### JEE Main 12 Jan 2019 Evening Question 23

##### Question: Molecules of benzoic acid $ (C _6H _5COOH) $ dimerise in benzene. V g of the acid dissolved in 30 g of benzene shows a depression in freezing point equal to 2 K. If the percentage association of the acid to form dimer in the solution is 80, then w is (Given that $ K _{f}=5,K,kg,mo{l^{-1}}, $ Molar mass of benzoic acid $ =122g,mo{l^{-1}} $ ) [JEE Main Online Paper Held On 12-Jan-2019 Evening]

#### Options:

A) 2.4 g

B) 1.8 g

C) 1.0 g

D) 1.5 g

## Show Answer

#### Answer:

Correct Answer: A

#### Solution:

- $ \Delta T _{f}=ik _{f}.m $ - (i) where, m = molality $ \Delta T _{f}= $ depression in freezing point i = van’t Hoff factor and $ m=\frac{w}{122}\times \frac{1000}{30} $

For association, $ i=1+( \frac{1}{2}-1 )0.8=0.6 $

So, form eqn. (i), $ 2=0.6\times 5\times \frac{w}{122}\times \frac{1000}{30} $ or $ w=\frac{122\times 2\times 30}{0.6\times 5\times 100}=2.44g $