MCQ Questions for Class 11 Medical Chemistry Hydrogen Quiz 2

The fuel used to produce electrical energy in the space rocket Saturn V, used by Neil Armstrong to reach the moon was:

0%
$CO+O_{2}$
0%
$F_{2}+O_{2}$
0%
$CH_{4}+O_{2}$
0%
liquid $H_{2}+O_{2}$

Explanation

The fuel used was a liquid mixture of

$H_{2}$ and

$O_{2}$ .

It was used for the following reasons:
1.

$H_{2}$ is a highly efficient fuel, as it has the highest calorific value.

2. Mixing

$O_{2}$ was required to support the combustion reaction:

$2H_{2}+O_{2}\rightarrow 2H_{2}O$ .

3. Further, the water produced at the end of the combustion reaction provided a good source of pure drinking water for the astronauts.

Metal hydride on treatment with water gives________.

0%
$H_{2}O_{2}$
0%
$H_{2}O$
0%
acid
0%
hydrogen

Explanation

Metal hydrides react with water to form hydrogen gas and the metal hydroxide.
Example:

$LiH+H_2O\rightarrow H_2+LiOH$ .

Hydrogen does not resemble alkali metals in which one of the following properties?

0%
Tendency to form cation
0%
Nature of oxide
0%
Combination with halogens
0%
Reducing character

Explanation

Hydrogen loses one electron and forms cation

$(H^+)$ .

Oxides of hydrogen are neutral but metal oxides are basic in nature.

Hydrogen combines with halogens

$(X)$ and forms

$HX$ molecule.

Hydrogen loses one electron and causes reduction. Hence, it acts as a reducing agent.

Hydrogen finds a place in VII A group of periodic table because_________.

0%
it is a light element
0%
$H -$ atom has only one electron in its valence shell
0%
it forms hydride of the type $NaH$
0%
it has $7$ isotopes

Explanation

Hydrogen finds a place in VII A group of periodic table because hydrogen forms a compound

$NaH$ like halogens

$NaX$ (

$X$ is halogen).

The lightest element in the periodic table is_________.

0%
lithium
0%
fluorine
0%
hydrogen
0%
helium

Explanation

The lightest element in the periodic table is hydrogen.
Its atomic weight is

$1.008$ amu.

$H_{2}O_{2}$ present in the atmosphere is formed by the action of sunlight on the atmospheric:

0%
$H_{2}$ and $O_{2}$
0%
$H_{2}O$ and $O_{3}$
0%
moisture and $O_{2}$
0%
$H_{2}O$ and $SO_{2}$

Explanation

$H_2O_2$ present in the atmosphere is formed by the action of sunlight on the atmospheric

$H_2O$ and

$O_3$ .

The reaction is as follows:

$H_2O + \dfrac 13O_3 + \text{hv} \rightarrow H_2O_2$

The peroxide linkage

$(-O-O-)$ is not present in:

0%
$H_{2}SO_{5}$
0%
$CrO_{5}$
0%
$HNO_{4}$
0%
$HClO_{4}$

Explanation

Hint:

$\text { A peroxide linkage is simply a } O-O \text { bond. }$

Correct Answer: Option $$D$$

Explanation for correct option:

We can find the

$O-O$ linkage in the given compounds by their structures.

$HClO_{4}$ structure has no such linkage, hence it is the correct option.

Explanation for incorrect options:

All other compounds have $O-O$ linkage as shown below:

Hence, these are incorrect options.

$H_{2}O_{2}$ forms prismatic crystals at:

0%
$0^{o}C$
0%
$2^{o}C$
0%
$2K$
0%
$-1.7^{o}C$

Explanation

On cooling,

$95-96\%$ , the peroxide in solid carbon dioxide and ether, or in methyl chloride at

$-23^oC$ , solidifies to a hard crystalline mass.

If a little of this solid is placed in the

$95\%$ and the solution is cooled to

$-10^oC$ , columnar prismatic crystals of pure solid hydrogen peroxide melting at

$-1.7^oC$ are obtained. These crystals explode with a trace of platinum black; alone, they are fairly stable.

The product obtained at anode when

$50\%\: H_{2}SO_{4}$ aqueous solution is electrolysed using platinum electrodes is :

0%
$H_{2}SO_{3}$
0%
$H_{2}S_{2}O_{8}$
0%
$O_{2}$
0%
$H_{2}$

Explanation

Hydrogen peroxide can be produced by electrolysis of sulfuric acid. At the anode the following reaction takes place:

$H_2SO_4^- \rightarrow S_2O_8^{2-} + 2H^+ + 2e^-$
The reaction of this ion with water serves as a commercial preparation of hydrogen peroxide:

$S_2O_8^{2-}(aq) + H_2O\rightarrow H_2O_2(aq) + 2HSO_4^-$

The melting point of most of the solid substances increases with an increase in pressure. However ice melts at a temperature lower than its usual melting points when pressure is increased. This is because:

0%
the density of ice is lesser than $H_{2}O$
0%
pressure generates heats
0%
the chemical bonds break under pressure
0%
ice is not a true solid

Electrolysis of

$50\%\: H_{2}SO_{4}$ produces:

0%
$H_{2}S_{2}O_{8}$ at anode
0%
$H_{2}$ at anode
0%
$H_{2}SO_{5}$ at cathode
0%
$H_{2}O_{2}$ at anode

Explanation

Hydrogen peroxide can be produced by electrolysis of sulfuric acid. At the anode the following reaction takes place:

$H_2SO_4^- \rightarrow S_2O_8^{2-} + 2H^+ + 2e^-$
The reaction of this ion with water serves as a commercial preparation of hydrogen peroxide:

$S_2O_8^{2-}(aq.) + H_2O\rightarrow H_2O_2(aq.) + 2HSO_4^-$ .

Catalytic union of

$H_{2}$ and

$O_{2}$ to get

$H_{2}O_{2}$ is:

0%
auto oxidation
0%
electrolysis of $50\%\:Na_{2}O_{2}$
0%
hydrolysis of aqueous solution of $\left ( NH_{4} \right )_{2}SO_{4}+H_{2}SO_{4}$
0%
treating $Na_{2}O_{2}$ with cold dil. $Na_{2}O_{2}$

Which of the following compound gives

$H_{2}O_{2}$ by the addition of

$H_{2}O$ ?

0%
$CO(NH_2)_2.H_2O_2$
0%
$Na_{2}HPO_{4}.H_{2}O_{2}$
0%
$(NH_{4})_{2}SO_{4}.H_{2}O_{2}$
0%
All of the above

Explanation

$CO(NH_{2})_{2}.H_{2}O_{2}\overset {water}{\longrightarrow} CO(NH_{2})_{2}+H_{2}O_{2}$

$Na_{2}HPO_{4}.H_{2}O_{2}\overset{water}{\longrightarrow} Na_{2}HPO_{4}+H_{2}O_{2}$

$(NH_{4})_{2}SO_{4}.H_{2}O_{2}\overset{water}{\longrightarrow}(NH_{4})_{2}SO_{4}+H_{2}O_{2}$

Hence, option D is correct,

Which one of the following compounds undergoes hydrolysis during distillations to yield hydrogen peroxide?

0%
$H_{2}S_{2}O_{8}$
0%
$H_{2}S_{2}O_{6}$
0%
$HNO_{3}$
0%
$H_{4}P_{2}O_{7}$

Explanation

One of the methods of preparation of

$H_2O_2$ is from sulphuric acid.
Electrolysis of acidic sulphate solutions leads to the formation of peroxodisulfuric acid which then further hydrolyses to give

$H_2O_2$ .

In the preparation of

$H_{2}O_{2}$ by auto-oxidation method, the starting substance is:

0%
2-ethylanthraquinone.
0%
2-ethylanthraquinol.
0%
p-benzoquinone.
0%
N-methylaniline.

Explanation

In the auto-oxidation method, 2-Ethylanthraquinone is reduced to 2-ethylanthraquinol by Hydrogen in the presence of palladium. This 2-ethylanthraquinol is oxidised by air to 2-ethylanthraquinone again.

Then

$H_{2}O_{2}$ forms. So, here the starting substance is 2-ethylanthraquinone.

Hence, the correct option is

$\text{A}.$

Assertion (A): Hydrogen shows resemblance with alkali metals as well as halogens.
Reason (R): Hydrogen exists in atomic form only at high temperature.

0%
Both A and R are true and R is the correct explanation of A
0%
Both A and R are true but R is not a correct explanation of A
0%
A is true but R is false
0%
A is false but R is true

In the hydrogen peroxide molecule :

0%
two hydrogen atoms are connected to one of the oxygen
0%
all the four atoms are in the same plane
0%
the four atoms are arranged in a non-linear and non-planar manner
0%
$O-H$ bonds are polar but molecule is non-polar

Explanation

Hydrogen peroxide

$(H_2O_2)$ is a nonplanar molecule with (twisted) symmetry. It is non–linear, non–planar molecule with a open book structure. The

$–O–O–$ linkage is called peroxy linkage. The structure is given in the figure above.

In case of

$H_{2}O_{2}$ in solid state, the angle between the planes containing

$H-$ atoms is:

0%
$100^{o}$
0%
$90^{o}$
0%
$109^{o}28^{'}$
0%
$180^{o}$

$H_{2}O_{2}$ molecule the O-O bond length is:

0%
$1.34 \mathring{A}$
0%
$1.48 \mathring{A}$
0%
$1.54 \mathring{A}$
0%
$1.20 \mathring{A}$

What is the solution used for the preparation of hydrogen peroxide by electrolytic procedure?

0%
$0.2 N NaOH$
0%
$50$ % $H_{2}SO_{4}$
0%
$20$ % $NaOH$
0%
$5$ % $NaCl$

$H_{2}O_{2}$ , the degree of hydrogen bonding is:

0%
same as in water
0%
more than in water
0%
less than in water
0%
zero

Explanation

Due to the presence of two oxygen atoms in

$H_2O_2$ , there would be four hydrogen bonds corresponding to each molecule. Whereas, in water there would be only three hydrogen bonds per molecule.

The angle between the planes of

$H_{2}O_{2}$ molecule in gaseous phase is:

0%
$101.5^{o}$
0%
$90^{o}$
0%
$111.5^{o}$
0%
$109^{o}28^{'}$

Explanation

$C$ is the correct option which can be seen from the diagram.

Which of the following is a linear molecule?

0%
$H_{2}O$
0%
$H_{2}O_{2}$
0%
$D_{2}O$
0%
$N_{2}O$

Explanation

$\textbf{Hint: The shape of a molecule depends on the number of lone pairs and the bond pairs present}\\\;\;\;\;\;\;\;\;\;\;\;\textbf{around the central atom.}$

$\textbf{Step-1: Find shape of }$ $\mathbf{H_2O}$

In $H_2O$ , the oxygen atom is the central atom which is surrounded with two hydrogen atoms and hence will

have two bond pairs. The oxygen atoms will also consist of two lone pairs in $$H_2O$.

Hence the shape of $H_2O$ will be bent and the geometry will be tetrahedral.

$\textbf{Step-2: Find shape of }$ $\mathbf{H_2O_2}$

A $H_2O_2$ molecule consists of two hydrogen atoms and two oxygen atoms. Each oxygen atom in $H_2O_2$ is

surrounded with two other atoms and two lone pairs. Hence the shape around each oxygen atom

in $H_2O_2$ will be bent.

$\textbf{Step-3: Find shape of }$ $\mathbf{D_2O}$

In $D_2O$ , the oxygen atom is the central atom which is surrounded with two deuterium

atoms and hence will have two bond pairs. The oxygen atoms will also consist of two lone

pairs in $D_2O$ . Hence the shape of $D_2O$ will be bent and the geometry will be tetrahedral.

$\textbf{Step-4: Find shape of }$ $\mathbf{N_2O}$

$N_2O$ , has below structure. The central nitrogen atom is forming two $\mathbf{\sigma}$ bonds and two

$\mathbf{\pi}$ bonds. So it is $\mathbf{sp}$ hybridized. So the shape is linear..

Hence, the correct answer is option $\textbf{(D)}$ $N_2O$ .

Hydrogen peroxide has a:

0%
linear structure
0%
closed chain structure
0%
closed book structure
0%
open book structure

Explanation

The structure of

$H_2O _2$ is not planar but open book structure. The O-O-H bond angle is 101.5

$^o$ and the O-O single bond distance is 1.48 A

$^o$ .

Hence, option D is correct.

Statement-1: Both sodium and hydrogen have outer electronic configuration

$ns^{1}$ but hydrogen is less reactive than sodium.
Statement-2: Hydrogen has greater ionisation energy.

0%
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1
0%
Statement-1 is true, Statement-2 is false
0%
Statement-1 is false, Statement-2 is true

Explanation

Both sodium and hydrogen is having same outer electronic configuration

$ns^1$ but hydrogen have less reactivity because the ionization potential of

$Na$ is less. Hence, it can loose its electron easily and participates in chemical reaction easily, so reason is the correct explanation for assertion so option A would be correct.

$30$ volume hydrogen peroxide means:

0%
$30\%\:H_{2}O_{2}$ solution
0%
$30\:\text{cm}^{3}$ of the solution contains $1$ g of $H_{2}O_{2}$
0%
$1\:\text{cm}^{3}$ of the solution liberates $30\:\text{cm}^{3}$ of $O_{2}$ at STP
0%
$30\:\text{cm}^{3}$ of the solution contains $1$ mole of $H_{2}O_{2}$

Explanation

$30$ volume hydrogen peroxide means,

$1$ ml of hydrogen peroxide solution liberates

$30$ ml of oxygen gas at STP

$1$ ml

$= 1\:\text{cm}^3$

Option C is correct.

One ml of

$H_{2}O_{2}$ solution gives

$50$ ml of

$O_{2}$ at NTP, so it is:

0%
$10$ vol
0%
$25$ vol
0%
$50$ vol
0%
$100$ vol

Explanation

$1$ ml of hydrogen peroxide gives

$50$ ml of oxygen at NTP means,

$50$ volume hydrogen peroxide.

$3.4$ g of

$H_{2}O_{2}$ decomposes, the weight of oxygen liberated from it is:

0%
$1.6$ g
0%
$2.24$ g
0%
$1.16$ g
0%
$3.2$ g

Explanation

$2{ H }_{ 2 }{ O }_{ 2 } \rightarrow 2{ H }_{ 2 }O +{ O }_{ 2 }$ .

$68$ g of

${ H }_{ 2 }{ O }_{ 2 }$ produce ---

$32$ g of

${ O }_{ 2 }$

$3.4$ g of

${ H }_{ 2 }{ O }_{ 2 }$ produce ---

$\dfrac { 32 }{ 68 } \times 3.4 = 1.6$ g of

${ O }_{ 2 }$ .

The volume of

$20\:V-H_{2}O_{2}$ solution, required to prepare

$5$ litres of oxygen at STP is:

0%
$125$ ml
0%
$250$ ml
0%
$100$ ml
0%
$50$ ml

Explanation

$20$ ml of

${ O }_{ 2 }$ is prepared by

$1$ ml of

${ H }_{ 2 }{ O }_{ 2 }$ at STP.

$1$ ml of

${ O }_{ 2 }$ is prepared by

$= \dfrac{1}{20}$ ml of

${ H }_{ 2 }{ O }_{ 2 }$ .

$5000$ ml of

${ O }_{ 2 }$ is prepared by =

$\dfrac { 1 }{ 20 } \times 5000 = 250$ ml of

${ H }_{ 2 }{ O }_{ 2 }$ .

Hence, option B is correct.

The volume of perhydrol which on decomposition gives 1.5 lit of

$O_{2}$ gas at STP is:

0%
$25$ ml
0%
$15$ ml
0%
$10$ ml
0%
$20$ ml

Explanation

As, it is a perhydrol sample,

$1$ ml of

$H_{2}O_{2}$ would give

$100$ ml of

$O_{2}$ at room temperature. Thus,

$1.5$ L of

$O_{2}$ would be given by

$15$ ml of

$H_{2}O_{2}$ .

The volume strength of

$15\%$ (w/v) solution of

$H_{2}O_{2}$ is :

0%
$50$
0%
$4.94$
0%
$494$
0%
$0.494$

Explanation

The solution is

$15\% (w/v)$ . It means

$100 \ ml$ of solution contains

$15$ g of hydrogen peroxide.
The concentration in g/L is

$\dfrac {15}{100} \times 1000 =150 g/L$ .
The relationship between the concentration in g/L and the volume strength

$\text{x}$ is
Conc. in g/L

$= \dfrac {680\times \text{(x)}}{224} g/L$

Thus,

$150 = \dfrac {680\times \text{(x)}}{224}$

$\text{x}=50$ . Thus, the volume strength of solution is

$50$ .

The volume (in ml) of oxygen liberated at STP when

$100$ ml of

$15\:V$

$H_{2}O_{2}$ decomposes is:

0%
$150$
0%
$300$
0%
$3000$
0%
$1500$

Explanation

$1$ ml

$H_2O_2$ of

$15$ vol liberates

$15$ ml of

$O_2$ at STP.
Hence,

$100$ ml

$H_2O_2$ of

$15$ vol liberates

$15\times100$ ml of

$O_2$ at STP or

$1500$ ml of

$O_2$ at STP.

Hence, option D is the correct answer.

The volume strength of

$3N\:H_{2}O_{2}$ solution is:

0%
$11.2\:\text{V}$
0%
$8.4\:\text{V}$
0%
$22.4\:\text{V}$
0%
$16.8\:\text{V}$

Explanation

$3\:N$ means

$3$ g eqwt. in

$1$ L solution.

$N=\dfrac{\text{weight}}{\text{eq. wt.}}$ .

$($ eq. wt. for hydrogen peroxide

$= 17$ g

$)$

Therefore,

$3 = \dfrac{\text{wt}}{17}$ or

$\text{wt} = 3 \times17 = 51$ g.

Now, from the equation, we know that

$68$ g of

$H_2O_2$ gives

$22.4$ L of Oxygen.

Therefore,

$51$ g of hydrogen peroxide gives

$22.4 \times \dfrac{51}{68} = 16.8$ L.

Thus, volume strength is

$16.8$ L, because volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.

The

$\%$ by weight of hydrogen in

$H_{2}O_{2}$ is :

0%
$50$
0%
$25$
0%
$6.25$
0%
$5.88$

Explanation

The molecular weight of

$H_2O_2$ is

$34$ g.

It has

$2$ g of

$H_2$ in it.
Therefore, percentage weight of

$H_2$

$=$

$\dfrac{2}{34}\times100 = 5.88\%$

Hence, the percentage weight of hydrogen in

$H_2O_2$ is

$5.88\%$ .

$H_{2}O_{2}$ turns blackened lead paintings to white colour. In this reaction it oxidizes

$PbS$ to

$PbSO_{4}$ . The number of moles of

$H_{2}O_{2}$ needed to oxidize

$0.1$ moles of

$PbS$ is:

0%
1 mole
0%
0.1 mole
0%
0.5 mole
0%
0.4 mole

Explanation

The balanced equation is:

$PbS+4H_{2}O_{2}\rightarrow PbSO_{4}+4H_{2}O$ .

The stiochiometric coefficient of

$PbS$ is

$1$ while that of

$H_{2}O_{2}$ is

$4$ in the balanced reaction.

So, 0.1 mol will be oxidised by 0.4 mol of

$H_2O_2$

The co-ordination number of

$Na$ in solid

$NaH$ is:

0%
$4$
0%
$6$
0%
$8$
0%
$12$

Explanation

Coordination number of

$Na$ in

$NaH$ is

$6$ as its shape is octahedral.

The reaction between

$H_2O_2$ and

$KMnO_4$ is

$2KMnO_4+3H_2SO_4+5H_2O_2 \rightarrow K_2SO_4+2MnSO_4+8H_2O+5O_2$ .

In a reaction excess of

$H_2O_2$ is added to 0.1 mole of acidified

$KMnO_4$ solution. Then the STP volume of

$O_2$ liberated is:

0%
5.6 L
0%
6.6 L
0%
11.2 L
0%
22.4 L

Explanation

$0.1$ mole of

$KMnO_{4}$ produces

$0.25$ moles of

$O_{2}$ which has a volume of 22.4 x 0.25 litres at STP.

The concentration of

$H_{2}O_{2}$ in a solution containing 34 gm in 500ml is given in List I and their corresponding values are given in List II. Then match List I with List II.

List I List II

A) Molarity	1) 68
B) Normality	2) 22.4
C) % w/v	3) 4
D) Volume strength	4) 2
	5) 10

0%
(A)-5, (B)-3, (C)-1, (D)-2
0%
(A)-4, (B)-3, (C)-1, (D)-2
0%
(A)-3, (B)-4, (C)-1, (D)-2
0%
(A)-1, (B)-2, (C)-3, (D)-4

Explanation

(A) The molar mass of hydrogen peroxide is 34 g/mol.

The molarity of 34 g of hydrogen peroxide in 500 ml is

$\displaystyle \frac {34g}{34g/mol} \times \frac {1000mL/L}{500mL} = 2$ M.

(B) The normality of hydrogen peroxide is twice its molarity. It is

$\displaystyle 2 \times 2 = 4$ N.

$\displaystyle \frac {34 g \times 1000mL/L}{500mL} = 68$ %

(D) Volume strength

$\displaystyle = 11.2 \times$ molarity

$\displaystyle = 11.2 \times 2 = 22.4$

Strength of 2N

$H_{2}O_{2}$ solution is approximately:

0%
10 volumes
0%
44.8 volumes
0%
22.4 volumes
0%
11.2 volumes

Explanation

volume strength is referred to the volume of oxygen gas liberated from 1 unit volume of

$2H_2O_2$ solution.

According to the reaction

$2H_2O_2 \rightarrow 2H_2O+O_2$

$1$ mol of

$H_2O_2$ produces

$0.5$ mol or

$11.2\ L$ of

$O_2$ .

The strength of the compound

$=11.2 \times$ molarity of the solution

We have,

$2N$

$H_{2}O_{2}$

$=1M$

$H_{2}O_{2}$

$\therefore$ strength of

$2N$

$H_{2}O_{2}=$ 11.2 volumes.

The number of electrons involved in the manufacture of 1 mole of

$H_{2}O_{2}$ from 50%

$H_{2}SO_{4}$ :

0%
2 moles
0%
3 moles
0%
1 mole
0%
4 moles

Explanation

The equations are:

$2H_{2}SO_{4}\rightarrow 2HSO_{4}^{-}+2H^{+}$

$2HSO_{4}^{-}\rightarrow H_{2}S_{2}O_{8}+2e^{-}$ At anode

$2H^{+}+2e^{-}\rightarrow H_{2}$ At cathode

$H_{2}S_{2}O_{8}+2H_{2}O\rightarrow 2H_{2}SO_{4}+H_{2}O_{2}$

Hence, two moles of electrons are transferred in the formation of 1 mole of

$H_{2}O_{2}$ .

The weight - volume percentage (w/v) of

$10\ V$

$H_{2}O_{2}$ solution is:

0%
$6$ %
0%
$2$ %
0%
$11$ %
0%
$3$ %

Explanation

The volume strength of

$H_2O_2$ can be converted into the weight - volume percentage (w/v) by the following expression:

The weight - volume percentage (w/v)

$= \dfrac {68x}{224}$ %

Where

$x$ represents volume strength which in this example is

$10 V$ .

The weight - volume percentage (w/v)

$= \dfrac {68 \times 10}{224}= 3$ %

Which ionic hydride is stable upto its melting point?

0%
$NaH$
0%
$CaH_{2}$
0%
$LiH$
0%
$BaH_{2}$

Explanation

All ionic hydrides of group

$I$ and group

$II$ decompose before their melting point except

$LiH$ .

$LiH$ is stable upto its melting point.
The ionic character of the bonds in hydrides increases from

$LiH$ to

$CsH$ and their stability decreases in the same order.

Weight of

$H_{2}O_{2}$ present in

$560$ ml of

$20$ volume

$H_{2}O_{2}$ solution is approximately:

0%
69 g
0%
34 g
0%
32 g
0%
3.4 g

Explanation

Volume strength is referred to the volume of oxygen gas liberated from 1 mL volume of

$2H_2O_2$ solution.

According to the reaction

$2H_2O_2 \rightarrow 2H_2O+O_2$

$1\ mol$ (

$34\ g$ ) of

$H_2O_2$ produces

$0.5$ mol or

$11200\ mL$ of

$O_2$ .

$20\ mL$ oxygen will be obtained from :

$\frac {34}{11200} \times 20=0.061\ g$

Therefore,

$0.061g$ of

$H_2O_2$ in

$1\ mL$ gives 20 volume

$2H_2O_2$

Therefore in

$560\ mL$ solution, amount of

$H_2O_2$ present =

$560 \times 0.061=34.16g$

Option B is correct.

The volume strength of

$1N$

$H_{2}O_{2}$ solution is:

0%
$11.2\: V$
0%
$22.4\: V$
0%
$1\: V$
0%
$5.6\: V$

Explanation

$1 N$ means

$1 g$ equivalent weight in

$1 L$ solution.

$N= \dfrac{weight}{eq.wt}$ (eq. wt for hydrogen peroxide

$=17g$ )

Therefore,

$1 = \dfrac{wt}{17}$ or

$wt = 1\times 17 = 17 g$

Now, from the reaction, we know that

$68 g$ of

$H_2O_2$ gives

$22.4 L$ of Oxygen.

Therefore,

$17 g$ of hydrogen peroxide gives

$\dfrac{22.4 \times 17}{68} = 5.6 L$ .

Thus, the volume strength is

$5.6L$ because volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.

$50$ ml of a certain sample of

$H_{2}O_{2}$ gives

$1200$ ml of

$O_{2}$ at STP. The volume strength of

$H_{2}O_{2}$ is:

0%
22.4 V
0%
20 V
0%
24 V
0%
11.2 V

Explanation

It is given that

$50$ ml of a certain sample of

$H_2O_2$ gives

$1200$ ml of

$O_2$ at STP.

Volume strength is the volume of oxygen at STP given by one volume sample of hydrogen peroxide on heating.

$50$ ml of

$H_2O_2$ gives

$1200$ ml of

$O_2$ , then volume strength will be

$= \dfrac {1200}{50}= 24\ V$

Hence, the volume strength of

$H_2O_2$ is

$24\ V$ .

Elements of which of the following groups do not form hydrides?

0%
$1,2,3$
0%
$3,4,5$
0%
$7,8,9$
0%
$6,7,8$

Explanation

The region of the periodic table from the group

$7$ to

$9$ is referred to as the hydride gap as they do not form hydrides.

Examples of such elements are Mn, Fe, Co, Ru etc. These elements do not form hydrides on account of low affinity for hydrogen in their normal oxidation states.

Hence, option C is correct.

Statement-1:

$CaH_{2}$ is used as main source to transport gas as 1 pound of it gives 16.5 cubic feet

$H_{2}$ .
Statement-2: It is difficult to transport

$H_{2}$ gas.

0%

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.
0%
Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1.
0%

Statement-1 is true, Statement-2 is false.
0%
Statement-1 is false, Statement-2 is true.

Explanation

Statement-1:

$CaH_{2}$ (also called hydrolith is used as main source to transport gas as 1 pound of it gives 16.5 cubic feet

$H_{2}$ .
Statement-2: It is difficult to transport

$H_{2}$ gas due to safety considerations as hydrogen is highly inflammable..

Hence, statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

An excess of acidified solution of

$KI$ is added to a

$25$ mL

$H_2O_2$ solution. The iodine liberated requires

$20$ mL of

$0.3\ N\ Na_2S_2O_3$ solution. Calculate the volume strength of

$H_2O_2$ :

0%
$1.144$
0%
$1.244$
0%
$1.344$
0%
$1.444$

Explanation

$I_{2}+2Na_{2}S_{2}O_{3}\rightarrow Na_{2}S_{4}O_{6}+2NaI$

$20$ ml of

$0.3M(=0.3N)$

$Na_{2}S_{2}O_{3}$ will react with

$\dfrac {(20)(0.3)}{(1000)(2)}$ moles of

$I_{2}$ which is

$0.003$ moles or

$0.006$ equivalents. Thus there are

$0.006$ equivalents in

$25$ ml of

$H_{2}O_{2}$ , hence, its normality is

$0.24$ N and thus, the volume strength is

$0.24 \times 5.6 =1.344$ .

Assertion: Reducing power of nascent hydrogen is more than atomic hydrogen.
Reason: Bond energy of hydrogen is maximum.

0%
Both Assertion & Reason are correct and Reason is correct explanation of Assertion
0%
Both Assertion & Reason are correct but Reason is not correct explanation of Assertion
0%
Assertion is correct but Reason is incorrect
0%
Assertion is wrong but Reason is correct

Excess of

$KI$ and dil.

$H_2SO_4$ , were mixed in

$50$ ml

$H_2O_2$ . The

$I_2$ liberated required

$20$ ml of

$0.1$ N

$Na_2S_2O_3$ . Find out the strength of

$H_2O_2$ in

$g/lit^{-1}$ ?

0%
$0.17 g/lit$
0%
$0.34 g/lit$
0%
$0.68 g/lit$
0%
$1.36 g/lit$

Explanation

$I_{2}+2Na_{2}S_{2}O_{3}->Na_{2}S_{4}O_{6}+2NaI$
20ml of 0.1M

$Na_{2}S_{2}O_{3}$ will react with

$\dfrac {(20)(0.1)}{(1000)(2)}$ moles of

$I_{2}$ which is 0.001 moles or 0.002 equivalents. Thus there are 0.002 equivalents in 50ml of

$H_{2}O_{2}$ . Hence, its strength is 0.002 x 17/0.05=0.68

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Hydrogen Quiz 2 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

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