Question

The vapour pressure of water at 20${}^{\circ }\text{C}$ is 17 mm Hg. Calculate the vapour pressure of a solution containing $2.8$ g of urea $\left(N{H}_{2}CON{H}_2\right)$ in $50$ g of water N = 14, C = 12, O = 16, H = 1 ($16.71$ mm Hg)

Answer 1

Vapour pressure of water= $P_1^o=17mmHg$

Weight of water= $W_1=50g$

Weight of urea= $W_2=2.8g$

Molar mass of water= $M_1=18g/mol\Rightarrow n_1=\frac{50}{18}=2.78$

Molar mass of water= $M_2=60g/mol\Rightarrow n_2=\frac{2.8}{60}=0.047$

$\therefore$ Vapour pressure of solution $P_1$

$\frac{P_1^o-P_1}{P_1^o}=\frac{n_2}{n_2+n_1}$

$\Rightarrow \frac{17-P_1}{17}=\frac{0.047}{2.78+0.047}=0.0166$

$\Rightarrow P_1=16.71mmHg$