In transforming $0.01 m o l e$ of $P b S$ to $P b S O_{4}$ , the volume of $10$ volume $H_{2} O_{2}$ required will be:

11.2 m L

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22.4 m L

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33.6 m L

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44.8 m L

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Solution

The correct option is D $44.8 m L$
$P b S + 4 H_{2} O_{2} \to P b S O_{4} + 4 H_{2} O$

$0.04 m o l$ of $H_{2} O_{2}$ is required to react with $0.01 m o l$ $P b S$ .

Molarity of $H_{2} O_{2}$ may be calculated as,

Volume strength $= M o l a r i t y \times 11.2$

$10 = M \times 11.2$

$M = 0.892$

Number of moles $= \frac{M V}{1000}$

$0.04 = \frac{0.892 \times V}{1000}$

$V = 44.8 m L$ .

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The volume of a "10 volume solution" of H₂O₂ required to liberate 500mL O₂ at N.T.P, is:

P b S + H_{2} O_{2} \to P b S O_{4} + H_{2} O

P b S + H_{2} O_{2} \to P b S O_{4} + H_{2} O

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