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

Answer

C

Explanation

We have
`kappa = (1)/(R)`. Cell constant
For `0.02 M` solution :
`1.4 S m^(-1) = (1)/(50 ohm)`. Cell contant
or Cell constant `= (1.4 S m^(-1)) (50 ohm)`
`= 70 m^(-1)`
For `0.5 M` solution
`kappa = ((1)/(280 ohm)) (70 m^(-1)) = 0.25 Sm^(-1)`
We have
`Lambda_(m) = (kappa)/("Molarity")`
`= ((0.25 S m^(-1))/(0.5 mol L^(-1)))`
`= (0.25 S m^(-1))/(0.5 mol(10^(-3) m^(3))^(-1))`
`= 0.5 xx 10^(-3) S m^(2) mol^(-1)`
`= 0.5 xx 10^(-4) S m^92) mol^(-1)`