Resistance of a conductivity cell filed with a solution of an cleetrolyto of concentration `0.1M` is `100 Omega.` The conductivity of this solution is `1.29 S m ^(-1).` Resistance of the same cell when filled with `0.2 M` of the same solution is `520 Omega.` The molar coonductivity of `0.2 M` solution os electrolyte will be

Options

A. `1.24xx10^(-4) Sm^(2) mol ^(-1)`
B. `12.4 xx10^(-4) Sm^(2) mol ^(-1)`
C. `124xx10 ^(-4) S m ^(2) mol ^(-1)`
D. `1240 xx10^(-4) S m ^(2) mol ^(-1)`

Answer

B

Explanation

`R= 100Omega , k =1/R ((l)/(A)),`
`l/A`(cell constnat `= 1.29xx10m^(1)`
Given, `R =520Omega , C=0.2M,`
`mu` (molar conductivity)= ?
`mu =k xxV ` (k can be calculated as `k=1/R ((l)/(A))`
now cell constnat is known. )
Here,
`mu =(1)/(520) xx129 xx(1000)/(0.2)xx10^(-6)m ^(3)`
`=12.4xx10^(-4) Sm^(2) mol ^(-1)`