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

`k = (1)/(R) xx(1)/(a)`
`1.3 = (1)/(50) xx (1)/(a)`
`:. (l)/(a) = 65 m^(-1)`
k of `0.4m` solution is
`k = (1)/(R) xx(1)/(a)`
`= (1)/(260) xx 65 = (1)/(4) ohm^(-1) m^(-1) = (1)/(4) Sm^(-1)`
`^^_(m) = (k)/("molarity") = (1)/(4 xx 0.4) S m^(-1) mol^(-1) sm^(3)`
`= (1)/(4 xx 0.4 xx 1000) Sm^(-) mol^(-1) m^(3)`
`=(1)/(1.6) xx 10^(-3)`
`=6.25 xx 10^(-4) S m^(2) mol^(-1)`