Question

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 in solution were treated with excess of $AgNO_{3}$ to give 4.78 g of AgCl (molar mass = $143.5 \,g\, mol^{-1})$. The formula of the complex is (Atomic mass of Ag = 108 u)

• 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 1

Answer: (A)

Mole of $CoCl_{3}.6NH_{3}=\frac{2.675}{267.5}=0.01\,mol$
$AgNO_{3}$ (aq)+$Cl^{-}$ (aq) $\to AgCl ↓ $ (white) $+NO^{-}_{3}$ (aq)
Mole of $AgCl=\frac{4.78}{143.5}=0.03\,mol$
$0.01\,mol\,CoCl_{3}. 6NH_{3}$ gives $= 0.03\,mol\,AgCl$
$\therefore 1\,mol\,CoCl_{3}. 6NH_{3}$ ionises to give = 3 mol $Cl^{-}$
Hence, the formula of compound is $[Co(NH_{3})_{6}]Cl_{3}$