Question

Acetic acid dimerizes when dissolved in benzene. As a result boiling point of the solution rises by $0.36^{\circ}C$, when $100\, g$ of benzene is mixed with "X" g of acetic acid. In this solution, if experimentally measured molecular weight of acetic acid is 117.8 and molar elevation constant of benzene is $2.57\, K\, kg\, mol^{-1}$, what is the weight % and degree of dissociation (in %) of acetic acid in benzene?

• A. 1.62 and 98.3
• B. 0.81 and 98.3
• C. 0.5 and 86
• D. 1 and 98.3

Answer 1

The Correct Option is A

Given $: \Delta T_{b}=0.36^{\circ} C$
$M_{A}$( weight of solvent )$=100\,g $
$K_{b}=257\,k \,kg \,mol ^{-1}$

Experimental molecular weight of acetic acid $=117.8 $

As, $i=\frac{\text { normal molar mass }}{\text { abnormal molar mass }} $
$\therefore i=\frac{60}{117.8}=0.51 $

Now, $ \Delta T_{b}=i K_{b} \times m$
$0.36=0.51 \times 257 \times \frac{x / 60 \times 1000}{100 \times} $
$x=1.65\, g$

Weight $\%$ of acetic acid

$=\frac{\text { weight of acetic acid }}{\text { weight of solution }} \times 100$
$=\frac{1.65}{101.65} \times 100=1.62$

Also, dimerisation of acetic acid is given as,

$2 CH _{3} COOH \rightleftharpoons\left( CH _{3} COOH \right)_{2}$
$\alpha($ degree of association $)=\frac{i-1}{\frac{1}{n}-1}=\frac{0.51-1}{\frac{1}{2}-1}$
$=.98$ or $98 \%$

Thus, weight $\%$ and degree of association of acetic acid is $1.62$ and $98 \%$ respectively.