Question

The equivalent conductance of $B{a}^{2+}$ and $C{l}^{-}$ are respectively $127$ and $76$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$ at infinite dilution. What will be the equivalent conductance of $BaC{l}_2$ at infinite dilution?

• A. $139.5$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$
• B. $203$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$
• C. $279$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$
• D. $101.5$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$

Answer 1

The correct option is A $139.5$ $oh{m}^{-1}$ $c{m}^2$ $e{q}^{-1}$

Kohlrausch law:

"At infinite dilution, when dissociation is complete, each ion makes a definite contribution towards equivalent conductance of the electrolyte irrespective of the nature of the ion with which it Is associated and the value of equivalent conductance at infinite dilution for any electrolyte is the sum of the contribution of its constituent ions", i.e., anions and cations.

The chemical equation is,

$BaC{l}_2\to B{a}^{2+}+2C{l}^{-}$

${\Lambda }_{BaC{l}_2}^0={\Lambda }_{B{a}^{2+}}^0+2C{l}^{-}=127+2\times 76=279$ $oh{m}^{-1}$$c{m}^2$$mo{l}^{-1}$

Equivalent conductivity $=\frac{279}{2}=139.5$ $oh{m}^{-1}$$c{m}^2$$e{q}^{-1}$