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)`.

Options

A. `124 xx 10^(-4) S m^(2) mol^(-1)`
B. `1240 xx 10^(-4) S m^(2) mol^(-1)`
C. `1.24 xx 10^(-4) S m^(2) mol^(-1)`
D. `12.4 xx 10^(-4) S m^(2) mol^(-1)`

Answer

A

Explanation

at `C = 0.1 M: (1)/(a) = kappaR = 1.29 xx 100`
at `C = 0.02 M, kappa = (1)/(a xx R) = (1.29 xx 100)/(520)`
`= 0.248 ohm^(-1) m^(-1)`
Also
`Lambda = (kappa)/(M("in" mol//L)) = (kappa)/(M xx 10^(3)("in" mol//m^(3)))`
`= (0.248)/(0.02 xx 10^(3) = 124 xx 10^(-4) Sm^(2) mol^(-1)`