In the complex with formula `MCl_(3).4H_(2)O` the co-ordination number of the metal M is six. And there is a no molecule of hydration in it. The volume of 0.1 M `AgNO_(3)` solution needed to precitate the free chloride ions in 200 ml of 0.01 M solution of the complex is

Options

A. 40mL
B. 20mL
C. 60mL
D. 80mL

Answer

B

Explanation

The coordination number of metal `M=6`
number of molecules of hydration `=0`
thus , the formula of complex `= [M (H_(2) O)_(4) Cl_(2) ]Cl`
` [M (H_(2) O_(4) Cl_(2)]Cl+AgNO_(3) to [M (H_(2)O)_(4) Cl_(2) ]Cl `
` Cl^(-) AgNO_(3) to NO_(3)^(-)+AgCl`
Let the volume of ` AgNO_(3) ` used = V mL
`:' M_(1) V_(1)=M_(2)V_(2)`
` ( " for " Ag NO_(3) ) (" for " Cl^(-))`
` therefore 0.1 xxV = 200 xx0.01 or V=20 mL `