The resistance of `0.01N` solution of an electrolysis is `210 Omega` at `298 K` with a cell constant of `0.88 cm^(-1)`. Calculate the conductivity and equivalent conducitivty of the solution.

Explanation

`R=210 ohm, (l)/(a)=0.88cm^(-1)`
`k=(1)/(R)xx(1)/(a)=(1)/(210)xx0.88=4.19ohm^(-1)cm^(-1)` or `S cm^(-1)`
`wedge_(eq)=(kxx1000)/(N)=(4.19xx10^(-3)xx1000)/(0.01)`
`=419.05 S ch^(2)eq^(-1)`