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 mol.L"^(-1)` The conductivity of 0.1(M) KCl solution is `"1.29S.m"^(-1).`

Explanation

Cell constant = conductivity `xx` resistance
`="1.29 S.m"^(-1)xx100Omega=129m^(-1)=1.29cm^(-1)`
So, the conductivity of `"0.02 mol.L"^(-1)KCl` solution
`=("cell constant")/("resistance")=(129)/(520)="0.248 S.m"^(-1)`
Concentration `="0.02 mol.L"^(-1)`
`=1000xx"0.02 mol.m"^(-3)="20 mol.m"^(-3)`
Hence, molar conductivity `=Lambda_(m)=(kappa)/(c)=(248xx10^(-3)S.m^(-1))/("20 mol.m"^(-3))`
`=124xx10^(-4)"S.m"^(2)."mol"^(-1)`
Alternatively, `kappa =("1.29 cm"^(-1))/(520 Omega)=0.248xx10^(-2)"S.cm"^(-1)` and
`Lambda_(m)=kappa xx1000cm^(3).L^(-1).M^(-1)`
`=(0.248xx10^(-2)"S. cm"^(2)."mol"^(-1))/("0.02 mol.L"^(-1))`
`="124 S.cm"^(2)."mol"^(-1)`