The vapour pressure of pure liquid A is 0.80 atm. On mixing a non-volatile B to A, its vapour pressure becomes 0.6 atm. The mole fraction of B in the solution is:

Options

A. 0.25
B. 1
C. 0.5
D. 0.75

Answer

To solve the problem of finding the mole fraction of non-volatile solute B in the solution, we can use the concept of relative lowering of vapor pressure. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- The vapor pressure of pure liquid A (P₀) = 0.80 atm
- The vapor pressure of the solution (P) = 0.60 atm

### Step 2: Calculate the Change in Vapor Pressure
The change in vapor pressure (ΔP) can be calculated as:
\[ \Delta P = P₀ - P = 0.80 \, \text{atm} - 0.60 \, \text{atm} = 0.20 \, \text{atm} \]

### Step 3: Use the Formula for Relative Lowering of Vapor Pressure
The relative lowering of vapor pressure is given by the formula:
\[ \frac{\Delta P}{P₀} = \frac{n_B}{n_A + n_B} \]
Where:
- \(n_B\) = moles of solute B
- \(n_A\) = moles of solvent A

Since we are looking for the mole fraction of B, we can express it as:
\[ x_B = \frac{n_B}{n_A + n_B} \]

### Step 4: Relate the Change in Pressure to Mole Fraction
From the relative lowering of vapor pressure:
\[ \frac{\Delta P}{P₀} = \frac{0.20 \, \text{atm}}{0.80 \, \text{atm}} = 0.25 \]
This means:
\[ \frac{n_B}{n_A + n_B} = 0.25 \]

### Step 5: Express the Mole Fraction of B
Rearranging the equation gives:
\[ n_B = 0.25(n_A + n_B) \]
This can be rewritten as:
\[ n_B = 0.25n_A + 0.25n_B \]
Now, isolating \(n_B\):
\[ n_B - 0.25n_B = 0.25n_A \]
\[ 0.75n_B = 0.25n_A \]
\[ \frac{n_B}{n_A} = \frac{0.25}{0.75} = \frac{1}{3} \]

### Step 6: Calculate the Mole Fraction of B
Now, we can find the mole fraction of B:
\[ x_B = \frac{n_B}{n_A + n_B} = \frac{n_B}{n_A + n_B} = \frac{n_B}{n_A + \frac{1}{3}n_A} = \frac{n_B}{\frac{4}{3}n_A} \]
Substituting \(n_B = \frac{1}{3}n_A\):
\[ x_B = \frac{\frac{1}{3}n_A}{\frac{4}{3}n_A} = \frac{1}{4} = 0.25 \]

### Final Answer
Thus, the mole fraction of B in the solution is:
\[ \boxed{0.25} \]