1.

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

Answer»

SOLUTION :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)`


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