Question

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)`.

Answer 1

The cell constant is given by the equation
cell constant `=G^(*)=`conductivity`xx`resistance
`=1.29S//mxx100Omega=129m^(-1)=1.29cm^(-1)`
conductivity of 0.02 mol `L^(-1)` KCl solution=cell constant/resistance ltbr `=(G^(*))/(R)=(129m^(-1))/(520Omega)=0.248Sm^(-1)`
Concentration `=0.02mol L^(-1)`
`=1000xx0.02mol" "m^(-3)=20mol" "m^(-3)`
Molar conductivity `=^^_(m)=(k)/(c)`
`=(248xx10^(-3)Sm^(-1))/(20mol" "m^(-3))=124xx10^(-4)S" "m^(2)mol^(-1)`
Alternatively, `k=(1.29cm^(-1))/(520Omega)=0.248xx10^(-2)S" "cm^(-1)`
and `^^_(m)=kxx1000cm^(3)L^(-1)" molarity"^(-1)`
`=(0.248xx10^(-2)Scm^(-1)xx1000cm^(3)L^(-1))/(0.02mol" "L^(-1))`
`=124S" "cm^(2)mol^(-1)`