Question

Calculate the relative lowering of vapour pressure if vapour pressure of pure solvent and vapour pressure of solution at 25°C are 32 and 30 mm Hg respectively.

• A. 0.0721
• B. 0.0552
• C. 0.0625
• D. 0.9375

Answer 1

Answer: (C)

0.0625

Answer 2

The relative lowering of vapour pressure is given by Raoult's Law:

$$\text{Relative Lowering} = \frac{P^0 - P}{P^0}$$

Where:

• $P^0$ is the vapor pressure of the pure solvent
• $P$ is the vapor pressure of the solution

Given:

• $P^0 = 32 \, \text{mm Hg}$
• $P = 30 \, \text{mm Hg}$

Substitute the values:

$$\text{Relative Lowering} = \frac{32 - 30}{32} = \frac{2}{32} = 0.0625$$