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

PbS + 4H2O2 → PbSO4 + 4H2O

1 mole - 4 mole

0.01 mole - ??

$\frac{0.01\times 4}{1}=0.04 \mathrm{moles} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2$

 $$\begin{gathered} \mathrm{n}=\frac{\mathrm{weight}}{\mathrm{Molecular} \mathrm{weight}} \\ \mathrm{Weight}=0.04\times 34=1.36 \mathrm{gr} \\ 34 \mathrm{gr} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution} \mathrm{contains}=1.36 \mathrm{gr} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \\ 10 \mathrm{vol} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2(3.03 \mathrm{gr})\to 100 \mathrm{ml} \\ 1.36 \mathrm{gr} \to ?? \\ \frac{1.36\times 100}{3.03}=44.8 \mathrm{ml} \end{gathered}$$