ 0.026
This positive value correctly represents the relative lowering.
💡 Prevention Tips:
  • Conceptual Clarity: Understand that non-volatile solutes decrease vapour pressure, making P⁰ > P.
  • Formula Mastery: Consistently use the formula Relative Lowering = (P⁰ - P) / P⁰.
  • Sign Check: Always verify that your calculated values for 'lowering' and 'relative lowering' are positive. A negative result indicates a calculation error.
  • JEE/CBSE Note: Both CBSE and JEE expect this fundamental understanding. A sign error can lead to a complete loss of marks for numerical problems.
CBSE_12th
Critical Approximation

❌ Misapplication of Dilute Solution Approximation: <span style='color: #e67e22;'>(P° - P) / P° ≈ n₂/n₁</span>

Students frequently approximate the relative lowering of vapour pressure formula from (P° - P) / P° = n₂ / (n₁ + n₂) to (P° - P) / P° ≈ n₂ / n₁. This approximation is valid *only* for very dilute solutions, where the moles of solute (n₂) are negligible compared to the moles of solvent (n₁). Applying this simplification to moderately concentrated solutions leads to significant inaccuracies in calculated molar masses or vapour pressures.
💭 Why This Happens:
  • Over-generalization: Students often assume 'dilute solution' applies to all problems without verifying the relative magnitudes of n₂ and n₁.
  • Confusion with other Colligative Properties: Some colligative properties (like elevation in boiling point or depression in freezing point) are frequently introduced in their approximate forms, valid for dilute solutions. Students mistakenly extend this level of approximation directly to RLVP without understanding its specific conditions.
  • Memorization without understanding: Relying on memorized formulas without a clear grasp of their derivation and conditions of applicability.
✅ Correct Approach:
  • The fundamental relationship for Relative Lowering of Vapour Pressure is given by Raoult's Law: (P° - P) / P° = X₂, where X₂ is the mole fraction of the solute.
  • Always use the precise definition of mole fraction: X₂ = n₂ / (n₁ + n₂).
  • Therefore, the exact formula for RLVP is (P° - P) / P° = n₂ / (n₁ + n₂).
  • The approximation (P° - P) / P° ≈ n₂ / n₁ is only valid when n₂ << n₁ (i.e., for very dilute solutions). For CBSE exams, it is generally safer to use the exact formula unless the problem explicitly states 'very dilute' and the calculated magnitudes clearly justify the approximation.
  • JEE Perspective: For JEE, understand when approximations are justified (e.g., if n₂ is less than 5-10% of n₁, approximation might be acceptable for MCQ options, but for numerical answers, precision is key unless specified).
📝 Examples:
❌ Wrong:

Problem: 20 g of a non-volatile solute (Molar Mass = 100 g/mol) is dissolved in 81 g of water (Molar Mass = 18 g/mol). Calculate the relative lowering of vapour pressure.

Incorrect Approach (using approximation):

  • Moles of water (n₁) = 81 g / 18 g/mol = 4.5 mol
  • Moles of solute (n₂) = 20 g / 100 g/mol = 0.2 mol
  • Incorrect Approximation: (P° - P) / P° ≈ n₂ / n₁ = 0.2 / 4.5 ≈ 0.0444
✅ Correct:

Correct Approach:

  • Moles of water (n₁) = 4.5 mol
  • Moles of solute (n₂) = 0.2 mol
  • Correct Formula: (P° - P) / P° = n₂ / (n₁ + n₂)
  • (P° - P) / P° = 0.2 / (4.5 + 0.2) = 0.2 / 4.7 ≈ 0.04255

Observation: The difference between 0.0444 and 0.04255 is significant (approximately 4.3% error), which can lead to incorrect final answers in examinations.

💡 Prevention Tips:
  • Always Start with Exact Formula: Begin all RLVP calculations with (P° - P) / P° = n₂ / (n₁ + n₂).
  • Check for Dilution Condition: Only use the approximation n₂ / n₁ if n₂ is extremely small (e.g., < 1% of n₁), or if the problem explicitly guides you to use an approximation for dilute solutions.
  • Differentiate Denominators: Clearly distinguish between (P° - P) / P° and (P° - P) / P. Remember that (P° - P) / P = n₂ / n₁ is an exact relationship derived from Raoult's Law, not an approximation. The approximation error discussed here pertains to the (P° - P) / P° form.
  • Practice Both Methods: Solve problems using both exact and approximate methods to develop an intuition for when the approximation introduces an unacceptable error.
CBSE_12th
Critical Other

❌ Confusing Mole Fraction of Solute vs. Solvent in Relative Lowering of Vapour Pressure

Students frequently interchange the mole fraction of the solvent (Xsolvent) with the mole fraction of the non-volatile solute (Xsolute) when applying Raoult's Law for relative lowering of vapour pressure. They also sometimes confuse the term 'lowering of vapour pressure' (ΔP) with 'relative lowering of vapour pressure' (ΔP/P°).
💭 Why This Happens:
This common error stems from an incomplete understanding of the derivation of Raoult's Law for non-volatile solutes. While Raoult's Law directly states that the vapour pressure of the solution (Psolution) is proportional to the mole fraction of the solvent (Psolution = Xsolvent * P°), students often directly apply Xsolvent to the relative lowering without going through the complete derivation. The 'relative' part, specifically 'relative to what', gets overlooked, leading to an incorrect substitution.
✅ Correct Approach:
The relative lowering of vapour pressure is defined as (P° - Psolution) / P°, where P° is the vapour pressure of the pure solvent and Psolution is the vapour pressure of the solution. For a non-volatile solute, this quantity is equal to the mole fraction of the non-volatile solute (Xsolute).
The derivation is as follows:
  • According to Raoult's Law: Psolution = Xsolvent * P°
  • Lowering of vapour pressure: ΔP = P° - Psolution = P° - (Xsolvent * P°) = P°(1 - Xsolvent)
  • Since Xsolvent + Xsolute = 1, then (1 - Xsolvent) = Xsolute
  • Therefore, ΔP = P° * Xsolute
  • And Relative Lowering of Vapour Pressure = ΔP/P° = Xsolute.
It is crucial to use Xsolute for the relative lowering.
📝 Examples:
❌ Wrong:
A student is asked to calculate the relative lowering of vapour pressure for an aqueous solution where the mole fraction of water (solvent) is 0.95. They might incorrectly state that the relative lowering is 0.95.
✅ Correct:
Given the mole fraction of the solvent (Xsolvent) = 0.95.
The mole fraction of the non-volatile solute (Xsolute) = 1 - Xsolvent = 1 - 0.95 = 0.05.
Therefore, the relative lowering of vapour pressure = Xsolute = 0.05.
💡 Prevention Tips:
  • Understand the Derivation: Always recall the full derivation from Psolution = Xsolvent * P° to ΔP/P° = Xsolute. This reinforces why Xsolute is used.
  • Mnemonic: Remember that 'Relative Lowering' is directly equal to the 'Mole Fraction of Solute'.
  • JEE Focus: While approximate formulas for dilute solutions (ΔP/P° ≈ nsolute / Nsolvent) are common, the fundamental identity ΔP/P° = Xsolute is key to avoid conceptual errors.
  • CBSE Focus: Be prepared to clearly state and explain Raoult's Law, define relative lowering, and show its relation to the mole fraction of the non-volatile solute.
CBSE_12th
Critical Conceptual

❌ Ignoring Van't Hoff Factor (i) for Electrolyte Solutes

Students frequently overlook the dissociation of electrolyte solutes when applying the formula for relative lowering of vapour pressure. This leads to an incorrect calculation of the effective mole fraction of the solute particles, thereby producing an erroneous result for the colligative property.
💭 Why This Happens:
  • Lack of clear understanding that colligative properties depend solely on the number of solute particles, not their chemical identity.
  • Tendency to apply the basic formula (P° - P)/P° = Xsolute, which is strictly for non-electrolytes, without considering the solute's nature.
  • Difficulty in identifying electrolytes or correctly determining their Van't Hoff factor (i).
✅ Correct Approach:
Always identify if the solute is an electrolyte (dissociates into ions) or a non-electrolyte (does not dissociate). If it's an electrolyte, calculate the Van't Hoff factor (i), which represents the number of particles formed per formula unit. The modified formula for electrolytes is: (P° - P) / P° = i × Xsolute, where P° is the vapour pressure of the pure solvent, P is the vapour pressure of the solution, and Xsolute is the mole fraction of the solute (calculated based on its initial molar amount).
📝 Examples:
❌ Wrong:
A student calculates the relative lowering of vapour pressure for a 0.1 molal aqueous solution of NaCl by directly using XNaCl (calculated from 0.1 mol NaCl and 1000g H₂O). They ignore that NaCl dissociates into Na⁺ and Cl⁻ ions, effectively doubling the number of solute particles.
✅ Correct:
For the 0.1 molal aqueous NaCl solution, the student should first recognize that NaCl is an electrolyte and dissociates as NaCl → Na⁺ + Cl⁻. Assuming complete dissociation, the Van't Hoff factor i = 2. Therefore, the relative lowering of vapour pressure must be calculated using (P° - P) / P° = 2 × XNaCl , where XNaCl is the mole fraction of NaCl calculated from its initial molar amount. This correctly accounts for the increased number of particles.
💡 Prevention Tips:
  • Critical Check: Before applying any colligative property formula, determine if the solute is an electrolyte or non-electrolyte.
  • For Electrolytes: Always find the Van't Hoff factor (i). Remember, for strong electrolytes, 'i' equals the number of ions produced per formula unit (e.g., NaCl → 2, CaCl₂ → 3). For non-electrolytes (like glucose, urea), i = 1.
  • Conceptual Clarity: Understand that colligative properties are *number-dependent*. Electrolytes contribute more 'effective' particles due to dissociation, directly impacting the magnitude of the property.
JEE_Main
Critical Other

❌ <span style='color: red;'>Ignoring van't Hoff factor (i) for Electrolytes or Associating Solutes</span>

Students frequently apply the formula for relative lowering of vapour pressure, (P0 - Ps)/P0 = χsolute, without considering the actual behavior of the solute in solution. If the solute is an electrolyte (e.g., NaCl, CaCl2) that dissociates, or an associating solute (e.g., acetic acid in benzene) that polymerizes, the effective number of solute particles changes. Neglecting the van't Hoff factor (i) for such solutes leads to significant errors in calculations for RLVP and derived quantities like molar mass or degree of dissociation/association.
💭 Why This Happens:
  • Lack of conceptual clarity: Not fully understanding that colligative properties depend on the number of solute particles, not just the initial moles added.
  • Over-reliance on basic formulas: Blindly applying the standard Raoult's Law (valid for non-electrolytes/non-associating solutes) without analyzing the solute's nature.
  • Overlooking subtle cues: Failing to identify keywords like "electrolyte," "dissociates," "associates," "ionic compound," or mentions of specific solvents in the problem statement.
✅ Correct Approach:
For electrolytes or associating solutes, the van't Hoff factor (i) must be incorporated into the RLVP formula. The corrected formula is: (P0 - Ps)/P0 = i × χsolute.
Here, χsolute = nsolute / (nsolute + nsolvent), where nsolute is the initial moles of solute added. The factor 'i' accounts for the actual number of particles formed in solution per molecule of solute added.
  • For dissociation: i = 1 + (N-1)α, where N is the number of ions produced per molecule, and α is the degree of dissociation.
  • For association: i = 1 + (1/N-1)α, where N is the number of molecules associating (e.g., 2 for dimerization), and α is the degree of association.
📝 Examples:
❌ Wrong:
A student calculates the relative lowering of vapour pressure for a 0.1 m aqueous solution of NaCl as if it were a non-electrolyte. They use (P0 - Ps)/P0 = χNaCl directly. This approach ignores that NaCl dissociates into Na+ and Cl- ions, effectively doubling the number of particles, thus yielding an approximately half-value of the actual RLVP.
✅ Correct:
For a 0.1 m aqueous solution of NaCl (a strong electrolyte), assuming complete dissociation (α=1), NaCl dissociates into 2 ions (N=2). Thus, the van't Hoff factor i = 1 + (2-1)×1 = 2. The correct relative lowering of vapour pressure must be calculated using (P0 - Ps)/P0 = i × χNaCl = 2 × χNaCl. If the problem asks for the experimental molar mass of the solute, neglecting 'i' would give a molar mass that is half the actual molar mass (for NaCl).
💡 Prevention Tips:
  • Always identify solute nature: Before applying any colligative property formula, determine if the solute is an electrolyte (strong/weak acid/base/salt) or a molecule known to associate (e.g., carboxylic acids in non-polar solvents).
  • Read questions carefully: Scrutinize problem statements for keywords like "dissociates," "associates," "electrolyte," "ionic compound," or any mention of degree of dissociation/association.
  • Incorporate van't Hoff factor: If it's an electrolyte or an associating solute, remember to multiply the calculated colligative property by 'i'. For non-electrolytes that don't associate, i=1.
  • Practice with varied problems: Solve numerical problems involving both ideal and non-ideal (due to i-factor) solutions for all colligative properties to reinforce the concept.
JEE_Advanced
Critical Approximation

❌ Misapplying Dilute Solution Approximation for Relative Lowering of Vapour Pressure

Students frequently use the approximate form (P0 - P) / P0 ≈ nsolute / nsolvent for relative lowering of vapour pressure, even when the solution is not sufficiently dilute. This error can lead to a significant deviation from the correct answer, which is crucial in JEE Advanced where options are often designed to catch such mistakes.
💭 Why This Happens:
This mistake arises from an incomplete understanding of Raoult's Law and the conditions under which its derived approximations are valid. The precise expression for relative lowering of vapour pressure is (P0 - P) / P0 = Xsolute = nsolute / (nsolute + nsolvent). The approximation (P0 - P) / P0 ≈ nsolute / nsolvent is valid *only* when the solution is very dilute (i.e., nsolute << nsolvent), allowing nsolute to be neglected in the denominator of the mole fraction. Students often overlook this critical condition.
✅ Correct Approach:
  • Always begin with the precise form of Raoult's Law for relative lowering of vapour pressure: (P0 - P) / P0 = Xsolute = nsolute / (nsolute + nsolvent).
  • Use the approximation (P0 - P) / P0 ≈ nsolute / nsolvent only if the problem explicitly states the solution is highly dilute, or if calculations clearly show that the mole fraction of the solute is very small (e.g., Xsolute < 0.01-0.05).
  • For moderately concentrated or non-dilute solutions, the exact mole fraction of the solute must be used without any approximation in the denominator.
📝 Examples:
❌ Wrong:

Wrong Calculation of Molar Mass from RLVAP:

Given: Relative lowering of vapour pressure = 0.02, mass of solute = 10g, mass of solvent (water) = 90g.

Incorrect Approach (assuming dilute solution directly):
(P0 - P) / P0 = (wsolute/Msolute) / (wsolvent/Msolvent)
0.02 = (10/Msolute) / (90/18)
0.02 = (10/Msolute) / 5
Msolute = 10 / (5 * 0.02) = 10 / 0.1 = 100 g/mol

(Here, Xsolute = 0.02 is not small enough to neglect 'nsolute' in the denominator in precise calculations for JEE Advanced.)

✅ Correct:

Correct Calculation of Molar Mass from RLVAP:

Using the same data:

Correct Approach (using the exact formula):
(P0 - P) / P0 = (wsolute/Msolute) / [(wsolute/Msolute) + (wsolvent/Msolvent)]
0.02 = (10/Msolute) / [(10/Msolute) + (90/18)]
Let nsolute = 10/Msolute. nsolvent = 5 mol.
0.02 = nsolute / (nsolute + 5)
0.02(nsolute + 5) = nsolute
0.02nsolute + 0.1 = nsolute
0.1 = 0.98nsolute
nsolute = 0.1 / 0.98 ≈ 0.102 mol
Msolute = 10 / 0.102 ≈ 98.04 g/mol

The difference (100 vs 98.04 g/mol) is significant and demonstrates how misapplying approximations can lead to incorrect answers in JEE Advanced.

💡 Prevention Tips:
  • Always Check Dilution: Before applying any approximation, carefully read the problem statement to confirm if the solution is genuinely dilute. If not, stick to the exact formula.
  • Understand the Denominator: Remember that the mole fraction of solute, Xsolute, is nsolute / (nsolute + nsolvent). The approximation comes from neglecting nsolute in the denominator.
  • JEE Advanced Alert: JEE Advanced problems are designed to test your conceptual clarity. Blindly using approximations without checking conditions is a common trap.
  • Practice Both Forms: Solve problems using both the exact and approximate formulas to develop an intuitive understanding of when each is appropriate and the magnitude of error introduced by approximation.
JEE_Advanced
Critical Unit Conversion

❌ Inconsistent Unit Usage for Pressure and Mass in Relative Lowering of Vapour Pressure Calculations

Students frequently make critical errors by using inconsistent units for pressure (e.g., mixing mmHg and atm) or mass/molar mass (e.g., using molar mass in kg/mol when other masses are in grams) when applying Raoult's Law for relative lowering of vapour pressure. This leads to incorrect numerical values, as the formula relies on a ratio of pressures or mole fractions derived from consistent mass units.
💭 Why This Happens:
This mistake primarily occurs due to a lack of meticulous attention to detail during problem-solving, especially in the high-pressure environment of the JEE Advanced exam. Questions are often designed with mixed units to test this specific understanding. Students might rush through the problem, assume units will cancel out, or fail to convert all given values to a common, consistent unit system before performing calculations. Sometimes, the molar mass is implicitly assumed in g/mol, but the given mass of solute/solvent might be in kg.
✅ Correct Approach:
The correct approach is to always convert all quantities to a consistent set of units before starting any calculations. For pressure, choose one unit (ee.g., Pascal, mmHg, atm) and convert all given pressures to that unit. For mass, ensure all masses (solute, solvent) are in the same unit (e.g., grams) when calculating moles, which means molar mass should be in g/mol. This ensures that ratios and differences are dimensionally consistent and yield the correct result. For JEE Advanced, explicitly writing down units with each value helps prevent oversight.
📝 Examples:
❌ Wrong:

Consider a problem where pure solvent vapour pressure (P°) is given as 760 mmHg and the solution vapour pressure (Ps) is given as 0.9 atm. A common mistake would be to calculate the relative lowering directly as:

(P° - Ps) / P° = (760 - 0.9) / 760

This calculation is fundamentally flawed because it subtracts a value in atmospheres from a value in millimeters of mercury, leading to a meaningless numerical result.

✅ Correct:

Using the same problem: P° = 760 mmHg, Ps = 0.9 atm.

First, convert all pressures to a consistent unit. Let's convert atm to mmHg (since 1 atm = 760 mmHg):

Ps = 0.9 atm * 760 mmHg/atm = 684 mmHg

Now, calculate the relative lowering of vapour pressure:

(P° - Ps) / P° = (760 mmHg - 684 mmHg) / 760 mmHg
= 76 / 760 = 0.1

This is the correct approach, ensuring consistent units throughout the calculation.

💡 Prevention Tips:
  • Read Carefully: Always pay close attention to the units specified for each quantity in the problem statement.
  • Standardize Units: Before any calculation, convert all given values to a single, consistent system of units (e.g., SI units, or a system chosen for convenience like mmHg for all pressures).
  • Write Units Explicitly: During calculations, write down the units alongside the numerical values. This visual aid helps in identifying inconsistencies.
  • Check Dimensionality: After setting up the formula, mentally or physically check if the units on both sides of the equation or within an operation are compatible.
JEE_Advanced
Critical Formula

❌ Incorrect application of Relative Lowering of Vapour Pressure (RLVP) formula

Students frequently make critical errors in understanding and applying the correct formula for Relative Lowering of Vapour Pressure. The most common mistake is confusing the mole fraction of the solute with that of the solvent, or using the vapor pressure of the solution (P) instead of the pure solvent (P°) in the denominator for the 'relative' aspect.
💭 Why This Happens:
This mistake primarily stems from a conceptual misunderstanding of what 'relative lowering' truly signifies. Many students memorize the formula without grasping that the relative lowering is proportional to the concentration of solute particles, expressed as its mole fraction. Additionally, confusing P and P° in the denominator indicates a lack of clarity regarding the reference state (pure solvent) for the 'relative' calculation. Hasty reading of questions or not paying attention to which component's mole fraction is being calculated also contributes.
✅ Correct Approach:
The correct formula for Relative Lowering of Vapour Pressure (RLVP), a colligative property, is:
(P° - P) / P° = X_solute
Where:
  • P° = Vapour pressure of the pure solvent
  • P = Vapour pressure of the solution
  • X_solute = Mole fraction of the solute (moles of solute / total moles in solution)
Key Point for JEE Advanced: If the solute is electrolytic (associating or dissociating), remember to incorporate the van't Hoff factor (i) into the mole fraction of the solute: (P° - P) / P° = i * X_solute. For CBSE, simple non-electrolytes are common, but JEE requires 'i' understanding.
📝 Examples:
❌ Wrong:
A student calculates the relative lowering as (P° - P) / P = X_solute, or mistakenly uses X_solvent in place of X_solute, leading to a completely incorrect result. For instance, if given the mole fraction of solvent (X_solvent), they might use it directly instead of calculating X_solute = 1 - X_solvent.
✅ Correct:
Consider a non-volatile solute in a solvent. If the vapor pressure of pure water (P°) is 760 mmHg and the solution's vapor pressure (P) is 750 mmHg, the correct relative lowering of vapor pressure is:
RLVP = (760 - 750) / 760 = 10 / 760 = 0.01316.
This value is equal to the mole fraction of the solute (X_solute).
If a student instead calculated (760-750)/750, they would get 10/750 = 0.01333, which is incorrect.
💡 Prevention Tips:
  • Thoroughly understand definitions: Clearly differentiate between lowering of vapor pressure (P° - P) and relative lowering of vapor pressure ((P° - P) / P°).
  • Identify components: Always remember that RLVP is a function of the mole fraction of the solute.
  • Check the denominator: The 'relative' aspect means dividing by the vapor pressure of the pure solvent (P°), not the solution.
  • Practice with 'i': For JEE Advanced, actively practice problems involving the van't Hoff factor 'i' to correctly calculate the effective mole fraction of solute.
  • Read carefully: Pay close attention to whether the problem provides the mole fraction of solute or solvent.
JEE_Advanced
Critical Calculation

❌ <span style='color: #FF0000;'>Misapplication of van't Hoff Factor ('i') and Mole Fraction in RLVP Calculations</span>

Students frequently make critical errors by either neglecting the van't Hoff factor 'i' for electrolytic solutes or misapplying it in the Relative Lowering of Vapour Pressure (RLVP) formula. A common mistake involves incorrectly formulating the mole fraction of the solute, especially when considering the 'very dilute' solution approximation.
💭 Why This Happens:
  • Conceptual Confusion: Forgetting that RLVP, a colligative property, depends on the total number of particles in solution. For electrolytes, this necessitates accounting for the van't Hoff factor 'i' due to dissociation or association.
  • Formula Misinterpretation: Confusion arises between the exact mole fraction formula for the solute, $chi_{solute} = frac{i cdot n_{solute}}{i cdot n_{solute} + n_{solvent}}$, and its common approximation for very dilute solutions, $chi_{solute} approx frac{i cdot n_{solute}}{n_{solvent}}$. Incorrectly choosing or applying these can lead to errors.
✅ Correct Approach:
  • The fundamental relationship for Relative Lowering of Vapour Pressure is $frac{P^0 - P_s}{P^0} = chi_{solute}$.
  • For non-volatile electrolytic solutes, always calculate the effective moles of solute: $n'_{solute} = i imes n_{solute}$.
  • The exact mole fraction of solute, which equals the RLVP, is then: $chi_{solute} = frac{n'_{solute}}{n'_{solute} + n_{solvent}}$.
  • JEE Advanced Tip: The approximation $frac{P^0 - P_s}{P^0} approx frac{i imes n_{solute}}{n_{solvent}}$ is only valid and should be used if the problem explicitly states 'very dilute solution' or if the context clearly implies such an approximation. Otherwise, always use the exact formula for precision, as JEE Advanced problems often require it.
📝 Examples:
❌ Wrong:

Calculate RLVP for a solution containing 0.1 mol NaCl (assuming i=2) in 10 mol water.

  • Common Error 1 (Ignoring 'i'): Treating NaCl as a non-electrolyte:

    $RLVP = chi_{solute} = frac{0.1}{0.1 + 10} = frac{0.1}{10.1} approx 0.0099$. (Incorrect)

  • Common Error 2 (Partial 'i' application): Applying 'i' only to the numerator, but not considering effective moles in the denominator:

    $RLVP = chi_{solute} = frac{0.1 imes 2}{0.1 + 10} = frac{0.2}{10.1} approx 0.0198$. (Incorrect, as total moles in solution should also reflect effective solute moles).

✅ Correct:

Using the same data (0.1 mol NaCl (i=2) in 10 mol water):

  • Correct Calculation (Exact Method):

    Effective moles of solute ($n'_{solute}$) = $i imes n_{NaCl} = 2 imes 0.1 = 0.2$ mol.

    Moles of solvent ($n_{water}$) = 10 mol.

    RLVP = $chi_{solute} = frac{n'_{solute}}{n'_{solute} + n_{solvent}} = frac{0.2}{0.2 + 10} = frac{0.2}{10.2} approx 0.0196$.

  • Correct Calculation (Very Dilute Approximation, if applicable):

    If the problem states 'very dilute', then $RLVP approx frac{n'_{solute}}{n_{solvent}} = frac{0.2}{10} = 0.02$. (This simplified form is common for ideal dilute solutions where $n'_{solute} ll n_{solvent}$).

💡 Prevention Tips:
  • Identify Solute Type: Always check if the solute is an electrolyte (dissociates/associates) or a non-electrolyte.
  • Calculate 'i' Accurately: If the solute is an electrolyte, determine the correct van't Hoff factor 'i' (considering degree of dissociation/association), if not provided.
  • Use Effective Moles Consistently: For electrolytes, use $n'_{solute} = i imes n_{solute}$ consistently in both the numerator and denominator of the mole fraction expression for all colligative property calculations.
  • Exact vs. Approximate: Always prefer the exact mole fraction formula for RLVP unless the problem explicitly states or the options clearly suggest a 'very dilute solution' approximation.
JEE_Advanced
Critical Conceptual

❌ Ignoring Van't Hoff Factor (i) for Electrolytic Solutes

Students frequently forget to account for the dissociation or association of electrolytic solutes when calculating the relative lowering of vapour pressure, treating them as non-electrolytes. This leads to an incorrect number of particles in the solution and thus, an inaccurate calculation of the solute's mole fraction.
💭 Why This Happens:
This error stems from a fundamental misunderstanding that colligative properties, including Relative Lowering of Vapour Pressure (RLVP), depend on the number of solute particles, not their chemical identity. While many problems for CBSE boards might focus on non-electrolytes (where i=1), JEE Advanced questions almost always involve electrolytes to test this conceptual depth. Students often mechanically apply Raoult's law without considering the effective number of particles.
✅ Correct Approach:
For electrolytic solutes (which dissociate into ions) or associating solutes (which form larger molecules), the Van't Hoff factor (i) must be incorporated into Raoult's law. The formula for RLVP becomes:
(P° - P) / P° = i * xsolute
where xsolute is the mole fraction of the solute calculated assuming no dissociation/association, and i accounts for the effective number of particles. For ideal electrolytes, i is typically the number of ions produced per formula unit (e.g., for NaCl, i=2; for CaCl₂, i=3). For real solutions, 'i' can be less due to incomplete dissociation, requiring further calculation.
📝 Examples:
❌ Wrong:
Calculating RLVP for a 0.1 M NaCl solution by taking xsolute = 0.1 / (0.1 + nsolvent) directly, effectively assuming i=1. This ignores that each NaCl unit produces two particles (Na⁺ and Cl⁻).
✅ Correct:
For a 0.1 M NaCl solution, the effective mole fraction of solute particles should be calculated using i=2. Thus, the effective number of moles of solute particles would be 2 * 0.1 = 0.2 moles, and the formula becomes (P° - P) / P° = 2 * (0.1 / (0.2 + nsolvent)). The Van't Hoff factor multiplies the mole fraction of the solute, effectively increasing the number of particles.
💡 Prevention Tips:
  • Always check the nature of the solute: Is it an electrolyte (ionic compound, strong acid/base) or a non-electrolyte (sugar, urea, ethanol)?
  • Calculate 'i' for electrolytes: Determine the number of ions formed upon dissociation (e.g., K₂SO₄ → 2K⁺ + SO₄²⁻, so i=3). If the degree of dissociation is given, use i = 1 + (n-1)α.
  • Remember the conceptual basis: Colligative properties depend on the number of particles, not their size or identity.
  • Practice JEE Advanced level problems: These almost always incorporate the Van't Hoff factor for electrolytes.
JEE_Advanced
Critical Calculation

❌ Incorrect Calculation of Mole Fraction of Solute (X_solute)

Students frequently make critical errors in calculating the mole fraction of the solute (Xsolute), which is directly equal to the relative lowering of vapour pressure according to Raoult's Law. This mistake often arises from either confusing the moles of solute with the moles of solvent, or incorrectly applying approximations in dilute solutions, leading to a wrong value for Xsolute and consequently an incorrect relative lowering or solute molar mass.
💭 Why This Happens:
This error stems from:
  • Conceptual Confusion: Not clearly distinguishing between moles of solute (nsolute) and moles of solvent (Nsolvent).
  • Formula Misapplication: Using the mole fraction of solvent (Xsolvent) instead of solute (Xsolute) for RLVP.
  • Approximation Errors: While for very dilute solutions, Xsolute ≈ nsolute/Nsolvent is sometimes used, students may apply this approximation incorrectly in moderately dilute or concentrated solutions, or misuse it when the exact formula is required, especially in molar mass calculations.
  • Carelessness: Simple arithmetic errors or swapping values during calculation.
✅ Correct Approach:

The relative lowering of vapour pressure (RLVP) is strictly equal to the mole fraction of the solute (Xsolute).
The exact formula must always be used unless explicitly stated otherwise or when solving for an approximate value in very dilute solutions.

  • Exact Formula: (P0 - Ps) / P0 = Xsolute = nsolute / (nsolute + Nsolvent)
    Where:
    P0 = Vapour pressure of pure solvent
    Ps = Vapour pressure of solution
    nsolute = Moles of solute = (mass of solute / molar mass of solute)
    Nsolvent = Moles of solvent = (mass of solvent / molar mass of solvent)
  • JEE Tip: For JEE, always aim for the exact formula unless the problem specifically guides towards a dilute solution approximation for simplifying calculations (e.g., when determining molar mass of non-volatile solute in very dilute solutions, where nsolute + Nsolvent ≈ Nsolvent). However, for RLVP itself, Xsolute = nsolute / (nsolute + Nsolvent) is precise.
📝 Examples:
❌ Wrong:

Problem: A solution contains 10 g of a non-volatile solute (Molar Mass = 100 g/mol) in 90 g of water (Molar Mass = 18 g/mol). Calculate the relative lowering of vapour pressure.

Wrong Calculation:
nsolute = 10 / 100 = 0.1 mol
Nsolvent = 90 / 18 = 5 mol
Student mistakenly calculates Xsolute as nsolute / Nsolvent = 0.1 / 5 = 0.02 (using the dilute solution approximation incorrectly or confusing the denominator).

✅ Correct:

Correct Calculation:
nsolute = 10 / 100 = 0.1 mol
Nsolvent = 90 / 18 = 5 mol
Using the exact formula for Xsolute:
Xsolute = nsolute / (nsolute + Nsolvent)
Xsolute = 0.1 / (0.1 + 5)
Xsolute = 0.1 / 5.1
Xsolute = 0.0196 (approx.)
Thus, the relative lowering of vapour pressure is 0.0196.

💡 Prevention Tips:
  • Understand Definitions: Clearly know what 'mole fraction of solute' means: moles of solute divided by total moles (solute + solvent).
  • Write Formulas: Always write down the complete formula for Raoult's Law (RLVP = Xsolute = nsolute / (nsolute + Nsolvent)) before substituting values.
  • Identify Variables: Explicitly identify nsolute and Nsolvent from the problem statement.
  • Check Dilution: Be cautious with dilute solution approximations. For JEE, unless the problem is specifically designed for such approximation, use the exact formula.
  • Practice Regularly: Solve a variety of problems to solidify your understanding and application of the formula.
JEE_Main
Critical Formula

❌ Incorrect Application of Relative Lowering of Vapour Pressure (RLVP) Formula

A common critical mistake is misapplying Raoult's Law for Relative Lowering of Vapour Pressure. Students frequently interchange the vapor pressure of the pure solvent (P°) with the vapor pressure of the solution (P) in the denominator of the RLVP expression. Additionally, they might incorrectly use the mole fraction of the solvent (χ_solvent) instead of the mole fraction of the non-volatile solute (χ_solute).
💭 Why This Happens:
This error stems from a fundamental misunderstanding of the term 'relative lowering' – it means relative to the *pure solvent's* vapour pressure, not the solution's. Algebraic manipulation errors during derivation or simply rote memorization without conceptual clarity often lead to using 'P' instead of 'P°' in the denominator. The confusion between solute and solvent mole fractions arises from not understanding that the *lowering* is caused by the *solute* particles.
✅ Correct Approach:
The correct formula for Relative Lowering of Vapour Pressure (RLVP) is derived from Raoult's Law for non-volatile solutes and is given by:
(P° - P) / P° = χ_solute
Where:
  • P° is the vapour pressure of the pure solvent.
  • P is the vapour pressure of the solution.
  • χ_solute is the mole fraction of the non-volatile solute.

Remember, the 'relative' part implies comparison to the initial (pure solvent) state.
📝 Examples:
❌ Wrong:
A student might incorrectly write the formula as:
(P° - P) / P = χ_solute
Or even more erroneously:
(P° - P) / P° = χ_solvent
This will lead to an incorrect calculated value for the molar mass of the solute or the vapour pressure of the solution.
✅ Correct:
Given pure solvent vapour pressure (P°) = 100 mmHg, solution vapour pressure (P) = 90 mmHg.
The correct Relative Lowering of Vapour Pressure is:
(P° - P) / P° = (100 - 90) / 100 = 10 / 100 = 0.1
This value is equal to the mole fraction of the solute (χ_solute).

JEE Main Tip: This formula is fundamental. Any algebraic error here will cascade, resulting in incorrect final answers, especially in problems involving molar mass determination or colligative properties.
💡 Prevention Tips:
  • Understand the Derivation: Don't just memorize. Understand how Raoult's Law (P = P°χ_solvent) leads to the RLVP formula.
  • Identify Variables Clearly: Always distinguish between P° (pure solvent) and P (solution) vapour pressures.
  • Focus on Solute: Relative lowering is directly caused by and proportional to the mole fraction of the *non-volatile solute*.
  • Practice: Solve various numerical problems, writing down the formula correctly each time until it's ingrained.
  • Cross-check Units: While not a formula error directly, ensure consistency in pressure units.
JEE_Main
Critical Sign Error

❌ Sign Error and Incorrect Denominator in Relative Lowering of Vapour Pressure

Students frequently make sign errors when calculating the 'lowering of vapour pressure' (P⁰ - P) by mistakenly subtracting the initial (solvent) from the final (solution) pressure (P - P⁰). Additionally, a common error in 'relative lowering of vapour pressure' is using the vapour pressure of the solution (P) in the denominator instead of the pure solvent (P⁰). These errors lead to incorrect signs or magnitudes, rendering the entire calculation wrong.
💭 Why This Happens:
  • Conceptual Confusion: A lack of clear understanding that adding a non-volatile solute *reduces* the vapour pressure, meaning P⁰ (pure solvent) is always greater than P (solution).
  • Memorization without Understanding: Rote learning of the formula without grasping the physical significance of 'lowering' (always a positive value) and 'relative' (referencing the original state).
  • Haste and Pressure: During exams, students might hastily swap terms or misrecall the denominator due to time constraints or stress.
✅ Correct Approach:
Always remember the fundamental principle: vapour pressure of the pure solvent (P⁰) is *always greater than* the vapour pressure of the solution (P).
  • Lowering of Vapour Pressure: This is defined as ΔP = P⁰ - P. This value must always be positive.
  • Relative Lowering of Vapour Pressure: According to Raoult's Law, this is given by (P⁰ - P) / P⁰ = x_solute. The denominator *must always be P⁰* (vapour pressure of the pure solvent).
📝 Examples:
❌ Wrong:
If P⁰ = 50 mmHg and P = 40 mmHg:
  • Incorrect Lowering: P - P⁰ = 40 - 50 = -10 mmHg (Incorrect sign).
  • Incorrect Relative Lowering: (P⁰ - P) / P = (50 - 40) / 40 = 10 / 40 = 0.25 (Incorrect denominator).
✅ Correct:
Using the same values: P⁰ = 50 mmHg and P = 40 mmHg:
  • Correct Lowering: P⁰ - P = 50 - 40 = 10 mmHg.
  • Correct Relative Lowering: (P⁰ - P) / P⁰ = (50 - 40) / 50 = 10 / 50 = 0.2.
This value (0.2) is then equal to the mole fraction of the solute (x_solute).
💡 Prevention Tips:
  • Check the Sign: The lowering of vapour pressure (P⁰ - P) must *always* yield a positive value. If it's negative, you've swapped P⁰ and P.
  • Remember the Reference: 'Relative' implies comparison to the original state. Hence, the denominator for relative lowering is always the vapour pressure of the *pure solvent* (P⁰).
  • Conceptual Understanding: Visualize the process. Solute particles occupy surface area, reducing the escape of solvent molecules, thus lowering vapour pressure. P⁰ > P is a fundamental outcome.
  • Practice with Simple Values: Work through problems by assigning simple numbers to P⁰ and P to reinforce the correct application of the formula.
JEE_Main
Critical Approximation

❌ Ignoring or Misapplying the Dilute Solution Approximation in RLVVP Calculations

Students frequently overlook the critical approximation for Relative Lowering of Vapour Pressure (RLVVP) when dealing with dilute solutions. The exact formula is (P° - P)/P° = nsolute / (nsolute + nsolvent). However, for dilute solutions, where the moles of solute (nsolute) are significantly less than the moles of solvent (nsolvent), the denominator can be approximated as nsolvent. Thus, the formula simplifies to (P° - P)/P° ≈ nsolute / nsolvent. Failing to use this approximation when appropriate often leads to unnecessarily complex calculations or, worse, incorrect answers in time-bound JEE Main exams.
💭 Why This Happens:
This mistake stems from a lack of deep understanding of the conditions under which Raoult's Law for colligative properties is most accurately applied, i.e., for ideal dilute solutions. Students might stick rigidly to the exact formula, unaware that the approximation is not only permissible but often intended in exam problems to simplify arithmetic. It can also be due to poor conceptual clarity on when a solution is considered 'dilute enough' for this approximation to be valid.
✅ Correct Approach:
For any problem involving RLVVP, first assess if the solution is dilute. Indications include a very small RLVVP value (e.g., less than 0.05 or 0.1), or explicitly stated low concentrations of solute. In such cases, confidently apply the dilute solution approximation: (P° - P)/P° = nsolute / nsolvent. This not only simplifies the algebra, preventing the need to solve complex quadratic-like equations for molar mass but also aligns with the typical expectations for JEE Main questions designed to test this understanding. This approximation is accurate enough for most JEE problems.
📝 Examples:
❌ Wrong:

Problem: A non-volatile solute 'X' causes a relative lowering of vapor pressure of 0.02 when 5g of it is dissolved in 100g of water. Calculate the molar mass of X (Mwater = 18 g/mol).

Wrong Approach: Using the exact formula:
0.02 = (5/M) / (5/M + 100/18)
0.02 = (5/M) / (5/M + 5.55)
0.02(5/M + 5.55) = 5/M
0.1/M + 0.111 = 5/M
0.111 = 4.9/M
M = 4.9 / 0.111 ≈ 44.14 g/mol. (More complex algebra, prone to calculation errors, and not the intended method for dilute solutions.)

✅ Correct:

Problem: A non-volatile solute 'X' causes a relative lowering of vapor pressure of 0.02 when 5g of it is dissolved in 100g of water. Calculate the molar mass of X (Mwater = 18 g/mol).

Correct Approach: Since RLVVP = 0.02 (which is small), the solution is dilute. Apply the approximation:
RLVVP ≈ nsolute / nsolvent
Moles of water (nsolvent) = 100g / 18 g/mol ≈ 5.55 mol
0.02 ≈ (5g / MX) / 5.55 mol
0.02 * 5.55 = 5 / MX
0.111 = 5 / MX
MX = 5 / 0.111 ≈ 45.05 g/mol. (Simpler calculation, leading to the expected answer in JEE. Note that the difference between 44.14 and 45.05 can lead to choosing a wrong option.)

💡 Prevention Tips:
  • Identify Dilute Solutions: Always check if the given RLVVP value is small or if the problem describes a low concentration of solute. This is your cue to use the approximation.
  • Understand the 'Why': Remember that in dilute solutions, nsolute is negligible compared to nsolvent, making (nsolute + nsolvent) ≈ nsolvent.
  • Practice Both Ways: For initial practice, solve some problems using both the exact and approximate methods to appreciate the simplification and the minimal difference in results for dilute solutions.
  • JEE Context: In JEE Main, if the problem can be significantly simplified by an approximation that doesn't drastically alter the result for a dilute solution, it's almost always the intended method.
JEE_Main
Critical Other

❌ Ignoring Van't Hoff Factor 'i' for Electrolytes and Associating Solutes in RLVP Calculations

Many students fail to account for the effective number of particles in solution when the solute is an electrolyte (dissociates) or an associating solute (associates). They often directly apply the simple Raoult's Law for Relative Lowering of Vapour Pressure (RLVP), Pinitial - Pfinal / Pinitial = χsolute, without considering the Van't Hoff factor (i). This leads to significantly incorrect values for RLVP or calculated molar masses.
💭 Why This Happens:
This critical mistake stems from a superficial understanding of colligative properties. Students often memorize the basic RLVP formula for non-volatile, non-electrolyte solutes without fully grasping that colligative properties depend on the number of solute particles, not just the number of moles added. They overlook the conditions under which the simple formula is applicable, especially the impact of dissociation or association on the particle count.
✅ Correct Approach:
Always analyze the nature of the solute. If it's an electrolyte (e.g., NaCl, CaCl2) or an associating solute (e.g., ethanoic acid in benzene), the Van't Hoff factor (i) must be incorporated into Raoult's Law. The correct modified formula for Relative Lowering of Vapour Pressure is:
(P° - P) / P° = i × χsolute
where 'i' accounts for the dissociation or association of the solute.
📝 Examples:
❌ Wrong:
Problem: Calculate the relative lowering of vapour pressure for a 0.1 molal (0.1 m) aqueous solution of NaCl.
Wrong Approach: Students might assume NaCl is a non-electrolyte and calculate χNaCl directly.
Moles of water in 1 kg = 1000/18 = 55.5 moles.
χNaCl = (0.1) / (0.1 + 55.5) ≈ 0.0018
So, RLVP ≈ 0.0018 (Incorrect)
✅ Correct:
Correct Approach: For NaCl, it dissociates into Na+ and Cl- ions. So, the Van't Hoff factor 'i' = 2 (assuming complete dissociation).
Moles of water in 1 kg = 1000/18 = 55.5 moles.
χNaCl = (0.1) / (0.1 + 55.5) ≈ 0.0018
Now, apply the corrected formula:
RLVP = i × χNaCl = 2 × 0.0018 = 0.0036 (Correct)
This shows the RLVP is twice the value obtained by ignoring 'i'.
💡 Prevention Tips:
  • Read Carefully: Always identify the solute's nature (electrolyte, non-electrolyte, associating) from the problem statement.
  • Conceptual Clarity: Understand that colligative properties depend on the *effective number of particles*.
  • Apply Van't Hoff Factor: For JEE Main, remember that 'i' is crucial for problems involving electrolytes. For strong electrolytes, 'i' often equals the number of ions produced per formula unit (e.g., i=2 for NaCl, i=3 for CaCl2).
  • Practice with Variety: Solve problems involving both electrolyte and non-electrolyte solutes to reinforce this distinction.
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Relative lowering of vapour pressure

Subject: Chemistry
Unit: 6 - SOLUTIONS
Sub-unit: 6.2 - Colligative Properties
Complexity: Mid
Syllabus: JEE_Main

Content Completeness: 55.6%

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