The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 20 — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

[Cr(H2O)5 Cl]Cl2 + 2AgNO3  [Cr(H2O)5 Cl](NO3)2 + 2AgCl 0.01 M × 20 ml 0.1 M × V = 0.2 mili mole 0.2 milimole [Cr(H2O)5 Cl]Cl2 requires 0.4 milimole AgNO3 0.1 × V = 0.4 V = . 4 . = 4 ml