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

`0.1456 amp`

Explanation

Cell constant `= (0.50)/(1.50) =(1)/(3)`
Specific conductance
`= ("equivalent conductance")/("volume (cc) containing 1 eq.")`
`=(97.1)/(10,000)` (for N solution `V = 10,000 "cc"`)
`= 0.0097 "mho cm"^(-1)`
Conductance = specific conductance/cell constant
`(0.00971)/(1//3)=0.02913 "mho"`
`:.` resistance `= (1)/(0.02913)"ohm"`
`:.` current in amp `= ("potential difference (volt)")/("resistance(ohm)")`
(Ohm's law) `= (5)/(1//0.02913)=0.1456` ampere.