When 0.01 mole of a cobalt complex is treated with excess silver … - Vidyadip

MCQ

When $0.01$ mole of a cobalt complex is treated with excess silver nitrate solution$, 4.305 g$ silver chloride is precipitated. The formula of the complex is:

✓
$ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3} $
B
$ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6 \mathrm{Cl}_3\right]} $
C
$ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2} $
D
$ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{NO}_3}$

✓

Answer

Correct option: A.

$ {\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_6\right] \mathrm{Cl}_3} $

$4.305g\ \text{AgCl} = \frac{4.305}{143.5}\ mol = 0.03\ mol.$
As $0.01$ mole of the complex gives $0.03$ mole of $\text{AgCI},$ this shows that there are three ionisable $Cl.$

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