Question

$2\, g$ of benzoic acid $(C_{6}H_{5}COOH)$ dissolved in $25\, g$ of benzene shows a depression in freezing point equal to $1.62\, K$. Molal depression constant for benzene is $4.9\, K\, kg\,mol^{-1}$. What is the percentage association of acid if it forms dimer in solution?

• A. $50\%$
• B. $70\%$
• C. $99\%$
• D. $35\%$

Answer 1

Answer: (C)

We know, $\Delta T = i \times K_{f} \times \frac{w_{2} \times 1000}{M_{2} \times w_{1}}$
Given : $\Delta T = 1.62\, K; K_{f}= 4.9\, kg\,$ mol $^{-1}$;
$w_{2} = 2 \,g; M_{2} = 122\, g$ mol $^{-1}; w_{1} = 25\, g$
Substituting these values in the above equation, we get
$1.62 = i \times 4.9 \times \frac{2 \times 1000}{122 \times 25}$
$i = 0.504$
For association
$\alpha = \frac{1 -i}{1 -1/n} n = 2$ for dimerization of benzoic acid
$\alpha = \frac{1 -0.504}{1 - 1/2} = 0.992$
$\therefore \%$ association $= 99.2$