Resistance of `0.2 M` of solution of an electrolysis is `50 Omega`. The specific conductance of the solution is `1.4 S m^(-1)`. The resistanceof `0.5 M` solution of the same electrolyte is `280 Omega`. The molar conductivity of `0.5 M` solution of the electrolysis in `Sm^(2) mol^(-1)` is :

Options

A. `5 xx 10^(3)`
B. `5 xx 10^(2)`
C. `5 xx 10^(-4)`
D. `5 xx 10^(-3)`

Answer

C

Explanation

`K = (1)/(R) xx (l)/(A)`
`(K_(1))/(K_(2)) = (R_(2))/(R_(1))`
`(1.4)/(K_(2)) = (280)/(50)`
`K_(2) = (7)/(28) = (1)/(4) Sm^(-1)`
`lambda_(m) =(K)/(M xx 1000) = (1)/(4) xx (1)/(0.5 xx 1000)`
`= 5 xx 10^(-4) Sm^(2) mol^(-1)`