Question

Taking the example of `Al_(2)(SO_(4))_(3)`, derive the relation between molar conductivity and equivalent conductivity.

Answer 1

Molar conductivity, `wedge_(m)=(kappaxx1000)/(c_(m)("Molarity"))`, Equivalent conductivity, `wedge_(eq)=(kappaxx1000)/(c_(eq)("Normality"))`
Eq. wt. of `Al_(2)(SO_(4))_(3)=(Mol." " wt.)/(6)` (Total +ve valency of Al in `Al_(2)(SO_(4))_(3)=6`) `therefore`Normality`=6xx`Molarity
Hence, `(wedge_(m))/(wedge_(eq))=("Normality")/("Molarity")=6` or `wedge_(m)=6wedge_(eq)`
Alternatively, `wedge_(eq)=(wedge_(m))/(v_(+)z_(+))=(wedge_(m))/(2xx3)=(wedge_(m))/(6)`