Question

Acetic acid dimerizes when dissolved in benzene. As a result boiling point of the solution rises by $0.36^{\circ }C$, when $100g$ 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 $2.57Kkgmo{l}^{-1}$, 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
• E. 1.4 and 99

Answer 1

Answer: (A)

Given $:\Delta T_b=0.36^{\circ }C$

$M_A$( weight of solvent )$=100g$

$K_b=257kkgmo{l}^{-1}$

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, $\Delta T_b=i{K}_b\times m$

$0.36=0.51\times 257\times \frac{x/60\times 1000}{100\times }$

$x=1.65g$

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,

$2C{H}_{3}COOH\rightleftharpoons {\left(C{H}_{3}COOH\right)}_2$

$\alpha ($ 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.