Question

When 12.2 g of benzoic acid is dissolved in 100 g of water, the freezing point of solution was found to be $-{0.93}^{o}C\left(K_f\right(H_{2}O)=1.86Kkgmo{l}^{-1})$. Then the number (n) of benzoic acid molecules associated (assuming 100% association) is

• A. 2.00
• B. 2
• C. 2.0

Answer 1

Answer: (A), (B), (C)

$nPhCOOH\to (PhCOOH{)}_n$

$\frac{1}{n}=i${As, $\alpha$=1}

$\Delta T_f=i\times k_f\times m$

$0.93=\frac{1}{n}\times 1.86\times \frac{12.2\times 1000}{122\times 100};n=2$