A conductance cell was filled with a 0.02 M KCl solution which has a specific conductance of `2.768xx10^(-3)ohm^(-1)cm^(-1)` . If its resistance is 82.4 ohm at `25^(@)` C the cell constant is:

Answer

`0.229 cm^(-1)`

Explanation

Cell constant `G^(**) = (kappa)/(C) = (2.78 xx 10^(-3))/(1//82.5)`
`= 0.229 cm^(-1)`