A solution containing 2.675 g of `CoCl_(3).6NH_(3)` (molar mass `= 267.5 g mol^(-1)`) is passed through a cation exchanger. The chloride ions obtained is solution were treated with excess of `AgNO_(3)` to give 4.73 g of `AgCl` (molar mass = `143.5 g mol^(-1)`). The formula of the complex is (At. mass of Ag = 108 u)

Options

A. `[CoCl_(3) (NH_(3))_(5)]`
B. `[CoCl(NH_(3))_(5)]Cl_(2)`
C. `[Co(NH_(3))_(6)]Cl_(3)`
D. `[CoCl_(2) (NH_(3))_(4)]Cl`

Answer

C

Explanation

Moles of the complex `=(2.675)/(267.5)=0.01`
moles of AgCl precipitated `=(4.48)/(143.5)=0.33`
Thus, 1 mole of the complex will precipitate `AgCl ""(0.033)/(0.01)=3" moles"`. That means 1 molecules of the complex contains 3 ionizable Cl.
Hence the formula is `[Co(NH_(3))_(6)]Cl_(3)`