When 100 mL of 0.2 M solution of \(\text{CoCl}_3 \cdot x\text{NH}_3\) is treated with excess of \(\text{AgNO}_3\) solution, \(3.6 \times 10^{22}\) ions are precipitated. The value of \(x\) is \((N = 6 \times 10^{23} \text{mol}^{-1})\)
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The number of moles of AgCl precipitated per mole of complex equals the number of chloride ions present in the ionization sphere (outside the square bracket). For Co(III) ammine complexes, coordination number is 6.
Step 1: Calculate moles of the complex:
Molarity (M) = 0.2 mol/L.
Volume (V) = 100 mL = 0.1 L.
Total moles of complex = \(M \times V = 0.2 \times 0.1 = 0.02\) moles.
Step 2: Calculate moles of Chloride ions precipitated:
Number of AgCl molecules/ions precipitated = \(3.6 \times 10^{22}\).
Avogadro's Number (\(N_A\)) = \(6 \times 10^{23}\).
Moles of precipitate = \(\frac{3.6 \times 10^{22}}{6 \times 10^{23}} = 0.6 \times 10^{-1} = 0.06\) moles.
Step 3: Determine number of ionizable Chlorines (n):
Ratio \(\frac{\text{Moles of AgCl}}{\text{Moles of Complex}} = n\).
\(n = \frac{0.06}{0.02} = 3\).
This means 3 \(\text{Cl}^-\) ions are outside the coordination sphere.
Step 4: Determine formula and x:
The coordination number of Cobalt(III) is typically 6.
The formula is written as \([\text{Co(NH}_3)_x]\text{Cl}_3\).
For the coordination number to be 6, \(x\) must be 6.
Formula: \([\text{Co(NH}_3)_6]\text{Cl}_3\).
Thus, \(x = 6\).
Final Answer: \(x = 6\).
