Question

Resistance of a conductivity cell filled with 0.1 mol `L^(-1)` KCl solution is `100Omega`. If the resistance of the same cell when filled with 0.02 mol `L^(-1)` KCl solution is `520Omega`, calculate the conductivity and molar conductivity of 0.02 M KCl solution. The conductivity of 0.1 M KCl solution is 1.29S/m.

Answer 1

(i) For 0.1 M KCl solution, `R=100Omega,kappa=1.29" S "m^(-1)` ,brgt Cell constant=Conductivity`xx`Resistance`=1.29" S "m^(-1)xx100Omega=129m^(-1)xx100Omega=129m^(-1)(S=Omega^(-1))=1.29cm^(-1)`
(ii) Conductivity of 0.02 M KCl solution `(kappa)=("Cell constant")/("Resistance")=(129m^(-1))/(520Omega)=0.248Omega^(-1)m^(-1)` or `0.248" S "m^(-1)`
Concentration of the solution `=0.02M=0.02" mol "L^(-1)=0.02xx10^(3)" mol "m^(-3)=20" mol "m^(-3)`
Molar conductivity`=(kappa)/(c_(m))=(0.248Sm^(-1))/("20 mol "m^(-3))=0.0124" S "m^(2)mol^(-1)=1.24xx10^(-2)Sm^(2)mol^(-1)`