The freezing point of benzene decreases by `0.45^(@)C` when `0.2 g` of acetic acid is added to `20 g` of benzene. IF acetic acid associates to form a dimer in benzene, percentage association of acetic acid in benzene will be
`(K_(f) "for benzene" = 5.12 K kg mol^(-1))`

Options

A. 0.646
B. 0.804
C. 0.746
D. 0.946

Answer

D

Explanation

In benzene
`underset(1-alpha)(2CH_3COOH) rArr underset(alpha//2)((CH_3COOH)_2)`
Here `alpha` is degree of association
`DeltaT_f = iK_fm`
`0.45 = ( 1-alpha/2)(5.12) ((0.2)/60)/(20/1000)`
`1-alpha/2 = 0.527`
`alpha = 0.945`
% degree of association = 94.6%