The vapour pressure of a solvent decreased by `10 mm` of `Hg` when a non-volatile solute was added to the solvent. The mole fraction of solute is `0.2`, what would be the mole fraction of solvent if the decrease in vapour pressure is `20 mm` of `Hg`.

Options

A. 0.8
B. 0.6
C. 0.4
D. 0.2

Explanation

To solve the problem, we will apply Raoult's law, which relates the vapor pressure of a solution to the mole fraction of the solvent and the vapor pressure of the pure solvent.

### Step-by-Step Solution:

1. **Understanding Raoult's Law**:
Raoult's law states that the vapor pressure of a solvent in a solution (P) is equal to the mole fraction of the solvent (X_solvent) multiplied by the vapor pressure of the pure solvent (P0):
\[ P = X_{\text{solvent}} \cdot P_0 \]
The decrease in vapor pressure when a non-volatile solute is added is given by:
\[ \Delta P = P_0 - P \]

2. **Setting Up the Initial Condition**:
From the problem, we know that when a non-volatile solute is added, the vapor pressure decreases by 10 mm Hg, and the mole fraction of the solute (X_solute) is 0.2. Therefore, the mole fraction of the solvent can be calculated as:
\[ X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - 0.2 = 0.8 \]

3. **Using the Decrease in Vapor Pressure**:
The relationship between the decrease in vapor pressure and the mole fraction of the solute can be expressed as:
\[ \Delta P = P_0 \cdot X_{\text{solute}} \]
For the first condition:
\[ 10 \, \text{mm Hg} = P_0 \cdot 0.2 \quad \text{(1)} \]

4. **Finding P0**:
From equation (1), we can solve for \(P_0\):
\[ P_0 = \frac{10 \, \text{mm Hg}}{0.2} = 50 \, \text{mm Hg} \]

5. **Setting Up for the Second Condition**:
Now, we need to find the mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg. Using the same relationship:
\[ \Delta P = P_0 \cdot X_{\text{solute}} \quad \text{(2)} \]
For the second condition:
\[ 20 \, \text{mm Hg} = 50 \, \text{mm Hg} \cdot X_{\text{solute}} \quad \text{(2)} \]

6. **Finding X_solute for the Second Condition**:
Rearranging equation (2) gives:
\[ X_{\text{solute}} = \frac{20 \, \text{mm Hg}}{50 \, \text{mm Hg}} = 0.4 \]

7. **Calculating the Mole Fraction of the Solvent**:
Now, we can find the mole fraction of the solvent:
\[ X_{\text{solvent}} = 1 - X_{\text{solute}} = 1 - 0.4 = 0.6 \]

### Final Answer:
The mole fraction of the solvent when the decrease in vapor pressure is 20 mm Hg is **0.6**.

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