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

Explanation

The cell constant is given by the equation :
Cell constant `=G^(**)=` conductivity `xx` resistance
`=1.29 S//m m 100Omega =129m^(-1)=1.29 cm^(-1)`
Conductivity of `0.02 mol L^(-1)` Kcl solution = cell constant /resistance
`=(G^(**))/(R)=(129m^(-1))/(520Omega)=0.248Sm^(-1)`
Concentration `=0.02 mol L^(-1)`
`=1000xx0.02 mol m^(-3) =20 mol m^(-3)`
Molar conductivity `=^^_(m)=k/c `
`(248xx10^(-3)Sm^(-1))/(20 mol m ^(-3))=124xx10^(-4) Sm^(2) mol ^(-1)`
Alternatively, `k=(1.29 cm^(-1))/(520Omega)=0.248xx10^(-2) S cm^(-1)`
`and ^^_(m) =k xx1000 cm^(3) L-1 "molarity"^(-1)`
`=(0.248 xx10^(-2) S cm^(-1)xx1000 cm ^(23) L^(-1))/(0.02 mol L^(-1))`
`=124S cm^(2) mol ^(-1)`