The vapour pressure of benzene at `90^@C` is 1020 torr. A solution of 15 g of a solute in 58.8 g benzene has a vapour pressure of 990 torr. The molecular weight of the solute is

Options

A. 78.2
B. 204.2
C. 148.2
D. 676.53

Answer

To find the molecular weight of the solute in the given problem, we can use Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the concentration of the solute.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Vapor pressure of pure benzene (P0) = 1020 torr
- Vapor pressure of the solution (Ps) = 990 torr
- Mass of solute (W2) = 15 g
- Mass of solvent (W1) = 58.8 g
- Molar mass of benzene (M1) = 78 g/mol (calculated from its molecular formula C6H6)

2. **Calculate the Change in Vapor Pressure**:
\[ \Delta P = P_0 - P_s = 1020 \, \text{torr} - 990 \, \text{torr} = 30 \, \text{torr} \]

3. **Apply Raoult's Law**:
According to Raoult's Law, the change in vapor pressure can be expressed as:
\[ \Delta P = \frac{W_2 \cdot M_1}{W_1 \cdot M_2} \]
Rearranging gives us:
\[ M_2 = \frac{W_2 \cdot M_1}{W_1 \cdot \Delta P} \]

4. **Substitute the Known Values**:
- \(W_2 = 15 \, \text{g}\)
- \(M_1 = 78 \, \text{g/mol}\)
- \(W_1 = 58.8 \, \text{g}\)
- \(\Delta P = 30 \, \text{torr}\)

Plugging in these values:
\[ M_2 = \frac{15 \, \text{g} \cdot 78 \, \text{g/mol}}{58.8 \, \text{g} \cdot 30 \, \text{torr}} \]

5. **Calculate Molar Mass of the Solute**:
\[ M_2 = \frac{1170}{1764} \approx 0.663 \, \text{g/mol}^{-1} \]
\[ M_2 \approx 676.53 \, \text{g/mol} \]

6. **Final Answer**:
The molecular weight of the solute is approximately **676.53 g/mol**.