Resistance of 0.2 M solution of an electrolyte is `50 Omega`. The specific conductance of the solution of 0.5 M solution of same electrolyte is `1.4 S m^(-1)` and resistance of same solution of the same electrolyte is `280 Omega`. The molar conductivity of 0.5 M solutions of the electrolyte is `5 m^(2) mol^(-1)` is

Options

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

Answer

A

Explanation

`x=1.4S//m`
`R=50Omega`
`M=0.2 `
`K=(1)/(R)xx(l)/(A)`
`implies(l)/(A)=1.4xx50m^(-1)`.
Now, new solution has M=0.5, `R=280Omega`
`impliesK=(1)/(R)xx(l)/(A)=(1)/(280)xx1.4xx50=(1)/(4)`
`implies lamda_(M)=(K)/(1000xxM)=((1)/(4))/(1000xx0.5)=(1)/(2000)=5xx10^(-4)`