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%
