Explanation

(d)
$PbS + 4H_{2}O_{2}\ \rightarrow PbSO_{4} + 4H_{2}O$
from the above equation
$\because 1\ mole\ of\ PbS\ required\ 4\ moles\ of\ H_{2}O_{2}$
0.01 moles of PbS required 0.04 mole of $H_{2}O_{2}$
Weight of 0.04 mole $H_{2}O_{2} = 1.36\ g$
10 volume of $H_{2}O_{2}$ means,
1mL of such solution of $H_{2}O_{2}$ on decomposition by heat produces 10mL of oxygen at NTP.
$H_{2}O_{2}$ decomposes as,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2H_{2}O_{2}\ \rightarrow 2H_{2}O + O_{2}$
Thus 1mL of 10 volume $H_{2}O_{2}$ solution contains
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{68}{22400} \times 10\ g\ of\ H_{2}O_{2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.030.35\ g\ of\ H_{2}O_{2}$
$\because 0.03035\ \ g\ of\ H_{2}O_{2}\$is present in 1 mL of 10 volume $H_{2}O_{2}.$
$\therefore 1.36\ g\ of\ H_{2}O_{2}\ present\ in\frac{1}{0.03035} \times 1.36\ mL\ of\ 10\ volume\ \ of\ H_{2}O_{2} = 44.81\ mL$