Question

When $2.44$ grams of benzoic acid $\left(C_6{H}_{5}COOH\right)$ dissolved in $25$ grams of benzene, it shows depression of freezing point equal to $2.2K$. Molal depression constant of benzene is $5.0Kkgmo{l}^{-1}$. What is the percentage association of acid, if it forms dimer in solution?

• A. $50\%$
• B. $77\%$
• C. $95\%$
• D. $90\%$

Answer 1

Answer: (D)

Given

Total number of moles of equilibrium $=1-\alpha +\alpha /2$
$=1-\alpha /2$
$\therefore i=\frac{\text{ Total moles at equilibrium }}{\text{ Initial moles }}=\frac{1-\alpha /2}{1}$
We know that, $\Delta T_f=i{K}_{f}m$
$2\cdot 2=\left(1-\frac{\alpha }{2}\right)\times 5.0kkgmo{l}^{-1}\times \frac{2.44}{122}\times \frac{1000}{25}$
$1-\frac{\alpha }{2}=\frac{2.2\times 122\times 25}{244\times 1000\times 5}$
$\Rightarrow 1-\frac{\alpha }{2}=0.55$
$\alpha =0.90$
Percentage degree of association $=90\%$