The vapour pressure of pure benzene at `88^(@)C` is `957mm` and that of toluene at the same temperature is `379.5mm`. Calculate the composition of a benzene-toluene mixture boiling at `88^(@)C`

Explanation

To solve the problem of finding the composition of a benzene-toluene mixture boiling at \(88^\circ C\), we can use Raoult's Law, which states that the vapor pressure of a component in a mixture is equal to the vapor pressure of the pure component multiplied by its mole fraction in the mixture.

### Step-by-Step Solution:

1. **Identify the given data:**
- Vapor pressure of pure benzene (\(P^0_{C_6H_6}\)) at \(88^\circ C\) = \(957 \, mmHg\)
- Vapor pressure of pure toluene (\(P^0_{C_7H_8}\)) at \(88^\circ C\) = \(379.5 \, mmHg\)

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