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

Answer

B

Explanation

Specific conductance = Observed conductance `xx` Cell constant
`k = (1)/(R)xx ` Cell constant
or Cell constant = `k xx R =1.3S m^(-1) xx 50 Omega = 65 m^(-1)`
For 0.4 M solution , specific conductance
`(1)/(R) xx ` Cell constant = `(1)/(260 Omega) xx 65 m^(-1) = (65)/(260) S m^(-1)`
`Lambda_(m) = (k (Sm^(-1)))/(1000 L m^(-3) xx "Molarity" (mol L^(-1)))`
`= (65//260)/(1000 xx0.4) S m^(2) mol^(-1) = 6.25 xx 10^(-4) S m^(2) mol^(-1)`.