A solution of two volatile liquids A and B obeys Raoult's law. At a certain temperature it is found that when the pressure above the mixture in equilibrium is 402.5 mm of Hg, the mole fraction of A in the vapour is 0.35 and in the liquid it is 0.65. What are the vapour pressures of two liquids at this temperature ?

Explanation

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

### Step-by-Step Solution:

1. **Identify Given Values:**
- Total pressure \( P = 402.5 \, \text{mmHg} \)
- Mole fraction of A in vapor \( y_A = 0.35 \)
- Mole fraction of A in liquid \( x_A = 0.65 \)
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