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. `1.24xx10^(-4)Sm^(2)mol^(-1)`
D. `12.4xx10^(-4)Sm^(2)mol^(-1)`

Answer

A

Explanation

At C = `0.1` M,
`(l)/(a)=KR=1.29xx100`
At C = `0.02` M
`K=(l)/(axxR)=(1.29xx100)/(520)`
`=0.248ohm^(-1)m^(-1)`
Also,
`Lambda=(K)/("M(in mol/L)")=(K)/(Mxx10^(3)("in mol"//m^(3)))`
`=(0.248)/(0.02xx10^(3))`
`=124xx10^(-4)"S m"^(2)mol^(-1)`