Question

For a dilute solution containing $2.5\, g$ of a non-volatile non-electrolyte solute in $100\,g$ of water, the elevation in boiling point at $1$ atm pressure is $2^{\circ}C$. Assuming concentration of solute is much lower than the concentration of solvent, the vapour pressure (mm of $Hg$) of the solution is (take $K_b = 0.76 \,K\, kg \,mol^{-1}).$

• A. 724
• B. 740
• C. 736
• D. 718

Answer 1

Answer: (A)

The elevation in boiling point is
$\Delta T_b=K_b \cdot m : m$ = molality $=\frac{n_2}{w_1} \times 1000$
$ [n_2 =$ Number of moles of solute , $ w_1 =$ Weight of solvent in gram]
$\Rightarrow \, 2=0.76 \times \frac{n_2}{100} \times 1000$
$\Rightarrow n_2 =\frac{5}{19}$
Also, from Raoult- law of lowering of vapour pressure :
$ \frac{-\Delta p}{p^{\circ}}=x_2 =\frac{n_2}{n_1+n_2}=\frac{n_2}{n_1} \, \, [\because \, n_1 > > n_2]$
$\Rightarrow - \Delta p =760 \times \frac{5}{19} \times \frac{18}{100}$
$ =36 \, mm$ of $Hg$
$\Rightarrow p=760-36=724 \, mm$ of $ Hg$