In transforming 0.01 mole of PbS to ${\mathrm{PbSO}}_4$, the volume of “10 volume” ${\mathrm{H}}_2{\mathrm{O}}_2$ required will be ____ ml (near integer)

$\mathrm{PbS}+4{\mathrm{H}}_2{\mathrm{O}}_2\to {\mathrm{PbSO}}_4+4{\mathrm{H}}_2\mathrm{O}$

$1\mathrm{mole}\mathrm{of}\mathrm{PbS}-4\mathrm{moles}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2$

$0.01\mathrm{mole}\mathrm{of}\mathrm{PbS}-0.04\mathrm{moles}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2$

Weight of 0.04 mole of ${\mathrm{H}}_2{\mathrm{O}}_2= 1.36 \mathrm{gm}$

10 volumes of $H_2{O}_2$ means

$2{\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}_2\mathrm{O}+{\mathrm{O}}_2$

$\frac{68}{22400}\times 10\mathrm{gm}\mathrm{of}{\mathrm{H}}_2{\mathrm{O}}_2$

$=0.03035 \mathrm{gm}{\mathrm{H}}_2{\mathrm{O}}_2$

1.36gm of ${\mathrm{H}}_2{\mathrm{O}}_2$ present in 

$\frac{1}{0.03035}\times 1.36$

$=44.8 \mathrm{ml}$