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)

$23.4\, mm \,Hg$ and $0.017$

Answer 2

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 $= 23.4\,mm\,Hg$