\[\underset{\begin{smallmatrix} 1\,mole \\ 0.01\,mole \end{smallmatrix}}{\mathop{PbS}}\,+\underset{\begin{smallmatrix} 4\,moles \\ 0.04\,mole \end{smallmatrix}}{\mathop{4{{H}_{2}}{{O}_{2}}}}\,\to PbS{{O}_{4}}+4{{H}_{2}}O\]
Weight of $$0.04\,mole$$   of $${H_2}{O_2} = 1.36\,g$$
Now, $${H_2}{O_2}$$  decomposes as : $$2{H_2}{O_2} \to 2{H_2}O + {O_2}$$
$$ \Rightarrow 1\,mL$$   of volume $${H_2}{O_2}$$  solution contains $$\frac{{68}}{{22400}} \times 10\,g = 0.03035\,g$$      of $${H_2}{O_2}.$$
$$ \Rightarrow 0.03035\,g$$   of $${H_2}{O_2}$$  is present in $$1\,mL$$  of 10 volume $${H_2}{O_2}$$
$$\therefore 1.36\,g$$   of $${H_2}{O_2}$$  will be present in $$\frac{1}{{0.03035}} \times 1.36 = 44.81\,mL$$      of 10 volume $${H_2}{O_2}.$$

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

$${K_2}C{r_2}{O_7}$$   an orange coloured solution, on acidification with $$dil\,{H_2}S{O_4}$$   is oxidised by $${H_2}{O_2}$$  to blue peroxide of chromium, $$Cr{O_5},$$  which is unstable.
$$\eqalign{ & {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to {K_2}S{O_4} + {H_2}C{r_2}{O_7} \cr & \left[ {{H_2}{O_2} \to {H_2}O + \left[ O \right]} \right] \times 4 \cr & \underline {{H_2}C{r_2}{O_7} + 4\left[ O \right] \to 2Cr{O_5} + {H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr & \underline {{K_2}C{r_2}{O_7} + {H_2}S{O_4} + 4{H_2}{O_2} \to 2Cr{O_5} + {K_2}S{O_4} + 5{H_2}O} \cr} $$

$${H_2}{O_2}$$  is unstable liquid and decomposes into water and oxygen either on standing or on heating.

$$Ba{O_2}.8{H_2}O + {H_2}S{O_4} \to BaS{O_4} + {H_2}{O_2} + 8{H_2}O$$

Presence of calcium and magnesium salts in the form of hydrogencarbonate, chloride and sulphate in water makes water 'hard'.

Reducing nature of $${H_2}{O_2}$$  means it reduces other substance and itself gets oxidised. In such reactions $${O_2}$$  is evolved.
In (A), $$F{e^{2 + }}$$ gets oxidised to $$F{e^{3 + }}.$$
In (B), $${I_2}$$  gets reduced to $${I^ - }.$$
In (C), $$M{n^{2 + }}$$ gets oxidised to $$M{n^{4 + }}.$$
In (D), $$Pb{S^{\left( {2 - } \right)}}$$  gets oxidised to $$P{b^{\left( { + 6} \right)}}S{O_4}.$$

$$Na{\text{ zeolite}} + CaC{l_2} \to Ca\,{\text{zeolite}}\, + 2NaCl$$

Anhydrous $$CuS{O_4}$$  is white which on reaction with $${H_2}O$$  forms hydrated $$CuS{O_4}$$  which is blue in colour.
$$\mathop {CuS{O_4}}\limits_{{\text{(anhydrous,}}\,{\text{white)}}} + 5{H_2}O \to \mathop {CuS{O_4}.5{H_2}O}\limits_{{\text{(hydrated,}}\,{\text{blue)}}} $$

$${\text{Volume}}\,{\text{strength}} = {\text{Normality}} \times 5.6 = 1.5 \times 5.6 = 8.4L$$