Problem 84 - A Complete Resource Book in Chemistry for JEE Main

Question

Two liquids having vapour pressures \(\mathrm{P}_{1}^{0}\) and \(\mathrm{P}_{2}^{0}\) in pure state in the ratio of \(2: 1\) are mixed in a molar ratio of \(1: 2\). The ratio of their moles in the vapour state can be (a) \(2: 1\) (b) \(1: 2\) (c) \(1: 1\) (d) \(3: 3\)

Step-by-Step Solution

Answer

The ratio of moles in the vapour state is \(1:1\), option (c).

Step 1: Understand Raoult's Law

Raoult's Law states that the partial vapour pressure of each component in an ideal solution is equal to the vapour pressure of the pure component multiplied by its mole fraction in the solution. For a binary mixture, this can be written as \( P_1 = x_1 P_1^0 \) and \( P_2 = x_2 P_2^0 \) where \( x_1 \) and \( x_2 \) are the mole fractions of components 1 and 2.

Step 2: Define the Given Ratios

It's given that the pure vapour pressures of the liquids \( P_1^0 \) and \( P_2^0 \) are in the ratio of \( 2:1 \). Thus, we can write \( P_1^0 = 2P_2^0 \). It is also given that the molar ratio of the liquids when mixed is \( 1:2 \), which implies the mole fractions are \( x_1 = \frac{1}{3} \) and \( x_2 = \frac{2}{3} \).

Step 3: Calculate Partial Pressures

Using Raoult's Law, calculate the partial pressures of each component: \( P_1 = \left(\frac{1}{3}\right)(2P_2^0) = \frac{2}{3}P_2^0 \) and \( P_2 = \left(\frac{2}{3}\right)(P_2^0) = \frac{2}{3}P_2^0 \).

Step 4: Determine Moles in Vapour Phase

The moles in the vapour state are proportional to the partial pressures: \( n_1 = \frac{P_1}{P_1 + P_2} \) and \( n_2 = \frac{P_2}{P_1 + P_2} \). Substituting the partial pressures, we get: \( n_1 = \frac{\frac{2}{3}P_2^0}{\frac{4}{3}P_2^0} = \frac{1}{2} \) and \( n_2 = \frac{\frac{2}{3}P_2^0}{\frac{4}{3}P_2^0} = \frac{1}{2} \).

Step 5: Conclude with the Ratio of Moles

Thus, the ratio of the moles of the liquids in the vapour state is \(1:1\). This corresponds to option (c) from the selections.

Key Concepts

Vapour Pressure

Vapour pressure is a fundamental concept in understanding the behavior of liquids, especially when they are part of a mixture. It represents the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. When a liquid is placed in a closed container, molecules escape from the liquid to the vapor phase until equilibrium is reached. The pressure exerted by these molecules is the vapour pressure.
In the context of Raoult's Law, which applies to ideal solutions, the vapour pressure of a component in a mixture is calculated based on its pure component vapour pressure and its mole fraction in the solution. This means, in an ideal solution:

• The contribution of each component to the total vapour pressure is directly proportional to its presence in the mixture.
• For two substances A and B, the overall vapour pressure can be given by: \[ P_{total} = x_A P_A^0 + x_B P_B^0 \]
• Where \( P_A^0 \) and \( P_B^0 \) are the vapour pressures of the pure components.

Understanding these relationships allows us to predict and explain the boiling points, volatility, and behavior of mixtures.

Binary Mixture

A binary mixture consists of two different substances combined together. In chemistry, analyzing such mixtures is key to understanding various properties like boiling point, vapour pressure, and phase behavior.
In the given exercise, we deal with a binary mixture made up of two liquids. The concept of a binary mixture simplifies the analysis as you only need to consider two components. Here, the liquids are in a molar ratio of 1 to 2, meaning every molecule of the first liquid is paired with two molecules of the second liquid. This molar ratio translates directly into mole fractions:

• For a binary mixture of components 1 and 2 with a molar ratio of 1:2, the mole fractions are \( x_1 = \frac{1}{3} \) and \( x_2 = \frac{2}{3} \).
• Mole fractions describe the proportion of each component's presence in the mixture.

By using these mole fractions alongside the vapour pressures of pure components, we can effectively apply Raoult's Law to determine the properties of the mixture when it is in equilibrium.

Ideal Solution

An ideal solution is a theoretical concept where the intermolecular interactions between the different components of a mixture are equal to those found within each pure component. This means:

• The enthalpy change of mixing is zero.
• The volume change upon mixing is also zero.
• The mixture obeys Raoult's Law perfectly.

In the case of an ideal solution, each component contributes to the total vapour pressure, and their influence is solely determined by their respective mole fractions and pure component vapour pressures. This assumption simplifies calculations, making ideal solutions a good starting point for understanding real systems.
While practical solutions often deviate from ideal behavior due to differences in molecular sizes, shapes, or interactions, ideal solutions provide insights into what to expect when complexities are introduced. Understanding the concept of an ideal solution and how it applies in calculating properties like the vapour pressure can clarify more complex behaviors seen in real-world mixtures.