Question

The vapour pressure of pure water at ${25}^{\circ }C$ is 23.76 torr. The vapour pressure of a solution containing 5.40 g of a nonvolatile substance in 90.0 g water is 23.32 torr. The molecular weight of the solute is:

• A. $97.24$ g/mol
• B. $24.29$ g/mol
• C. $50.44$ g/mol
• D. $57.24$ g/mol

Answer 1

Answer: (D)

$57.24$ g/mol

Let $M$ be the molecular weight of solute.

Number of moles of solute $=\frac{5.40}{M}$

Number of moles of water $=\frac{90.0}{18}=5$

Mole fraction of solute $=\frac{\frac{5.40}{M}}{\frac{5.40}{M}+5}=\frac{5.40}{5.40+5M}$

Relative lowering in the vapour pressure $=\frac{P^0-P}{P^0}=\frac{23.76-23.32}{23.76}=0.01852$

The relative lowering in the vapour pressure of the solution is equal to the mole fraction of solute.

$0.01852=\frac{5.40}{5.40+5M}$

$⟹5.40+5M=291.6$

$5M=286.2$

$⟹M=57.24g/mol$

Hence, the correct option is $\text{D}$