Question

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

• A. $23.8\, mm \,Hg$ and $0.16$
• B. $25.4\, mm \,Hg$ and $0.02$
• C. $30.2\, mm \,Hg$ and $0.020$
• D. $23.4\, mm \,Hg$ and $0.017$

Answer 1

Answer: (D)

Given, $p^{\circ} = 23.8 \,mm \,Hg$
$w_{2} = 50 \,g, M^{2} \left(urea\right) = 60\, g\, mol^{-1}$, $p_{s} =?$,
$\frac{p^{\circ}-p_{s}}{p^{\circ}} = ?$
$w_{1} = 850 \, g, M_{1} \left(water\right) = 18 \, g\, mol^{-1}$
$\therefore n_{2} = \frac{50}{60} = 0.83$,
$n_{1} = \frac{850}{18} = 47.22$
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, $\frac{\Delta p}{p^{\circ}} = 0.017$
$\therefore \Delta p = 0.017 \times 23.8$
$p^{\circ} -p_{s} = 0.017 \times 23.8$
$p_{s} = 23.8 - 0.017 \times 23.8$
$p_{s} = 23.4\,mm\,Hg$
Thus, vapour pressure of water in the solution
$= 23.4\,mm\,Hg$