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 $\mathrm{Co}{\left({\mathrm{NH}}_{3}\right)}_{5}{\mathrm{Cl}}_{3}$

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 and an electrolytic solution that must contain $\left[\mathrm{Co}{\left({\mathrm{NH}}_{3}\right)}_{5}\mathrm{Cl}{\right]}^{2+}$ and two Cl- as constituent ions. Thus, it is a 1:2 electrolyte.

The reaction is as follows:

Hence, option (b) is correct.