NEET Practice Questions (MCQs) with Answers & Solutions

(C). Commercially ${\mathrm{H}}_2{\mathrm{O}}_2$ is prepared by electrolysis of $50\%$ sulphuric acid.

Hydrogen peroxide is prepared by the electrolysis of 30% to 50%  ice-cold H₂SO₄. When acidified sulfate solution is electrolyzed at high current density, peroxodisulphate is obtained. Peroxodisulphate is then hydrolyzed to get hydrogen peroxide.

2HSO^–₄(aq) [Electrolysis] → HO₃SOOSO₃H(aq) [Hydrolysis] → 2HSO^–₄(aq)+2H+(aq)+H₂O₂(aq)

(C).  ${\mathrm{H}}_2{\mathrm{O}}_2$ does not give silver peroxide with moist silver iodine.

${\mathrm{Ag}}_2\mathrm{O}+{\mathrm{H}}_2{\mathrm{O}}_2\to 2\mathrm{Ag}+{\mathrm{O}}_2+{\mathrm{H}}_2\mathrm{O}$

$$\begin{gathered} \left(\mathrm{A}\right). 2\mathrm{Na}+2{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2+2\mathrm{NaOH} \\ \mathrm{A} \mathrm{B} \mathrm{C} \\ \mathrm{and} \mathrm{Zn}+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{ZnO}}_2+{\mathrm{H}}_2 \\ \mathrm{D} \mathrm{C} \mathrm{B} \end{gathered}$$

(A). 136 ppm ${\mathrm{CaSO}}_4$ equivalent to 100 ppm ${\mathrm{CaCO}}_3$ equivalent to = 95 ppm ${\mathrm{MgCl}}_2$.

       Thus 68 ppm ${\mathrm{CaSO}}_4$ = 100 × 68 / 136 = 50 ppm ${\mathrm{CaCO}}_3$.

       19 ppm ${\mathrm{MgCl}}_2$ = 100 × 19/95 = 20 ppm ${\mathrm{CaCO}}_3$.

       Total hardness = 50 + 20 = 70 ppm

(A). We know that strength = Normality × Eq. wt. and Eq. wt. of H₂O₂ = 17

       $\therefore$  Strength of $1.6 \mathrm{N} {\mathrm{H}}_2{\mathrm{O}}_2$ solution = 1.6 × 17 g/l

       Now 68g of ${\mathrm{H}}_2{\mathrm{O}}_2$ gives 22400 ml ${\mathrm{O}}_2$ at N.T.P.

        $\therefore$  1.6 × 17 g of ${\mathrm{H}}_2{\mathrm{O}}_2$ will give

$\frac{22400}{68}\times 1.6\times 1.7=8960 \mathrm{ml} \mathrm{of} {\mathrm{O}}_2$ at N.T.P.

       But 1.6 × 17 g of ${\mathrm{H}}_2{\mathrm{O}}_2$ are present in 1000 ml of ${\mathrm{H}}_2{\mathrm{O}}_2$ solution.

       Hence 1000 ml of ${\mathrm{H}}_2{\mathrm{O}}_2$ solution gives 8960 ml of ${\mathrm{O}}_2$ at N.T.P.

       $\therefore$1 ml of ${\mathrm{H}}_2{\mathrm{O}}_2$ solution will give $=\frac{8960}{1000}=8.96$ ml of ${\mathrm{O}}_2$ at N.T.P.

        Hence the volume strength of 1.6N ${\mathrm{H}}_2{\mathrm{O}}_2$ solution = 8.96 volume.

$$\begin{gathered} \left(\mathrm{C}\right). \mathrm{Mol} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{in} 2\mathrm{L} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution} ={\mathrm{V}}_{\left(\mathrm{L}\right)}\times \mathrm{M}=2\times 4=8 \mathrm{mols} \\ \therefore \mathrm{Mass} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{present}=8\times \mathrm{GMM}=8\times 34=272 \mathrm{g} \\ \mathrm{Mol} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{in} 400 {\mathrm{cm}}^3 \mathrm{of} \mathrm{solution}=400\times {10}^{-3}\times 4=1.6 \mathrm{mol} \\ 2{\mathrm{H}}_2{\mathrm{O}}_2 \to 2{\mathrm{H}}_2\mathrm{O}+ {\mathrm{O}}_2 \\ 2 \mathrm{mol} 22.4 \mathrm{L} \mathrm{at} \mathrm{STP} \\ 2 \mathrm{mol} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{gives} {\mathrm{O}}_2=22.4 \mathrm{L} \mathrm{at} \mathrm{STP} \\ 1.6 \mathrm{mol} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{gives} {\mathrm{O}}_2=\frac{22.4\times 1.6}{2}=17.92 \mathrm{L} \end{gathered}$$

(B). For ${\mathrm{H}}_2{\mathrm{O}}_2, {\mathrm{V}}_1=30 \mathrm{ml}, {\mathrm{N}}_1=?$

      $$\begin{gathered} \mathrm{For} {\mathrm{KMnO}}_4, {\mathrm{V}}_2=30 \mathrm{ml}, {\mathrm{N}}_2=\mathrm{N}/10 \\ \mathrm{Applying} \mathrm{normality} \mathrm{equation}, \\ {\mathrm{N}}_1{\mathrm{V}}_1={\mathrm{N}}_2{\mathrm{V}}_2 \mathrm{i}.\mathrm{e}., 30\times {\mathrm{N}}_1=30\times 1/10 \\ \therefore {\mathrm{N}}_1=0.1 \mathrm{N} \\ \mathrm{Thus}, \mathrm{the} \mathrm{normality} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution}=0.1 \mathrm{N}. \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ {\mathrm{H}}_2{\mathrm{O}}_2\to 2{\mathrm{H}}^{+}+{\mathrm{O}}_2+2{\mathrm{e}}^{-} \\ \therefore \mathrm{eq}. \mathrm{wt} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2=34/2=7 \\ \mathrm{Hence} \mathrm{strength} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution}=\mathrm{Normality}\times \mathrm{eq}. \mathrm{wt}.=0.1\times 17=1.7 \mathrm{g}/\mathrm{litre} \\ \therefore \%\mathrm{strength} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2=\frac{1.7\times 100}{1000}=0.17\% \\ \mathrm{Consider} \mathrm{the} \mathrm{chemical} \mathrm{equation}, \\ 2{\mathrm{H}}_2{\mathrm{O}}_2\stackrel{ }{\to }2{\mathrm{H}}_2\mathrm{O}+ {\mathrm{O}}_2 \\ 68\mathrm{g} 22400 \mathrm{ml} \mathrm{at} \mathrm{N}.\mathrm{T}.\mathrm{P}. \\ \mathrm{Now} 68 \mathrm{g} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{give} {\mathrm{O}}_2 \mathrm{at} \mathrm{N}.\mathrm{T}.\mathrm{P}.=22400 \mathrm{ml} \\ 1.7 \mathrm{g} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{will} \mathrm{give} {\mathrm{O}}_2=\frac{22400}{68}\times 1.7=560 \mathrm{ml} \\ \mathrm{But} 1.7 \mathrm{g} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{are} \mathrm{present} \mathrm{in} 1000 \mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution}. \\ \mathrm{Hence} 1000 \mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution} \mathrm{gives} 560 \mathrm{ml} \mathrm{of} {\mathrm{O}}_2 \mathrm{at} \mathrm{N}.\mathrm{T}.\mathrm{P}. \\ \therefore 1 \mathrm{ml} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution} \mathrm{will} \mathrm{give} =\frac{560}{1000}=0.56 \mathrm{mol} \mathrm{of} {\mathrm{O}}_2 \mathrm{at} \mathrm{N}.\mathrm{T}.\mathrm{P}. \\ \therefore \mathrm{Volume} \mathrm{strength} \mathrm{of} {\mathrm{H}}_2{\mathrm{O}}_2 \mathrm{solution}=0.56. \end{gathered}$$ 

(B). ${\mathrm{D}}_2\mathrm{O}$ is called heavy water. Due to stronger $\mathrm{D}–\mathrm{O}$ bonds, the degree of dissociation of ${\mathrm{D}}_2\mathrm{O}$ is

       lower than that of ${\mathrm{H}}_2\mathrm{O}$.

(B).  The extent of H-bonding in ${\mathrm{H}}_2{\mathrm{O}}_2$ is higher than that in ${\mathrm{H}}_2\mathrm{O}$.

(C). The electronegativity difference between $\mathrm{Be}\left(1.5\right)$ and $\mathrm{H}\left(2.1\right)$ is small.