The vapour pressure of pure solvent is 0.8 mm of Hg at a particular temperature. On addition of a non-volatile solute ‘A’ the vapour pressure of solution becomes 0.6 mm of Hg. The mole fraction of component ‘A’ is:

Options

A. 0.25
B. 0.75
C. 0.5
D. 0.35

Answer

To find the mole fraction of the non-volatile solute 'A' in the solution, we can use Raoult's Law, which relates the vapor pressure of a solution to the vapor pressure of the pure solvent and the mole fraction of the solute.

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- Vapor pressure of pure solvent (P₀) = 0.8 mm Hg
- Vapor pressure of the solution (Pₛ) = 0.6 mm Hg

**Step 2: Calculate the relative lowering of vapor pressure.**
The relative lowering of vapor pressure can be calculated using the formula:
\[ \text{Relative Lowering of Vapor Pressure} = \frac{P₀ - Pₛ}{P₀} \]

Substituting the values:
\[ \text{Relative Lowering of Vapor Pressure} = \frac{0.8 \, \text{mm Hg} - 0.6 \, \text{mm Hg}}{0.8 \, \text{mm Hg}} = \frac{0.2 \, \text{mm Hg}}{0.8 \, \text{mm Hg}} = 0.25 \]

**Step 3: Relate the relative lowering of vapor pressure to the mole fraction of the solute.**
According to Raoult's Law:
\[ \text{Relative Lowering of Vapor Pressure} = x_A \]
where \( x_A \) is the mole fraction of the solute 'A'.

From our calculation in Step 2:
\[ x_A = 0.25 \]

**Step 4: Conclusion.**
The mole fraction of component 'A' is 0.25.

### Final Answer:
The mole fraction of component 'A' is 0.25.

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