Benzene and toluene form nearly ideal solutions. 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 `78` g of benzene and `46` g of toluene in torr is

Options

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

Explanation

To find the partial vapor pressure of benzene in a solution containing 78 g of benzene and 46 g of toluene at 20°C, we will use Raoult's Law. Here's a step-by-step solution:

### Step 1: Calculate the number of moles of benzene and toluene.
- **Molecular weight of benzene (C₆H₆)** = 78 g/mol
- **Molecular weight of toluene (C₇H₈)** = 92 g/mol

**Moles of benzene:**
\[ \text{Moles of benzene} = \frac{\text{mass of benzene}}{\text{molecular weight of benzene}} = \frac{78 \text{ g}}{78 \text{ g/mol}} = 1 \text{ mol} \]

**Moles of toluene:**
\[ \text{Moles of toluene} = \frac{\text{mass of toluene}}{\text{molecular weight of toluene}} = \frac{46 \text{ g}}{92 \text{ g/mol}} = 0.5 \text{ mol} \]

### Step 2: Calculate the total number of moles in the solution.
\[ \text{Total moles} = \text{Moles of benzene} + \text{Moles of toluene} = 1 \text{ mol} + 0.5 \text{ mol} = 1.5 \text{ mol} \]

### Step 3: Calculate the mole fraction of benzene.
\[ \text{Mole fraction of benzene} (X_B) = \frac{\text{Moles of benzene}}{\text{Total moles}} = \frac{1 \text{ mol}}{1.5 \text{ mol}} = \frac{2}{3} \]

### Step 4: Use Raoult's Law to find the partial vapor pressure of benzene.
According to Raoult's Law:
\[ \text{Partial vapor pressure of benzene} (P_B) = X_B \times P^0_B \]
where \( P^0_B \) is the vapor pressure of pure benzene.

Given:
- \( P^0_B = 75 \text{ torr} \)

Substituting the values:
\[ P_B = \left(\frac{2}{3}\right) \times 75 \text{ torr} = 50 \text{ torr} \]

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

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