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