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 $-0.93^{\circ} \mathrm{C}\left(\mathrm{K}_{f}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.86 \mathrm{~K} \mathrm{~kg}\right.$ $\mathrm{mol}^{-1}$ ). The number $(\mathrm{n})$ of benzoic acid molecules associated (assuming $100 \%$ association) is_____.
Solution:
$\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{i} \times \mathrm{k}_{\mathrm{f}} \times \mathrm{m}$
$0-(-0.93)=\mathrm{i} \times 1.86 \times \frac{12.2}{122 \times 100} \times 1000$
$\mathrm{i}=\frac{0.93}{1.86}=0.5$
$\mathrm{i}=1+\left(\frac{1}{\mathrm{n}}-1\right) \alpha$
$\frac{1}{2}=1+\left(\frac{1}{n}-1\right) \times 1$
$\mathrm{n}=2$