Resistance of a conductivity cell filled with 0.1 mol `L^(-1)` KCl solution is 100 ohms. If the resistance of the same cell filled with 0.02 mol `L^(-1)` KCl solution is 520 ohms, calculate the conductivity and molar conductivity of 0.02 mol `L^(-1)` KCl solution. The conductivity of 0.1 mol `L^(-1)` KCl solution is `1.29 xx 10^(-2) S cm^(-1)`

Explanation

Cell constant, `G^(@)` = Conductivity `xx` Resonance
`=1.29 xx 10^(-2) S cm^(-1) xx 100` ohm
`=1.29 cm^(-1)`
Conductivity of 0.02 mol `L^(-1)` KCl `=G^(@)/("Resistance") = (1.29 cm^(-1))/(520 ohm) = 0.00248 S cm^(-1)`
`Lambda_(m) = (k xx 1000)/("concentration") = (2.48 xx 10^(-3) xx 1000)/0.02 = 124 S cm^(2) mol^(-1)`