Resistance of ` 0.2 M` solution of an electrolue is ` 50 Omega`. The specific conductance of the solution is ` 1.4 S m^^(-1)`. The resistance of `0.5` M solution of the same electrolyte is `280. Omega`. The molar conducitivity of ` 0.5 M` solution of the electrolyte is ` S m^2 "mol"^(-1)` is.

Options

A. ` 5 xx10^(-4)`
B. `5 xx10^(-3)`
C. `5 xx10^3`
D. `5 xx10^2`

Answer

A

Explanation

For `0.2 M` solution , ` R= 50 Omega`
` k= 14 S m^(-1) = 1.4 xx 10^(-2) S cm^(-1)`
The resistivity of the solution is related to specific cnductance by
` rho = 1/k = 1/(1.4 xx 10^(-2)) Omega cm`
It is known that , `R= rho 1/a`
It is kinown tht , `R= rho = 1/a`
`rArr l/a = R xx 1/(rho) = 50 xx 1.4 xx 10^(-2) cm`
for `0.5 M` solution,
`R= 280 Omega`
` k=? `
`l/a = 50 xx 1.4 xx 10^(-2) xm`
The specific conductance of the solution is given by
` k= 1/(rho) = 1/R xx l/a`
` k= 1/(280) xx 50 xx 1.4 xx 10^(-2)`
`2.5 xx 10^(-3) S cm^(-1)`
Now , the molar conductivity of the solution is given by
`Lambda_m = (K xx 1000)/M = (2.5 xx 10^(-3) xx 1000)/(0.5)`
` 5.5 cm^2 "mol"^(-1)`
` = 5 xx 10^(-4) S m^2 "mol"^(-1)`
Hence, the molar conductivity of `0.5 M` solution of the electrolyte is `5xx 10^(-4) Sm^2 "mol"^(-1)`.