10 gram of a solute was dissolved in 80 g of acetone at 30ºC to form an ideal solution. The vapour pressure was found to be 271 mm of Hg. Vapour pressure of pure acetone at `30^(@)` C is 283 mm Hg. The molecular weight of the solute is

Options

A. 100.5
B. 120.5
C. 150.5
D. 163.5

Explanation

To solve the problem, we will use Raoult's law, which relates the vapor pressure of a solvent in a solution to the vapor pressure of the pure solvent and the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Mass of solute (A) = 10 g
- Mass of solvent (B, acetone) = 80 g
- Vapor pressure of the solution (P_solution) = 271 mm Hg
- Vapor pressure of pure acetone (P°_B) = 283 mm Hg

2. **Use Raoult's Law:**
According to Raoult's law, the vapor pressure of the solution can be expressed as:
\[ P_{\text{solution}} = P^\circ_B \cdot X_B \]
where \(X_B\) is the mole fraction of the solvent (acetone).

3. **Calculate the Mole Fraction of the Solvent:**
Rearranging the equation gives:
\[ X_B = \frac{P_{\text{solution}}}{P^\circ_B} \]
Substituting the values:
\[ X_B = \frac{271 \, \text{mm Hg}}{283 \, \text{mm Hg}} \approx 0.958 \]

4. **Calculate the Mole Fraction of the Solute:**
The mole fraction of the solute (A) can be calculated as:
\[ X_A = 1 - X_B = 1 - 0.958 = 0.042 \]

5. **Calculate Moles of Solvent (B):**
The molecular weight of acetone (C₃H₆O) is 58 g/mol. Therefore, the number of moles of acetone (B) is:
\[ n_B = \frac{\text{mass of B}}{\text{molecular weight of B}} = \frac{80 \, \text{g}}{58 \, \text{g/mol}} \approx 1.379 \, \text{mol} \]

6. **Relate Moles of Solute (A) to Mole Fraction:**
The mole fraction of the solute can also be expressed in terms of moles:
\[ X_A = \frac{n_A}{n_A + n_B} \]
Rearranging gives:
\[ n_A = X_A \cdot (n_A + n_B) \]
Substituting \(X_A\) and \(n_B\):
\[ n_A = 0.042 \cdot (n_A + 1.379) \]

7. **Solve for Moles of Solute (A):**
Rearranging the equation:
\[ n_A - 0.042 n_A = 0.042 \cdot 1.379 \]
\[ n_A (1 - 0.042) = 0.058058 \]
\[ n_A \approx \frac{0.058058}{0.958} \approx 0.0607 \, \text{mol} \]

8. **Calculate the Molecular Weight of the Solute:**
The molecular weight (M_A) of the solute is given by:
\[ M_A = \frac{\text{mass of A}}{n_A} = \frac{10 \, \text{g}}{0.0607 \, \text{mol}} \approx 164.3 \, \text{g/mol} \]

### Final Answer:
The molecular weight of the solute is approximately **164.3 g/mol**.