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 need to follow these steps:

### Step 1: Understand the relationship between vapor pressure and mole fraction
The relative lowering of vapor pressure can be expressed using the formula:
\[ \frac{P_0 - P}{P_0} = x_{solute} \]
where \(P_0\) is the vapor pressure of the pure solvent, \(P\) is the vapor pressure of the solution, and \(x_{solute}\) is the mole fraction of the solute.

### Step 2: Analyze the first scenario
In the first scenario, we know:
- The decrease in vapor pressure (\(P_0 - P\)) is 10 torr.
- The mole fraction of the solute (\(x_{solute}\)) is 0.1.

Using the formula:
\[ \frac{10}{P_0} = 0.1 \]
From this, we can solve for \(P_0\):
\[ P_0 = \frac{10}{0.1} = 100 \text{ torr} \]

### Step 3: Analyze the second scenario
In the second scenario, we are told:
- The decrease in vapor pressure (\(P_0 - P\)) is now 20 torr.

We can use the same formula:
\[ \frac{20}{P_0} = x_{solute} \]
Substituting \(P_0 = 100\) torr:
\[ \frac{20}{100} = x_{solute} = 0.2 \]

### Step 4: Calculate the mole fraction of the liquid
The mole fraction of the solvent (liquid) can be calculated using the relationship:
\[ x_{solvent} + x_{solute} = 1 \]
Substituting the value of \(x_{solute}\):
\[ x_{solvent} = 1 - x_{solute} = 1 - 0.2 = 0.8 \]

### Final Answer
The mole fraction of the liquid (solvent) when the decrease in vapor pressure is 20 torr is:
\[ \boxed{0.8} \]