Question

When 0.1 mol $CoC{l}_3(N{H}_3{)}_5$ is treated with excess of $AgN{O}_3,$ 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to
(a) 1:3 electrolyte
(b) 1:2 electrolyte
(c) 1:1 electrolyte
(d) 3:1 electrolyte

Answer 1

One mol of $AgN{O}_3$ precipitates one mol of chloride ion. In the above reaction, when 0.1 mol $CoC{l}_3(N{H}_3{)}_5$ is treated with an excess of $AgN{O}_3$, 0.2 moles of AgCl is obtained thus, there must be two free chloride ions in the solution of electrolyte.
So, molecular formula of complex will be $\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2$ and electrolyte solution must contain $\left[Co\right(N{H}_3{)}_{5}Cl{]}^{2+}$ and two $C{l}^{-}$ as constituent ions. Thus, it is 1:2 electrolyte.
$\left[Co\right(N{H}_3{)}_{5}Cl]C{l}_2\to [Co\left(N{H}_3{)}_{5}Cl{]}^{2+}\right(aq)+2C{l}^{-}(aq)$
Hence, option (b) is the correct.