The vapour pressure at a given temperature of an ideal solution containing `0.2 mol` of non-volatile solute and `0.8 mol` of a solvent is `60 mm` of `Hg`. The vapour pressure of the pure solvent at the same temperature will be

Options

A. `120 mm Hg`
B. `150 mm Hg`
C. `60 mm Hg`
D. `75 mm Hg`

Explanation

To solve the problem, we need to determine the vapor pressure of the pure solvent given the vapor pressure of an ideal solution containing a non-volatile solute.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Moles of non-volatile solute (n_solute) = 0.2 mol
- Moles of solvent (n_solvent) = 0.8 mol
- Vapor pressure of the solution (P_solution) = 60 mm Hg

2. **Calculate Total Moles**:
- Total moles (n_total) = n_solute + n_solvent
- \( n_{total} = 0.2 \, \text{mol} + 0.8 \, \text{mol} = 1.0 \, \text{mol} \)

3. **Calculate Mole Fraction of Solvent**:
- Mole fraction of solvent (X_solvent) = \( \frac{n_{solvent}}{n_{total}} \)
- \( X_{solvent} = \frac{0.8 \, \text{mol}}{1.0 \, \text{mol}} = 0.8 \)

4. **Use Raoult’s Law**:
- According to Raoult’s Law, the vapor pressure of the solution (P_solution) is given by:
\[ P_{solution} = P^0_{solvent} \times X_{solvent} \]
- Where \( P^0_{solvent} \) is the vapor pressure of the pure solvent.

5. **Rearranging the Equation**:
- We can rearrange the equation to find \( P^0_{solvent} \):
\[ P^0_{solvent} = \frac{P_{solution}}{X_{solvent}} \]

6. **Substituting the Values**:
- Substitute the known values into the equation:
\[ P^0_{solvent} = \frac{60 \, \text{mm Hg}}{0.8} \]

7. **Calculate \( P^0_{solvent} \)**:
- Perform the calculation:
\[ P^0_{solvent} = \frac{60}{0.8} = 75 \, \text{mm Hg} \]

### Final Answer:
The vapor pressure of the pure solvent at the same temperature is **75 mm Hg**.

---