On treating 100 mL of 0.1 M aqueous solution of the complex CrCl ₃ 6 H ₂ O with excess of AgNO ₃, 2.86 ~g of AgCl was obtained. The complex is

Step-by-step solution

Given, Molarity of the complex =0.1 M Volume of the complex =100 ~mL Mass of AgCl obtained =2.86 ~g Now, no. of moles in CrCl ₃ 6 H ₂ O = molarity volume 1000 = 0.1 100 1000 =0.01 moles No. of moles of AgCl = 2.86 143 =0.02 moles Number of Cl ⁻ ions present in the ionisation sphere = Moles of ions precipitated with excess AgNO ₃ Moles of complex = 0.02 0.01 =2 It means two chlorine ions must be present in the solution. So, the complex will be [ Cr ( H ₂ O )₅ Cl ⁻ ] Cl ₂ H ₂ O