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. `5xx10^(3)`
B. `5xx10^(2)`
C. `5xx10^(-4)`
D. `5xx10^(-3)`

Answer

c

Explanation

`R=(1)/(k)xx(l)/(a) ((l)/(a)=` cell constant `)`
For `0.2M` solution
`50=(1)/(k)xx(l)/(a)implies50=(1)/(1.4)xx(l)/(a)`
`(l)/(a)=70m^(-1)`
For `0.5M` solution
`280=(1)/(k) xx70`
`k=(1)/(4)Sm^(-1)`
`^^_(m)((kxx1000)/(M))(10^(-2)m)^(3)`
`^^_(m)=(1)/(4)xx((1000)/(M))(10^(-2)m)^(3)`
`=(1)/(4)xx(1000)/(0.5)xx10^(-6)`
`=500xx10^(-6)=5xx10^(-4)Sm^(2) mol^(-1)`