JEE Main 2016 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

22 g of benzoic acid (M=122) dissolved in 100 g of benzene lowers its freezing point by 0.32 K (Kf=5.12 K kg mol^-1). If the acid associates…

Single correct option

1.22\ \mathrm{g}

of benzoic acid (

M=122

) dissolved in

100\ \mathrm{g}

of benzene lowers its freezing point by

0.32\ \mathrm{K}

(

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

). If the acid associates completely only to the extent of forming dimers,

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

, the percentage of acid molecules that are associated is

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Answer: (A) $75\%$

Molality

m=(1.22/122)/0.100=0.100

mol/kg, so the expected depression is

5.12\times0.100=0.512

K. Van't Hoff factor

i=0.32/0.512=0.625

. For dimerisation with fraction

\alpha

associated, total particles

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

. So

1-\alpha/2=0.625

\alpha=0.75

, i.e.

75\%

. The

62.5\%

option is

i

itself reported as a percentage, and

37.5\%

(1-i)

without the factor of 2 for dimer formation.

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