Problem 163 - A Complete Resource Book in Chemistry for JEE Main

Question

At \(80^{\circ} \mathrm{C}\), the vapour pressure of pure liquid 'A' is 520 \(\mathrm{mm} \mathrm{Hg}\) and that of pure liquid ' \(\mathrm{B}\) ' is \(1000 \mathrm{~mm} \mathrm{Hg}\). If a mixture solution of 'A' and 'B' boils at \(80^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) pressure, the amount of 'A' in the mixture is (1 \(\mathrm{atm}=\) \(760 \mathrm{~mm} \mathrm{Hg}) .\) (a) \(52 \mathrm{~mol}\) per cent (b) 34 mol per cent (c) 48 mol per cent (d) \(50 \mathrm{~mol}\) per cent

Step-by-Step Solution

Answer

The amount of 'A' in the mixture is 50 mol percent.

Step 1: Understanding Raoult's Law

To find the amount of liquid 'A' in the mixture, we use Raoult's Law, which states that the partial vapor pressure of each component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

Step 2: Applying Raoult's Law to the Mixture

Given that the total vapor pressure of the mixture equals atmospheric pressure at 80°C (760 mm Hg), we can write this as: \( P_A + P_B = 760 \, \text{mm Hg} \), where \( P_A \) and \( P_B \) are partial pressures of A and B, respectively.

Step 3: Determining Partial Pressures

From Raoult's Law, the partial pressure of A, \( P_A = x_A \times P^0_A \), and for B, \( P_B = x_B \times P^0_B \). Here, \( P^0_A = 520 \, \text{mm Hg} \) and \( P^0_B = 1000 \, \text{mm Hg} \). Also, \( x_A + x_B = 1 \).

Step 4: Setting up Equations

Substitute the expressions for \( P_A \) and \( P_B \) into the total pressure equation: \( x_A \times 520 + (1 - x_A) \times 1000 = 760 \).

Step 5: Solving for Mole Fraction of A

Solve the equation \( x_A \times 520 + (1 - x_A) \times 1000 = 760 \):1. Expand: \( 520x_A + 1000 - 1000x_A = 760 \)2. Combine like terms: \( -480x_A = -240 \)3. Solve for \( x_A \): \( x_A = \frac{1}{2} \) or \( x_A = 0.5 \).

Step 6: Calculating the Mole Percent

The mole fraction, \( x_A = 0.5 \), translates to 50% when converted to mole percent, as \( \text{Mole percent} = x_A \times 100 \).

Key Concepts

Vapor Pressure

Vapor pressure is a fundamental concept in chemistry that describes 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, some molecules at the surface will escape into the gas phase, creating vapor. The pressure these vapor molecules exert is known as vapor pressure.

This pressure depends on the nature of the liquid and the temperature:

• Pure substances have specific vapor pressures at given temperatures.
• The vapor pressure increases with temperature because more molecules have sufficient energy to escape into the vapor phase.
• In a mixture, each component will contribute to the total vapor pressure based on its own properties.

Understanding vapor pressure is crucial for predicting how substances behave under various conditions and is a key factor in boiling, as we will explore further.

Mole Fraction

Mole fraction is a way of expressing the concentration of a component in a mixture. It represents the ratio of the number of moles of one component to the total number of moles in the mixture:

\[x_i = \frac{n_i}{n_{total}}\]

This concept is essential when applying Raoult's Law:

• The mole fraction of each component must add up to 1, such that \( x_A + x_B = 1 \) for a binary mixture.
• Mole fraction can never exceed 1 and is dimensionless, making it a convenient way to describe mixture compositions.

Mole fraction plays a vital role in defining how much impact each component has on properties like vapor pressure and boiling point.

Partial Vapor Pressure

Partial vapor pressure refers to the pressure contributed by each component of a mixture to the total vapor pressure. According to Raoult's Law, the partial vapor pressure for a component is calculated by multiplying the vapor pressure of the pure component by its mole fraction in the mixture. This gives us:

• \( P_A = x_A \times P^0_A \)
• \( P_B = x_B \times P^0_B \)

Where:

• \( P_A \) and \( P_B \) are the partial pressures of components A and B.
• \( P^0_A \) and \( P^0_B \) are the vapor pressures of the pure components.

The sum of all partial pressures in a mixture equals the total pressure, providing a way to understand how each component contributes to the behavior of the whole system. Calculating partial vapor pressures helps predict how mixtures will behave, particularly in separation processes or when forming azeotropes.

Boiling Point

The boiling point of a liquid is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. At this point, the liquid becomes a vapor:

• When the vapor pressure reaches ambient pressure, the liquid begins to vaporize vigorously, resulting in boiling.
• This boiling continues as long as the heat is supplied, the external pressure remains constant, and the liquid is present.

In a mixture, the boiling point is influenced by the vapor pressures of its components:

• The total vapor pressure must reach the ambient pressure for the mixture to boil.
• Raoult's Law governs how the components' contributions to vapor pressure affect the boiling behavior.

Understanding boiling points is essential for applications such as distillation and lyophilization, where phase changes are controlled for specific outcomes.