The vapour pressure of a pure liquid `A` is `40 mm Hg` at `310 K`. The vapour pressure of this liquid in a solution with liquid `B` is `32 mm Hg`. The mole fraction of `A` in the solution, if it obeys Raoult's law, is:

Options

A. 0.8
B. 0.5
C. 0.2
D. 0.4

Explanation

To find the mole fraction of liquid A in the solution, we can use Raoult's law, which states that the vapor pressure of a solution is directly proportional to the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Vapor pressure of pure liquid A, \( P^0_A = 40 \, \text{mm Hg} \)
- Vapor pressure of the solution, \( P_A = 32 \, \text{mm Hg} \)

2. **Apply Raoult's Law:**
According to Raoult's law, the vapor pressure of the solution can be expressed as:
\[ P_A = X_A \cdot P^0_A \]
where \( X_A \) is the mole fraction of liquid A in the solution.

3. **Rearrange the Equation:**
To find the mole fraction \( X_A \), rearrange the equation:
\[ X_A = \frac{P_A}{P^0_A} \]

4. **Substitute the Values:**
Substitute the known values into the equation:
\[ X_A = \frac{32 \, \text{mm Hg}}{40 \, \text{mm Hg}} \]

5. **Calculate the Mole Fraction:**
Perform the division:
\[ X_A = 0.8 \]

### Final Answer:
The mole fraction of liquid A in the solution is \( X_A = 0.8 \).

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