Question

$2g$ of benzoic acid $(C_6{H}_{5}COOH)$ dissolved in $25g$ of benzene shows a depression in freezing point equal to $1.62K$. Molal depression constant for benzene is $4.9Kkgmo{l}^{-1}$. 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 : $\Delta T=1.62K;K_f=4.9kg$ mol ${}^{-1}$;
$w_2=2g;M_2=122g$ mol ${}^{-1};w_1=25g$
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 $=99.2$