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Byju's Answer

Standard XII

Chemistry

Stoichiometric Calculations

A solution co...

Question

A solution containing $2.675$ $g$ of $C o {C l}_{3} .6 N H 3$ $(m o l a r m a s s = 267.5 g m o 1^{- 1})$ is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $A g N O_{3}$ to give $4.78$ $g$ of $A g C l$ $(m o l a r m a s s = 143.5 g m o 1^{- 1})$ . The formula of the complex is:(At. Mass of $A g$ $= 108 u$ )

[C o (N H_{3})_{6}] {C l}_{3}

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[C o {C l}_{2} (N H_{3})_{4}] C l

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[C o {C l}_{3} (N H_{3})_{3}]

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[C o C l (N H_{3})_{5}] {C l}_{2}

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Solution

The correct option is A $[C o (N H_{3})_{6}] {C l}_{3}$
The ionization of 1 mole of the complex will give x moles of chloride ions
$C o {C l}_{3} .6 N H 3 \to x {C l}^{-}$
x moles of chloride ions will give x moles of AgCl ppt. This corresponds to 1 mole of the complex.
${x C l}^{-} + x A g N O_{3} \to x A g C l (p p t) + x N O_{3}^{-}$
$m o l e s o f A g C l = x \times m o l e s o f C o {C l}_{3} .6 N H 3$
$\frac{4.78}{143.5} = x \frac{2.675}{267.5}$
$0.0331 = x 0.01$
$∴ x = 3$ . So, it has 3 ionizable chloride ions.

$∴$ The complex is $[C o (N H_{3})_{6}] C l_{3}$ .

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Similar questions

Q. A solution containing 2.675 g of

C o C l_{3} .6 N H_{3}

(Molar mass = 267.5 g/mol) is passed through a cation exchanger. The chloride ions obtained in the solution is treated with an excess of

A g N O_{3}

to give 4.78 g of

A g C l

(Molar mass = 143.5 g/mol). The formula of the complex is:
(Atomic mass of Ag = 108 u)

Q. A solution containing

2.675 g

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Q. A solution containing

2.675 g

C o C l_{3} .6 N H_{3}

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The chloride ions obtained in solution were treated with excess of

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Q. A mixture contains

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A solution containing 2.675 g of CoCl3.6NH3 (molar mass=267.5gmo1−1) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78 g of AgCl (molar mass=143.5gmo1−1). The formula of the complex is:(At. Mass of Ag =108u)