The vapour pressure of a liquid decreases by 10 torr when a non-volatile solute is dissolved. The mole fraction of the solute in solution is 0.1. What would be the mole fraction of the liquid if the decrease in vapour pressure is 20 torr, the same solute being dissolved?

Options

A. 0.2
B. 0.9
C. 0.8
D. 0.6

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand Raoult's Law
According to Raoult's Law, the relative lowering of vapor pressure is directly related to the mole fraction of the solute in the solution. The formula is given by:

\[ \frac{\Delta P}{P_0} = X_{\text{solute}} \]

where:
- \(\Delta P\) = lowering of vapor pressure
- \(P_0\) = vapor pressure of the pure solvent
- \(X_{\text{solute}}\) = mole fraction of the solute

### Step 2: Use the First Set of Data
From the problem, we know that when a non-volatile solute is dissolved, the vapor pressure decreases by 10 torr, and the mole fraction of the solute is 0.1. We can set up the equation:

\[ \frac{10 \text{ torr}}{P_0} = 0.1 \]

### Step 3: Solve for the Vapor Pressure of Pure Solvent
Rearranging the equation to find \(P_0\):

\[ P_0 = \frac{10 \text{ torr}}{0.1} = 100 \text{ torr} \]

### Step 4: Use the Second Set of Data
Now, we need to find the mole fraction of the liquid when the vapor pressure decreases by 20 torr. We can set up a similar equation:

\[ \frac{20 \text{ torr}}{P_0} = X_{\text{solute}} \]

Substituting \(P_0 = 100 \text{ torr}\):

\[ \frac{20 \text{ torr}}{100 \text{ torr}} = X_{\text{solute}} \implies X_{\text{solute}} = 0.2 \]

### Step 5: Calculate the Mole Fraction of the Solvent
The sum of the mole fractions of the solute and solvent must equal 1:

\[ X_{\text{solvent}} + X_{\text{solute}} = 1 \]

Substituting \(X_{\text{solute}} = 0.2\):

\[ X_{\text{solvent}} + 0.2 = 1 \implies X_{\text{solvent}} = 1 - 0.2 = 0.8 \]

### Final Answer
The mole fraction of the liquid (solvent) when the decrease in vapor pressure is 20 torr is:

\[ \boxed{0.8} \]

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