The vapour pressure of acetone at `20^(@)C` is 185 torr. When `1.2 g` of non-volatile substance was dissolved in `100g` of acetone at `20^(@)C` its vapour pressure was 183 torr. The molar mass `(g mol^(-1))` of the substance is:

Options

A. 64
B. 128
C. 488
D. 32

Answer

To solve the problem, we need to determine the molar mass of a non-volatile substance that is dissolved in acetone, given the vapor pressures before and after the solute is added. We will use Raoult's law, which relates the vapor pressure of a solvent in a solution to the mole fraction of the solvent.

### Step-by-Step Solution:

1. **Identify Given Data:**
- Vapor pressure of pure acetone (P₀) = 185 torr
- Vapor pressure of the solution (Pₛ) = 183 torr
- Mass of the non-volatile solute (m₂) = 1.2 g
- Mass of acetone (m₁) = 100 g

2. **Calculate the Relative Lowering of Vapor Pressure:**
\[ \text{Relative lowering of vapor pressure} = \frac{P₀ - Pₛ}{P₀} = \frac{185 - 183}{185} = \frac{2}{185} \]

3. **Use Raoult's Law:**
According to Raoult's law:
\[ \frac{P₀ - Pₛ}{P₀} = \frac{n₂}{n₁ + n₂} \]
Where:
- \( n₁ \) = number of moles of acetone
- \( n₂ \) = number of moles of the solute

4. **Calculate Moles of Acetone (n₁):**
The molar mass of acetone (C₃H₆O) is:
\[ \text{Molar mass of acetone} = 12 \times 3 + 1 \times 6 + 16 = 58 \text{ g/mol} \]
Now, calculate the number of moles of acetone:
\[ n₁ = \frac{m₁}{\text{Molar mass of acetone}} = \frac{100 \text{ g}}{58 \text{ g/mol}} \approx 1.724 \text{ mol} \]

5. **Set Up the Equation:**
Since \( n₂ \) (moles of solute) is much smaller than \( n₁ \), we can approximate \( n₁ + n₂ \approx n₁ \):
\[ \frac{2}{185} = \frac{n₂}{n₁} \]
Rearranging gives:
\[ n₂ = n₁ \cdot \frac{2}{185} \]

6. **Calculate Moles of Solute (n₂):**
Substitute \( n₁ \):
\[ n₂ = 1.724 \cdot \frac{2}{185} \approx 0.0186 \text{ mol} \]

7. **Calculate Molar Mass of the Solute (M):**
The number of moles of solute is given by:
\[ n₂ = \frac{m₂}{M} \]
Rearranging gives:
\[ M = \frac{m₂}{n₂} = \frac{1.2 \text{ g}}{0.0186 \text{ mol}} \approx 64.52 \text{ g/mol} \]

8. **Final Answer:**
Rounding gives the molar mass of the non-volatile substance as approximately **64 g/mol**.