# 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%

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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