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. `[Co(NH_3)_6] Cl_3`
B. `[CoCl_2 (NH_3)_4] Cl`
C. `[CoCl_3 (NH_3)_3]`
D. `[CoCl(NH_3)_5] Cl_2`

Answer

A

Explanation

`underset(2.675g)(CoCl_3. 6NH_3) to xCl^-`
`x Cl^(-) + AgNO_3 to underset(4.78g)(x AgCl darr)`
Number of moles of the comples `=(2.675)/(267.5) = 0.01 ` moles
Number of moles of AgCl obtained `=(4.78)/(143.5)=0.03` moles
`therefore ` No. of moles of AgCl obtained `=3 xx ` No. of moles of complex
`therefore n =(0.03)/(0.01)=3`
`therefore 3Cl^-` ions are precipitable.
Hence the formula of the complex is `[Co(NH_3)_6]Cl_3`