JEE Main 2025 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

22 g of benzoic acid (M=122) dissolved in 50 g of benzene (Kf=5.12 K kg mol^-1) lowers the freezing point by 0.64 K. If the acid associates into…

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1.22\ \mathrm{g}

of benzoic acid (

M=122

) dissolved in

50\ \mathrm{g}

of benzene (

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

) lowers the freezing point by

0.64\ \mathrm{K}

. If the acid associates into dimers, the percentage association is

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Answer: (D) $75\%$

Molality

=0.01/0.050=0.2\ \mathrm{mol\,kg^{-1}}

, so the calculated

\Delta T_f = 5.12\times0.2=1.024

K. Then

i=0.64/1.024=0.625

. For dimerisation

2\mathrm{A}\rightleftharpoons \mathrm{A_2}

the van't Hoff factor is

i=1-\alpha/2

, so

\alpha = 2(1-0.625)=0.75

, i.e.

75\%

. The value

62.5\%

is just

i

and

37.5\%

1-i

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