Question

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

• A. 6 mL
• B. 7 mL
• C. 8 mL
• D. 9 mL

Answer 1

The correct option is A 6 mL

mmol of the complex $=30\times 0.01=0.3$ mmol

$\left[Cr\right(H_{2}O{)}_5]C{l}_2\to [Cr(H_{2}O{)}_{5}Cl{]}^{2+}+2C{l}^{-}$

$\because$ 1 mole of the complex gives two moles of chloride ions when dissolved in a solution.
$\therefore$ 3 mmol of complex produces 6 mmol of $C{l}^{-}$.
Hence, 0.6 mmol of $A{g}^{+}$ would be required for precipitation.

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