When a certain conductance cell was filled with 0.1 mol `L^(-1)`KCl, it has a resistance of 85`Omega` at `25^(@)`C. When the same cell was filled with an aqueous solution of 0.052 mol `L^(-1)` of an electrolyte solution, the resistance was 96 `Omega` . Calculate the molar conductivity of the electrolyte at this concentration (Conductivity of 0.1 `mol L^(-1)` KCl solution is `1.29 xx 10^(-2) S `cm^(-)`)

Answer

`219.65 S cm^(2) mol^(-1)`

Explanation

`kappa = G xx G^(**) or G^(**) = (kappa)/(G)`
where G is conductance and `G^(**)` is cell constant `G^(**) = (1.29 xx 10^(-2) S cm^(-1))/(1//85) = 1.0965 cm^(-1)`
`kappa` for `0.052 mol L^(-1)` solution .
`kappa = G xx G^(**) = (1)/(96) xx 1.0965 = 1.142 xx 10^(-2) S cm^(-1)`
`Lambda = (kappa xx 1000)/(M)`
`= (1.142 xx 10^(-2) xx 1000)/(0.052)`
`= 219.6 S cm^(2) mol^(-1)`