$$\eqalign{ & \because \,\,22.4\,{\text{litre}}\,\,{O_2}\,{\text{at }}N.T.P.{\text{ obtained by }}68{\text{ }}gm\,\,{\text{of}}\,{H_2}{O_2} \cr & \therefore \,\,1\,{\text{litre}}\,\,{O_2}\,{\text{at}}\,N.T.P.{\text{ obtained by}} \cr & \frac{{68}}{{22.4}}gm\,\,{\text{of}}\,{H_2}{O_2} \cr & \therefore \,\,20\,{\text{litre}}\,\,{O_2}\,{\text{at}}\,N.T.P.{\text{ obtained by}} \cr & \frac{{68}}{{22.4}} \times 20\,gm\,\,{\text{of}}\,{H_2}{O_2} = 60.71\,gm\,\,{\text{of}}\,{H_2}{O_2} \cr & \therefore \,\,1000\,ml\,{O_2}\,{\text{at }}N.T.P.{\text{ obtained by}} \cr & = 60.71\,gm\,\,{\text{of}}\,{H_2}{O_2} \cr & \therefore \,{\kern 1pt} 100\,ml\,{O_2}\,{\text{at }}N.T.P.{\text{ obtained by}} \cr & {\text{ = }}\frac{{60.71}}{{1000}} \times 100 = 6.071\% \cr} $$

$${H_2}{O_2}$$  is an unstable liquid and decomposes into water and oxygen. It is more stable in acidic medium than in basic medium.

Occlusion is the phenomenon of adsorption of hydrogen by metal.

Oxidising nature of $${H_2}{O_2}$$  means it oxidises other substance and itself gets reduced. In such reactions $${O_2}$$  is not evolved.
In (A), $$MnO_4^ - $$  gets reduced from $$M{n^{7 + }}$$  to $$M{n^{2 + }}.$$
In (B), $$F{e^{3 + }}$$ gets reduced to $$F{e^{2 + }}.$$
In (C), $${I^{-}}$$ gets oxidised to $${I_2}.$$
In (D), $$KI{O_4}$$  gets reduced from $${I^{7 + }}$$  to $${I^{5 + }}.$$

No explanation is given for this question. Let's discuss the answer together.

Decomposition of $${H_2}{O_2}$$  is catalysed by metals, rough surfaces, sunlight, dust particles, glass and alkalies.

TIPS/Formulae :
A diamagnetic substance contains no unpaired electron.
$${H_2}$$  is diamagnetic as it contains all paired electrons.
$$\mathop {{H_2} = \sigma _b^2}\limits_{\left( {{\text{diamagnetic}}} \right)} ,\,\,\,\mathop {H_2^ + = \sigma _b^1,}\limits_{\left( {{\text{paramagnetic}}} \right)} \,\,\mathop {H_2^ - = \sigma _b^2,}\limits_{\left( {{\text{paramagnetic}}} \right)} \,\,\,\mathop {\sigma _a^{*1};He_2^ + }\limits_{\left( {\operatorname{paramagnetic} } \right)} = \mathop {\sigma _b^2,\sigma _a^{*1}}\limits_{\left( {{\text{paramagnetic}}} \right)} $$

In the reaction $$A{g_2}O$$  is reduced to $$Ag$$  hence $${H_2}{O_2}$$  acts as a reducing agent.
$$A{g_2}O + {H_2}{O_2} \to 2Ag + {H_2}O + {O_2}$$

Hydrogen directly combines with highly reactive metals to give hydrides.
$$Ca + {H_2} \to Ca{H_2}$$

Fire due to action of water on saline hydrides cannot be extinguished with water or $$C{O_2}.$$  These hydrides can reduce $$C{O_2}$$  at high temperature to produce $${O_2}.$$