Question

$0.5$ molal solution of a solute in benzene shows a depression in freezing point equal to $2\, K$. Molal depression constant for benzene is $5\, K\, kg\, mol ^{-1}$. If the solute forms dimer in benzene, what is the $\%$ association?

• A. 40
• B. 50
• C. 60
• D. 80

Answer 1

Answer: (A)

Given, molality of solution, $m=0.5\, m$
$\Delta T_{f}=2\, K$
$K_{f}=5\, K\, kg\, mol ^{-1}$
of association, $\alpha=$ ?
Step I Calculation of van't Hoff factor (i)
$\Delta T_{f}=i K_{f} M$
$i=\frac{\Delta T_{f}}{K_{f} M}$
$i=\frac{2 K}{5 K kg mol ^{-1} \times 0.5\, m }$
$i=\frac{4}{5}$
Step II Calculation of $\%$ association of acid $(\alpha)$.
As solute form dimer in benzene, i.e. $2\, A\, A_{2}$
Therefore, value of $x=2$
$\alpha=\frac{i-1}{1 / x-1}$
$\alpha=\frac{4 / 5-1}{1 / 2-1}=0.4$
$\therefore \%$ of $\alpha=0.4 \times 100=40 \%$