Coordination CompoundHard
Question
Co-ordination number of Cr in CrCl3.5H2O is six. The maximum volume of 0.1 N AgNO3 needed to precipitate the chlorine in outer sphere in 200 ml of 0.01 M solution of the complex is/are:
Options
A.140 ml
B.40 ml
C.80 ml
D.20 ml
Solution
(B) [Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl] (NO3)2
number of mole of complex = 200 × 0.01 = 2
required milli mole of AgNO3 = 4
milli mole = M × Vml ⇒ 4 = 0.1 × Vml = 40 ml
(D) [Cr(H2O)5Cl2]Cl + AgNO3 → AgCl + [Cr(H2O)5Cl2] (NO3)
number of mole of complex = 200 × 0.01 = 2
required milli mole of AgNO3 = 2
milli mole = M × Vml ⇒ 2 = 0.1 × Vml = 20 ml.
number of mole of complex = 200 × 0.01 = 2
required milli mole of AgNO3 = 4
milli mole = M × Vml ⇒ 4 = 0.1 × Vml = 40 ml
(D) [Cr(H2O)5Cl2]Cl + AgNO3 → AgCl + [Cr(H2O)5Cl2] (NO3)
number of mole of complex = 200 × 0.01 = 2
required milli mole of AgNO3 = 2
milli mole = M × Vml ⇒ 2 = 0.1 × Vml = 20 ml.
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