Question

For a dilute solution containing $2.5g$ 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 $K_b=0.76Kkgmo{l}^{-1}).$

• 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 :
$\frac{-\Delta p}{p^{\circ }}=x_2=\frac{n_2}{n_1+n_2}=\frac{n_2}{n_1}[\because n_1>>n_2]$
$\Rightarrow -\Delta p=760\times \frac{5}{19}\times \frac{18}{100}$
$=36mm$ of $Hg$
$\Rightarrow p=760-36=724mm$ of $Hg$