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
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a
0.003
b
0.01
c
0.001
d
0.002
answer is C.
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Detailed Solution
CrH2O4Cl2Cl+AgNO3→CrH2O4Cl2NO3+AgClppt No. of mole=0.01×1001000=10−3 So, mole of AgCl=0.001