If 0.01 M solution of an electrolyte has a resistance of 40 ohms in a cell having a cell constant of 0.4 `cm^(–1)` then its molar conductance in `ohm^(–1) cm^2 mol^(– 1)` will be

Options

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

Answer

D

Explanation

Molarity = 0.01 M , Resistance = 40 ohm,
Cell constant `(1)/(a) = 0.4 cm^(-1)` .
Specific conductivity (k) = `("cell constant")/("resistance") = (0.4)/(40) = 0.01 ohm^(-1) cm^(-1)`
Molar conductance `(Lambda_(m)) = (1000 k)/("Molarity")`
`= (1000 xx 0.1)/(0.01) = 10^(3) ohm^(-1) cm^(2) mol^(-1)`