Explanation

(b)
$"10\ volume\ H_{2}O_{2}"\$means 1mL of its solution on decomposition at NTP, give 10 mL oxygen gas. Volume of oxygen formed from 100 mL of solution at NTP = 1000 ML.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2H_{2}O_{2}\ \ \ \ \ \ \ \ \ \ \rightarrow \ \ \ \ \ 2H_{2}O + O_{2}$
2 moles 1 mole
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2 \times 34\ g\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 22400\ mL.$
$\because 22400\ mL\ O_{2}$ formed at NTP by decomposition of 68 g $H_{2}O_{2}.$
$\therefore\ \ 1\ mL\ O_{2}\ formed\ at\ NTP\ from\frac{68}{22400}\ of\ H_{2}O_{2}$
$\therefore 1000\ mL\ O_{2}\ formed\ at\ NTP\ from\$
$\ \ \ \ \ \ \ \ \ \frac{68 \times 1000}{22400}\ g\ H_{2}O_{2} = 3.035\ g\ H_{2}O_{2}$
So, concentration of “10 volume $H_{2}O_{2}$”
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 3.0\%$ approximately