Benzene and toluene form nearly ideal solution. At `20^(@)C` the vapour pressure of benzene is 75 torr and that of toluene is 22 torr. The partial vapour pressure of benzene at `20^(@)C` for a solution containing 78g of benzene and 46 g of toluene in torr is-

Options

A. 50
B. 25
C. 37.5
D. 53.5

Answer

To solve the problem of finding the partial vapor pressure of benzene in a solution containing 78 g of benzene and 46 g of toluene at 20°C, we will follow these steps:

### Step 1: Calculate the number of moles of benzene and toluene.

1. **Molar mass of benzene (C6H6)**: 78 g/mol
2. **Molar mass of toluene (C7H8)**: 92 g/mol

**Number of moles of benzene**:
\[ \text{Moles of benzene} = \frac{\text{Mass of benzene}}{\text{Molar mass of benzene}} = \frac{78 \, \text{g}}{78 \, \text{g/mol}} = 1 \, \text{mol} \]

**Number of moles of toluene**:
\[ \text{Moles of toluene} = \frac{\text{Mass of toluene}}{\text{Molar mass of toluene}} = \frac{46 \, \text{g}}{92 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 2: Calculate the mole fraction of benzene.

The mole fraction of benzene (\(X_{C6H6}\)) is calculated using the formula:
\[ X_{C6H6} = \frac{\text{Moles of benzene}}{\text{Moles of benzene} + \text{Moles of toluene}} = \frac{1}{1 + 0.5} = \frac{1}{1.5} = \frac{2}{3} \]

### Step 3: Use Raoult's Law to calculate the partial vapor pressure of benzene.

Raoult's Law 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:
\[ P_{C6H6} = P^0_{C6H6} \times X_{C6H6} \]
Where:
- \(P^0_{C6H6}\) = vapor pressure of pure benzene = 75 torr
- \(X_{C6H6}\) = mole fraction of benzene = \(\frac{2}{3}\)

Substituting the values:
\[ P_{C6H6} = 75 \, \text{torr} \times \frac{2}{3} = 50 \, \text{torr} \]

### Final Answer:
The partial vapor pressure of benzene at 20°C for the solution is **50 torr**.

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