On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decrea

Q: Why can't we just use the approximation $n \ll N$ for Raoult's Law?

While the approximation $\frac{P^\circ - P_s}{P^\circ} \approx \frac{n}{N}$ is useful for very dilute solutions, it introduces a small error. In JEE Advanced, unless the calculation is impossible, always use the exact relation $\frac{P^\circ - P_s}{P_s} = \frac{n}{N}$ to ensure your answer falls perfectly within the accepted range.

Q: How did we get the formula $\frac{P^\circ - P_s}{P_s} = \frac{n}{N}$?

Start with the standard Raoult's Law: $\frac{P^\circ - P_s}{P^\circ} = \frac{n}{n+N}$. Subtract both sides from 1 to get $\frac{P_s}{P^\circ} = \frac{N}{n+N}$. Invert this to get $\frac{P^\circ}{P_s} = \frac{n+N}{N} = \frac{n}{N} + 1$. Finally, subtract 1 from both sides to arrive at the exact and much simpler relation!

Q: Why do we use the mass of the solvent in kilograms for molality?

Molality is defined as the number of moles of solute per kilogram of solvent. This ensures that the concentration term is independent of temperature, which is crucial when we are studying temperature-dependent phenomena like freezing point depression.

Q: Does the nature of the non-volatile solute affect the answer?

No! Colligative properties like relative lowering of vapor pressure and depression in freezing point depend only on the *number* of solute particles, not their chemical identity. That's why the problem doesn't even name the solute!

Q: Why did we not convert the vapor pressure from mm Hg to atm or bar?

In the formula for relative lowering of vapor pressure, we are taking a ratio of pressures: $\frac{P^\circ - P_s}{P_s}$. The units in the numerator and denominator cancel out perfectly, so as long as they are the same, you don't need to convert them!

Question

On dissolving 0.5 g of a non-volatile non-ionic solute to 39 g of benzene, its vapor pressure decreases from 650 mm Hg to 640 mm Hg. The depression of freezing point of benzene (in K) upon addition of the solute is _______.
(Given data : Molar mass and the molal freezing point depression constant of benzene are $78 	ext{ g mol}^{-1}$ and $5.12 	ext{ K kg mol}^{-1}$, respectively)

Sigma Percentile · Accepted Answer

### The Physical Setup: A Tale of Two Properties

Imagine you are standing in a laboratory, holding a beaker filled with exactly $39 	ext{ g}$ of pure benzene. At a given temperature, the molecules of benzene are constantly escaping into the vapor phase, creating a vapor pressure of $650 	ext{ mm Hg}$.

Now, you drop $0.5 	ext{ g}$ of a mysterious, non-volatile, non-ionic solute into the beaker. As it dissolves, the solute particles occupy space at the surface of the liquid, physically blocking some of the benzene molecules from escaping. Consequently, the vapor pressure drops to $640 	ext{ mm Hg}$.

This phenomenon is known as the **Relative Lowering of Vapor Pressure**, a classic colligative property. But the problem doesn't stop there. It asks us to find out how this addition affects another colligative property: the **Depression in Freezing Point**.

### The Master Equation: Raoult's Law and the Elite Shortcut

To connect the macroscopic drop in vapor pressure to the microscopic amount of solute added, we rely on Raoult's Law. The standard textbook formula states that the relative lowering of vapor pressure is equal to the mole fraction of the solute:

$$ \frac{P^\circ - P_s}{P^\circ} = \frac{n}{n + N} $$

Where $P^\circ$ is the vapor pressure of the pure solvent, $P_s$ is the vapor pressure of the solution, $n$ is the moles of solute, and $N$ is the moles of solvent.

However, solving this equation for $n$ can be algebraically tedious because $n$ appears in both the numerator and the denominator. In the high-pressure environment of JEE Advanced, we need a faster, more elegant weapon. By mathematically rearranging the standard formula (subtracting both sides from 1 and inverting), we arrive at an exact, highly powerful shortcut:

$$ \frac{P^\circ - P_s}{P_s} = \frac{n}{N} $$

Notice the subtle difference? The denominator on the left is now $P_s$ (the solution's vapor pressure), and the denominator on the right is simply $N$. The annoying $n$ is completely isolated! This is not an approximation; it is an exact mathematical identity.

### Executing the First Phase: Finding the Moles

Before we unleash our shortcut, let's quickly calculate the moles of our solvent, benzene ($C_6H_6$). Given its mass is $39 	ext{ g}$ and its molar mass is $78 	ext{ g mol}^{-1}$:

$$ N = \frac{39}{78} = 0.5 	ext{ mol} $$

Now, let's substitute our known values into the elite shortcut formula. The drop in pressure ($P^\circ - P_s$) is $650 - 640 = 10 	ext{ mm Hg}$. The final pressure $P_s$ is $640 	ext{ mm Hg}$.

$$ \frac{10}{640} = \frac{n}{0.5} $$

Cross-multiplying gives us the moles of the solute instantly:

$$ n = \frac{10 	imes 0.5}{640} = \frac{5}{640} 	ext{ mol} $$

**Pro-Tip:** Do not convert this fraction into a decimal yet. Keeping it as a fraction will allow for beautiful cancellations in the next step.

### The Second Phase: The Freezing Point Drop

With the moles of the solute secured, we transition to the second colligative property. The addition of a solute increases the entropy of the liquid phase, making it harder for the solvent to freeze into an ordered solid. This results in a depression of the freezing point, governed by the equation:

$$ \Delta T_f = K_f 	imes m $$

Here, $K_f$ is the molal freezing point depression constant (given as $5.12 	ext{ K kg mol}^{-1}$), and $m$ is the molality of the solution. Molality is defined as the moles of solute per kilogram of solvent:

$$ m = \frac{n}{W_{	ext{solvent}}(	ext{in kg})} = \frac{n}{W_{	ext{solvent}}(	ext{in g})} 	imes 1000 $$

### The Final Calculation

Let's substitute all our values into the freezing point depression formula. We have $n = \frac{5}{640}$ and the mass of benzene is $39 	ext{ g}$.

$$ \Delta T_f = 5.12 	imes \frac{\frac{5}{640}}{\frac{39}{1000}} $$

Rearranging this into a single, clean fraction:

$$ \Delta T_f = 5.12 	imes \frac{5}{640} 	imes \frac{1000}{39} $$

$$ \Delta T_f = \frac{5.12 	imes 5000}{640 	imes 39} $$

$$ \Delta T_f = \frac{25600}{24960} $$

Dividing these numbers yields:

$$ \Delta T_f = 1.0256... 	ext{ K} $$

Rounding off to two decimal places, we get our final answer: **$1.03 	ext{ K}$**.

### A Word of Caution on Approximations

Many textbooks suggest using the approximation $n \ll N$ for dilute solutions, which simplifies Raoult's Law to $\frac{P^\circ - P_s}{P^\circ} \approx \frac{n}{N}$. If we had used this approximation, we would have calculated $n = \frac{5}{650}$, leading to a $\Delta T_f$ of approximately $1.01 	ext{ K}$.

While $1.01$ is close to $1.03$, in the highly competitive arena of JEE Advanced, precision is paramount. The exact rearranged formula $\frac{P^\circ - P_s}{P_s} = \frac{n}{N}$ takes exactly the same amount of time to compute but guarantees zero approximation error. Always choose mathematical rigor when the path is clear!

Enter Numerical Value:

Visualized Solution

Physical Setup

Relative Lowering of Vapor Pressure

The Elite Shortcut

Moles of Solvent (Benzene)

Solving for Moles of Solute

Depression in Freezing Point

Substituting Values

Final Calculation

Final Answer

The Way Forward

The Sigma Insight: Colligative Properties

The Physical Setup

The Master Equation

Executing the First Phase

The Second Phase

The Final Calculation

A Word of Caution on Approximations

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