Question

The ratio of amounts of $H_{2}S$ needed to precipitate all the metal ions from 100 ml of 1 M $AgN{O}_3$ and 100 ml of 1 M $CuS{O}_4$ will be

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. None of these

Answer 1

The correct option is B 1 : 2

$AgN{O}_3=2A{g}^{+}+S^{2-}\to A{g}_{2}S$
$\left(H_{2}S\right)$
$\because$ 2 mole $\to$ 1 mole $[100\times 1=100millimole]$
$\therefore$ 100 miliimole $\to$ 50 millimole $H_{2}S$ required
$CuS{O}_4\equiv C{U}^{+2}+S^{2-}\to Cus$
$\left(H_{2}S\right)$
$\because$ 1 mole $\to$ 1 mole $[100\times 1=100millimole]$
$\therefore$ 100 miliimole $\to$ 100 millimole $H_{2}S$ required
Ratio $\frac{50}{100}=\frac{1}{2}$