The vapour pressure of pure liquid `A` at `310^(@)`C is `120` torr. The vapour pressure of this liquid in solution with liquid `B` is `72` torr. Calculate the mole of fraction of `A` in solution if the mixture obeys Raoult's law.

Options

A. 0.06
B. 0.9
C. 0.2
D. 0.6

Explanation

To solve the problem, we will use Raoult's Law, which states that the partial vapor pressure of a component in a solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.

### Step-by-Step Solution:

1. **Identify the given data**:
- Vapor pressure of pure liquid A, \( P_{A}^{0} = 120 \) torr
- Vapor pressure of liquid A in the solution, \( P_{A} = 72 \) torr

2. **Apply Raoult's Law**:
According to Raoult's Law:
\[ P_{A} = P_{A}^{0} \cdot X_{A} \]
where \( P_{A} \) is the vapor pressure of A in the solution, \( P_{A}^{0} \) is the vapor pressure of pure A, and \( X_{A} \) is the mole fraction of A in the solution.

3. **Rearranging the equation to find \( X_{A} \)**:
We can rearrange the equation to solve for the mole fraction \( X_{A} \):
\[ X_{A} = \frac{P_{A}}{P_{A}^{0}} \]

4. **Substituting the values**:
Now, substitute the values of \( P_{A} \) and \( P_{A}^{0} \):
\[ X_{A} = \frac{72 \text{ torr}}{120 \text{ torr}} \]

5. **Calculating \( X_{A} \)**:
Performing the calculation:
\[ X_{A} = \frac{72}{120} = 0.6 \]

6. **Conclusion**:
The mole fraction of liquid A in the solution is \( 0.6 \).

### Final Answer:
The mole fraction of A in the solution is \( 0.6 \).

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