Question

When 0.1 mol \[\ce{CoCl3 (NH3)5}\] is treated with excess of \[\ce{AgNO3}\], 0.2 mol of \[\ce{AgCl}\] are obtained. The conductivity of solution will correspond to ______.

Options

• 1:3 electrolyte

• 1:2 electrolyte

• 1:1 electrolyte

• 3:1 electrolyte

Answer 1

When 0.1 mol \[\ce{CoCl3 (NH3)5}\] is treated with excess of \[\ce{AgNO3}\], 0.2 mol of \[\ce{AgCl}\] are obtained. The conductivity of solution will correspond to 1:2 electrolyte.

Explanation:

One mole of \[\ce{AgNO3}\] precipitates one mole of chloride ion. In the above reaction, when 0.1 mole \[\ce{CoCl3 (NH3)5}\] is treated with excess of \[\ce{AgNO3}\], 0.2 mol of \[\ce{AgCl}\] are obtained thus, there must be two free chloride ions in the solution of electrolyte.

So, molecular formula of complex will be \[\ce{[CO(NH3)5Cl]Cl2}\] and electrolytic solution must contain \[\ce{[CO(NH3)5Cl]^{2+}}\] and two CT as constituent ions. Thus, it is 1:2 electrolyte.

\[\ce{[Co(NH3)5Cl]Cl2 -> [Co(NH3)5Cl]^{2+} (aq) + 2Cl^{-} (aq)}\]