`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

0.1834

Explanation

Number of milliequivalents of acid= Number fo milliequivalent of base
`(0.5xx"basicity")/("m"_("acid"))xx1000=37.5xx(1)/(10)`
`(0.5xx1xx1000)/("m"_("acid"))=3.75`
`"m"_("acid")=(500)/(3.75)`
`Delta"T"_("f")=[1+("n"-1)alpha]"K"_("f")("W"_("solute")xx1000)/("m"_("solute")xx"W"_("solvent"))`
`0.165=(1+alpha)(1.86xx1.5xx1000)/((500)/(3.75)xx150)`
`alpha=0.1828`
`%"dissociation=18.28"%`