20 volumes H₂O₂ solution has a strength of about

${\text{∵\hspace{0.17em}\hspace{0.17em}22.4\hspace{0.17em}\hspace{0.17em}litres\hspace{0.17em}\hspace{0.17em}O}}_2\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{at}\mathrm{\hspace{0.17em}}\mathrm{N}.\mathrm{T}.\mathrm{P}.\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{obtained}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{by}\mathrm{\hspace{0.17em}}68\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{g}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{of}\mathrm{\hspace{0.17em}}{\mathrm{H}}_2{\mathrm{O}}_2$

${\text{∵\hspace{0.17em}1\hspace{0.17em}\hspace{0.17em}litre\hspace{0.17em}\hspace{0.17em}O}}_2\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{at}\mathrm{\hspace{0.17em}}\mathrm{N}.\mathrm{T}.\mathrm{P}.\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{obtained}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{by}\mathrm{\hspace{0.17em}}\frac{68}{22.4}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{g}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{of}\mathrm{\hspace{0.17em}}{\mathrm{H}}_2{\mathrm{O}}_2$

$\therefore {\text{\hspace{0.17em}20\hspace{0.17em}\hspace{0.17em}litre\hspace{0.17em}\hspace{0.17em}O}}_2\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{at}\mathrm{\hspace{0.17em}}\mathrm{N}.\mathrm{T}.\mathrm{P}.\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{obtained}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{by}\mathrm{\hspace{0.17em}}\frac{68}{22.4}\mathrm{\times 20 \hspace{0.17em}\hspace{0.17em}}\mathrm{g}\mathrm{\hspace{0.17em}\hspace{0.17em}}\mathrm{of}\mathrm{\hspace{0.17em}}{\mathrm{H}}_2{\mathrm{O}}_2 = 60.71 \mathrm{g} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2$

$\therefore 1000 ml{O}_{2}atN.T.P.obtainedby\text{\hspace{0.17em}60.71 \hspace{0.17em}\hspace{0.17em}}gof{H}_2{O}_2 = 60.71 g of H_2{O}_2$

$\therefore 100ml{O}_{2}atN.T.P.obtainedby\text{\hspace{0.17em}=}\frac{60.71}{1000}\text{× 100 } = 6.71\%$