Question

The freezing point of a solution containing $0.2g$ of acetic acid in $20g$ of benzene is lowered by ${0.45}^{\circ }C.$. Choose the correct option(s) :
Given :
$(K_f{)}_{\text{Benzene}}=5.12Kkgmo{l}^{-1}$

• A. $\text{Degree of association for acetic acid is}0.55$
• B. $\text{Degree of association for acetic acid is}0.94$
• C. $\text{Van't hoff factor for acetic acid is}0.53$
• D. $\text{Van't hoff factor for acetic acid is}0.92$

Answer 1

Answer: (B), (C)

$\text{Van't hoff factor for acetic acid is}0.53$

$\text{Weight of acetic acid}\left(w\right)=0.2g\text{Weight of benzene}\left(W\right)=20g=\frac{20}{1000}kg\Delta T_f={0.45}^{\circ }C$

$\text{Let}{m}_{obs}\text{be observed molecular mass of acetic acid}$

$\Delta T_f=K_f\times m$

$\text{m is molality}$

$m_{obs}=\frac{1000\times K_f\times w}{W\times \Delta T_f}=\frac{1000\times 5.12\times 0.2}{20\times 0.45}=113.78gmo{l}^{-1}$

$\text{Theoretical molecular mass of acetic acid}=60gmo{l}^{-1}$

$\text{van't Hoff factor}=\frac{\text{Theoretical molecular mass}}{\text{Observed molecular mass}}i=\frac{60}{113.78}\approx 0.53$

$2C{H}_{3}COOH\rightleftharpoons (C{H}_{3}COOH{)}_2\begin{array}{ccc}\text{Before association}& C& 0\\ \text{After association}& (C-C\beta )& \frac{C\beta }{2}\end{array}$

$i=\frac{C-C\beta +\frac{C\beta }{2}}{C}0.53=1-\frac{\beta }{2}$
$\beta =0.94$