An excess of AgNO3 is added to 100 mL of a 0.05 M solution of tetraaquadichloridochromium (III) chloride. The number of moles of AgCl precipitated will be 10-3. (Nearest integer)
Answer
5
Solution
Tetraaquadichloridochromium(III) chloride = [Cr(H2O)4Cl2]Cl
[Cr(H2O)4Cl2]Cl [Cr(H2O)4Cl2]+ + Cl- ⇒ 1 ionisable Cl- per formula unit
Moles of complex = 0.100 L × 0.05 M = 5 10-3 mol
Ag+ + Cl- AgCl ⇒ moles of AgCl = 5 10-3
Answer: 5