A solution containing 2.675 g of ${CoCl}_{3} . 6 {NH}_{3}$ (molar mass = 267.5 g $m o l^{- 1}$ ) 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)

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} Mole of {CoCl}_{3} . 6 {NH}_{3} = \frac{2 .675}{267 .5} = 0 .01 mol \\ {AgNO}_{3} (aq) + {Cl}^{-} (aq) \to AgCl ↓ (white) + {NO}_{3}^{-} (aq) \\ Mole of AgCl = \frac{4 .78}{143 .5} = 0 .03 mol \\ 0 .01 mol {CoCl}_{3} . 6 {NH}_{3} gives = 0.03 mol AgCl \\ ∴ 1 mol {CoCl}_{3 .} 6 {NH}_{3} ionises to give = 3 mol {Cl}^{-} \\ Hence, the formula of compound is [Co {({NH}_{3})}_{6}] {Cl}_{3} \end{array}$