When 1 mol `CrCl_(3).6H_(2)O` is treated with excess of `AgNO_(3)`, 3 mol of `AgCl` are obtained. The formula of the coplex is

Options

A. `[CrCl_(3)(H_(2)O)_(3)].3H_(2)O`
B. `[CrCl_(2)(H_(2)O)_(4)]Cl.2H_(2)O`
C. `[CrCl(H_(2)O)_(5)]Cl_(2).H_(2)O`
D. `[Cr(H_(2)O)_(6)]Cl_(3)`

Answer

4

Explanation

3 mol of AgCl means `3Cl^(-)` are given in the solution hence, the formula of the complex will be `[Cr(H_(2)O)_(6)]Cl_(3)`