The ratio of amounts of $H_{2} S$ needed to precipitate all the metal ions from 100 ml of 1 M $A g N O_{3}$ and 100 ml of 1 M $C u S O_{4}$ will be

1 : 1

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1 : 2

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2 : 1

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None of these

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Solution

The correct option is B 1 : 2
$A g N O_{3} = 2 A g^{+} + S^{2 -} \to A g_{2} S$
$(H_{2} S)$
$∵$ 2 mole $\to$ 1 mole $[100 \times 1 = 100 m i l l i m o l e]$
$∴$ 100 miliimole $\to$ 50 millimole $H_{2} S$ required
$C u S O_{4} \equiv C U^{+ 2} + S^{2 -} \to C u s$
$(H_{2} S)$
$∵$ 1 mole $\to$ 1 mole $[100 \times 1 = 100 m i l l i m o l e]$
$∴$ 100 miliimole $\to$ 100 millimole $H_{2} S$ required
Ratio $\frac{50}{100} = \frac{1}{2}$

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