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

Explanation

`^^_(m)=(kxx1000)/("molarity")=(0.248cm^(-1)xx1000cm^(3)L^(-1))/(0.20" mol "L^(-1))`
`=124" S "cm^(2)" "mol^(1)`