Question

3g of urea is dissolved in 45 g of $H_2O$. The relative lowering of vapour pressure is

• A. $0.05$
• B. $0.04$
• C. $0.02$
• D. $0.01$

Answer 1

Answer: (C)

Relative lowering of vapour pressure

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}$

where $p^{\circ}=$ vapour pressure of the pure solvent.

$p_{s}=$ vapour pressure of the solution

$n_{2}=$ number of moles of the solute

$n_{1}=$ number of moles of the solvent

$n_{2}= \frac{w_{2}}{M_{2}}=\frac{3}{60}=\frac{1}{20}=0.05 $

$n_{1}=\frac{w_{1}}{M_{1}}=\frac{45}{18}=2.5 $

$\therefore \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}=\frac{0.05}{2.5+0.05}=0.02$