JEE Main 2021 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

22 g of benzoic acid (molar mass 122 g mol^-1) is dissolved in 100 g of benzene, where 80% of the acid exists as dimers. The depression in freezing…

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1.22\ \mathrm{g}

of benzoic acid (molar mass

122\ \mathrm{g\,mol^{-1}}

) is dissolved in

100\ \mathrm{g}

of benzene, where

80\%

of the acid exists as dimers. The depression in freezing point is (

K_f

of benzene

=5.12\ \mathrm{K\,kg\,mol^{-1}}

)

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Answer: (D) $0.307\ \mathrm{K}$

Moles of acid

=0.01

, so molality

m=0.01/0.100=0.10\ \mathrm{mol\,kg^{-1}}

. Dimerisation:

2\mathrm{C_6H_5COOH}\rightleftharpoons(\mathrm{C_6H_5COOH})_2

. If

\alpha=0.8

is the fraction associated, moles of particles per mole of acid

=1-\alpha+\alpha/2=1-\alpha/2=0.6

, so

i=0.6

. Then

\Delta T_f=iK_fm=0.6\times5.12\times0.10=0.307\ \mathrm{K}

. The value

0.717

comes from wrongly using

i=1+\alpha/2

as if the solute dissociated, and

0.512

ignores association altogether.

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