`1.5g` of monobasic acid when dissolved in `150g` of water lowers the freezing point by `0.165^(@)C.0.5g` of the same acid when titrated, after dissolution in water, requires `37.5 mL` of `N//10` alkali. Calculate the degree of dissociation of the acid `(K_(f)` for water `= 1.86^(@)C mol^(-1))`.

Answer

`18.27%`

Explanation

`(0.5)/(M) = 3.75 xx 10^(-3) rArr M = 133.33`
`0.165 = (1-alpha) xx (1.86 xx 1.5 xx 1000)/(133.33 xx 150)`
`1 + alpha =1.1827`
`alpha = 0.1827 = 18.27%`