MCQ
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$1:1$
- ✓$1:2$
- C$2:1$
- DNone of these
✓
Answer
Correct option: B.
$1:2$
b
(b) $AgN{O_3} \equiv 2A{g^ + } + \mathop {{S^{2 - }}}\limits_{({H_2}S)} \to A{g_2}S$
(b) $AgN{O_3} \equiv 2A{g^ + } + \mathop {{S^{2 - }}}\limits_{({H_2}S)} \to A{g_2}S$
$2$ mole $ \to $ $1$ mole [$100 \times 1 =100$ millimole]
$\therefore $ $100$ miliimole $ \to $ $50$ millimole ${H_2}S$ required
$CuS{O_4} \equiv C{u^{ + 2}} + \mathop {{S^{2 - }}}\limits_{({H_2}S)} \to CuS$
$1$ mole $ \to $ $1$ mole [$100 \times 1 =100$ millimole]
$\therefore $ $100$ millimole $ \to $ $100$ millimole ${H_2}S$ required
Ratio $\frac{{50}}{{100}} = \frac{1}{2}$.
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