Core Logic
Since the complex behaves as a 1:2 electrolyte, it must dissociate into 3 ions: 1 complex cation and 2 counter anions.
This means two chloride ions (Cl^-$Cl^-$) must reside outside the coordination sphere.
The structural formula is: [Cr(H₂O)₅Cl]Cl₂ ₂O$[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O}$.
Reaction with silver nitrate:
[Cr(H₂O)₅Cl]Cl₂ ₂O + 2AgNO₃ arrow 2AgCl + complex nitrate$[\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2}\cdot\mathrm{H}_{2}\mathrm{O} + 2\mathrm{AgNO}_{3} \rightarrow 2\mathrm{AgCl} \downarrow + \text{complex nitrate}$
Step 1: Calculate Moles of Complex
Molar mass of CrCl₃· 6H₂O = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5 g/mol$CrCl_3\cdot 6H_2O = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5 \text{ g/mol}$.
Moles of the complex = (2.75)/(266.5) 0.01031 moles$\frac{2.75}{266.5} \simeq 0.01031 \text{ moles}$.
Step 2: Calculate Mass of Precipitate
Since each mole of complex gives 2 moles of AgCl$AgCl$:
Moles of AgCl$AgCl$ formed = 2 × 0.01031 = 0.02063 moles$2 \times 0.01031 = 0.02063 \text{ moles}$.
Molar mass of AgCl = 108 + 35.5 = 143.5 g/mol$AgCl = 108 + 35.5 = 143.5 \text{ g/mol}$.
Mass of AgCl = 0.02063 × 143.5 2.96 gm$AgCl = 0.02063 \times 143.5 \simeq 2.96 \text{ gm}$.
Rounding to the nearest integer yields 3.
Pattern Recognition
A 1:2 electrolyte contains 3 total ions. For CrCl₃ · 6H₂O$CrCl_3 \cdot 6H_2O$, this strictly dictates 2 Cl^-$2 Cl^-$ are outside the coordination sphere. Hence n$n$ factor for chloride precipitation is exactly 2.
Chapter Mix
Class 12 Chemistry: Coordination Compounds
Class 11 Chemistry: Some Basic Concepts of Chemistry