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On treatment of 100 ml of 0.1 M solution of the complex CrCl 3 .6H 2 O with excess of AgNO 3 , 4.305 g of AgCl was obtained. The complex is

Options

  1. A[Cr(H 2 O) 3 Cl 3 ] . 3H 2 O
  2. B[Cr(H 2 O) 4 Cl 2 ] Cl.2H 2 O
  3. C[Cr(H 2 O) 5 Cl)Cl 2 .H 2 O
  4. D[Cr(H 2 O) 6 ]Cl 3

Correct answer

D. [Cr(H 2 O) 6 ]Cl 3

Step-by-step solution

Mol of AgCl = 4 . 3 0 5 1 4 3 . 5 = 0 . 0 3 = mol of Cl - given by the complex. Mol of the complex = 100 x 10 -3 x 0.1 = 0.01 ; Cr H 2 O 6 Cl 3 ⟶ Cr H 2 O 6 3 + + 3 Cl - 0 . 0 1 mol 0 . 0 1 mol 0 . 0 3 mol

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