Question

The volume (in $\text{mL}$) of $0.1\text{M}AgN{O}_3$ required for the complete precipitation of chloride ions present in $30\text{mL}$ of $0.01\text{M}$ solution of $\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2$, as silver chloride is close to:

• A. $6\text{mL}$
• B. $7\text{mL}$
• C. $8\text{mL}$
• D. $9\text{mL}$

Answer 1

The correct option is A $6\text{mL}$

mmol of the complex $=30\times 0.01=0.3\text{mmol}$

$\left[Cr\right(H_{2}O{)}_{5}Cl]C{l}_2\to [Cr(H_{2}O{)}_{5}Cl{]}^{2+}+2C{l}^{-}$

$\because$ $1$ mole of the complex gives two moles of chloride ions when dissolved in a solution.
$\therefore$ $0.3\text{mmol}$ of complex produces $0.6\text{mmol}$ of $C{l}^{-}$.
Hence, $0.6\text{mmol}$ of $A{g}^{+}$ would be required for precipitation.

Volume of $AgN{O}_3$ required $=\frac{0.6}{0.1}=6\text{mL}$