Question

A solution is made by mixing one mole of volatile liquid A with 3 moles of volatile liquid B. The vapor pressure of pure A is 200 mm Hg and that of the solution is 500 mm Hg. The vapor pressure of pure B and the least volatile component of the solution, respectively, are:

• A. 1400 mm Hg, A
• B. 1400 mm Hg, B
• C. 600 mm Hg, B
• D. 600 mm Hg, A

Hint:

Raoult's Law is useful for determining the vapor pressure of components in an ideal solution. It states that the vapor pressure is proportional to the mole fraction of the component in the solution.

Answer 1

The Correct Option is D

The relation between vapor pressures is given by Raoult's law: \[ P_A = P_A^0 \cdot X_A, \quad P_B = P_B^0 \cdot X_B \] Given: \[ P_A + P_B = 500 \, \text{mm Hg}, \quad P_A^0 = 200 \, \text{mm Hg}, \quad P_B^0 = 600 \, \text{mm Hg} \] Using the mole fractions and Raoult's law: \[ P_A = 200 \times \frac{1}{4}, \quad P_B = 600 \times \frac{3}{4} \] Thus, the answer is \( P_A = 600 \, \text{mm Hg}, P_B = 600 \, \text{mm Hg} \).

Answer 2

This problem deals with the vapor pressure of a solution containing two volatile liquids, which can be described by Raoult's Law. We are asked to find the vapor pressure of pure liquid B and identify the least volatile component of the solution.

Concept Used:

Raoult's Law for Ideal Solutions: For a solution of volatile liquids, the partial vapor pressure of each component in the solution is directly proportional to its mole fraction. The total vapor pressure of the solution is the sum of the partial vapor pressures of all components.

The total vapor pressure, \( P_{\text{total}} \), is given by:

\[ P_{\text{total}} = P_A + P_B = P_A^\circ x_A + P_B^\circ x_B \]

where:

• \( P_A^\circ \) and \( P_B^\circ \) are the vapor pressures of the pure components A and B, respectively.
• \( x_A \) and \( x_B \) are the mole fractions of components A and B in the solution.

Volatility: A component's volatility is directly related to its vapor pressure in the pure state. The component with a lower pure vapor pressure is considered less volatile.

Step-by-Step Solution:

Step 1: List the given information.

• Moles of liquid A, \( n_A = 1 \) mole
• Moles of liquid B, \( n_B = 3 \) moles
• Vapor pressure of pure A, \( P_A^\circ = 200 \) mm Hg
• Total vapor pressure of the solution, \( P_{\text{total}} = 500 \) mm Hg

Step 2: Calculate the mole fractions of A (\( x_A \)) and B (\( x_B \)).

The total number of moles in the solution is:

\[ n_{\text{total}} = n_A + n_B = 1 + 3 = 4 \, \text{moles} \]

The mole fraction of A is:

\[ x_A = \frac{n_A}{n_{\text{total}}} = \frac{1}{4} \]

The mole fraction of B is:

\[ x_B = \frac{n_B}{n_{\text{total}}} = \frac{3}{4} \]

Step 3: Apply Raoult's Law to find the vapor pressure of pure B (\( P_B^\circ \)).

Substitute the known values into Raoult's Law equation:

\[ P_{\text{total}} = P_A^\circ x_A + P_B^\circ x_B \] \[ 500 \, \text{mm Hg} = (200 \, \text{mm Hg}) \left( \frac{1}{4} \right) + P_B^\circ \left( \frac{3}{4} \right) \]

Step 4: Solve the equation for \( P_B^\circ \).

\[ 500 = 50 + \frac{3}{4} P_B^\circ \] \[ 500 - 50 = \frac{3}{4} P_B^\circ \] \[ 450 = \frac{3}{4} P_B^\circ \] \[ P_B^\circ = 450 \times \frac{4}{3} \] \[ P_B^\circ = 150 \times 4 = 600 \, \text{mm Hg} \]

Step 5: Identify the least volatile component.

Compare the vapor pressures of the pure components:

• \( P_A^\circ = 200 \) mm Hg
• \( P_B^\circ = 600 \) mm Hg

Since the vapor pressure of pure A is lower than that of pure B (\( P_A^\circ < P_B^\circ \)), component A is the least volatile.

The vapor pressure of pure B is 600 mm Hg and the least volatile component of the solution is A.