The resistance of cell containing 0.1N KCl solution and 0.1N `AgNO_(3)` solution was 337.62 and 362.65 ohms respectively at 298 K. The conductivity of 0.1 N KCl is 0.01286 `"ohm"^(-)cm^(-)` at 298 K. Find the cell constant and equivalent conductance of 0.1 N `AgNO_(3)` solution.

Explanation

We know that
(a) Cell constant `=("Conductivity of 0.1 N KCl")/("Observed conductance") `
= Conductivity of 0.1 N KCl `xx` Resistance
`=0.01286 xx 337.62 =4.342cm^(-1)`
(b) `wedge` for 0.01 N `AgNO_(3)=` Conductivity of 0.1 N `AgNO_(3) xx ` Volume of 0.1 N `AgNO_(3)` containing 1 gram equivalent of electrolyte (in mL)
= [Observed conductance `xx` cell constant] `xx (10 xx 1000)`
`=(1)/(362.65) xx 4.342 xx 10xx1000`
`=119.73"mho cm"^(2)" (gm.equiv)"^(-)`