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

Answer

`124s cm^(2)mol^(-1)`

Explanation

`wedge_(m)=(kxx1000)/("Molarity")`
`=(0.0248Scm^(-1)xx1000cm^(2)L^(-1))/(0.20mL^(-1))`
`=124 S cm^(2) mol^(-1)`