Question

If the resistance of a conductivity cell containing 0.01 mol/L KCl solution at 298 K is 1000 Ω, then the conductivity of the cell is [cell constant = 0.219 cm^-1]

Answer 1

Correct option is (c)

As we know  =1/resistance x cell constant

= 0.219 cm^-1 / 1000Ω

= 2.19 x 10^-4 5cm^-1

Conductivity of 0.01 mol/ l kcl solution will be 

= 2.19 x 10^-4 / 0.01 x 1000

= 21.9 S cm^-1