Question:
When $12.2 \mathrm{~g}$ of benzoic acid is dissolved in $100 \mathrm{~g}$ of water, the freezing point of solution was found to be $-0.93^{\circ} \mathrm{C}\left(\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right)$. The number $(\mathrm{n})$ of benzoic acid molecules associated (assuming $100 \%$ association ) is
Solution:
(12)
n PhCOOH $\rightarrow(\mathrm{PhCOOH})_{\mathrm{n}}$
$N=\frac{1}{X}=\mathrm{i}\{$ As $\quad \alpha=1\}$
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{k}_{\mathrm{f}} \times \mathrm{m}$
$0.93=\frac{1}{n} \times 1.86 \times \frac{12.2 \times 1000}{122 \times 100}$
$n=2$