Question

In the complex with formula $MCl_3 \cdot 4H_2O,$ the coordination number of the metal M is six and there is no molecule of hydration in it. The volume, if 0.1 M $AgNO_3$ 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
• E. 10 mL

Answer 1

Answer: (B)

The coordination number of metaI M = 6
Number of molecules of hydration = 0
Thus, the formula of complex = $[ M (H_2O)_4 Cl_2 ]Cl $
$[M(H_2O)_4Cl_2]Cl+AgNO_3 \longrightarrow [M(H_2O)_4Cl_2]NO_3 +AgCI $
$\, \, \, \, \, \, \, \, \, \, \, \, Cl^- + AgNO_3 \longrightarrow NO^-_3 + AgCl$
Let the volume of $AgNO_3$ used = V
$\because \, \, \, \, \, \, \, \, \, \, \,{}^{M_1VC_1}_{for\, AgNO_3}=\,{}^{M_2VC_2}_{for\, Cl^-}$
$\therefore \, \, \, \, \, \, \, \, \, \, \, 0. 1 \times V = 200\times 0.01 $
$or \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, V=20 mL$