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

Answer

D

Explanation

`k=-(1)/(R)((I)/(a))=(1)/(520)xx1.29xx10 m^(-1)`
Now, we have `mu=kxxv`
Hence,
`mu=(1)/(520)xx129xx(100)/(0.2)xx10^(-6) m^(2)`
`=12.4xx10^(-4) S ^(2)// "mol"`