Resistance of a conductvity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 `Omega`. The conductivity of this solution is 1.29 `Sm^(-1)`. Resistance of the same cell when filled with 0.02M of the same solution is `520 Omega`. the molar conductivity of 0.02M solution of the electrolyte will be:

Options

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

Answer

D

Explanation

`C=0.1M,R=100Omega`
`K=1.29Sm^(-1)=(1)/(100)xx(l)/(A)`
`0.02M,R=520Omega`
`K=(1)/(520)xx129`
`Å_(M)=((1)/(520)xx129)/(1000xx0.02)=124xx10^(-4)Sm^(2)mol^(-1)`
or `K=Gsigma=(r)/(R)=1,sigma=KR`
`sigma=1.29xx100=129`
for other concentration
`r=129`
`K'=(sigma)/(R)=(129)/(520),^^_(M)=(1000K)/(M)=(1000xx129)/(0.02xx500)=5Cm^(2)mol^(-1)`