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 find the mole fraction of component B in the vapor phase above an equimolar mixture of two pure liquids A and B at 300 K, we can follow these steps:

### Step 1: Understand the given data
We have:
- Vapor pressure of pure liquid A, \( P^0_A = 150 \, \text{mm Hg} \)
- Vapor pressure of pure liquid B, \( P^0_B = 100 \, \text{mm Hg} \)
- The mixture is equimolar, meaning the mole fraction of A and B in the liquid phase is equal.

### Step 2: Determine the mole fractions in the liquid phase
Since the mixture is equimolar:
- Mole fraction of A, \( X_A = 0.5 \)
- Mole fraction of B, \( X_B = 0.5 \)

### Step 3: Calculate the total vapor pressure using Dalton's Law
According to Dalton's Law of Partial Pressures:
\[ P_{\text{total}} = P^0_A \cdot X_A + P^0_B \cdot X_B \]
Substituting the values:
\[ P_{\text{total}} = (150 \, \text{mm Hg} \cdot 0.5) + (100 \, \text{mm Hg} \cdot 0.5) \]
\[ P_{\text{total}} = 75 + 50 = 125 \, \text{mm Hg} \]

### Step 4: Calculate the partial pressure of component B
The partial pressure of component B can be calculated as:
\[ P_B = P^0_B \cdot X_B \]
Substituting the values:
\[ P_B = 100 \, \text{mm Hg} \cdot 0.5 = 50 \, \text{mm Hg} \]

### Step 5: Calculate the mole fraction of B in the vapor phase
The mole fraction of B in the vapor phase, \( Y_B \), is given by:
\[ Y_B = \frac{P_B}{P_{\text{total}}} \]
Substituting the values:
\[ Y_B = \frac{50 \, \text{mm Hg}}{125 \, \text{mm Hg}} = 0.4 \]

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