The volume (in `mL`) of `0.1M Ag NO_(3)` required for complete precipitation of chloride ions present in `30 mL` of `0.01M` solution of `[Cr(H_(2)O)_(5)Cl]Cl_(2)`, as silver chloride is close to:

Answer

6

Explanation

`[Cr(H_(2)O)_(5)Cl]Cl_(2)+2 AgNO_(3)rarr 2 AgCl+[Cr(H_(2)O)_(5)Cl](NO_(3))_(2)`
Number of ionizable `Cl^(-)` ions in the complex = 2 Millimoles of `Cl^(-)` ions = Molarity `xx` volume `xx 2 = 0.01 xx 30 xx 2 = 0.6`
Therefore, `Ag^(+)` required for complete precipitation of `Cl^(-)` ions = 0.6 millimoles
As Millimoles = Molarity `xx` V ml
0.6 = `0.1xxV`
V = 6 ml.