The ratio of the molar amounts of $H_{2} S$ needed to precipitate the metal ions from $20 m L$ each of 1M $C d (N O_{3})_{2}$ and 0.5 M $C u S O_{4}$ is

1 : 1

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2 : 1

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1 : 2

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Indefinite

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Solution

The correct option is B 2 : 1
$C d^{+ 2} + S^{2 -} \to C d S$
$20 \times 1 = 20$
$C u^{+ 2} + S^{2 -} \to C u S$
$20 \times 0.5 = 10$
Ratio $= 2 : 1$

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The ratio of the molar amounts of $H_{2} S$ needed to precipitate the metal ions from $20 m L$ each of 1M $C d (N O_{3})_{2}$ and 0.5 M $C u S O_{4}$ is

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Q. The ratio of amounts of

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needed to precipitate all, the metal ions from 100 ml of

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and 100 ml of

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will be:

The ratio of the molar amounts of $H_{2} S$ needed to precipitate the metal ions from $20 m L$ each of 1M $C d (N O_{3})_{2}$ and 0.5 M $C u S O_{4}$ is

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The ratio of the molar amounts of H2S needed to precipitate the metal ions from 20 mL each of 1M Cd(NO3)2 and 0.5 M CuSO4 is

The correct option is B 2 : 1Cd+2+S2−→CdS 20×1=20 Cu+2+S2−→CuS 20×0.5=10 Ratio=2:1

The ratio of the molar amounts of $H_{2} S$ needed to precipitate the metal ions from $20 m L$ each of 1M $C d (N O_{3})_{2}$ and 0.5 M $C u S O_{4}$ is

The correct option is B 2 : 1
$C d^{+ 2} + S^{2 -} \to C d S$
$20 \times 1 = 20$
$C u^{+ 2} + S^{2 -} \to C u S$
$20 \times 0.5 = 10$
Ratio $= 2 : 1$