Question

The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by 0.45 $^{\circ}$C. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and $K_f$ for benzene = $5.12\, K \,kg \,mol^{-1}$) $M_{observed}$ of acetic acid = 113.78

• A. 0.945
• B. 0.549
• C. 0.782
• D. 1

Answer 1

The Correct Option is A

Given : w$_{2}$ = 0.2 g, w$_{1}$ = 20 g, $\Delta T_{f }= 0.45 ^{\circ}C$ $\Delta T_{ f} =\frac{1000\times K_{ f} \times w_{2}}{w_{1}\times M} \Rightarrow 0.45=\frac{1000\times5.12\times0.2}{20\times M}$ $\therefore\quad M_{\left(observed\right)}=113.78\left(acetic acid\right)$ $\quad\quad\quad2CH_{3}COOH{\rightleftharpoons} \left(CH_{3}COOH\right)_{2}\quad$ Before association $\quad\quad\quad\quad1\quad\quad\quad\quad0$ After association$\quad\quad\quad\quad1-\alpha\quad\quad\quad\alpha/2$ $\quad\quad\quad\quad$ (where a is degree of association) Molecular weight of acetic acid = 60 $i=\frac{Normal molecular mass}{Observed molecular mass}$ $\therefore\quad \frac{M_{\left(normal\right)}}{M_{\left(observed\right)}}=1-\alpha+\frac{\alpha}{2}$ or,$\quad \frac{60}{113.78}=1-\alpha+\frac{\alpha}{2} \therefore \alpha$ = 0.945 or 94.5%