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. `80.4%`
B. `74.6%`
C. `94.6%`
D. `64.6%`

Answer

C

Explanation

`DeltaT_f =0.45`
`m= (((0.2)/60)xx1000)/(20)=1/6`
`K_f=5.12 K kg //mol`
`i=1+(1/n-1)beta(n=2)`
`=1-beta/2`
Now , `DeltaT_f=iK_fm`
`0.45 =(1-beta/2)(5.12)(1/6)`
`impliesbeta=0.94`
`:. %` Association `~~94%`