A conductivity cell when filled with 0.1 (N) KCl at `25^(@)C` the resistance at `25^(@)C` becomes 307.62 ohm, and when filled in with `0.1 (N)AgNO_(3)` solution, the resistance becomes 362.65 ohm at the same temperature. At `25^(@)C`, the specific conductance of 0.1 (N) KCl solution is `"0.01286 ohm"^(-1)."cm"^(-1)`. From these data, calculate (i) cell constant (ii) equivalent conductivity of `0.1 (N)AgNO_(3)` solution.

Explanation

`"For KCl soluion, "kappa =(l)/(A)xx(1)/(R)" "therefore" "0.001406=(l)/(A)xx(1)/(25)`
`therefore" "(l)/(A)="0.03515 ohm"^(-1)."cm"^(-1)`
`"For NaOH solution, "kappa =(l)/(A)xx(1)/(R)`
`therefore" "kappa =0.03515xx(1)/(14.8)=2.375xx10^(-3)" ohm"^(-1)."cm"^(-1)`
`Lambda =kappa xx(1000)/(N)=2,375xx10^(-3)xx(1000)/(0.01)`
`="237.5 ohm"^(-1)."cm"^(2)."g- equive"^(-1)`