At 300 K two pure liquids A and B have vapour pressures respectively 150 mm Hg and 100 mm Hg. In a equimolar liquid mixture of A and B, the mole fraction of B in the vapour phase above the solution at this temperature is:

Options

A. 0.6
B. 0.5
C. 0.8
D. 0.4

Answer

To solve the problem, we need to determine the mole fraction of component B in the vapor phase above an equimolar mixture of liquids A and B at 300 K. Here are the steps to find the solution:

### Step 1: Identify the given data
- Vapor pressure of pure liquid A, \( P_{A}^{0} = 150 \, \text{mm Hg} \)
- Vapor pressure of pure liquid B, \( P_{B}^{0} = 100 \, \text{mm Hg} \)
- The mixture is equimolar, meaning the mole fraction of A (\( X_A \)) and B (\( X_B \)) in the liquid phase is \( 0.5 \) each.

### Step 2: Calculate the vapor pressures of A and B in the solution
Using Raoult's Law, the vapor pressure of each component in the solution can be calculated as follows:

- Vapor pressure of A in the solution:
\[ P_{A} = X_{A} \cdot P_{A}^{0} = 0.5 \cdot 150 \, \text{mm Hg} = 75 \, \text{mm Hg} \]

- Vapor pressure of B in the solution:
\[ P_{B} = X_{B} \cdot P_{B}^{0} = 0.5 \cdot 100 \, \text{mm Hg} = 50 \, \text{mm Hg} \]

### Step 3: Calculate the total vapor pressure of the solution
The total vapor pressure of the solution (\( P_S \)) is the sum of the individual vapor pressures:
\[ P_S = P_{A} + P_{B} = 75 \, \text{mm Hg} + 50 \, \text{mm Hg} = 125 \, \text{mm Hg} \]

### Step 4: Calculate the mole fraction of B in the vapor phase
The mole fraction of B in the vapor phase (\( y_B \)) can be calculated using the formula:
\[ y_B = \frac{P_B}{P_S} \]
Substituting the values:
\[ y_B = \frac{50 \, \text{mm Hg}}{125 \, \text{mm Hg}} = 0.4 \]

### Final Answer
The mole fraction of B in the vapor phase above the solution at 300 K is \( 0.4 \).

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