1.

When a certain conductance cell was filled with 0.1 mol L^(-1)KCl, it has a resistance of 85Omegaat 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»


SOLUTION :`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)`


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