Calculating conductivity and molar conductivity: Resistance of a conductivity cell filled with `0.1 M KCl` solution is `100 Omega`. If the resistance of the same cell when filled with `0.02 M KCl` solutions `520 Omega` and the conductivity of `0.1 KCl` solution is `1.29 S m`, calculate the conductivity and moalr conductivity of `0.02 M KCl` solution.
Strategy : Calculate the cell constant with the help of `0.01 M KCl` solution (both `R` and `kappa` are Known). Use the cell constant to determine the conductivity of `0.02 M KCl` solution and finally find its molar conductivity using the molarity

Explanation

Consider `0.01 M KCl` solution
Cell constant `(G^(**))` = conductivity `xx` resistance
`= (1.29 S m^(-1)) xx (100 Omega)`
`= 1.29 Sm^(-1)`
Now consider `0.02 MKCl` solution :
Conductivity `(kappa)` = Cell constant `(G^(**))`/resistance `(R)`
`= 129 m^(-1)//520Omega`
`= 0.248 S m^(-1)`
Molar concentration `= 0.02 mol L^(-1)`
`= ((1000 L)/(m^(3))) xx (0.02 mol L^(-1))`
`= 20 mol m^(-3)`
Therefore, molar conductivity is given as
`Lambda_(m) = (kappa)/(C)`
`= (0.248 S m^(-1))/(20 mol m^(-3))`
`= (248 xx 10^(-3) S m^(-1))/(2 xx 10^(1) mol m^(-3))`
`= 124 xx 10^(-4) s m^(2)mol^(-1)`