Question

An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Determine the lowering in vapour pressure at 298 K.

[Given : $K_f=1.86$ K kg mol${}^{-1}$, $K_b=0.512$ K kg mol${}^{-1}$ and vapour pressure of water at $298K=23.756$ mm of Hg].

• A. $0.13$
• B. $0.15$
• C. $0.16$
• D. $0.1378$

Answer 1

The correct option is D $0.1378$

Lowering in vapour pressure
According to Raoult's law

$\frac{P^o-P}{P^o}={\chi }_B$

Now,

${\chi }_B=\frac{n_B}{n_A+n_B}$

For dilute solution

$n_A+n_B\approx n_A$

$X_B=\frac{n_B}{n_A}=\frac{n_B}{W_A}\times M{w}_A$ ...(i)

and

$molality=\frac{n_B\times 1000}{W_A}$ ....(ii)

Dividing equation (i) by (ii), we get

$\frac{{\chi }_B}{m}=\frac{M{w}_A}{1000}\Rightarrow {\chi }_B=\frac{m\times M{w}_A}{1000}$

$\therefore {\chi }_B=\frac{0.322\times 18}{1000}=0.00580$

Now $\frac{P^o-P}{P^o}=0.00580$

or $P^o-P=P^o\times 0.00580=23.756\times 0.00580=0.1378mm$

$\therefore$ Lowering in vapour pressure =0.1378 mm Hg