Question

For a dilute solution containing $3.2 \,g$ of a non-volatile nonelectrolyte solute in $100 \,g$ water, the elevation in boiling point at $2$ atm pressure is $3^{\circ} C$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of $Hg$ ) of the solution is (Take, $\left.K _{ b }=0.76 \,K\, mol ^{-1} kg \right)$

• A. 724
• B. 740
• C. 742
• D. 718

Answer 1

Answer: (C)

$ \Delta T_{b}=\frac{1000 K_{b} w_{1}}{m_{1} w_{2}} $

$3=1000 \times 0.76 \frac{w_{1}}{m_{1} w_{2}} $

$\therefore \frac{w}{m_{1}}-=\frac{3}{1000 \times 0.76}=\frac{3}{760}$

By Raoult's law,

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\chi_{\text {solute }}=\frac{n_{1}}{n_{1}+n_{2}}$

But, $ n_{2} > > n_{1} $

Thus, $ \frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{n_{1}}{n_{2}}=\frac{w_{1} m_{2}}{m_{1} w_{2}}$

$\frac{p^{\circ}-p_{s}}{p^{\circ}}=\frac{18}{760} $

$(m_{2}$ molar mass of solvent $=18)$

$p^{\circ}-p_{s} =\frac{18}{760} \times 760=18\, mm \,Hg $

$\therefore p_{s}=p^{\circ}-18=760-18=742 \,mm \,Hg$