The conductivity of a 0.12 N solution of an electrolyte of the type `A^+B^-` is 0.024 S `cm^(-1)` . Calculate its (i) equivalent, (ii) molar conductivities.

Explanation

Soin.: Normality of a solution is given by the number of equivalent per `dm^3` (or per litre) of the solution. Therefore, concentration of the given solution = 0.12 equiv/ `dm^3` conductivity of the solution, (K) = 0.024 S `cm^(-1)` Then,
(i) The equivalent conductivity is given by,
`A_(eq)=(1000k)/C_(eq)=(1000xx0.024)/0.12` S `cm^2 "equiv"^(-1)`
= 200 S `cm^2` `"equiv"^(-1)`
(ii) The given electrolyte is of the type `A^+B^-` This means thateach cation (or anion) carries a charge equivalent to the charge carried by one electron. So, Z=1.
Then, `A_m=1xxA_(eq)=1xx200` S `cm^2mol^(-1)`
=200 S `cm^2mol^-1`