The volume (in mL) of 0.1 M AgNO_3 required | vedclass

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(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water to release $2$ moles of ionizable $Cl^-$ ions per mole of the complex.Number of millimoles o

Vedclass · Accepted Answer

$6$

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(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water to release $2$ moles of ionizable $Cl^-$ ions per mole of the complex.Number of millimoles of $[Cr(H_2O)_5Cl]Cl_2 = 30 \ mL 	imes 0.01 \ M = 0.3 \ mmol$.Since each mole of complex provides $2$ moles of $Cl^-$, the total millimoles of $Cl^- = 0.3 	imes 2 = 0.6 \ mmol$.For complete precipitation, $Ag^+$ reacts with $Cl^-$ in a $1:1$ molar ratio $(Ag^+ + Cl^- ightarrow AgCl(s))$.Therefore, millimoles of $AgNO_3$ required $= 0.6 \ mmol$.Using the formula $M = \frac{n}{V(in \ L)}$ or $M = \frac{mmol}{V(in \ mL)}$, we have $0.1 = \frac{0.6}{V}$.$V = \frac{0.6}{0.1} = 6 \ mL$.

List-$I$	List-$II$
$I$. $[Cr(CN)_6]^{4-}$	$P$. $t_{2g}$ orbitals contain $4$ electrons
$II$. $[RuCl_6]^{2-}$	$Q$. $\mu$ (spin-only) $= 4.9 \ BM$
$III$. $[Cr(H_2O)_6]^{2+}$	$R$. low spin complex ion
$IV$. $[Fe(H_2O)_6]^{2+}$	$S$. metal ion in $4+$ oxidation state
	$T$. $d^4$ species

The volume (in $mL$) of $0.1 \ M$ $AgNO_3$ required for complete precipitation of chloride ions present in $30 \ mL$ of $0.01 \ M$ solution of $[Cr(H_2O)_5Cl]Cl_2$, as silver chloride is close to

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