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. `0.150`
B. `0.25`
C. `0.50`
D. `0.75`

Answer

To find the mole fraction of component B in the solution, we can use the concept of relative lowering of vapor pressure. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Vapor pressure of pure liquid A (P₀) = 0.80 atm
- Vapor pressure of the solution (P₁) = 0.60 atm

2. **Calculate the Lowering of Vapor Pressure:**
- The lowering of vapor pressure (ΔP) can be calculated as:
\[ ΔP = P₀ - P₁ = 0.80 \, \text{atm} - 0.60 \, \text{atm} = 0.20 \, \text{atm} \]

3. **Use the Formula for Relative Lowering of Vapor Pressure:**
- The formula for relative lowering of vapor pressure is:
\[ \frac{ΔP}{P₀} = X_B \]
where \(X_B\) is the mole fraction of the solute (component B).

4. **Substitute the Values into the Formula:**
- Substitute the values we have into the formula:
\[ \frac{0.20 \, \text{atm}}{0.80 \, \text{atm}} = X_B \]

5. **Calculate the Mole Fraction of B:**
- Performing the calculation:
\[ X_B = \frac{0.20}{0.80} = 0.25 \]

6. **Conclusion:**
- The mole fraction of B in the solution is:
\[ X_B = 0.25 \]

### Final Answer:
The mole fraction of B in the solution is **0.25**.

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