Resistance of `0.2 M` solution of an electrolyte is `50 ohm`. 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 conductivity of `0.5 M` solution of the electrolyte in `S m^(2) mol^(-1)` id

Options

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

Answer

A

Explanation

For `0.2` M solution, R = `50Omega`
`k=1.4Sm^(-1)=1.4xx10^(-2)"S cm"^(-1)`
`rArrrho=(1)/(K)=(1)/(1.4xx10^(-2))Omegacm`
Now, `R=rho(l)/(a)`
`rArr(l)/(a)=(R)/(rho)=50xx1.4xx10^(-2)`
For `0.5` M solution, R = `280Omega`
K = ?
`(l)/(a)=50xx1.4xx10^(-2)`
`rArrR=rho(l)/(a)rArr(1)/(rho)=(1)/(R)xx(l)/(a)`
`rArrK=(1)/(280)xx50xx1.4xx10^(-2)`
`=(1)/(280)xx70xx10^(-2`
`=2.5xx10^(-3)"S cm"^(-1)`
Now, `Lambda_(m)=(sigmaxx1000)/(M)=(2.5xx10^(-3)xx1000)/(0.54)`
`"5 S cm"^(2)mol^(-1)=5xx10^(-4)"S m"^(2)mol^(-1)`