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

Explanation

Step I : Let us first calculate the cell constant,
Cell constant, `(G^(**))=` Conductivity `(K) xx` Resistance (R)
Resistance of 0.1 M KCl solution
`=100 Omega`
Conductivity of 0.1 M KCl solution
`=1.29 Sm^(-1)`
Cell constant `=1.29 (Sm^(-1))xx100 Omega`
`=129 m^(-1)`
`=1.29 cm^(-1)`
Step II : Calculation of conductivity of 0.02 M KCl solution
Resistance of solution `=520 Omega`
Cell constant `(G^(**))=1.29 cm^(-1)`
Conductivity, `kappa=("Cell constant")/("Resistance")`
`=(1.29 cm^(-1))/(520 Omega)`
`=0.248xx10^(-2) S cm^(-1)`
Step III : Calculation of molar conductivity
`lambda_(m)=(1000xx kappa)/C`
`C=0.02 M, kappa =0.248xx10^(-2) S cm^(-1)`
`lambda_(m)=(1000xx0.248xx10^(-2))/(0.02)`
`=124 S cm^(2) mol^(-1)`