For vessel \(A\),
\(10 \mathrm{vol} \mathrm{H}_2 \mathrm{O}_2\) means \(1 \mathrm{~mL}\) of \(\mathrm{H}_2 \mathrm{O}_2\) solution gives \(10 \mathrm{mLO}_2\).
\(\begin{array}{cc}2 \mathrm{H}_2 \mathrm{O}_2 & \longrightarrow 2 \mathrm{H}_2 \mathrm{O}+\mathrm{O}_2 \\ 2 \mathrm{~mol} & 1 \mathrm{~mol} \\ =68 \mathrm{~g} & 22400 \mathrm{~mL} \text { at } \mathrm{NTP}\end{array}\)
\(\therefore 22400 \mathrm{~mL}_2\) is obtained by \(68 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2\).
\(\because 10 \mathrm{~mL} \mathrm{O}_2\) is obtained by \(\frac{68 \times 10}{22400}=0.03 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2\)
Vessel \(B\),
\(30 \mathrm{vol} \mathrm{H}_2 \mathrm{O}_2\) means \(1 \mathrm{~mL}\) of \(\mathrm{H}_2 \mathrm{O}_2\) solution gives \(30 \mathrm{mLO}_2\).
Means, \(22400 \mathrm{~mL}\) of \(\mathrm{O}_2\) is obtained by \(68 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2\).
\(\because 30 \mathrm{~mL} \mathrm{O}_2\) is obtained by \(\frac{68 \times 30}{22400}=0.09 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2\)
Vessel \(C\),
Given, it is filled with \(250 \mathrm{~mL}\) of \(2 \mathrm{M} \mathrm{H}_2 \mathrm{O}_2\).
Molarity \(=\frac{\text { number of moles of } \mathrm{H}_2 \mathrm{O}_2}{\text { volume of sol in } \mathrm{L}}\)
\(\therefore \quad 2=\frac{x / 34}{250 / 1000}\) or \(x=17 \mathrm{~g} \mathrm{H}_2 \mathrm{O}_2\)
\(\therefore\) The weight (in g) of \(\mathrm{H}_2 \mathrm{O}_2\) present in vessels follows order :
\(C > B > A\)