In the complex with formula MCl3 ⋅ 4H2O, the coordination number of the metal M is six and there is no molecule of hydration in it. The volume, if 0.1 M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex, is

Question

In the complex with formula MCl3 ⋅ 4H2O, the coordination number of the metal M is six and there is no molecule of hydration in it. The volume, if 0.1 M AgNO3 solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex, is (A) 40 mL (B) 20 mL (C) 60 mL (D) 80 mL

Tardigrade Admin · Accepted Answer

The coordination number of metaI M = 6
Number of molecules of hydration = 0
Thus, the formula of complex =

[M (H_{2} O)_{4} C l_{2}] Cl

[M (H_{2} O)_{4} C l_{2}] Cl + A g N O_{3} ⟶ [M (H_{2} O)_{4} C l_{2}] N O_{3} + A g C I

C l^{-} + A g N O_{3} ⟶ N O_{3}^{-} + A g Cl

Let the volume of

A g N O_{3}

used = V

∵_{f or A g N O_{3}}^{M_{1} V C_{1}} =_{f or C l^{-}}^{M_{2} V C_{2}}

∴ 0.1 \times V = 200 \times 0.01

or V = 20 m L

Q. In the complex with formula $MC l_{3} \cdot 4 H_{2} O,$ the coordination number of the metal M is six and there is no molecule of hydration in it. The volume, if 0.1 M $A g N O_{3}$ solution needed to precipitate the free chloride ions in 200 mL of 0.01 M solution of the complex, is

40 mL

20 mL

60 mL

80 mL

10 mL