Question

Dry air is passed through a solution containing $10\,g$ of a solute in $90\,g$ of water and then through pure water. The loss in weight of solution is $2.5\,g$ and that of pure solvent is $0.05 \,g$ . Calculate the molecular weight of the solute.

• A. 50
• B. 180
• C. 102
• D. 25

Answer 1

Answer: (C)

102

Answer 2

According to Raoults law $\frac{p-{{p}_{s}}}{p}=\frac{n}{n+N}=\frac{0.05}{2.5+0.05}$
$=\frac{0.05}{2.55}=\frac{1}{51}$
Weight of solute $=\frac{w}{W}\times M\times \frac{p}{p-{{p}_{s}}}$
$=\frac{10\times 18}{90}\times 51$
$=102g$