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

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(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water as: $[Cr(H_2O)_5Cl]Cl_2 \rightarrow [Cr(H_2O)_5Cl]^{2+} + 2Cl^-$. Each mole of the complex p

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$4$

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(A) The complex $[Cr(H_2O)_5Cl]Cl_2$ dissociates in water as: $[Cr(H_2O)_5Cl]Cl_2 ightarrow [Cr(H_2O)_5Cl]^{2+} + 2Cl^-$. Each mole of the complex provides $2$ moles of ionizable chloride ions $(Cl^-)$. Number of millimoles of the complex $= Molarity 	imes Volume (mL) = 0.01 \, M 	imes 20 \, mL = 0.2 \, mmol$. Since $1$ mole of complex gives $2$ moles of $Cl^-$,$0.2 \, mmol$ of complex gives $0.4 \, mmol$ of $Cl^-$. The precipitation reaction is: $Ag^+ + Cl^- ightarrow AgCl(s)$. Thus,$0.4 \, mmol$ of $AgNO_3$ is required to precipitate $0.4 \, mmol$ of $Cl^-$. Using $Molarity = \frac{millimoles}{Volume (mL)}$,we get $0.1 = \frac{0.4}{V}$. $V = \frac{0.4}{0.1} = 4 \, mL$.

The volume (in $mL$) of $0.1 \, M \, AgNO_3$ required for complete precipitation of chloride ions present in $20 \, mL$ of $0.01 \, M$ solution of $[Cr(H_2O)_5Cl]Cl_2$ as silver chloride is $..........$.

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