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 solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.

### Step-by-Step Solution:

1. **Identify the given values:**
- Vapor pressure of pure liquid A, \( P_{A}^{0} = 40 \, \text{mm Hg} \)
- Vapor pressure of the solution, \( P_{A} = 32 \, \text{mm Hg} \)

2. **Apply Raoult's Law:**
According to Raoult's law:
\[ P_{A} = P_{A}^{0} \times X_{A} \]
where \( X_{A} \) is the mole fraction of liquid A in the solution.

3. **Rearrange the equation to find \( X_{A} \):**
\[ X_{A} = \frac{P_{A}}{P_{A}^{0}} \]

4. **Substitute the known values:**
\[ X_{A} = \frac{32 \, \text{mm Hg}}{40 \, \text{mm Hg}} \]

5. **Calculate \( X_{A} \):**
\[ X_{A} = 0.8 \]

6. **Conclusion:**
The mole fraction of liquid A in the solution is \( 0.8 \).

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

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