A $0.0100 M$ KCl solution has conductivity $1.41 \times 1 0^{- 3} S c m^{- 1}$ at 298 K. When this solution is placed in a conductivity cell, the measured conductance is $6.13 \times 1 0^{- 4} S$ . The cell constant of the cell in $c m^{- 1}$ is (two decimals)

Numerical answer type, answer in cm^-1, tolerance 0.02.

Answer: 2.3 cm^-1

Cell constant

G^{*} = κ / G = κ R

G^{*} = \frac{1.41 \times 1 0 ^{- 3}}{6.13 \times 1 0 ^{- 4}} = 2.30 c m^{- 1}