The vapour pressure of pure liquid solvent `A` is `0.80` atm. When a non-volatile substance `B` is added to the solvent, its vapour pressure drops to `0.60` atm, the mole fraction of component `B` in the solution is

Options

A. 0
B. 0.25
C. 2. 0
D. 3. 0

Explanation

To solve the problem, we will use Raoult's Law, which states that the vapor pressure of a solvent in a solution is equal to the vapor pressure of the pure solvent multiplied by its mole fraction in the solution.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Vapor pressure of pure solvent A, \( P^0_A = 0.80 \) atm
- Vapor pressure of the solution, \( P_t = 0.60 \) atm

2. **Apply Raoult's Law:**
According to Raoult's Law, the vapor pressure of the solution can be expressed as:
\[ P_t = P^0_A \cdot x_A \]
where \( x_A \) is the mole fraction of solvent A.

3. **Rearranging the Equation:**
We can rearrange the equation to find \( x_A \):
\[ x_A = \frac{P_t}{P^0_A} \]

4. **Substituting the Values:**
Substitute the known values into the equation:
\[ x_A = \frac{0.60 \, \text{atm}}{0.80 \, \text{atm}} = 0.75 \]

5. **Finding the Mole Fraction of Component B:**
Since the sum of the mole fractions in a solution is equal to 1, we can find \( x_B \):
\[ x_B = 1 - x_A = 1 - 0.75 = 0.25 \]

6. **Final Answer:**
The mole fraction of component B in the solution is:
\[ x_B = 0.25 \]