Question

A solution containing 2.675 g of CoCl3 . 6 NH3 (molar mass = 267.5 g mol–1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass = 143.5 g mol–1). The formula of the complex is (At. mass of Ag = 108 u)

• A. [Co(NH3)6 ] Cl3
• B. [CoCl2 (NH3)4] Cl
• C. [CoCl3(NH3)3]
• D. [CoCl(NH3)5] Cl2

Answer 1

Answer: (A)

$Co(NH3)6$ Cl3

Answer 2

Mole of CoCl3 . 6NH3 = $\frac{2.675}{267.5}$ = 0.01

AgNO3 (aq) + Cl– (aq) $\longrightarrow$ AgCl Æ (white)

Mole of AgCl = $\frac{4.78}{143.5}$ = 0.03

0.01 mole of CoCl3 . 6NH3 gives 0.03 mole of AgCl

1 mole of CoCl3 . 6NH3 ionises to give 3 moles of Cl–.

Hence the formula of compound is [Co(NH3)6] Cl3 .