Resistance of `0.2 M` solution of an electrolyte is `50 Omega`. The specific conductance of the solution is ` 1.3 S m^(-1)`. If resistance of the `0.4 M` solution of the same electrolyte is `260 Omega`, its molar conductivity is .

Options

A. `6.25 xx 10^(-4) S m^2 mol^(-1)`
B. `625 xx 10^(-4) S m^2 mol^-1`
C. `62.5 S m^(2) mol^-1`
D. `6250 S m^(2) mol^(-1)`

Answer

A

Explanation

`kappa = 1.3 S m^(-1) = 1.3 xx 10^-2 S cm^(-1)`
`kappa = 1/R ((l)/(a))`
`1.3 xx 10^(-2) = (1)/(50) ((l)/(a))`
`therefore ((l)/(a)) = 1.3 xx 10^(-2) xx 50 = 65.0 xx 10^(-2) cm^(-1)`
`Lambda_(m) = (kappa xx 1000)/(M)`
`kappa = 1/R ((l)/(a)) = 1/260 xx 65.0 xx 10^-2`
`Lambda_m = 1/260 xx (65.0 xx 10^(-2) xx 1000)/(0.4)`
`6.25 S mol^(-1) cm^(2)`
`= 6.25 xx 10^(4) S mol^(-1) m^(2)`.