Explanation

(a)
$\left\lbrack H_{2}O_{2} \rightarrow H_{2}O + \frac{1}{2}O_{2} \right\rbrack \times 2$
$2H_{2}O_{2} \rightarrow 2H_{2}O + \ \ \ \ \ \ \ \ \ \ \ \ \ O_{2}\$
$\ \ 68\ g\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 22.4\ L\ at\ NTP$
$\because 22.4\ L\ O_{2}\ at\ NTP\ is\ obtained\ by\ 68\ g\ of\ H_{2}O_{2}$
$\therefore 20\ L\ O_{2}\ at\ NTP\ will\ be\ obtained\ by\ H_{2}O_{2}$
$= \frac{68}{22.4} \times 20 = 60.7\ g/L$
$\therefore 1000\ mL\ O_{2}\ at\ NTP\ is\ obtained\ by\ H_{2}O_{2} = 60.7\ g$
$\therefore Percentage\ strength = \frac{60.7 \times 100}{1000} = 6.07\ g$