The volume (in mL) of 0.125; M; AgNO_{3} requ | vedclass

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Question

(C) The chemical formula of the complex is $[Co(NH_{3})_{6}]Cl_{3}$.Each mole of $[Co(NH_{3})_{6}]Cl_{3}$ contains $3$ moles of ionizable $Cl^{-}$ ion

Vedclass · Accepted Answer

$26.92$

Vedclass · Answer

(C) The chemical formula of the complex is $[Co(NH_{3})_{6}]Cl_{3}$.Each mole of $[Co(NH_{3})_{6}]Cl_{3}$ contains $3$ moles of ionizable $Cl^{-}$ ions.The reaction with $AgNO_{3}$ is: $[Co(NH_{3})_{6}]Cl_{3} + 3AgNO_{3} ightarrow [Co(NH_{3})_{6}](NO_{3})_{3} + 3AgCl$.Number of moles of $[Co(NH_{3})_{6}]Cl_{3} = \frac{0.3 \; g}{267.46 \; g/mol} \approx 0.0011216 \; mol$.Since $1$ mole of complex reacts with $3$ moles of $AgNO_{3}$,moles of $AgNO_{3}$ required $= 3 	imes 0.0011216 = 0.0033648 \; mol$.Using the molarity formula $M = \frac{n}{V(L)}$,we have $V(L) = \frac{n}{M} = \frac{0.0033648}{0.125} = 0.0269184 \; L$.Converting to $mL$,$V = 0.0269184 	imes 1000 = 26.92 \; mL$.

List-$I$	List-$II$
$(A). [Ag(CN)_2]^-$	$1. \text{Square planar, } 1.73 \, B.M.$
$(B). [Cu(CN)_4]^{3-}$	$2. \text{Linear, } 0 \, B.M.$
$(C). [Cu(CN)_6]^{4-}$	$3. \text{Octahedral, } 0 \, B.M.$
$(D). [Cu(NH_3)_4]^{2+}$	$4. \text{Tetrahedral, } 0 \, B.M.$
$(E). [Fe(CN)_6]^{4-}$	$5. \text{Octahedral, } 1.73 \, B.M.$

The volume (in $mL$) of $0.125\; M\; AgNO_{3}$ required to quantitatively precipitate chloride ions in $0.3\; g$ of $[Co(NH_{3})_{6}]Cl_{3}$ is (Given: $M_{[Co(NH_{3})_{6}]Cl_{3}} = 267.46\; g/mol$,$M_{AgNO_{3}} = 169.87\; g/mol$).

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