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

m moles of `[Cr(H_(2)O)_(6)Cl]Cl_(2) = 0.01 xx 30 = 0.3`
m moles of `Cl^(-) = 0.3 xx 2 = 0.6`
[1 mole of complex gives 2 `Cl^(-)` ions]
m moles of `Ag^(+)`= m moles of `Cl^(-)`
`0.1 xx V = 0.6`
V = 6mL