Resistance of 0.2 M solution of an electrolyte is `50Omega` . The specific conductance of the solution is `1.4 Sm^(-1)`. The resistance of 0.5 M solution of the same electrolyte is :

Options

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

Answer

A

Explanation

Given for 0.2 M solution
`R = 50 Omega`
`k = 1.4 S m^(-1) = 1.4 xx 10^(-2) S cm^(-1)`
Now , `R = rho "" (t)/(a) = (1)/(k) xx (l)/(a)`
`implies (l)/(a) = R xx k = 50 xx 1.4 xx 10^(-2)`
For 0.5 M solution R = `200 Omega , k =` ?
`(l)/(a) = 50 xx 1.4 xx 10^(-2)`
`implies R = rho""(l)/(a) = (1)/(k) xx (l)/(a)`
`implies k = (1)/(280) xx 50 xx 1.4 xx 10^(-2)`
`= (1)/(280) xx 70 xx 10^(-2) = 2.5 xx 10^(-3) S cm^(-1)`
Now `Lambda_(m) = (k xx 1000)/(M) = (2.5 xx 10^(-3) xx 1000)/(0.5)`
`= 5.5 cm^(3) mol^(-1) = 5 xx 10^(-4) S m^(2) mol^(-1)`