The vapour pressure `(VP)` of a dilute solution of non-volatile solute is `P` and the `VP` of a pure solvent is `P^(@)`. The lowering of the `VP` is

Options

A. `+ve`
B. `-ve`
C. `P//P^(@)`
D. `P^(@)//P`

Explanation

To find the lowering of vapor pressure when a non-volatile solute is added to a solvent, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Let \( P^0 \) be the vapor pressure of the pure solvent.
- Let \( P \) be the vapor pressure of the solution containing the non-volatile solute.

2. **Identify the Lowering of Vapor Pressure**:
- The lowering of vapor pressure can be defined as:
\[ \Delta P = P^0 - P \]
- Here, \( \Delta P \) represents the decrease in vapor pressure due to the addition of the solute.

3. **Apply Raoult's Law**:
- According to Raoult's Law, the vapor pressure of the solution is directly proportional to the mole fraction of the solvent in the solution. The relationship can be expressed as:
\[ P = P^0 \cdot X_{\text{solvent}} \]
- Where \( X_{\text{solvent}} \) is the mole fraction of the solvent.

4. **Relate Mole Fraction to Lowering of Vapor Pressure**:
- The mole fraction of the solute can be expressed as:
\[ X_{\text{solute}} = 1 - X_{\text{solvent}} \]
- Therefore, the lowering of vapor pressure can also be expressed in terms of the mole fraction of the solute:
\[ \frac{\Delta P}{P^0} = X_{\text{solute}} \]

5. **Conclusion**:
- The lowering of vapor pressure is directly proportional to the mole fraction of the solute in the solution. Thus, we can conclude that:
\[ \Delta P = P^0 - P \]
- This shows that the vapor pressure of the solution is always less than that of the pure solvent when a non-volatile solute is added.

### Final Answer:
The lowering of vapor pressure is given by \( \Delta P = P^0 - P \).

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