Hydrogen MCQ Questions & Answers in Inorganic Chemistry | Chemistry
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171.
In transforming $$0.01\,mole$$ of $$PbS$$ to $$PbS{O_4}$$ the volume of '10 volume' $${H_2}{O_2}$$ required will be
A
11.2 $$mL$$
B
22.4 $$mL$$
C
33.6 $$mL$$
D
44.8 $$mL$$
Answer :
44.8 $$mL$$
\[\underset{\begin{smallmatrix}
1\,mole \\
0.01\,mole
\end{smallmatrix}}{\mathop{PbS}}\,+\underset{\begin{smallmatrix}
4\,moles \\
0.04\,mole
\end{smallmatrix}}{\mathop{4{{H}_{2}}{{O}_{2}}}}\,\to PbS{{O}_{4}}+4{{H}_{2}}O\]
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 $${H_2}{O_2}.$$
172.
Last traces of water is removed from $${H_2}{O_2}$$ by
A
electrolysis
B
crystallisation
C
condensation
D
evaporation
Answer :
crystallisation
No explanation is given for this question. Let's discuss the answer together.
173.
An orange coloured solution acidified with $${H_2}S{O_4}$$ and treated with a substance $$'X'$$ gives a blue coloured solution of $$Cr{O_5}.$$ The substance $$'X'$$ is
A
$${H_2}{O_2}$$
B
$${H_2}O$$
C
$$dil\,HCl$$
D
$$conc\,HCl$$
Answer :
$${H_2}{O_2}$$
$${K_2}C{r_2}{O_7}$$ 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.
$$\eqalign{
& {K_2}C{r_2}{O_7} + {H_2}S{O_4} \to {K_2}S{O_4} + {H_2}C{r_2}{O_7} \cr
& \left[ {{H_2}{O_2} \to {H_2}O + \left[ O \right]} \right] \times 4 \cr
& \underline {{H_2}C{r_2}{O_7} + 4\left[ O \right] \to 2Cr{O_5} + {H_2}O\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \cr
& \underline {{K_2}C{r_2}{O_7} + {H_2}S{O_4} + 4{H_2}{O_2} \to 2Cr{O_5} + {K_2}S{O_4} + 5{H_2}O} \cr} $$
174.
$${H_2}{O_2}$$ is always stored in black bottles because
A
It is highly unstable
B
Its enthalpy of decomposition is high
C
It undergoes auto-oxidation on prolonged standing
D
None of these
Answer :
It undergoes auto-oxidation on prolonged standing
$${H_2}{O_2}$$ is unstable liquid and decomposes into water and oxygen either on standing or on heating.
175.
$$HCl$$ is added to following oxides. Which one would give $${H_2}{O_2}$$ ?
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 $$P{b^{\left( { + 6} \right)}}S{O_4}.$$
178.
When zeolite ( hydrated sodium aluminium silicate ) is treated with hard water, the sodium ions are exchanged with
A
adding a drop to anhydrous copper sulphate which changes its colour from white to blue
B
by boiling and testing for the presence of $${H_2}$$ and $${O_2}$$
C
by seeing its colour and transparency
D
by checking the production of lather when mixed with soap
Answer :
adding a drop to anhydrous copper sulphate which changes its colour from white to blue
Anhydrous $$CuS{O_4}$$ is white which on reaction with $${H_2}O$$ forms hydrated $$CuS{O_4}$$ which is blue in colour.
$$\mathop {CuS{O_4}}\limits_{{\text{(anhydrous,}}\,{\text{white)}}} + 5{H_2}O \to \mathop {CuS{O_4}.5{H_2}O}\limits_{{\text{(hydrated,}}\,{\text{blue)}}} $$
180.
The volume strength of $$1.5\,N{H_2}{O_2}$$ solution is