Ncert Exercises & Solutions

When 0.1 mol CoCl₃(NH₃)₅ is treated with an excess of AgNO₃, 0.2 mol of AgCl are obtained. The conductivity of the solution will correspond to-

1. 1:3 electrolyte

2. 1:2 electrolyte

3. 1:1 electrolyte

4. 3:1 electrolyte

Hint: Two chlorine atoms are present outside the complex.

Step 1:

One mole of AgNO₃ precipitates one mole of chloride ion. ln the above reaction, when 0.1 moles $Co {({NH}_{3})}_{5} {Cl}_{3}$

is treated with an excess of AgNO₃ 0.2 mole of AgCl are obtained thus, there must be two free chloride ions in the

solution of electrolyte.

Step 2:

So, the molecular formula of the complex will be $[Co {({NH}_{3})}_{5} Cl] {Cl}_{2}$ and an electrolytic solution that must contain $[Co {({NH}_{3})}_{5} Cl]^{2 +}$ and two Cl^- as constituent ions. Thus, it is a 1:2 electrolyte.

The reaction is as follows:

$[Co {({NH}_{3})}_{5} Cl] {Cl}_{2}_{(s)} \to {[Co {({NH}_{3})}_{5} Cl]^{2 +}}_{(aq)} + 2 {Cl}^{-}_{(aq)}$

Hence, option (b) is correct.

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