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, 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 (W_solute) = 15 g
- Mass of solvent (W_solvent) = 58.8 g
- Molecular weight of benzene (M_benzene) = 78 g/mol

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

3. **Use Raoult's Law:**
According to Raoult's law, the change in vapor pressure is given by:
\[ \frac{\Delta P}{P_s} = \frac{W_{\text{solute}}}{M_{\text{solute}}} \cdot \frac{M_{\text{solvent}}}{W_{\text{solvent}}} \]
Rearranging gives:
\[ M_{\text{solute}} = \frac{W_{\text{solute}} \cdot M_{\text{solvent}} \cdot P_s}{\Delta P} \]

4. **Substituting the Values:**
\[ M_{\text{solute}} = \frac{15 \, \text{g} \cdot 78 \, \text{g/mol} \cdot 990 \, \text{torr}}{30 \, \text{torr}} \]

5. **Calculating the Molecular Weight:**
\[ M_{\text{solute}} = \frac{15 \cdot 78 \cdot 990}{30} \]
\[ M_{\text{solute}} = \frac{1155300}{30} = 38510 \, \text{g/mol} \]
\[ M_{\text{solute}} = 676.5 \, \text{g/mol} \]

6. **Final Answer:**
The molecular weight of the solute is approximately **676.5 g/mol**.