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

Answer»

SOLUTION :* Molar conductivity `Lamda_(m)`:
Molar conductivity `(Lamda_(m))=(kxx1000)/(c)`
Where, conductivity `k=0.0248" S "cm^(-1)`
`therefore Lamda_(m)=(kxx1000)/("Molarity")`
`=((0.0248" S "cm^(-1))xx(1000cm^(3)DM^(-3)))/(0.20" mol "dm^(-3))(OR(m)L^(-1))/(OR" mol "L^(-1))`
`=124.0S" "cm^(2)mol^(-1)`


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