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 (c) 1:1.

Step 1: Understand the Given Information

We have two liquids, Liquid 1 and Liquid 2. The vapour pressures of these liquids in their pure states are \(\mathrm{P}_{1}^{0}\) and \(\mathrm{P}_{2}^{0}\) respectively, with a ratio of \(2:1\). Also, the liquids are mixed in a molar ratio of \(1:2\). Our goal is to find the ratio of their moles in the vapour state.

Step 2: Apply Raoult's Law

According to Raoult's Law, the partial vapour pressure of each component in a solution is given by the product of its mole fraction and its pure component vapor pressure. Using this, the partial pressures \(\mathrm{P}_1\) and \(\mathrm{P}_2\) for liquids 1 and 2 are \(x_1 \, \mathrm{P}_{1}^{0}\) and \(x_2 \, \mathrm{P}_{2}^{0}\) respectively, where \(x_1\) and \(x_2\) are the mole fractions of liquid 1 and liquid 2 in the solution.

Step 3: Calculate Mole Fractions

The molar ratio of the liquids is given as \(1:2\). Therefore, the mole fractions are \(x_1 = \frac{1}{3}\) and \(x_2 = \frac{2}{3}\). These fractions are calculated by taking the ratio of individual moles to the total moles, \(x_1 = \frac{1}{1+2}\) and \(x_2 = \frac{2}{1+2}\).

Step 4: Calculate Partial Vapour Pressures

Given that \(\mathrm{P}_{1}^{0}: \, \mathrm{P}_{2}^{0} = 2:1\), we can write \(\mathrm{P}_{1}^{0} = 2\mathrm{P}_{2}^{0}\). The partial pressures are \(P_1 = x_1 \, \mathrm{P}_{1}^{0} = \frac{1}{3} \, (2 \, \mathrm{P}_{2}^{0}) = \frac{2}{3} \, \mathrm{P}_{2}^{0}\) and unalterably, \(P_2 = x_2 \, \mathrm{P}_{2}^{0} = \frac{2}{3} \, \mathrm{P}_{2}^{0}\).

Step 5: Determine the Ratio of Moles in Vapour State

In the vapour state, the ratio of the moles of the liquids is equal to the ratio of their partial pressures. Thus, we have the ratio \(P_1:P_2 = \frac{2}{3} \, \mathrm{P}_{2}^{0} : \frac{2}{3} \, \mathrm{P}_{2}^{0} = 1:1\).

Key Concepts

Vapour Pressure

Vapour pressure is a critical concept in understanding how components behave in a mixture of liquids. It describes the pressure exerted by the vapor that is in equilibrium with its liquid phase at a given temperature. Imagine that above any liquid, there is a tiny cloud of vapor particles created from some of the liquid molecules escaping to the gas phase.

These escaping molecules create a pressure atop the liquid, known as the vapour pressure. The higher the vapour pressure of a liquid, the more volatile it is, meaning it can vaporize more quickly than substances with lower vapour pressures.

• Higher temperature typically increases vapour pressure.
• Different substances have distinct vapour pressures even at the same temperature.
• In mixtures, Raoult's Law helps to predict the behaviour of each component based on their pure vapour pressures.

Understanding vapour pressure is crucial for grasping how different substances contribute to the vapour phase in a mixture. This understanding leads us to our next topic, which is key in applying Raoult's Law to varieties of liquid solutions.

Mole Fraction

Mole fraction is a dimensionless quantity that describes the ratio of the moles of a specific component divided by the total moles of all components in a mixture. This measure tells us how much of each component is present in the mixture, independent of the size of the system.

Mole fractions are particularly important when using Raoult's Law to determine the partial pressures in a solution. Suppose you have a mixture of two liquids, A and B; knowing their mole fractions allows you to calculate how each contributes to the overall vapour pressure. The formula is:

• For component A:
  \( x_A = \frac{{ ext{moles of A}}}{{ ext{total moles}}} \)
• For component B:
  \( x_B = \frac{{ ext{moles of B}}}{{ ext{total moles}}} \)

The sum of mole fractions for all components in a mixture will always equal 1. This concept helps simplify complex problems, turning potentially overwhelming data sets into manageable fractions of the whole. Its importance is underscored when applying Raoult's Law, making mole fraction more than just a mathematical convenience.

Partial Pressure

The term partial pressure refers to the pressure exerted by a single component of a mixture of substances in the gas phase. This idea springs from Dalton's Law of Partial Pressures, which states that the total pressure exerted by a gas mixture is equal to the sum of the partial pressures of each individual component.

When implementing Raoult's Law, the partial pressure of a component is crucial. This is determined by the product of its mole fraction in the solution and its pure component vapour pressure. Mathematically, it can be described by:

• Partial pressure of component A:
  \( P_A = x_A \cdot P_A^0 \)
• Partial pressure of component B:
  \( P_B = x_B \cdot P_B^0 \)

These calculations allow you to predict how a mixture behaves, particularly in terms of which component will be more prevalent in the vapor phase. Knowing this is critically beneficial in chemical processes like distillation and other separation techniques, where control over the gaseous mix’s composition is vital.