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

0.001

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0.002

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0.003

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0.01

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Solution

The correct option is A 0.001
The molecular formula of dichlorotetraaquachromimum (III) chloride can be represented as $[C r (H_{2} O)_{4} C l_{2}] C l,$ . Thus, it contains one ionizable $C l^{-}$ .
Hence, an excess of $A g N O_{3}$ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromimum (III) chloride. The number of moles of $A g C l$ precipitated would be 0.001.

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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromimum (III) chloride. The number of moles of AgCl precipitated would be:

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