The freezing point of benzene decreases by 0.450C when 0.2 g of acetic acid is added to 20 g of benzene.  If acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be : (Kf for benzene=5.12 K kg mol−1)
Option: 1  74.6%
Option: 2 94.6
Option: 3  64.6%
Option: 4 80.4%  
 

Answers (1)

In benzene ,

    2 CH_{3}COOH \rightleftharpoons \left ( CH_{3}COOH \right )_{2}

      t = 0                   1                                 -

     t = t                      1 - \beta                          \frac{\beta }{2}

Given:

   w = 0.2g                W = 20 g                  \Delta T = 0.45 K

As we know , \Delta T = \frac{1000 \times K_{f} \times w}{M \times W }

\Delta T = \frac{1000 \times 5.12 \times 0.2}{M \times 20 }= 0.45

 

\therefore , observed M = 113.78 (acetic acid)

Molecular weight of acetic acid = 60

i =\frac{normal \; molecular \; mass}{observed \; molecular \; mass}

\therefore \frac{m_{normal}}{m_{observed}} = 1 - \beta + \frac{\beta }{2}

\frac{60}{113.78} = 1 - \beta + \frac{\beta }{2}

\Rightarrow \beta = 0.945

% degree of association = 94.5 %

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