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Depression in Freezing Point

The minimum q...

Question

The minimum quantity in grams of $H_{2} S$ needed to precipitate $63.5$ g of ${C u}^{2 +}$ will be nearly:
${C u}^{+ 2} + H_{2} S \to C u S + H_{2}$

63.5

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31.75

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34

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20

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Solution

The correct option is C $34$ g
$C u^{+ 2} + H_{2} S ⟶ C u S + H_{2}$
$1$ mole of $C u^{+ 2}$ requires $1$ mole of $H_{2} S$
Moles of $C u^{+ 2} = \frac{63.5}{63.5} = 1$ mole
$∴$ Mass of $H_{2} S$ = moles of $H_{2} S \times M_{H_{2} S}$
= $1 \times 34$
= $34 g$

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