The equivalent conductance of `0.10 N` solution of `MgCI_(2)` is 97.1 mho `cm^(2) eq^(-1)`. A cell electrodes that are `1.50 cm^(2)` in surface are and `0.50` cm apart is filled with `0.1N MgCI_(2)` solution. How much current will flow when the potential difference between the electrodes is 5 volts?

Answer

`0.1456amp`

Explanation

`^^_(eq)=97.1Scm^(2)eq^(-1),C=0.1N` ,brgt `A=1.5cm^(2),I=0.5cm`
`^^_(eq)=(1000xx((1)/(R)xx(l)/(A)))/(C)implies97.1=(1000)/(0.1)xx(1)/(R)xx(0.5)/(1.5)`
`R=34.33Omegaimpliesi=(V)/(R)=(5)/(34.33)=0.1456amp`