Benzene and toluene form an ideal solution. The vapour pressure of benzene and toluene are respectively `75mm` and `22mm` at `20^(@)C`. If the mole fractions of benzene and toluene in vapour are `0.63` and `0.37` respectively, calculate the vapour pressure of the ideal mixture.

Explanation

To calculate the vapor pressure of the ideal mixture of benzene and toluene, we can use Raoult's Law, which states that the vapor pressure of each component in an ideal 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 Given Data:**
- Vapor pressure of pure benzene, \( P^0_{B} = 75 \, \text{mm Hg} \)
- Vapor pressure of pure toluene, \( P^0_{T} = 22 \, \text{mm Hg} \)
- Mole fraction of benzene in vapor, \( y_{B} = 0.63 \)
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