### JEE Main Solved Paper 2017 Question 23

##### Question: The freezing point of benzene decreases by $ 0.45{{,}^{o}}C $ 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 :- ( $ K _{f} $ for benzene $ =5.12,K,kg,mo{l^{-1}} $ ) [JEE Main Solved Paper-2017]

#### Options:

A) 64.6%

B) 80.4%

C) 74.6%

D) 94.6%

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

In benzene $ 2CH _3COOH\rightarrow {{(CH _3COOH)}_2} $

$ i=1+( \frac{1}{2}-1 )\alpha $ $ i=1-\frac{\alpha }{2} $

Here $ \alpha $ is degree of association $ \Delta T _{f}=iK _{f}m $
$ 0.45=( 1-\frac{\alpha }{2} )(5.12)\frac{( \frac{0.2}{60} )}{\frac{20}{1000}} $

$ 1-\frac{\alpha }{2}=0.527 $ $ \alpha =0.945 $ % degree of association =94.5%