(i) Define specific and equivalent conductance. How are they related?
(ii) 0.05 M NaOH solution offered a resistance of `31.6 Omega` in a conductivity cell at 298 K. If the cell constant of the cell is `0.376cm^(-1)`, calculate molar conductance of NaOH solution.

Explanation

Given that,
Molarity of NaOH solution = 0.05 M
Resistance = `31.6Omega`
Conductance = `(1)/("Resistance")=(1)/(31.6)=0.03164Omega^(-1)`
From the equation,
Specific conductance `(kappa)`
= Conductance `xx` Cell constant
`=0.03164Omega^(-1)xx0.376cm^(-1)`
`=0.011896Omega^(-1)cm^(-1)`
As we know that
`wedge_(m)=(1000xxkappa)/("Molarity")=(1000xx0.011896)/(0.05)=237.92Omega^(-1)mol^(-1)`