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

0.003

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0.01

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0.001

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0.002

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Solution

The correct option is C $0.001$
$[C r (H_{2} O)_{4} C l_{2}] C l + A g N O_{3} \to [C r (H_{2} O)_{4} C l_{2}] N O_{3} + A g C l p p t$
No. of millimoles of solution = $100 mL \times 0.01 M$
$= 1 millimole$
$= 10^{- 3} mole$

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

The correct option is C 0.001[Cr(H2O)4Cl2]Cl+AgNO3→[Cr(H2O)4Cl2]NO3+AgClppt No. of millimoles of solution =100 mL×0.01 M =1 millimole =10−3 mole

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

The correct option is C $0.001$
$[C r (H_{2} O)_{4} C l_{2}] C l + A g N O_{3} \to [C r (H_{2} O)_{4} C l_{2}] N O_{3} + A g C l p p t$
No. of millimoles of solution = $100 mL \times 0.01 M$
$= 1 millimole$
$= 10^{- 3} mole$