The specific conductance of 0.01 (N) KCl solution at `25^(@)C,` is `"0.00140 ohm"^(-1)."cm"^(-1)`. The same conductivity cell when filled with `((N)/(100))NaOH` solution, the resistance of the solution was 14.8 ohm. If the resistance of the KCl solution is 25 ohm, what will be the equivalent conductivity of NaOH at `25^(@)C`?

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)`