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 find the vapor pressure of the pure solvent at the same temperature, we can use Raoult's Law, which states that the vapor pressure of a solvent in an ideal solution is directly proportional to the mole fraction of the solvent.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- 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 mmHg

2. **Calculate the Total Moles in the Solution:**
\[ n_{total} = n_{solute} + n_{solvent} = 0.2 \, \text{mol} + 0.8 \, \text{mol} = 1.0 \, \text{mol} \]

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

4. **Use Raoult’s Law to Relate the Vapor Pressure of the Solution to the Vapor Pressure of the Pure Solvent:**
\[ P_{solution} = X_{solvent} \cdot P_{0} \]
Rearranging gives:
\[ P_{0} = \frac{P_{solution}}{X_{solvent}} \]

5. **Substitute the Known Values:**
\[ P_{0} = \frac{60 \, \text{mmHg}}{0.8} = 75 \, \text{mmHg} \]

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