Question

3g of urea is dissolved in 45 g of $H_{2}O$. 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

$n_2=\frac{w_2}{M_2}=\frac{3}{60}=\frac{1}{20}=0.05$

$n_1=\frac{w_1}{M_1}=\frac{45}{18}=2.5$

$\therefore \frac{p^{\circ }-p_s}{p^{\circ }}=\frac{n_2}{n_1+n_2}=\frac{0.05}{2.5+0.05}=0.02$