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Q.

The coordination number of metal M in a complex having molecular formula MCl₃.(H₂O)₄ is 6 and there is no molecule of water of crystallization in it. The volume of 0.1M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M complex is

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a

40 mL

b

80 mL

c

60 mL

d

20 mL

answer is B.

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Detailed Solution

Metal coordination number M = 6
Number of Hydration Molecules = 0
As a result, the complex formula = $[M {(H_{2} O)}_{4} {Cl}_{2}] Cl [M {(H_{2} O)}_{4} {Cl}_{2}] Cl ⟶ [M {(H_{2} O)}_{4} {Cl}_{2}] {NO}_{3} + {AgCICl}^{-} + {AgNO}_{3} ⟶ {NO}_{3}^{-} + AgCl$
Let the volume of AgNO₃ used equal V.
M1xC₁ = M1xC₂
0.1×V=200×0.01
V=20mL

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