JEE Main 2018 · Chemistry · Solutions, Solid State and Surface Chemistry · medium

2 g of ethanoic acid (M=60) dissolved in 100 g of benzene lowers its freezing point by 0.640 K (Kf=5.12 K kg mol^-1). If the acid dimerises in…

Single correct option

1.2\ \mathrm{g}

of ethanoic acid (

M=60

) dissolved in

100\ \mathrm{g}

of benzene lowers its freezing point by

0.640\ \mathrm{K}

(

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

). If the acid dimerises in benzene, the percentage of acid molecules that exist as dimers is

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Answer: (D) $75\%$

Molality

=\dfrac{1.2/60}{0.100}=0.2

. Expected

\Delta T_f=5.12\times0.2=1.024

K, so

i=0.640/1.024=0.625

. For

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

with fraction

\alpha

associated, total moles

=1-\alpha/2

, so

0.625=1-\alpha/2

and

\alpha=0.75

, i.e.

75\%

. Using

i=1-\alpha

would give the incorrect

37.5\%

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