The freezing point of solution containing `0.2 g` of acetic acid in `20.0 g` of benzene is lowered by `0.45^(@)C`. Calculate the degree of association of acetic acid in benzene.
`(K_(f)=5.12 K^(@) mol^(-1) kg^(-1))`

Explanation

Let the observed molecular mass of acetic acid be `m_(obs)`
`m_(obs)=(1000xxK_(f)xxw)/(WxxDeltaT)`
`=(1000xx5.12xx0.2)/(20xx0.45)=113.78`
Normal molecular mass of acetic acid = 60
`{:(,2CH_(3)COOH,hArr,(CH_(3)COOH)_(2)),("Before association",1,,0),("After association",(1-alpha),,alpha//2):}`
van't Hoff factor `= ("Normal mol. mass")/("Obs. mol. mass")`
`=(1-alpha+alpha//2)/(1)`
`(60)/(113.78)=1-(alpha)/(2)`
or `alpha = 0.945`
or `94.5%` associated.