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

Explanation

Step I . Let us first calculate the cell constant ,
Cell constant , `G^(**)` = Conductivity `(kappa) xx ` Resistance (R)
Resistance of 0.1 M KCl solution = `100 Omega`
Conductivity of 0.1 M KCl solution = `1.29 S m^(-1)`
`therefore " "` Cell constant = `1.29 (S m^(-1)) xx 100 Omega`
`= 129 m^(-1)`
or `" " = 1.29 cm^(-1)`
Step II. Calculation of conductivity of 0.02 M KCl solution .
Resistance of solution = `520 Omega`
Cell constant `(G^(**)) = 1.29 cm^(-1)`
Conductivity `, kappa = ("Cell constant")/("Resistance")`
`= (1.29 cm^(-1))/(520 Omega)`
`= 0.248 xx 10^(-2) S cm^(-1)`
Step III . Calculation of molar conductivity .
`Lambda_(m) = (1000 xx kappa)/(C)`
`C = 0.02 M , kappa = 0.248 xx 10^(-2) S cm^(-1)`
`Lambda_(m) = (1000 xx 0.248 xx 10^(-2))/(0.02)`
`=124 S cm^(2) mol^(-1)`