Question

A potential difference of 20 volts applied to the ends of a column of `N/10` AgNO₃ solution, 4 cm in diameter and 12 cm in length, gives a current of 1.198 amperes. Calculate the specific and equivalent conductivity of the solution.

Answer 1

Given: 

Voltage V = 20 V

Current I = 1.198 A

Diameter of the column = 4 cm → Radius r = 2 cm

Length l = 12 cm

Normality = `N/10` = 0.1 N

Resistance `R = V/I`

= `20/1.198`

= 16.69 ohm

Area A = πr²

= 3.14 × (2)²

= 3.14 × 4

= 12.56 cm²

Specific conductivity `kappa = l/(R xx A)`

= `12/(16.69 xx 12.56)`

= 0.057 ohm⁻¹ cm⁻¹

Equivalent conductivity `Lambda_"eq" = (kappa xx 1000)/N`

= `(0.057 xx 1000)/0.1`

= 570 ohm⁻¹ cm² eq⁻¹