If `P^(@)` and `P_(s)` are vapour pressure of solvent and its solution, respectively, `chi_(1)` and `chi_(2)` are mole fractions of solvent and solute, respectively, then

Options

A. `P_(s)=P^(@)//chi_(2)`
B. `P^(@)-P_(s)=P^(@)chi_(2)`
C. `P_(s)=P^(@)chi_(2)`
D. `(P^(@)-P_(s))/P_(s)=chi_(1)/(chi_(1) + chi_(2))`

Explanation

To solve the problem, we will use the concept of relative lowering of vapor pressure (RLVP) and the definitions of vapor pressure and mole fractions.

### Step-by-Step Solution:

1. **Understanding the Variables**:
- Let \( P^0 \) = Vapor pressure of the pure solvent.
- Let \( P_s \) = Vapor pressure of the solution.
- Let \( \chi_1 \) = Mole fraction of the solvent.
- Let \( \chi_2 \) = Mole fraction of the solute.

2. **Using the RLVP Concept**:
The relative lowering of vapor pressure is given by the formula:
\[ \frac{P^0 - P_s}{P^0} = \chi_2 \]
This equation states that the relative lowering of vapor pressure (the difference between the vapor pressure of the pure solvent and the vapor pressure of the solution, divided by the vapor pressure of the pure solvent) is equal to the mole fraction of the solute.

3. **Rearranging the Equation**:
From the above equation, we can rearrange it to find \( P_s \):
\[ P^0 - P_s = P^0 \cdot \chi_2 \]
This implies:
\[ P_s = P^0 - P^0 \cdot \chi_2 \]
Simplifying further:
\[ P_s = P^0 (1 - \chi_2) \]

4. **Relating Mole Fractions**:
Since \( \chi_1 + \chi_2 = 1 \) (the sum of the mole fractions of solvent and solute must equal 1), we can express \( \chi_1 \) as:
\[ \chi_1 = 1 - \chi_2 \]
Therefore, we can also write:
\[ P_s = P^0 \cdot \chi_1 \]

5. **Final Equation**:
Thus, we have two important equations:
- \( P_s = P^0 \cdot \chi_1 \)
- \( P^0 - P_s = P^0 \cdot \chi_2 \)

6. **Identifying the Correct Option**:
From the options provided:
- Option B states: \( P^0 - P_s = P^0 \cdot \chi_2 \)
This is consistent with our derived equation, confirming that Option B is correct.

### Conclusion:
The correct answer is:
\[ P^0 - P_s = P^0 \cdot \chi_2 \]