Question

A solution containing $2.675\, g$ of $CoCl_3. 6\, NH_3$ (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 (At. 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)

$[Co(NH_3)_6]Cl_3$

Answer 2

${CoCl3. 6NH3 ->[AgNO_3] x AgCl ?}$ $n(AgCl) = x\, n(CoCl_3. 6NH_3)$ $\frac{4.78}{143.5} = x \frac{2.675}{267.5}\quad\therefore x = 3$ $\therefore$ The complex is $\left[Co\left(NH_{3}\right)_{6}\right]Cl_{3}$