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

Explanation

For 0.2 M sollutions, `R=50 Omega`
`K=14 S m^(-1)=1.4xx10^(-2) S cm^(-1)`
The resistivity of the solution is related to specific conductance by
`rho1/K=1/(1.4xx10^(-2))Omegam`
It is known that `R=rho1/a`
`implies R/(rho)=1/a implies 1/a=Rxx1/(rho)=50xx1.4xx10^(-2)cm`
For 0.5 M solution,
`R=280Omega`
`K=?`
`1/a=50xx1.4xx10^(-2)cm`
the specific conductance of the solution is given by
`1/(rho)=1/Rxx1?a`
`implies R/(rho)=1/a implies 1/a=Rxx1/(rho)=50xx1.4xx10^(-2)cm`
Now, mollar conductivity of the solution is given by
`Lambda_m=(kxx1000)/M=(2.5xx10^(-3)xx1000)/(0.5)=5.5 cm^(2) mol ^(-1) =5xx10^(-4) S ^(2) mol ^(-1)`
Hence, the molar conductivity of 0.5 M solution of the electrolyte is `5xx10^(-4) S m^(2) m o l^(-1)`