An excess of $A g N O_{3}$ is added to $100 m L$ of a $0.01 M$ solution of dichlorotetraaquachromium(III) chloride. The number of moles of $A g C l$ precipitated would be :-

0.01

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0.001

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0.02

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0.003

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Solution

The correct option is B 0.001
$[C r (H_{2} O)_{4} C l_{2}] C l \to [C r (H_{2} O)_{4} C l_{2}] + C l^{-}$
Initial: $100 \times 0.01$ $0$ $0$
moles $= 1 millimol$
As per the equation 1 mol of $[C r (H_{2} O)_{4} C l_{2}] C l$ furnishes 1 mol of $C l^{-}$ ions. Therefore, 1 millimol or $0.001 mol$ of $[C r (H_{2} O)_{4} C l_{2}] C l$ will furnish $0.001 mol$ of $C l^{-}$ ions.
Now, the solution of $[C r (H_{2} O)_{4} C l_{2}] C l$ is made to react with $A g N O_{3}$ and the reaction that takes place is:
$[C r (H_{2} O)_{4} C l_{2}] C l + A g N O_{3} \to [C r (H_{2} O)_{4} C l_{2}] N O_{3} + A g C l (↓)$
Initially: $0.001 mol$ excess
$∴$ $[C r (H_{2} O)_{4} C l_{2}] C l$ will act as the limiting reagent and will decide the moles of $A g C l$ precipitated.
Now, 1 mol of $[C r (H_{2} O)_{4} C l_{2}] C l$ precipitates 1 mol of $A g C l$ .
$∴$ $0.001 mole$ of $[C r (H_{2} O)_{4} C l_{2}] C l$ will precipitate $0.001 mole$ of $A g C l$ .

Hence, the correct answer is option (b).

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