When 0.1 mol `CoCl_(3)(NH_(3))_(5)` is treated with excess of `AgNO_(3)`, 0.2 mole of AgCl are obtained. The conductivity of solution will correspond to

Options

A. `1 : 3` electrolyte
B. `1 : 2` electrolyte
C. `1 : 1` electrolyte
D. `3 : 1` electrolyte

Explanation

To solve the problem, we need to analyze the coordination compound and the reaction that occurs when it is treated with excess AgNO3.

### Step-by-Step Solution:

1. **Identify the Coordination Compound**: The given compound is CoCl3(NH3)5. This indicates that cobalt (Co) is coordinated with 5 ammonia (NH3) ligands and 3 chloride (Cl) ligands.

2. **Reaction with AgNO3**: When the coordination compound is treated with excess AgNO3, the AgNO3 reacts with the chloride ions (Cl-) to form AgCl precipitate. The reaction can be represented as:
\[ Ag^+ + Cl^- \rightarrow AgCl(s) \]

3. **Determine the Moles of AgCl Formed**: According to the problem, 0.2 moles of AgCl are formed when 0.1 moles of CoCl3(NH3)5 are treated with AgNO3.

4. **Relate AgCl to Cl- Ions**: Since 1 mole of Ag+ reacts with 1 mole of Cl-, the formation of 0.2 moles of AgCl means that there must be 0.2 moles of Cl- ions available for the reaction.

5. **Calculate the Number of Cl- Ions**: The coordination compound CoCl3(NH3)5 contains 3 chloride ions. However, since we have 0.2 moles of Cl- ions reacting, we can conclude that there are additional Cl- ions that must be coming from outside the coordination sphere.

6. **Dissociation of the Complex**: When CoCl3(NH3)5 dissociates, it can be represented as:
\[ Co(NH3)5Cl^{2+} + 2Cl^- \rightarrow Co(NH3)5^{2+} + 3Cl^- \]
Here, 2 of the Cl- ions are free and available for reaction with AgNO3.

7. **Electrolyte Ratio**: The dissociation of the complex results in 1 mole of the complex ion (Co(NH3)5Cl^{2+}) and 2 moles of Cl- ions. Therefore, the electrolyte ratio is:
\[ 1 \text{ (complex ion)} : 2 \text{ (Cl- ions)} \]

8. **Conclusion**: The conductivity of the solution corresponds to the total number of ions produced when the complex dissociates. Thus, for every 1 mole of CoCl3(NH3)5, we get 1 mole of Co(NH3)5^{2+} and 2 moles of Cl-, leading to a total of 3 moles of ions.

### Final Answer:
The conductivity of the solution will correspond to a total of 3 moles of ions from the dissociation of 0.1 moles of CoCl3(NH3)5.