'V' ml of 0.25 M AgNO3 required for complete precipitation of chloride ions present in 50 ml of 0.02 M solution of [Cr(H2O)6]Cl3.'V' = ----------------------------------- ml

No. of milli moles of complex = 50 ×0.02 = 1[Cr(H2O)6]Cl3 ⇌[Cr(H2O)6]+3 + 3 Cl--No. of milli moles of Cl-- = No . of  milli moles of Ag+ = 1×3 = 3 ( 1 mole of complex gives 3 mole of Cl-)V (ml) ×0.25 = 3V(ml) =30.25 = 12