JEE Advanced · Chemistry · 23. Coordination Compounds
The volume (in \(\mathrm{mL}\) ) of \(0.1 \mathrm{M} \mathrm{AgNO}_3\) required for complete precipitation of chloride ions present in \(30 \mathrm{~mL}\) of \(0.01 \mathrm{M}\) solution of \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}_{)_5} \mathrm{Cl}\right] \mathrm{Cl}_2\right.\), as silver chloride is close to
- A 2
- B 4
- C 6
- D 8
Answer & Solution
Correct Answer
(C) 6
Step-by-step Solution
Detailed explanation
\(\mathrm{mmol}\) of complex \(=30 \times 0.01=0.3\) Also, 1 mole of complex \(\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2\) gives only two moles of chloride ion when dissolved in solution
\(
\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}^{2+}+2 \mathrm{Cl}^{-}\right.
\)
\(\Rightarrow \mathrm{mmol}\) of \(\mathrm{Cl}^{-}\)ion produced from its \(0.3 \mathrm{mmol}=0.6\)
Hence, \(0.6 \mathrm{mmol}\) of \(\mathrm{Ag}^{+}\)would be required for precipitation.
\(
\begin{aligned}
& \Rightarrow 0.60 \mathrm{mmol} \text { of } \mathrm{Ag}^{+}=0.1 \mathrm{M} \times \mathrm{V}(\text { in } \mathrm{mL}) \\
& \Rightarrow \quad V=6 \mathrm{~mL} .
\end{aligned}
\)
\(
\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right] \mathrm{Cl}_2 \longrightarrow\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}^{2+}+2 \mathrm{Cl}^{-}\right.
\)
\(\Rightarrow \mathrm{mmol}\) of \(\mathrm{Cl}^{-}\)ion produced from its \(0.3 \mathrm{mmol}=0.6\)
Hence, \(0.6 \mathrm{mmol}\) of \(\mathrm{Ag}^{+}\)would be required for precipitation.
\(
\begin{aligned}
& \Rightarrow 0.60 \mathrm{mmol} \text { of } \mathrm{Ag}^{+}=0.1 \mathrm{M} \times \mathrm{V}(\text { in } \mathrm{mL}) \\
& \Rightarrow \quad V=6 \mathrm{~mL} .
\end{aligned}
\)
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