An ideal solution was found to have a vapour pressure of 80 torr when the mole fraction of a non-volatile solute was 0.2. What would be the vapour pressure of the pure solvent at the same temperature?

Options

A. 64 torr
B. 80 torr
C. 100 torr
D. 400 torr

Answer

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by the mole fraction of the solvent in the solution.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of the solution (\(P_{solution}\)) = 80 torr
- Mole fraction of the non-volatile solute (\(X_{solute}\)) = 0.2

2. **Calculate the Mole Fraction of the Solvent:**
- The total mole fraction of the solution is equal to 1, which is the sum of the mole fractions of the solute and the solvent.
- Therefore, the mole fraction of the solvent (\(X_{solvent}\)) can be calculated as:
\[ X_{solvent} = 1 - X_{solute} = 1 - 0.2 = 0.8 \]

3. **Apply Raoult's Law:**
- According to Raoult's Law:
\[ P_{solution} = X_{solvent} \times P_{pure\ solvent} \]
- Rearranging this formula to find the vapor pressure of the pure solvent (\(P_{pure\ solvent}\)):
\[ P_{pure\ solvent} = \frac{P_{solution}}{X_{solvent}} \]

4. **Substitute the Values:**
- Now substitute the known values into the equation:
\[ P_{pure\ solvent} = \frac{80\ torr}{0.8} \]

5. **Calculate \(P_{pure\ solvent}\):**
- Performing the division:
\[ P_{pure\ solvent} = 100\ torr \]

### Final Answer:
The vapor pressure of the pure solvent at the same temperature is **100 torr**.

---