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In transforming 0.01 mole of PbS to $P b S O_{4},$ the volume of ‘10 volume $H_{2} O_{2}'$ required will be (at NTP)

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11.2 mL

22.4 mL

33.6 mL

44.8 mL

answer is D.

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Detailed Solution

$\underset{0.01 m o l e}{P b S} + \underset{0.04 m o l e}{4 H_{2} O_{2}} \to P b S O_{4} + 4 H_{2} O$

$\underset{0.04 m o l e}{4 H_{2} O_{2}} \to 4 H_{2} O + \underset{0.02 \times 22400 m L}{2 O_{2}}$ at N.T.P

Vol. of $O_{2}$ at N.T.P. = 0.02 × 22400 = 448 mL
Volume of 10 volume $H_{2} O_{2}$ solution
= 448/10 = 44.8 mL

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