The vapour pressure of two pure liquids A and B are 200 and 400 tor respectively at 300K. A liquid solution (ideal) of A and B for which the mole fraction of A is 0.40 is contained in a cylinder. The composition of components A and B in vapour phase after equilibrium is reached between vapour & liquid phase, respectively is

Options

A. `X_A = 0.62`, `X_B = 0.38`
B. `X_A = 0.50`, `X_B = 0.50`
C. `X_A = 0.25`, `X_B= 0.75`
D. `X_A = 0.30`, `X_B= 0.70`

Answer

To solve the problem, we need to find the mole fractions of components A and B in the vapor phase after equilibrium is reached between the vapor and liquid phases. We will use Raoult's Law and Dalton's Law of Partial Pressures.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure liquid A, \( P^0_A = 200 \, \text{Torr} \)
- Vapor pressure of pure liquid B, \( P^0_B = 400 \, \text{Torr} \)
- Mole fraction of A in the liquid phase, \( X_A = 0.40 \)

2. **Calculate the Mole Fraction of B:**
\[ X_B = 1 - X_A = 1 - 0.40 = 0.60 \]

3. **Apply Raoult's Law to Calculate Partial Pressures:**
- Partial vapor pressure of A, \( P_A = X_A \cdot P^0_A \)
\[ P_A = 0.40 \cdot 200 = 80 \, \text{Torr} \]
- Partial vapor pressure of B, \( P_B = X_B \cdot P^0_B \)
\[ P_B = 0.60 \cdot 400 = 240 \, \text{Torr} \]

4. **Calculate Total Vapor Pressure of the Solution:**
\[ P_T = P_A + P_B = 80 + 240 = 320 \, \text{Torr} \]

5. **Calculate Mole Fraction of A in the Vapor Phase:**
- Using Dalton's Law of Partial Pressures:
\[ Y_A = \frac{P_A}{P_T} = \frac{80}{320} = 0.25 \]

6. **Calculate Mole Fraction of B in the Vapor Phase:**
\[ Y_B = 1 - Y_A = 1 - 0.25 = 0.75 \]

### Final Answer:
- Mole fraction of A in the vapor phase, \( Y_A = 0.25 \)
- Mole fraction of B in the vapor phase, \( Y_B = 0.75 \)

### Conclusion:
The composition of components A and B in the vapor phase after equilibrium is reached is:
- \( Y_A = 0.25 \)
- \( Y_B = 0.75 \)