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 reaction between the complex \( CoCl_3(NH_3)_5 \) and excess \( AgNO_3 \), and how it relates to the conductivity of the solution.

### Step-by-Step Solution:

1. **Identify the Complex**: We start with the complex \( CoCl_3(NH_3)_5 \). This indicates that there are 3 chloride ions (Cl\(^-\)) coordinated to the cobalt ion and 5 ammonia (NH\(_3\)) ligands.

2. **Reaction with AgNO3**: When \( CoCl_3(NH_3)_5 \) is treated with excess \( AgNO_3 \), the chloride ions will react with silver nitrate to form silver chloride (AgCl). The reaction can be represented as:
\[ CoCl_3(NH_3)_5 + 3AgNO_3 \rightarrow Co(NH_3)_5 + 3AgCl + 3NO_3^- \]

3. **Moles of AgCl Formed**: According to the problem, 0.2 moles of AgCl are formed. This indicates that all 3 chloride ions from the complex have reacted, and the excess AgNO\(_3\) ensures that all Cl\(^-\) ions are precipitated as AgCl.

4. **Determine the Chloride Ions**: Since 0.2 moles of AgCl are produced, and each mole of AgCl corresponds to one mole of Cl\(^-\), it implies that 0.2 moles of Cl\(^-\) were present in the solution.

5. **Calculate the Total Ions in Solution**: The complex \( CoCl_3(NH_3)_5 \) dissociates in solution to give:
- 1 \( Co^{3+} \) ion
- 3 \( Cl^- \) ions (which are precipitated as AgCl)
- 5 \( NH_3 \) molecules (which do not contribute to conductivity as they are neutral)

Hence, the total ions present in the solution before the reaction are:
\[ 1 (Co^{3+}) + 3 (Cl^-) = 4 \text{ ions} \]

6. **Conductivity Contribution**: After the reaction, the \( Co^{3+} \) remains in the solution, and the \( Cl^- \) ions are precipitated out as AgCl. Therefore, the ions contributing to conductivity in the solution are:
- 1 \( Co^{3+} \) ion
- 0 \( Cl^- \) ions (since they are precipitated)
- 5 \( NH_3 \) molecules (neutral, do not contribute)

Thus, the total ions contributing to conductivity after the reaction is 1.

7. **Final Conductivity Ratio**: The final ratio of ions contributing to conductivity is:
\[ 1 \text{ (from } Co^{3+}) : 0 \text{ (from } Cl^-) = 1:0 \]

### Conclusion:
The conductivity of the solution corresponds to the presence of 1 ion from \( Co^{3+} \) and no contributing ions from \( Cl^- \) after the reaction.