The freezing point of a solution containing 0.2 g of acetic acid in 20.0 g benzene is lowered by `0.45^@C`. The degree of association of acetic acid in benzene is (Assume acetic acid dimerises in benzene and `K_f` for benzene = 5.12 K kg `"mol"^(-1)` ) ?

Options

A. `94.6%`
B. `54.9%`
C. `78.2%`
D. `100%`

Answer

A

Explanation

Given : `w_2=0.2 g , w_1=20 g , DeltaT_f =0.45^@C`
`DeltaT_f = (1000xxK_fxxw_2)/(w_1xxM) rArr 0.45=(1000xx5.12xx0.2)/(20xxM)`
`therefore M_"(observed)"`=113.78 (acetic acid )
As acetic acid dimerises in benzene , so,
`{:(,2CH_3COOH hArr, (CH_3COOH)_2),("Before association",1,0),("After association", 1-alpha,alpha//2):}`
(where `alpha` is degree of association)
Molecular weight of acetic acid =60
`i="Normal molecular mass"/"Observed molecular mass"`
`therefore M_"(normal)"/M_"(observed)"=1-alpha + alpha/2`
`60/113.78 = 1-alpha + alpha/2 therefore alpha` = 0.946 or 94.6%