Question

A solutions contains $2.675$g of $CuCl_{3},6NH_{3}$ (molar mass $= 2.67.5gmol^{- 1}$) is passed through a cation exchanger. The chloride ions obtained in solutions were treated with excess of $AgNO_{3}$to give $4.78g$of AgCl (molar mass = $143.5gmol^{- 1}$) The formula of the complex is (At. Mass of Ag = 108 u)

• A. $\lbrack CoCl(NH_{3})_{5}\rbrack Cl_{2}$
• B. $\lbrack Co(NH_{3})_{6}\rbrack Cl_{3}$
• C. $\lbrack CoCl_{2}(NH_{3})_{4}\rbrack Cl$
• D. $\lbrack CoCl_{3}(NH_{3})\rbrack_{3}$

Answer 1

Answer: (B)

$\lbrack Co(NH_{3})_{6}\rbrack Cl_{3}$

Answer 2

No. of moles of $CoCl_{3}.6NH_{3} = \frac{2.675}{267.5} = 0.01$

No. of moles of AgCl$= \frac{4.78}{143.5} = 0.03$

Since $0.01$moles of the complex $CoCl_{3}.6NH_{3}$gives $0.03$moles of AgCl on treatment with $AgNO_{3}.$ it implies the 3 chloirde ion are ionisable, in the complex. Thus, that formula of the complex is $\lbrack Co(NH_{3})_{6}\rbrack Cl_{3}.$