Question

Vapour pressure of pure water at $298\, K$ is $23.8\, mm \,Hg.$
$ \,50\, g$ of urea $(NH_{2}CONH_{2})$ is dissolved in $850\, g$ of water. The vapour pressure of water in the solutions is

• A. $43.4\, mm$
• B. $13.5\, mm$
• C. $25\, mm$
• D. $23.4\, mm$

Answer 1

Answer: (D)

Given, $P^{\circ} = 23.8\, mm$
$w_{2} = 50\, g, M_{2}$ (urea) $= 60\, g$ mol $^{-1}, P_{s} = $?
$\frac{P^{\circ} - P_{s}}{P^{\circ}} =$?
$w_{1} = 850\, g, M_{1}$ (water )$ = 18 g\, mol^{-1}$
$\therefore n_{2} = \frac{50}{60} = 0.83$
$\therefore n_{1} = \frac{850}{18} = 47.22$
Applying Raoult’s law, $\frac{P^{\circ} - P_{s}}{P^{\circ}} =\frac{n_{2}}{n_{1} + n_{2}}$
or $\frac{P^{\circ} - P_{s}}{P^{\circ}} = \frac{0.83}{47.22 + 0.83}$
$= \frac{0.83}{48.05} = 0.017$
$P^{\circ} - P_{s} = 0.017 \times 23.8$
or $P_{s} = 23.8 - 0.017 \times 23.8$
or $P_{s} = 23.4\, mm$