1.2 g of a non-volatile solute is added to 320 g of methyl alcohol at certain temperature. The vapour pressure is decreased from 400 mm to 399.2 mm Hg. Calculate the molecular weight of solute.

Answer

To solve the problem of calculating the molecular weight of a non-volatile solute added to methyl alcohol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Data**:
- Mass of solute (non-volatile) = 1.2 g
- Mass of solvent (methyl alcohol) = 320 g
- Initial vapor pressure of methyl alcohol (P₀) = 400 mm Hg
- Final vapor pressure (P) = 399.2 mm Hg

2. **Calculate the Change in Vapor Pressure (ΔP)**:
\[ ΔP = P₀ - P = 400 \, \text{mm Hg} - 399.2 \, \text{mm Hg} = 0.8 \, \text{mm Hg} \]

3. **Calculate the Relative Lowering of Vapor Pressure**:
\[ \text{Relative lowering} = \frac{ΔP}{P₀} = \frac{0.8 \, \text{mm Hg}}{400 \, \text{mm Hg}} = 0.002 = 2 \times 10^{-3} \]

4. **Calculate the Number of Moles of Solvent (Methanol)**:
- Molar mass of methanol (CH₃OH) = 12 (C) + 3(1) (H) + 16 (O) = 32 g/mol
\[ n_2 = \frac{\text{mass of solvent}}{\text{molar mass of solvent}} = \frac{320 \, \text{g}}{32 \, \text{g/mol}} = 10 \, \text{mol} \]

5. **Let the Molar Mass of the Solute be M**:
- Number of moles of solute (n₁):
\[ n_1 = \frac{1.2 \, \text{g}}{M} \]

6. **Calculate the Mole Fraction of Solute**:
- Mole fraction of solute (x₁):
\[ x_1 = \frac{n_1}{n_1 + n_2} \approx \frac{n_1}{n_2} \quad \text{(since } n_2 \gg n_1\text{)} \]
\[ x_1 = \frac{\frac{1.2}{M}}{10} = \frac{1.2}{10M} \]

7. **Set Up the Equation Using Raoult's Law**:
- From the relative lowering of vapor pressure:
\[ \frac{ΔP}{P₀} = x_1 \Rightarrow 2 \times 10^{-3} = \frac{1.2}{10M} \]

8. **Solve for M**:
\[ 2 \times 10^{-3} = \frac{1.2}{10M} \Rightarrow 2 \times 10^{-3} \times 10M = 1.2 \]
\[ 2 \times 10^{-2}M = 1.2 \Rightarrow M = \frac{1.2}{2 \times 10^{-2}} = 60 \, \text{g/mol} \]

9. **Conclusion**:
- The molecular weight of the solute is **60 g/mol**.