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 ^(-4) "mol"^(-1)`
C. ` 1. 24xx 10^(-4) S m^2 "mol"^(-1)`
D. ` 12.4 xx 10^(-4) S m^(2) "mol"^(-1)`

Answer

A

Explanation

`C= 0 . 1 M, R = 100 oh m`
`k= 1.29 ohm^(-1) m^(-1)`
` :. K= 1/R xx 1/a`
`:. 1/a = k. R= 1.29 = 1.29 xx 100` ohm
`C= 0.02 m, R= 530`
`k= 1/R xx l/a = 1/(520) xx 129 = 0.248 "ohm"^(-1) m^(-1)`
k decreases eith dilution .
`:. C = 0.02` and not `0.2`
Also `Lambda=k xx 1/(M("in" m//L))`
`K+ 1/(M xx 10^(-3) (m//m^3)) =(0.2 48 xx 1)/( 0.02 xx 10^(-3))`
` = 12 .4 xx 10^(-3) Sm^2 "mol"^(-1) = 124 xx 10^(-4) S m^2 "mol"^(-1)`.