The conductivity of 0.20 M solution of KCl at 298 K is `0.025 S cm^(-1)`?. Calculate its molar conductivity.

Explanation

Given that k = `0.025 S cm^(-1)`
`C = 0.2 "moles" L^(-1) = 0.2` moles `(1000 cm^(3))^(-1) =(0.2 "moles")/(1000 cm^(3))`
`Lambda_(m) = k/C = (0.025 S cm^(-1))/(0.2 "moles") xx 1000 cm^(3) = 125 S cm^(2) "mol"^(-1)`