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

Answer

A

Explanation

Cell constant`=("Conductivity (I)")/("Conductance (I)")=("Conductivity (II)")/("Conductance (II)")`
Conductivity (II)=`(129)/(520)Sm^(-1)`
0.2M=200 mol `m^(-3)`.
Molar conductivity of 0.2 M solution`=(129//520)/(200)`
`=12.4xx10^(-4)S" "m^(2)mol^(-1)`.
Note: From given it is not possible to calculate molar conductivity of 0.02M solution exactly. however, it is sure that its molar conductivity will be slightly greater than `12.4xx10^(-4)Sm^(2)mol^(-1)`. Assuming linear inverse variation of molar conductivity with concentration we can say that
`therefore`Molar conductivyt of 0.02M solution`=1.24xx10^(-3)xx10=124xx10^(-4)S" "m^(2)mol^(-1)`.