Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 `Omega`. The conductivity of this solution is 1.29 `Sm^(-1)`. Resistance of the same cell when filled with 0.02M of the same solution is `520 Omega`. the molar conductivity of 0.02M solution of the electrolyte will be:

Options

A. `1.24xx10^(4)"S m"^(2)mol^(-1)`
B. `12.4xx10^(4)"S m"^(2)mol^(-1)`
C. `124xx10^(4)"S m"^(2)mol^(-1)`
D. `1240xx10^(4)"S m"^(2)mol^(-1)`

Answer

A

Explanation

`K=(l)/(R.a),"cell constt."((l)/(a))=1.29xx100=126`
Again conductivity of `0.02` M solution
`x=(1)/(520)xx129`
`^^_(m)=(x xx1000)/(M)=(129)/(520)xx(1000)/(0.02)=1.24xx10^(4)Sm^(2)mol^(-1)`