The conductivity of pure water in a conductivity cell with electrodes of cross-sectional area `4cm^(2)` placed at a distance `2cm` apart is `8 xx 10^(-7) S cm^(-1)`. Calculate.
(a) The resistance of water.
(b) The current that would flow through the cell under the applied potential difference of 1 volt.

Explanation

Cell constant `= (l)/(a) = (2)/(4) = (1)/(2) cm^(-1)`
(a) Also, `k = (1)/(R) xx (l)/(a)`
`R = (1)/(k) xx (l)/(a) = (1)/(8 xx 10^(-7)) xx (1)/(2) = 6.25 xx 10^(5) ohm`
(b) From Ohm's law, `(V)/(i) = R`
`:. i = (1)/(6.25 xx 10^(5)) = 1.6 xx 10^(-6)` ampere