In transforming 0.01mole of PbS to PbSO4 the volume of ‘10volume’ H2O2 required will be

PbS + 4H2O2 → PbSO4 + 4H2O1 mole - 4 mole0.01 mole - ??0.01×41=0.04 moles of H2O2 n=weightMolecular weight Weight=0.04×34=1.36 gr 34 gr of H2O2 solution contains=1.36 gr of H2O2 10 vol of H2O2(3.03 gr)→100 ml 1.36 gr →?? 1.36×1003.03=44.8 ml

In transforming 0.01mole of PbS to PbSO4 the volume of ‘10volume’ H2O2 required will be

PbS + 4H2O2 → PbSO4 + 4H2O1 mole - 4 mole0.01 mole - ??0.01×41=0.04 moles of H2O2 n=weightMolecular weight Weight=0.04×34=1.36 gr 34 gr of H2O2 solution contains=1.36 gr of H2O2 10 vol of H2O2(3.03 gr)→100 ml 1.36 gr →?? 1.36×1003.03=44.8 ml

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In transforming 0.01mole of PbS to PbSO₄ the volume of ‘10volume’ H₂O₂ required will be

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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

PbS + 4H2O2 → PbSO4 + 4H2O

1 mole - 4 mole

0.01 mole - ??

\frac{0.01 \times 4}{1} = 0.04 moles of H_{2} O_{2}

$n = \frac{weight}{Molecular weight} Weight = 0.04 \times 34 = 1.36 gr 34 gr of H_{2} O_{2} solution contains = 1.36 gr of H_{2} O_{2} 10 vol of H_{2} O_{2} (3.03 gr) \to 100 ml 1.36 gr \to ? ? \frac{1.36 \times 100}{3.03} = 44.8 ml$