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 AgNO₃to give 4.78 g of AgCl (molar mass = 143.5 gmol^-1). The formula of the complex is
(At. mass of Ag = 108 u)

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$[CoCl {({NH}_{3})}_{5}] {Cl}_{2}$

$[{CoCl}_{3} {({NH}_{3})}_{3}]$

$[{CoCl}_{2} {({NH}_{3})}_{4}] Cl$

$[Co {({NH}_{3})}_{6}] {Cl}_{3}$

answer is A.

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Detailed Solution

$\underset{2.675 g}{{CoCl}_{3} \cdot 6 {NH}_{3}} ⟶ {xCl}^{-}$
${xCl}^{-} + {AgNO}_{3} ⟶ \underset{4.78 g}{xgCl ↓}$
Number of moles of the complex
$= \frac{2 .675}{267 .5} = 0 .01 moles$
Number of moles of AgCl obtained
$= \frac{4.78}{143.5} = 0.03 moles$
$∴$ No. of moles of AgCl obtained
$= x \times No. of moles of complex$
$∴ x = \frac{0.03}{0.01} = 3$
$Hence, the formula of complex is [Co {({NH}_{3})}_{6}] {Cl}_{3}$