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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachrominum (III) chloride.
The number of moles of AgCl precipitated would be-
  • a)
    0.001
  • b)
    0.002
  • c)
    0.003
  • d)
    0.01
Correct answer is option 'A'. Can you explain this answer?
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Calculation of moles of dichlorotetraaquachrominum (III) chloride

Moles = Molarity * Volume (in L)
Moles = 0.01 M * 0.1 L
Moles = 0.001 mol

Excess AgNO3 is added to the solution, which reacts with the chloride ions to form insoluble AgCl precipitate.

AgNO3 + Cl- → AgCl + NO3-

As AgNO3 is in excess, all the chloride ions will react to form AgCl precipitate.

Calculation of moles of AgCl precipitated

Moles of AgCl = Moles of Cl-

Moles of Cl- = Moles of dichlorotetraaquachrominum (III) chloride = 0.001 mol

Therefore, Moles of AgCl precipitated = 0.001 mol

As the volume of the solution is not changing during the reaction, the concentration of the resulting solution will be the same as the initial solution. Therefore, the concentration of AgCl precipitate will be:

Concentration of AgCl = Moles of AgCl / Volume of solution
Concentration of AgCl = 0.001 mol / 0.1 L
Concentration of AgCl = 0.01 M

Therefore, the correct answer is option 'A' (0.001 mol) as it represents the moles of AgCl precipitated.
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An excess of AgNO3 is added to 100 mL of a 0.01 M solution of dichlorotetraaquachrominum (III) chloride.The number of moles of AgCl precipitated would be-a)0.001b)0.002c)0.003d)0.01Correct answer is option 'A'. Can you explain this answer?
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