An excess of ${AgNO}_{3}$ is added to 100 mL of a 0.01 M solution of dichlorotetraaquachromium (III) chloride. The number of moles of $AgCl$ precipitated would be

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0.01

0.001

0.002

0.003

answer is A.

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Detailed Solution

$[Cr {(H_{2} O)}_{4} {Cl}_{2}] Cl + {AgNO}_{3} \to [Cr {(H_{2} O)}_{4} {Cl}_{2}] {NO}_{3} + \underset{ppt}{AgCl} No . of moles = 0 .01 \times \frac{100}{1000} = 10^{- 3} So, moles of AgCl = 0 .001$

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