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Step-by-Step Solution
Step 1: Identify the Relevant Species and Their Stoichiometry
The coordination complex is given as [\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2} .
In this complex, the coordination sphere is [\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]^{2+} , and there are 2 chloride ions ( \mathrm{Cl}^{-} ) outside the coordination sphere for each formula unit of the complex.
These 2 chloride ions are the ones that will precipitate with silver nitrate ( \mathrm{AgNO}_{3} ).
Step 2: Calculate the Moles of the Complex
The volume of the complex solution is 20 \,\mathrm{mL} which is 0.020 \,\mathrm{L} ,
and its molarity is 0.01\,\mathrm{M} .
The moles of the complex are:
\text{Moles of complex} \;=\; 0.01 \times 0.020
\;=\; 2.0 \times 10^{-4}\,\text{moles.}
Step 3: Find the Moles of Chloride Ions
Each mole of [\mathrm{Cr}(\mathrm{H}_{2}\mathrm{O})_{5}\mathrm{Cl}]\mathrm{Cl}_{2} contributes 2 moles of free chloride ions ( \mathrm{Cl}^{-} ).
Therefore, the total moles of \mathrm{Cl}^{-} are:
\text{Moles of }\mathrm{Cl}^{-} \;=\; 2 \times 2.0 \times 10^{-4}
\;=\; 4.0 \times 10^{-4}\,\text{moles.}
Step 4: Relate Chloride Ions to Silver Nitrate
To precipitate \mathrm{Cl}^{-} as \mathrm{AgCl} , each mole of \mathrm{Cl}^{-} requires 1 mole of \mathrm{AgNO}_{3} .
Hence, the moles of \mathrm{AgNO}_{3} needed are:
\text{Moles of }\mathrm{AgNO}_{3} \;=\; 4.0 \times 10^{-4}\,\text{moles.}
Step 5: Calculate the Required Volume of \mathrm{AgNO}_{3} Solution
The concentration of \mathrm{AgNO}_{3} solution is 0.1\,\mathrm{M} .
The volume of this solution required in liters is given by:
\text{Volume (L)}
\;=\; \frac{ \text{Moles of }\mathrm{AgNO}_{3} }{ \text{Molarity of }\mathrm{AgNO}_{3} }
\;=\; \frac{4.0 \times 10^{-4}}{0.1}
\;=\; 4.0 \times 10^{-3}\,\text{L.}
Converting liters to milliliters:
\text{Volume (mL)}
\;=\; 4.0 \times 10^{-3} \times 1000
\;=\; 4\,\mathrm{mL}.
Final Answer
The volume of 0.1\,\mathrm{M} \mathrm{AgNO}_{3} required is 4 mL.
