When 1 mol $CrCl_3.6H_2O$ is treated with excess of $AgNO_3$, 3 mol of $AgCl$ are obtained. The conductivity of the solution will correspond to

1:3 electrolyte

The correct answer is Option (4) → 1:3 electrolyte

$\text{AgNO}_3$ precipitates only free (ionizable) $\text{Cl}^-$ ions present outside the coordination sphere. If $3 \text{ mol AgCl}$ form, all three $\text{Cl}^-$ ions are outside the complex bracket.

So the complex is:

$[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3$

Dissociation in water

$[\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3 \rightarrow [\text{Cr}(\text{H}_2\text{O})_6]^{3+} + 3\text{Cl}^-$

Total ions produced = 1 complex cation + 3 anions = 4 ions

Electrolyte Type

This corresponds to one cation and three anions $\rightarrow$ 1 : 3 electrolyte.

Why others are wrong

• 1:1 electrolyte $\rightarrow$ would give 2 ions only (like $\text{NaCl}$).
• 1:2 electrolyte $\rightarrow$ gives 3 ions (like $\text{CaCl}_2$).
• 2:1 electrolyte $\rightarrow$ gives 3 ions but two cations + one anion (not the case).