Explanation

(a)
1mL $H_{2}O_{2}$ solution gives 11.2 mL $O_{2}$ at NTP
$\therefore 100\ mL\ H_{2}O_{2}\ solution\ gives\ O_{2} = 100 \times 11.2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1120.0\ mL\ O_{2}\ at\ NTP$
$H_{2}O_{2}$ decomposes as
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2H_{2}O_{2}\ (l)\ \rightarrow 2H_{2}O\ (l) + O_{2}(g)$
$\therefore\ \ 22400\ mL\ O_{2}\ at\ NTP\ is\ obtained\ from\ 68g\ H_{2}O_{2}$
$\therefore 1\ mL\ O_{2}\ at\ NTP\ is\ obtined\ from$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{68}{22400}\ g\ H_{2}O_{2}$
$\therefore\ \ 1120\ mL\ O_{2}\ at\ NTP\ is\ obtained\ from\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{68}{22400} \times 1120$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 34\ g$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ w = \frac{M \times m \times V}{1000}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ M = 1.0$