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 $K_b$=0.76K kg $mo{l}^{-1}$).

• A. 724
• B. 740
• C. 736
• D. 718

Answer 1

The correct option is A

724

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 solvent in gram]
$\Rightarrow 2=0.76\times \frac{n_2}{100}\times 1000$
$\Rightarrow n_2=\frac{5}{19}$
Also, from relative lowering of vapour pressure:
$\frac{-\Delta p}{p^0}=x_2=\frac{n_2}{n_1+n_2}\approx \frac{n_2}{n_1}[\because n_1>>n_2]$
$\Rightarrow -\Delta P=760\times \frac{5}{19}\times \frac{18}{100}$
= 36 mm of Hg
$\Rightarrow$ p = 760 – 36 = 724 mm of Hg