When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $-$0.93$^\circ$C (K_f(H₂O) = 1.86 K kg mol^$-$1). The number (n) of benzoic acid molecules associated (assuming 100% association) is ___________.

$\underset{\text{Benzoic acid}}{n \mathrm{PhCOOH}} \stackrel{\text { Association }}{\longrightarrow}(\mathrm{PhCOOH})_n$<br/><br/> Assuming $100 \%$ association ( $\alpha=1$ ),<br/><br/> $\Rightarrow i=1-\alpha\left(1-\frac{1}{n}\right)=\frac{1}{n}[\because \alpha+1]$ <br/><br/> Now, $\Delta T_f=K_f \times m \times i$<br/><br/> $0-(0.93)=1.86 \times \frac{w_B \times 1000}{w_A \times M_B} \times \frac{1}{n}$<br/><br/> $\left[\because w_B=\right.$ mass of $\mathrm{PhCOOH}=12.2 \mathrm{~g}$<br/><br/> $w_A=$ mass of $\mathrm{H}_2 \mathrm{O}=100 \mathrm{~g}$<br/><br/> $M_B=$ molar mass of $\left.\mathrm{PhCOOH}\right]$<br/><br/> $=122 \mathrm{~g} \mathrm{~mol}^{-1}$<br/><br/> $=1.86 \times \frac{12.2 \times 1000}{100 \times 122} \times \frac{1}{n}$<br/><br/> $$ \begin{aligned} \Rightarrow n &=\frac{1.86 \times 12.2 \times 1000}{0.93 \times 100 \times 122}=2 \end{aligned} $$<br/><br/> $\therefore$ Number of benzoic acid molecules associated, $n=2$