Multiple choice

The amount of 0.02 M AgNO3 volume required for complete precipitation of chloride ions present in 30 mL of 0.02 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to:

  1. 4 mL

  2. 5 mL

  3. 6 mL

  4. 7 mL

  5. 8 mL
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

[Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl](NO32
Number of ionisable chloride ions in complex [Cr(H2O)5Cl]Cl2 = 2 Millimoles = Molarity x Volume (mL) x 2 = 0.02 x 30 x 2 = 1.2 Therefore, required Ag+ ions = 1.2 millimoles According to Millimoles = Molarity x V (mL) 1.2 = 0.2 x VV = 6 mL

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