Mole fraction of vapour of `A` above solution in mixture of `A` and `B(X_(A) = 0.4)` will be `(P_(A)^(@) = 100mm, P_(B)^(@) = 200 mm)`:

Options

A. (a) `0.4`
B. (b) `0.8`
C. ( c) `0.25`
D. (d) None of these

Explanation

To solve the problem of finding the mole fraction of vapor of component A above a solution of A and B, we can follow these steps:

### Step 1: Identify Given Information
- Mole fraction of A in the liquid phase, \( X_A = 0.4 \)
- Mole fraction of B in the liquid phase, \( X_B = 1 - X_A = 0.6 \)
- Vapor pressure of pure A, \( P_A^0 = 100 \, \text{mm} \)
- Vapor pressure of pure B, \( P_B^0 = 200 \, \text{mm} \)

### Step 2: Calculate Total Vapor Pressure
To find the total vapor pressure (\( P_{\text{total}} \)) above the solution, we use Raoult's Law:
\[ P_{\text{total}} = P_A^0 \cdot X_A + P_B^0 \cdot X_B \]
Substituting the values:
\[ P_{\text{total}} = (100 \, \text{mm} \cdot 0.4) + (200 \, \text{mm} \cdot 0.6) \]
Calculating each term:
\[ P_{\text{total}} = 40 \, \text{mm} + 120 \, \text{mm} = 160 \, \text{mm} \]

### Step 3: Calculate Mole Fraction of A in Vapor Phase
Using Raoult's Law again, we can express the partial vapor pressure of A in the vapor phase:
\[ P_A = P_A^0 \cdot X_A \]
Substituting the values:
\[ P_A = 100 \, \text{mm} \cdot 0.4 = 40 \, \text{mm} \]

Now, we can find the mole fraction of A in the vapor phase (\( X_{A,\text{vapor}} \)):
\[ X_{A,\text{vapor}} = \frac{P_A}{P_{\text{total}}} \]
Substituting the values:
\[ X_{A,\text{vapor}} = \frac{40 \, \text{mm}}{160 \, \text{mm}} = \frac{1}{4} = 0.25 \]

### Step 4: Conclusion
The mole fraction of vapor of A above the solution is \( 0.25 \).

### Summary
- Mole fraction of A in vapor phase \( X_{A,\text{vapor}} = 0.25 \)