The vapour pressure of a solvent decreases by 5.4 torr when a non-volatile solute is added. In this solution, mole fraction of solute is 0.2. What would be mole fraction of the solvent if decreases in vapour pressure is 16.2 torr ?

Options

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

Answer

To solve the problem step by step, we will use the concept of relative lowering of vapor pressure.

### Step 1: Understand the relationship between vapor pressure lowering and mole fraction
According to Raoult's law, the decrease in vapor pressure (ΔP) of a solvent when a non-volatile solute is added is related to the mole fraction of the solute (χ_B) as follows:

\[ \frac{\Delta P}{P_0} = \chi_B \]

Where:
- ΔP = decrease in vapor pressure
- P_0 = vapor pressure of the pure solvent
- χ_B = mole fraction of the solute

### Step 2: Set up the equations for the first case
For the first case, we have:
- ΔP₁ = 5.4 torr
- χ_B₁ = 0.2

Using the formula, we can write:

\[ \frac{5.4}{P_0} = 0.2 \quad \text{(Equation 1)} \]

### Step 3: Set up the equations for the second case
For the second case, we have:
- ΔP₂ = 16.2 torr
- χ_B₂ = ? (unknown)

Using the same formula, we can write:

\[ \frac{16.2}{P_0} = \chi_B \quad \text{(Equation 2)} \]

### Step 4: Divide the two equations
Now, we divide Equation 1 by Equation 2:

\[ \frac{\frac{5.4}{P_0}}{\frac{16.2}{P_0}} = \frac{0.2}{\chi_B} \]

The \(P_0\) cancels out:

\[ \frac{5.4}{16.2} = \frac{0.2}{\chi_B} \]

### Step 5: Solve for χ_B
Now, we can solve for χ_B:

\[ \chi_B = \frac{0.2 \times 16.2}{5.4} \]

Calculating the right side:

\[ \chi_B = \frac{3.24}{5.4} = 0.6 \]

### Step 6: Find the mole fraction of the solvent (χ_A)
Since the sum of the mole fractions of solute and solvent must equal 1:

\[ \chi_A + \chi_B = 1 \]

Substituting the value of χ_B:

\[ \chi_A + 0.6 = 1 \]

Now, solving for χ_A:

\[ \chi_A = 1 - 0.6 = 0.4 \]

### Final Answer
The mole fraction of the solvent when the decrease in vapor pressure is 16.2 torr is:

\[ \chi_A = 0.4 \]