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 of finding the mole fraction of B in the vapor phase above an equimolar mixture of liquids A and B, we can follow these steps:

### Step 1: Determine the mole fractions of A and B in the liquid phase.
Given that the mixture is equimolar, we can assume:
- Moles of A = x
- Moles of B = x

The total moles in the mixture = moles of A + moles of B = x + x = 2x.

Now, we can calculate the mole fractions:
- Mole fraction of A (X_A) = moles of A / total moles = x / (2x) = 0.5
- Mole fraction of B (X_B) = moles of B / total moles = x / (2x) = 0.5

### Step 2: Calculate the total vapor pressure of the solution.
Using Raoult's Law, the total vapor pressure (P_T) can be calculated as:
\[ P_T = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]

Where:
- \( P_A^0 \) = vapor pressure of pure A = 150 mm Hg
- \( P_B^0 \) = vapor pressure of pure B = 100 mm Hg

Substituting the values:
\[ P_T = (150 \, \text{mm Hg} \cdot 0.5) + (100 \, \text{mm Hg} \cdot 0.5) \]
\[ P_T = 75 \, \text{mm Hg} + 50 \, \text{mm Hg} = 125 \, \text{mm Hg} \]

### Step 3: Calculate the mole fraction of B in the vapor phase.
Using the formula for the mole fraction of B in the vapor phase (Y_B):
\[ Y_B = \frac{P_B^0 \cdot X_B}{P_T} \]

Substituting the known values:
\[ Y_B = \frac{100 \, \text{mm Hg} \cdot 0.5}{125 \, \text{mm Hg}} \]
\[ Y_B = \frac{50}{125} = 0.4 \]

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