Q. 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 (Atomic mass of Ag=108 u)
A) $\left\lbrack Co\left( {NH}_{3} \right)_{6} \right\rbrack{Cl}_{3}$
B) $\lbrack{CoCl}_{2}\left( {NH}_{3} \right)_{4}\rbrack Cl$
C) $\lbrack{CoCl}_{3}\left( {NH}_{3} \right)_{3}\rbrack$
D) $\left\lbrack CoCl\left( {NH}_{3} \right)_{5} \right\rbrack{Cl}_{2}$
Explanation
(a)
Mole of ${CoCl}_{3}.6{NH}_{3}$=$\frac{2.675}{267.5}$=0.01
${AgNO}_{3}(aq) + {Cl}^{-}(aq)\ \longrightarrow AgCl \downarrow$ (white)
Moles of AgCl=$\frac{4.78}{143.5}$=0.03.
mole ${CoCl}_{3}.6{NH}_{3}$ gives =0.03 mol AgCl
$\therefore$ 1 mole ${CoCl}_{3}.6{NH}_{3}$ ionizes to gives =3 mol ${Cl}^{-}$
Hence, the formula of compound is $\left\lbrack Co\left( {NH}_{3} \right)_{6} \right\rbrack{Cl}_{3}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.