Question

When $0.1molCoC{l}_3(N{H}_3{)}_5$ is treated with an excess of $AgN{O}_3,0.2\text{mol of}AgCl$ is obtained. The conductivity of the solution will correspond to

• A. $3:1$ electrolyte
• B. $1:1$ electrolyte
• C. $1:3$ electrolyte
• D. $1:2$ electrolyte

Answer 1

The correct option is D $1:2$ electrolyte

Moles of $C{l}^{-}$ precipitated as per reaction.


Dissociation of electrolyte

$0.1\text{mol of}CoC{l}_3(N{H}_3{)}_5$ dissociated to from $0.2\text{mol chorilde ions as}AgCl$.
This can be represented as $\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2$:
$\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2\to [Co(N{H}_3{)}_{5}Cl{]}^{2+}+2C{l}^{-}$
Thus,$0.1\text{mole of}CoC{l}_3(N{H}_3{)}_5$ in electrolytic solution must contain $\left[CoC{l}_3\right(N{H}_3{)}_{5}Cl{]}^{2+}$ and two $C{l}^{-}$ as constitutent ions.
Thus, it is a $1:2$ electrolyte.

Hence, the correct answer is option $\left(b\right)$.