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))`

Answer

`94.6%`

Explanation

`M_(B)` (Observed ) `= (5.12 xx 0.2 xx 1000)/(20 xx 0.45) = 113.78`
` i= (60)/(113.78) = 0.527`
`2CH_(3) COOH hArr (CH_(3) COOH)_(2)`
`{:(,"Initial",1,0),(,"After dissociation",1-alpha,alpha//2):}`
`i= (1-alpha + alpha//2)/(1) = (1-alpha //2)/(1) = 0.527`
`alpha = 94.6%`