The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to

Question

The volume (in mL) of 0.1 M AgNO3 required for complete precipitation of chloride ions present in 30 mL of 0.01 M solution of [Cr(H2O)5Cl]Cl2, as silver chloride is close to (A) 6 (B) 4 (C) 8 (D) 2

Tardigrade Admin · Accepted Answer

[Cr(H2O)5Cl]Cl2 + 2AgNO3 → 2AgCl + [Cr(H2O)5Cl](NO3)2 Number of ionisable chloride ions in complex [Cr(H2O)5Cl] Cl2 = 2 Millimoles = Molarity × Volume (mL) × 2 = 0.01 × 30 × 2 = 0.6 Therefore, required Ag+ = 0.6 millimoles Millimoles = Molarity × V (in mL) 0.6 = 0.1 × V; V= 6 mL

Q. The volume (in mL) of 0.1MAgNO3​ required for complete precipitation of chloride ions present in 30mL of 0.01M solution of [Cr(H2​O)5​Cl]Cl2​, as silver chloride is close to

6

4

8

2

[Cr(H2​O)5​Cl]Cl2​+2AgNO3​→ 2AgCl+[Cr(H2​O)5​Cl](NO3​)2​ Number of ionisable chloride ions in complex [Cr(H2​O)5​Cl] Cl2​=2 Millimoles = Molarity × Volume (mL)×2 =0.01×30×2=0.6 Therefore, required Ag+=0.6 millimoles Millimoles = Molarity ×V (in mL) 0.6=0.1×V;V=6mL

Q. The volume (in mL) of $0.1 M A g N O_{3}$ required for complete precipitation of chloride ions present in $30 m L$ of $0.01 M$ solution of $[C r (H_{2} O)_{5} Cl] C l_{2}$ , as silver chloride is close to