Question

For a dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100g 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)

724

Answer 2

The elevation in boiling point is
$\Delta T_b=K_b.m:m=molality=\frac{n_2}{w_1} \times 1000$
$[n_2 =Number \, of \, moles \, of \, solute , \, w_1 =$ Weight of
\hspace50mm 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 :
$\frac{-\Delta p}{p^{\circ}}=x_2 =\frac{n_2}{n_1+n_2}=\frac{n_2}{n_1} \, \, \, \, [\because \, n_1 > > n_2]$
$\Rightarrow =\Delta p =760 \times \frac{5}{19} \times \frac{18}{100}$
\hspace30mm =36 \, mm \, of \, Hg
$\Rightarrow p=760-36=724 \, mm \, of \, Hg$