Question

Vapour pressure of pure water at $298K$ is $23.8mmHg$. $50g$ of urea is dissolved in $850g$ of water. The vapour pressure of water for this solution and its relative lowering are respectively.

• A. $23.8mmHg$ and $0.16$
• B. $25.4mmHg$ and $0.02$
• C. $30.2mmHg$ and $0.020$
• D. $23.4mmHg$ and $0.017$

Answer 1

Answer: (D)

Given, $p^{\circ }=23.8mmHg$
$w_2=50g,M^2\left(urea\right)=60gmo{l}^{-1}$, $p_s=?$,
$\frac{p^{\circ }-p_s}{p^{\circ }}=?$
$w_1=850g,M_1\left(water\right)=18gmo{l}^{-1}$
$\therefore n_2=\frac{50}{60}=0.83$,
$n_1=\frac{850}{18}=47.22$
Applying Raoult s law, $\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{n_2}{n_1+n_2}$
or, $\frac{p^{\circ }-p_s}{p^{\circ }}=\frac{0.83}{47.22+0.83}$
$=\frac{0.83}{48.05}=0.017$
Thus, relative lowering of vapour pressure $=0.017$
Again, $\frac{\Delta p}{p^{\circ }}=0.017$
$\therefore \Delta p=0.017\times 23.8$
$p^{\circ }-p_s=0.017\times 23.8$
$p_s=23.8-0.017\times 23.8$
$p_s=23.4mmHg$
Thus, vapour pressure of water in the solution
$=23.4mmHg$