The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S `cm^(-1)` . Calculate its molar conductivity.

Explanation

Given `k=2.50xx10^(-2),S^(-1)cm^(-1),M=0.20molL^(-1)`
`A_(m)=(kxx1000)/(M)=(1000xx2.50xx10^(-2))/(0.20)`
`=125.S^(-1)cm^(2)mol^(-1)`