Resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is `(100Omega)` . The conductivity of this solution is `1.29sm^(-1)`. Resistance of the same cell when filled with 0.2 M of the same solution is `520Omega` . The molar conductivity of (0.02)M solution of the electrolyte will be :

Options

A. ` 124xx10^(-4) sm^(2) mol^(-1)`
B. `1240xx10^(-4) Sm^(2) mol^(-1) `
C. `1.242xx10^(-4)S m ^(2) mol^(-1)`
D. ` 12.4xx10^(-4) sm ^(2) mol^(-1)`

Answer

D

Explanation

`K+ GxxG^ast`
For 0.1M solution `G^ast= (1.29)/(G) = 129m^(-1)` For 0.2M solution `K= Gxx(1)/(R ) = (129)/(520) sm^(-1)`
If we dilute 0.2 M solution by 10 times, its concentration becomes 0.02 M. Thus number of ions per unit volume is `(1)/(10)`th of initial one. So K for this solution is `(129)/(520xx10) SM^(-1)`