Question

On treatment of 100 ml of 0.1 M solution of the complex CrCl₃.6H₂O with excess of AgNO₃, 4.305 g of AgCl was obtained. The complex is

• A. [Cr(H₂O)₃ Cl₃] . 3H₂O
• B. [Cr(H₂O)₄Cl₂] Cl.2H₂O
• C. [Cr(H₂O)₅Cl)Cl₂ .H₂O
• D. [Cr(H₂O)₆]Cl₃

Answer 1

Answer: (D)

Mol of AgCl $=\frac{4\text{.}305}{143\text{.}5}=0\text{.}03$ = mol of Cl^- given by the complex.

Mol of the complex = 100 x 10^-3 x 0.1 = 0.01 ;