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JEE Main 201912 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 12 January
Shift: Evening Shift
Subject: Chemistry
Chapter: Solutions

JEE Main 2019 · 12 January · Evening Shift · Chemistry · Solutions

Molecules of benzoic acid ( C _ 6 H _ 5 COOH) dimerise in 30g of...

Molecules of benzoic acid

\left({C}_{6}{H}_{5}COOH\right)

dimerise in

30g

of benzene. '

w

'

g

of benzoic acid shows a depression in freezing point equal to

2K

. If the percentage association of the acid to form dimer in the solution is

80

, then

w

is:

(

Given that

{K}_{f} = 5{Kmol}^{- 1}

, molar mass of benzoic acid

= 122{gmol}^{- 1} )

Options

A $1.0g$
B $2.4g$
C $1.8g$
D $1.5g$

Correct answer

B. $2.4g$

Step-by-step solution

{\Delta T}_{f} = {k}_{f} \times m \times i

\ldots ( i )

Molecular weight of

{C}_{6}{H}_{5}COOH = 112

n = 2 because benzoic acid is dimerised, β = 80 % = 0.8

Here

i = \left(\dfrac{1}{n} - 1\right)\beta + 1

i = \left(\dfrac{1}{2} - 1\right)0.8 + 1

= 0.6

Put the value in equation

( i )

\text{2=0}\text{.6 \times}\dfrac{\dfrac{\text{wt}\text{.}\text{of benzoic acid}}{\text{112}}}{\dfrac{\text{30}}{\text{1000}}}\text{\times 5}

weight of benzoic acid

= 2.4g

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