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

Cell constant, `G**=` Conductivity `xx` Resistance
`=1.29 S m^(-1) xx 100 Omega`
`=129 m^(-1)=1.29cm^(-1)`
Conductivity `(k)` of `0.02 M KCl ` solution `=(G**)/(R)`
`=(129m^(-1))/(520Omega)`
`=0.248m^(-1)`
Concentration `=0.02 M =1000xx 0.02 mol m^(-3)`
`=20 mol m^(-3)`
Molar conductivity `(wedge_(m))=(k)/(c)=(0.248S m^(-1))/(20 mol m^(-3))`
`=124xx10^(-4)m^(2)mol ^(-1)`
Alternatively
`k=(1.29cm^(-1))/(520Omega)=0.248xx10^(-2)S cm^(-1)`
and `wedge _(m)=(kxx1000cm^(3)L^(-1)M^(-1))/(M)`
`=(0.248xx10^(-2)S cm^(-1)xx1000cm^(3)L^(-1))/(0.02M)`
`=124 S cm^(2)mol ^(-1)`