A chromium complex with a formula $$CrCl_{3}.6H_{2}O$$ has a spin only magnetic moment value of 3.87 BM and its solution conductivity corresponds to 1:2 electrolyte. 2.75 g of the complex solution was initially passed through a cation exchanger. The solution obtained after the process was reacted with excess of $$AgNO_{3}$$. The amount of AgCI formed in the above process is __ g. (Nearest
integer)
[Given: Molar massing $$mol^{-1}$$ Cr:52; Cl:35.5, Ag:108, 0:16, H:1]

We need to find the number of moles of AgCl precipitated when the solution from a cation exchanger is treated with AgNO₃.

Determine the structure of the complex.

Formula: CrCl₃·6H₂O

Spin-only magnetic moment = 3.87 BM → $$\mu = \sqrt{n(n+2)}$$ → $$3.87 = \sqrt{n(n+2)}$$ → $$n = 3$$ (unpaired electrons)

Cr³⁺ (3d³) has 3 unpaired electrons. ✓

1:2 electrolyte → Complex dissociates into 3 ions: one cation giving 2+ charge and 2 anions.

The formula that fits: $$[Cr(H_2O)_5Cl]Cl_2 \cdot H_2O$$

This gives: $$[Cr(H_2O)_5Cl]^{2+} + 2Cl^-$$ (1:2 electrolyte) ✓

Find moles of complex.

Molar mass of CrCl₃·6H₂O = 52 + 3(35.5) + 6(18) = 52 + 106.5 + 108 = 266.5 g/mol

Moles = 2.75/266.5 ≈ 0.01032 mol

After cation exchange.

When passed through a cation exchanger, the cation $$[Cr(H_2O)_5Cl]^{2+}$$ is retained. The solution contains 2 Cl⁻ ions per formula unit.

Moles of Cl⁻ = 2 × 0.01032 = 0.02064 mol

When treated with excess AgNO₃:

AgCl precipitated = 0.02064 mol

Mass of AgCl = 0.02064 × 143.5 = 2.96 g ≈ 3 g

The answer is 3.