1 . 22g of an organic acid is separately dissolved in 100g of benzene ( K _ b = 2 . 6Kkg mol ^ - 1 ) and 100g of acetone ( K _ b = 1 . 7Kkg mol ^ - 1 ) . The acid is known to dimerize in benzene but remain as a monomer

Correct answer: 0 Δ T _ b = i × k _ b × molality ; m = w _ 1 GMM _ 1 × 1000 w _ 2 For Acetone solution 0 . 17 = 1 × 1 . 7 × (1 . 22)/(GMM) × (1000)/(100) GMM of substance = 122gm / mol For Benzene solution Δ T _ b = i × k _ b × m = (1)/(2) × 2 . 6 × (1 . 22)/(122) × (1000)/(100) = 0 . 13 = 13 × 10 ^ - 2 = 13

JEE Main 202131 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 31 August
Shift: Evening Shift
Subject: Chemistry
Chapter: Solutions

JEE Main 2021 · 31 August · Evening Shift · Chemistry · Solutions

1 . 22g of an organic acid is separately dissolved in 100g of...

1 . 22g

of an organic acid is separately dissolved in

100g

of benzene \left({K}_{b} = 2 . 6Kkg{mol}^{- 1}\right) and 100g

of acetone \left({K}_{b} = 1 . 7Kkg{mol}^{- 1}\right) .

The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by

0 . 17 ° C .

The increase in boiling point of solution in benzene in

C ∘

x \times {10}^{- 2} .

The value of

x

is (Nearest integer)
[Atomic mass :

C = 12 . 0, H = 1 . 0, O = 16 . 0

]

Correct answer

Step-by-step solution

{\Delta T}_{b} = i \times {k}_{b} \times

molality ;

m = \dfrac{{w}_{1}}{{GMM}_{1}} \times \dfrac{1000}{{w}_{2}}

For Acetone solution

0 . 17 = 1 \times 1 . 7 \times \dfrac{1 . 22}{GMM} \times \dfrac{1000}{100}

GMM of substance

= 122gm / mol

For Benzene solution

{\Delta T}_{b} = i \times {k}_{b} \times m = \dfrac{1}{2} \times 2 . 6 \times \dfrac{1 . 22}{122} \times \dfrac{1000}{100} = 0 . 13 = 13 \times {10}^{- 2}

= 13

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