NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen 2021

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

Edited By Sumit Saini | Updated on Sep 10, 2022 05:35 PM IST

NCERT exemplar Class 11 Chemistry solutions chapter 9 provides the students with a compact and efficiently designed format for studying about the copious amounts of facts and information available about the element hydrogen which in many ways is the most important element in the periodic table and is essential for the students to be well versed with in order to have a strong foundation in the subject and implementing the knowledge learned about hydrogen. NCERT Exemplar Class 11 Chemistry chapter 9 solutions are around hydrogen and its various peculiar properties which explain why the element reacts in certain ways in chemical reactions and how its isotopic and chemical properties pave the way for the other elements in reactions.
Also read - NCERT Solutions for Class 11 Chemistry Chapter 9 Hydrogen

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 1)
Question:1
NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 2)
NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Short Answer Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Matching Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Assertion and Reason Type
NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Long Answer Type
Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen
What The Students Will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen
NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise
Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 1)

Question:1

Hydrogen resembles halogens in many respects for which several factors are responsible. Of the following factors which one is most important in this respect?
(i) Its tendency to lose an electron to form a cation.
(ii) Its tendency to gain a single electron in its valence shell to attain stable electronic configuration.
(iii) Its low negative electron gain enthalpy value.
(iv) Its small size.

Answer:

The answer is the option (ii) Its tendency to gain a single electron in its valence shell to attain stable electronic configuration.
Hydrogen resembles halogens as halogens with the configuration of $ns^{2} np^5$ in the seventeenth group, also harbour the ability to accept one electron and configure as an inert gas. Similarly, Hydrogen also can accept an electron to configure itself like helium.

$H \rightarrow He$
$1s^1$ $1s^2$

Question:2

Why does $H^+$ ion always get associated with other atoms or molecules?
(i) Ionisation enthalpy of Hydrogen resembles that of alkali metals.
(ii) Its reactivity is similar to halogens.
(iii) It resembles both alkali metals and halogens.
(iv) Loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to the small size, it cannot exist free.
Answer:

The answer is the option (iv) Loss of an electron from hydrogen atom results in a nucleus of very small size as compared to other atoms or ions. Due to the small size, it cannot exist freely.
A positive hydrogen ion is extremely small in size and cannot exist as a single alone atom but instead only exists as an association with other elements.

$H \rightarrow H^{+} + e^{-}$

Question:3

Metal hydrides are ionic, covalent or molecular in nature. Among LiH, NaH, KH, RbH, CsH, the correct order of increasing ionic character is
$(i) LiH > NaH > CsH > KH>RbH$
$(ii) LiH < NaH < KH < RbH < CsH$
$(iii) RbH > CsH > NaH > KH > LiH$
$(iv) NaH > CsH > RbH > LiH > KH$
Answer:

The answer is the option $(ii) LiH < NaH < KH < RbH < CsH$
This happens due to the reason that as the atomic size increases, the ionic; electropositive character of the metal hybrides also increases.

Question:4

Which of the following hydrides is electron-precise hydride?
$(i) B_{2}H_{6}$
$(ii) NH_{3}$
$(iii) H_{2}O$
$(iv) CH_{4}$
Answer:

The answer is the option $(iv) CH_{4}$
$CH_{4}$ has the precise number of electrons to form normal covalent bonds and is a precise electron hydride, in order to figure their own Lewis structures.

Question:5

Radioactive elements emit $\alpha, \beta \;and\; \gamma$ rays and are characterised by their half-lives. The radioactive isotope of Hydrogen is
(i) Protium
(ii) Deuterium
(iii) Tritium
(iv) Hydronium
Answer:

The answer is the option (iii) Tritium
Tritium is radioactive as the neutron-proton ratio is more than 1.5, and tritium has n=2 and p=1.

Question:6

Consider the reactions

(A) $H_{2}O_{2}+2HI\rightarrow I_{2}+2H_{2}O$ (B) $HOCl+H_{2}O_{2}\rightarrow H_{3}O^{+}+Cl^{-}+O_{2}$
Which of the following statements is correct about $H_{2}O_{2}$ with reference to these reactions? Hydrogen peroxide is ________.
(i) an oxidising agent in both (A) and (B)
(ii) an oxidising agent in (A) and reducing agent in (B)
(iii) a reducing agent in (A) and oxidising agent in (B)
(iv) a reducing agent in both (A) and (B)
Answer:

The answer is the option (ii) an oxidizing agent in (A) and reducing agent in (B)

$H_{2}O_{2}+2HI\rightarrow I_{2}+2H_{2}O$
$HOCl+H_{2}O_{2}\rightarrow H_{3}O^{+}+Cl^{-}+O_{2}$

Hydrogen peroxide in (A) is an oxidising agent as the oxygen in equation (a) gets reduced from $H_{2}O_{2} \;to\; H_{2}O$

$H_{2}O_{2}+2HI\rightarrow I_{2}+2H_{2}O$
+1 -1 +1 -1 0 +1 -2
Hydrogen peroxide in (B) is a reducing agent as oxygen increases from $H_{2}O_{2} \;to\;O_{2}$

Question:7

The oxide that gives $H_{2}O_{2}$ on treatment with dilute $H_{2}SO_4$ is —
$(i) PbO_{2}$
$(ii) BaO_{2}. 8H_{2}O + O_{2}$
$(iii) MnO_{2}$
$(iv) TiO_{2}$
Answer:

The answer is the option $(ii) BaO_{2}. 8H_{2}O + O_{2}$
This oxide has peroxide linkage $(O^{2-}_{2})$ when reacted with dilute $H_{2}SO_4$ , it produces $H_2O$ , however, dioxides do not produce the same products and the metal atom doesn’t give out water on treatment with dilute $H_{2}SO_4$ .

Question:8

Which of the following equations depict the oxidising nature of $H_{2}O_{2}$ ?

$(i)2MnO_{4}^{-}+6H^{+}+5H_{2}O_{2}\rightarrow 2Mn^{2+}+8H_{2}O+5O_{2}$
$(ii)2Fe^{3+}+2H^{+}+H_2O_{2}\rightarrow 2Fe^{2+}+2H_{2}O+O_{2}$
$(iii)2I^{-}+2H^{+}+H_{2}O_{2}\rightarrow I_{2}+2H_{2}O$
$(iv)KIO_{4}+H_{2}O_{2}\rightarrow KIO_{3}+H_{2}O+O_{2}$

Answer:

The answer is the option $(iii)2I^{-}+2H^{+}+H_{2}O_{2}\rightarrow I_{2}+2H_{2}O$
Iodine ions have a negative charge and are oxidised to form $I_{2}$ . Therefore, $H_{2}O_{2}$ is the oxidizing agent.

Question:9

Which of the following equation depicts reducing nature of $H_2O_2$ ?
$(i) \;2\left [ Fe(CN) _{6}\right ]^{4-}+2H^{+} + H_{2}O_{2}\rightarrow 2\left [Fe \left ( CN \right )_{ 6}\right ]^{3-}+2H_{2}O$
$(ii)I_{2}+H_{2}O_{2}+2OH^{-}\rightarrow 2I^{-}+2H_{2}O+O_{2}$
$(iii)Mn^{2+}+H_{2}O_{2}\rightarrow Mn^{4+}+2OH^{-}$
$(iv)PbS+4H_{2}O_{2}\rightarrow PbSO_{4}+4H_{2}O$

Answer:

The answer is the option $(ii)I_{2}+H_{2}O_{2}+2OH^{-}\rightarrow 2I^{-}+2H_{2}O+O_{2}$
Iodine in the form of $I_{2}$ is reduced to $I^{-}$ and thus $H_2O_2$ acts as the reducing agent.

Question:10

Hydrogen peroxide is _________.
(i) an oxidising agent
(ii) a reducing agent
(iii) both an oxidising and a reducing agent
(iv) neither oxidising nor reducing agent
Answer:

The answer is the option (iii) both an oxidizing and a reducing agent
We have seen examples of $H_2O_2$ acting as both the reducing and oxidising agent.

Question:11

Which of the following reactions increases the production of dihydrogen from synthesis gas?
$(i)CH_{4}(g)+H_{2}O (g)\xrightarrow[Ni]{1270K} CO(g)+3H_{2}(g)$
$(ii)C(s)+H_{2}O (g)\overset{1270K}{\rightarrow}CO(g)+H_{2}(g)$
$(iii)CO(g)+H_{2}O (g)\xrightarrow[catalyst]{673K}CO_{2}(g)+H_{2}(g)$
$(iv)C_{2}H_{6}+2H_{2}O \xrightarrow[Ni]{1270K}2CO+5H_{2}$
Answer:

The answer is the option $(iii)CO(g)+H_{2}O (g)\xrightarrow[catalyst]{673K}CO_{2}(g)+H_{2}(g)$
Carbon monoxide can be reacted with syngas mixtures using steam and iron chromate as enhancers to produce extra dihydrogen.

Question:12

When sodium peroxide is treated with dilute sulphuric acid, we get ______.
(i) sodium sulphate and water
(ii) sodium sulphate and oxygen
(iii) sodium sulphate, Hydrogen and oxygen
(iv) sodium sulphate and hydrogen peroxide
Answer:

The answer is the option (iv) sodium sulphate and hydrogen peroxide.
$Na_{2}O_{2} + H_{2}SO_{4}\rightarrow Na_{2}SO_{4} + H_{2}O_{2}$

Question:13

Hydrogen peroxide is obtained by the electrolysis of ______.
(i) water
(ii) sulphuric acid
(iii) hydrochloric acid
(iv) fused sodium peroxide
Answer:

The answer is the option (ii) sulphuric acid
$H_{2}SO_{4}\rightarrow H^{+} + HSO^{-4}$
$2 HSO{-4} \rightarrow on electrolysis HO_{3}SOOSO_{3}�H + 2e^{-}$
$HO_{3}SOOSO_{3}�H + 2H_2O (hydrolysis) \rightarrow 2 H_{2}SO_{4} + H_{2}O_{2}$

Question:14

Which of the following reactions is an example of the use of water gas in the synthesis of other compounds?

$(i)CH_{4}(g)+H_{2}O (g)\xrightarrow[Ni]{1270K} CO(g)+H_{2}(g)$
$(ii)CO(g)+H_{2}O (g)\xrightarrow[catalyst]{673K}CO_{2}(g)+H_{2}(g)$
$(iii)C_{n}H_{2n+n}+nH_{2}O(g)\xrightarrow[Ni]{1270K}nCO+(2n+1)H_{2}$
$(iv)CO(g)+2H_{2}(g)\xrightarrow[catalyst]{cobalt}CH_{3}OH(l)$
Answer:

The answer is the option $(iv)CO(g)+2H_{2}(g)\xrightarrow[catalyst]{cobalt}CH_{3}OH(l)$
Methanol used water gas for synthesis of other compounds.

Question:15

Which of the following ions will cause hardness in the water sample?
$(i) Ca^{2+}$
$(ii) Na^{+}$
$(iii) Cl^{-}$
$(iv) K^{+}$
Answer:

The answer is the option $(i) Ca^{2+}$
These ions often in the common forms of $Ca (HCO_3)$ or $CaCl_2$ will have an effect on the water by making it harsh and hard as the salts of calcium are soluble in water making it hard.

Question:16

Which of the following compounds is used for water softening?
$(i) Ca_3 (PO_4)_2$
$(ii) Na_{3}PO_{4}$
$(iii) Na_{6}P_{6}O_{18}$
$(iv) Na_{2}HPO_{4}$
Answer:

The answer is the option $(iii) Na_{6}P_{6}O_{18}$
Commercially known as Calgon, sodium hexametaphosphate is used to treat water and make it soft.

$2CaCl_{2} + Na_{2}[Na_{4}(P0_{3})_{6}] \rightarrow Na_{2}[Ca_{2}(P0_{3})_{6}] + 4NaCl$

Question:17

Elements of which of the following group(s) of periodic table do not form hydrides.
(i) Groups 7, 8, 9
(ii) Group 13
(iii) Groups 15, 16, 17
(iv) Group 14
Answer:

The answer is the option (i) Group 7, 8, 9
These elements are unable to form hydrides.

Question:18

Only one element of ________ forms hydride.
(i) group 6
(ii) group 7
(iii) group 8
(iv) group 9

Answer:

The answer is the option (i) Group 6
Only Chromium (Cr) is the element in group 6, capable of forming a hydride.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 2)

Question:19

Which of the following statements are not true for Hydrogen?
(i) It exists as a diatomic molecule.
(ii) It has one electron in the outermost shell.
(iii) It can lose an electron to form a cation which can freely exist
(iv) It forms a large number of ionic compounds by losing an electron.
Answer:

The answer is the option (iii) & (iv), i.e. it can lose an electron to form a cation which can freely exist, and it forms a large number of ionic compounds by losing an electron.
Hydrogen cannot form ionic compounds by giving an electron but instead forms many covalent bonds by the sharing of electrons.

Question:20

Dihydrogen can be prepared on a commercial scale by different methods. In its preparation by the action of steam on hydrocarbons, a mixture of CO and H₂ gas is formed. It is known as ____________.
(i) Water gas
(ii) Syngas
(iii) Producer gas
(iv) Industrial gas
Answer:

The answer is the option (i) and (ii), i.e. water gas and syngas.
Synthesis gas is a combination of CO and H₂, also known as water gas.

Question:21

Which of the following statement(s) is/are correct in the case of heavy water?
(i) Heavy water is used as a moderator in a nuclear reactor.
(ii) Heavy water is more effective as a solvent than ordinary water.
(iii) Heavy water is more associated than ordinary water.
(iv) Heavy water has a lower boiling point than ordinary water.
Answer:

The answer is the option (i) and (iii), i.e. Heavy water is used as a moderator in a nuclear reactor, and Heavy water is more associated than ordinary water.
It is associated with water as it has a higher mass as well as facilitates as a moderator in exchange reactions.

Question:22

Which of the following statements about Hydrogen are correct?
(i) Hydrogen has three isotopes, of which protium is the most common.
(ii) Hydrogen never acts as cation in ionic salts.
(iii) Hydrogen ion, $H^+$ , exists freely in solution.
(iv) Dihydrogen does not act as a reducing agent.
Answer:

The answer is the option (i) and (ii), i.e. Hydrogen has three isotopes of which protium is the most common and Hydrogen never act as a cation in ionic salts.
Protium, one of Hydrogen’s 3 isotopes is the most common as well as due to the small atomic size, it doe does not act as a cation but is associated with other molecules and compounds.

Question:23

Some of the properties of water are described below. Which of them is/are not correct?
(i) Water is known to be a universal solvent.
(ii) Hydrogen bonding is present to a large extent in liquid water.
(iii) There is no hydrogen bonding in the frozen state of water.
(iv) Frozen water is heavier than liquid water.
Answer:

The answer is the option (iii) and (iv). There is no hydrogen bonding in the frozen state of water, and Frozen water is heavier than liquid water.
Water exhibits different properties in different states due to hydrogen bonding, which exists in a water molecule. Ice is lighter than water due to the fact that there are empty spaces in the tetrahedrons of the hydrogen bonds.

Question:24

Hardness of water may be temporary or permanent. Permanent hardness is due to the presence of
(i) Chlorides of Ca and Mg in water
(ii) Sulphates of Ca and Mg in water
(iii) Hydrogen carbonates of Ca and Mg in water
(iv) Carbonates of alkali metals in water
Answer:

The answer is the option (i) and (ii) Chlorides of Ca and Mg in water and sulphate of Ca and Mg in water.
Two salts of calcium and magnesium when found as compounds of carbonate, chloride and sulphate dissolve in the water and make it hard.

Question:25

Which of the following statements is correct?
(i) Elements of group 15 form electron deficient hydrides.
(ii) All elements of group 14 form electron precise hydrides.
(iii) Electron precise hydrides have tetrahedral geometries.
(iv) Electron rich hydrides can act as Lewis acids.
Answer:

The answer is the option (ii) and (iii) All elements of group 14 form electron precise hydrides and electron precise hydrides have tetrahedral geometries.
All group 14 elements are tetrahedral in terms of geometry and form their own Lewis structures. They are electron precise hydrides which have enough number of electrons.

Question:26

Which of the following statements is correct?
(i) Hydrides of group 13 act as Lewis acids.
(ii) Hydrides of group 14 are electron deficient hydrides.
(iii) Hydrides of group 14 act as Lewis acids.
(iv) Hydrides of group 15 act as Lewis bases.
Answer:

The answer is the option (i) and (iv), i.e. Hydrides of group 13 act as Lewis acids and Hydrides of group 15 act as Lewis bases.
All the elements of group 13 are hydrides which act as Lewis acids since they form electron-deficient compounds. The group 14 elements are known as electron-rich hydrides and thus have extra electrons which are present in the form of lone pairs. Groups 15 to 17 have elements which form compounds that have 1 to 3 lone pairs and therefore act as Lewis bases.

Question:27

learn.careers360.com/ncert/question-which-of-the-following-statements-is-correct-i-metallic-hydrides-are-deficient-of-hydrogen/

(i) Metallic hydrides are deficient of Hydrogen.
(ii) Metallic hydrides conduct heat and electricity.
(iii) Ionic hydrides do not conduct electricity in solid-state.
(iv) Ionic hydrides are very good conductors of electricity in solid-state.

Answer:

The answer is the option (i), (ii), and (iii) Metallic hydrides are deficient of Hydrogen, Metallic hydrides conduct heat and electricity, and Ionic hydrides do not conduct electricity in solid-state.
Hydride and not volatile or conductive in the solid-state but are crystalline instead. While metallic hydrides are non-stoichiometric hydrides. They only conduct electricity in their molten state.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Short Answer Type

Question:28

How can the production of Hydrogen from water gas be increased by using a water gas shift reaction?
Answer:

The production of water gas is a chemical reaction. It is produced by superheated steam is passed over coal, nickel acts as the catalyst. The by-products are carbon dioxide and Hydrogen.
$CO + H_{2}O + H_{2} \xrightarrow[catalyst]{673K} CO_{2} + 2H_{2}$

Question:29

What are metallic/interstitial hydrides? How do they differ from molecular hydrides?
Answer:

Metallic hydrides are usually formed by d and f block elements. These hydrides are good conductors of heat and electricity; they lack Hydrogen, which makes them non-stoichiometric. These differ from molecular hydrides as they are formed by s, p block elements while the metals of group 7, 8, 9 do not form hydrides. Molecular hydrides are not good conductors of electricity or heat, unlike the metallic hydrides. They are also volatile compounds which have low melting and boiling points, but metallic hydrides, on the other hand, are hard in texture and have a certain metallic lustre.

Question:30

Name the classes of hydrides to which $H_2O$ , $B_2H_6$ and $NaH$ belong.
Answer:

$H_2O$ – Is a molecular hydride which is covalent and electron-rich.
$B_2H_6$ - Is a molecular hydride which is deficient in electrons.
$NaH$ – Ionic hydride

Question:31

If the same mass of liquid water and a piece of ice is taken, then why is the density of ice less than that of liquid water?
Answer:

Ice and water have different structures and since water expands on freezing the volume of the same amount to ice is more than that of liquid water. Therefore, the density of water is much higher than that of ice and therefore, ice will float on water.

Question:32

Complete the following equations:
$(i)PbS(s)+H_{2}O_{2}(aq)\rightarrow$
$(ii)CO(g)+2H_{2} (g)\xrightarrow[catalyst]{cobalt}$
Answer:

$(i)PbS(s)+H_{2}O_{2}(aq)\rightarrow PbSO_{4}+2H_{2}O$
$(ii)CO(g)+2H_{2}(g)\xrightarrow[catalyst]{cobalt} CH_{3}OH$

Question:33

Give reasons:
(i) Lakes freeze from the top towards the bottom.
(ii) Ice floats on water.

Answer:

The lake freezes from top to bottom because the temperature during winter keeps on decreasing, and the movement of water happens in such a way that the cold water is heavier and so it sinks to the bottom. While warm water replaces it by coming on the surface. The process repeats until the temperature decreases below 4 degrees and the lake keep on freezing from top to bottom.
Density of ice is less than water due to its structure which forms empty spaces between the water molecules, 4 atoms of Hydrogen surround 1 of oxygen, and therefore make ice float on water

Question:34

What do you understand by the term ‘auto protolysis' of water? What is its significance?
Answer:

Autoprotolysis is the process when two similar molecules react with each other to produce products that are called ions with Proton transfer. Autoprotolysis of water means the transfer of one Proton from a certain molecule to another. This explains the ability of water act as both acid and base. Therefore, making it amphoteric in nature.
$H_2O (l) + H_2O (l) \overset{reversible}{\rightarrow} H_{3}O^{+} (aq) + OH^{-} (aq)$

Question:35

Discuss briefly de-mineralisation of water by ion exchange resin.
Answer:

Demineralization of water means that all the soluble salts present in water are removed through cation and anion exchange. In the cation exchange process, cations of sodium, calcium and magnesium replace those of Hydrogen. While the anion exchange process exchanges OH. This both combine to produce water.
$H^{+} + OH^{-} \rightarrow H_{2}O$

Question:36

Molecular hydrides are classified as electron-deficient, electron precise and electron-rich compounds. Explain each type with two examples.
Answer:

There are three major types of molecular hydrides.
Electron precise hydrides are the ones that have just enough exact number of electrons to facilitate normal covalent bonds. These are the hydrides which are primarily comprised of group 14 elements: $CH_{4}$ .
Electron deficient hydrides are the ones which do not possess enough number of electrons to facilitate normal covalent bonds. Hydrides of group 13 a prime examples; $B_2H_6$ .
Electron rich hydrides are ones which have an excess number of electrons remaining after normal covalent bonds. Elements of group 15 16 17 are prime examples; $NH_3$ .

Question:37

How is heavy water prepared? Compare its physical properties with those of ordinary water.
Answer:

Heavy water is prepared by the exhaustive electrolysis of water.

S.RNo.	Property	$H_{2}O$	$D_{2}O$
(i)	Molecular mass (g/mol)	18.015	20.027
(ii)	Melting point (K)	273.0	276.8
(iii)	Boiling point (K)	373.0	374.4
(iv)	Density(298)g/cm	1.0000	1.1059
(v)	Enthalpy of vaporization(kJ/mol)	40.6	41.61

Question:38

Write one chemical reaction for the preparation of $D_2O_2$ .
Answer:

$D_{2}SO_{4}$ , when reacted in water over $BaO_{2}$ produces $D_{2}O_{2}$ .
$BaO_{2} + D_{2}SO_{4}\rightarrow BaSO_{4} + D_{2}O_{2}$

Question:39

Calculate the strength of 5 volume $H_{2}O_{2}$ solution.
Answer:

5 volume $H_{2}O_{2}$ solution signifies that 1 L of 5 volume $H_{2}O_{2}$ when decomposed, gives 5L of $O_2$ at NTP.
$H_{2}O_{2} \rightarrow 2H_{2}O + O_2$

$H_{2}O_{2}$ - 68g
$2H_{2}O + O_2$ – 22.4l at NTP
22.4L of $O_2$ at NTP is gotten from $H_{2}O_{2}$ is 68g

5L of $O_2$ at NTP is produced from $H_{2}O_{2}$ = $\frac{68}{22.4}\times5=15.18g$
But 5L of $O_2$ at NTP is from 5 volume of 1L $H_{2}O_{2}$

Strength of $H_{2}O_{2}$ in 5 volume $H_{2}O_{2}$ = 15.18L
% strength of $H_{2}O_{2}$ sol = $\frac{15.18}{1000}\times100=1.518$ %

Question:40

(i) Draw the gas phase and solid phase structure of $H_2O_2$ .
(ii) $H_2O_2$ is a better oxidising agent than water. Explain.
Answer:

(i) The structure of differ $H_2O_2$ s according to its physical state, as explained below: -

ii) $H_2O_2$ is a better oxidising agent than water as it works in both acidic and alkaline mediums. Water as an oxidising agent is reduced to $H_{2}$ and reacts with only active metals with other conditions such as the electrode potential being less than -0.83V.

Question:41

Melting point, enthalpy of vapourisation and viscosity data of $H_{2}O$ and $D_{2}O$ are given below:

Property	$D_{2}O$	$H_{2}O$
Melting point / K	373.0	374.4
Enthalpy of vapourisation at (373 K)/ kJ mol^-1	40.66	41.61
Viscosity/centipoise	0.8903	1.107

On the basis of this data explain in which of these liquids intermolecular forces are stronger?

Answer:

The given properties such as boiling point, enthalpy of vaporization and viscosity are all dependent on the intermolecular forces of the liquids. Water has a lower intermolecular force of attraction as compared to $D_{2}O$ , and thus the values are lower for water.

Question:42

Dihydrogen reacts with dioxygen $(O_{2})$ to form water. Write the name and formula of the product when the isotope of Hydrogen, which has one proton and one neutron in its nucleus is treated with oxygen. Will the reactivity of both the isotopes be the same towards oxygen? Justify your answer.
Answer:

Deuterium (D) is the isotope of Hydrogen which contains one proton and one neutron. Thus, on the reaction of water with deuterium, heavy water (deuterium oxide) is produced. Deuterium’s bond is stronger than the normal hydrogen bond, and thus $H_{2}$ reacts more with deuterium than oxygen.

$2D_{2} + O_{2} \rightarrow (heat) 2D_{2}O$

Question:43

Explain why HCl is a gas and HF is a liquid.
Answer:

F is a stronger electronegative than chlorine, and so forms a stronger bond with Hydrogen than chlorine. In order to break HF bonds, more energy is required as compared to break HCl bonds. This is the reason why HF has a higher boiling point than HCl and is a liquid at room temperature.

Question:44

When the first element of the periodic table is treated with dioxygen, it gives a compound whose solid-state floats on its liquid state. This compound has an ability to act as an acid as well as a base. What products will be formed when this compound undergoes autoionisation ?
Answer:

The first element is Hydrogen, and its molecular form is dihydrogen, so when it reacts with oxygen, it forms water, which is physical solid-state is ice and has a density lower than water, so it floats on water. Water has an amphoteric nature which acts as a base went around acids and vice versa.
$H_{2}O (as\;an\; acid) + NH_{3} \rightarrow NH4^{+} + OH^{-} (aq)$
$H_{2}O (as \;a\; base) + H_2S \rightarrow H_{3}O^{+} + HS^{-}$

Question:45

Rohan heard that instructions were given to the laboratory attendant to store a particular chemical, i.e., keep it in the darkroom, add some urea in it, and keep it away from dust. This chemical acts as an oxidising as well as a reducing agent in both acidic and alkaline media. This chemical is important for use in the pollution control treatment of domestic and industrial effluents.
(i) Write the name of this compound.
(ii) Explain why such precautions are taken for storing this chemical.

Answer:

$H_{2}O_{2}$
The compound is used in many industries such as textile and paper, bleaching agent and is composed of light and dust particles. It is widely employed to curb pollution by reducing it through oxidised action of harmful cyanides and reducing the effluents. It can function as both, an oxidising and reducing agent and thus is stored in the dark, away from dust as these can cause harmful chemical reactions.

Question:46

Give reasons why Hydrogen resembles alkali metals?
Answer:

Hydrogen has a specific unique electronic configuration which is similar to alkali metals but belongs to group 1 in the periodic table. Its strong nature like the alkali metals due to its configuration makes it give one electron to combine to form unipositive ions.

Question:47

Hydrogen generally forms covalent compounds. Give reason.
Answer:

Due to its configuration, Hydrogen also resembles certain halogens, given its ionization enthalpy. Therefore, when acting as a halogen, it forms hydrides by combining with the elements to form a diatomic molecule along with covalent compounds.

Question:48

Why is the ionisation enthalpy of Hydrogen higher than that of sodium?
Answer:

Hydrogen has a higher ionization enthalpy than sodium because sodium has a different configuration that is in the last shell, the electron is in 3s, which if lost, the configuration resembles neon, which is the noble gas. However, with Hydrogen, the case is different, and the electron is in the 1s orbital which on losing doenot amount to attainement of noble gas configuation.

Question:49

The basic principle of a hydrogen economy is transportation and storage of energy in the form of liquid or gaseous Hydrogen. Which property of Hydrogen may be useful for this purpose? Support your answer with the chemical equation if required.
Answer:

Hydrogen is an excellent source of energy. It can be alternate only used as fuel to empower automobiles and generate nuclear energy. It is looked at as a possible option and economy are the run using Hydrogen as a fuel can be called as a hydrogen economy.
Hydrogen is freely abundant and replaceable and so is regarded as a natural source since it also occurs as water, therefore can be largely utilised.
If Hydrogen is combusted, then it produces water and is not polluting for the environment.
A cell containing Hydrogen will give more power than the normal ones and as is looked at as the better alternative.
Hydrogen also acts as a great reducing agent and can be a substitute for the industries in place of carbon.

Question:50

What is the importance of heavy water?
Answer:

Heavy water is composed by electrolysis of water or through the fertilizer industries. It is a the great catalyst in nuclear reactors and can be used to study reactions.

Question:51

Write the Lewis structure of hydrogen peroxide.
Answer:

Question:52

An acidic solution of hydrogen peroxide behaves as an oxidising as well as the reducing agent. Illustrate it with the help of a chemical equation.
Answer:

1. Reactions which justify oxidising nature are-
H₂O₂ + 2KI $\rightarrow$ 2KOH + I₂
2. Reactions which justify Reducing nature are-
H₂O₂ + Cl₂ $\rightarrow$ 2HCl + O₂

Question:53

With the help of suitable examples, explain the property of $H_{2}O_{2}$ that is responsible for its bleaching action?
Answer:

$H_{2}O_{2}$ serves as a great bleaching agent and disinfectant, commercially sold as perhydrol and disintegrates on exposure to direct light.
The bleaching action of H2O2 is due to labile oxygen which it releases on decomposition.
H₂O₂→ H₂O+[O]
The nascent oxygen combines with coloured material and makes it colourless. It can bleach feather, silk, wool, paper, etc.

Question:54

Why is water molecule polar?
Answer:

The polar nature of the water molecule is due to the structure of the bent molecule and bond length of 95.7pm and the angle of 104.5.

Question:55

Why does the water show a high boiling point as compared to hydrogen sulfide? Give reasons for your answer.
Answer:

Water has a higher boiling point than hydrogen sulfide because the hydrogen bonding in water is quite different, and hydrogen sulfide does not have hydrogen bonding at all.

Question:56

Why can dilute solutions of hydrogen peroxide not be concentrated by heating? How can a concentrated solution of hydrogen peroxide be obtained?
Answer:

Hydrogen peroxide causes burns when it is heated, so it should not be concentrated. You should use other methods other than heating to make it concentrated, such as distillation with water under pressure. Further distillation under pressure and careful extraction can give you pure hydrogen peroxide.

Question:57

Why is hydrogen peroxide stored in wax-lined bottles?
Answer:

Hydrogen peroxide is highly reactive, and that is stored in dark coloured bottles because it disintegrates on exposure to direct light.
$2 H_{2}O_{2} \rightarrow 2 H_{2}O + O_{2}$

Question:58

Why does hard water not form lather with soap?
Answer:

Hard water contains soluble salts which make it rough and form a precipitate when used along with soap. Due to harsh chemicals dissolve in it, hard water is unsuitable for washing and laundry. The accumulation of salts makes it unsuitable for boilers is well.

Question:59

Phosphoric acid is preferred over sulphuric acid in preparing hydrogen peroxide from peroxides. Why?
Answer:

Phosphoric acid is preferred as it does not dissolve, but sulphuric acid can react with hydrogen peroxide to disintegrate and decompose.

Question:60

How will you account for $104.5^{\circ}$ bond angle in the water?
Answer:

$Sp^{3}$ hybridisation is present in oxygen along with a 109-degree angle, but the repulsive bond decreases the angle to a 104 degrees.

Question:61

Write redox reaction between fluorine and water?
Answer:

$2F_{2} (g) + 2H_{2}O \rightarrow 4H^{+} (aq) + 4F^{-} (aq) + O_{2}$

Question:62

Write two reactions to explain the amphoteric nature of water?

Answer:

$H_{2}O (l) + NH_{3} \rightarrow OH^{-} (aq) + NH_{4} (aq)$
$H_{2}O (l) + H_{2}S (aq) \rightarrow H_{2}O^{+} + HS^{-}$

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Matching Type

Question:63

Correlate the items listed in Column I with those listed in Column II. Find out as many correlations as you can.

Column I	Column II
(i) Synthesis gas	(a) $Na_{2}\left [ Na_{4}(PO_{3})_{6} \right ]$
(ii) Dihydrogen	(b) Oxidizing agent
(iii) Heavy water	(c) Softening agent
(iv) Calgon	(d) Reducing agent
(v)Hydrogen Peroxide	(e) Stochiometric compounds of s-blocks elements
(vi) Salt - like hydrides	(f) Prolonged electrolysisof water
	(g)Zn + NaOH
	(h) $Zn+dil. H_{2}SO_{4}$
	(i) Synthesis of menthol
	(j) Mixture of CO and H₂

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Answer:

(i) → (i), (j)
(ii) → (d), (e), (g), (h), (i)
(iii) → (f)
(iv) → (a), (c)
(v) → (b), (d)
(vi) → (e)

Question:64

Match Column I with Column II for the given properties/applications mentioned therein.

Column I	Column II
(i) H	(a) Used in the name perhydrol
(ii) $H_{2}$	(b) Can be reduced to dihydrogen by NaH.
(iii) $H_{2}O$	(c) Can be used in hydroformylation of olefin.
(iv) $H_{2}O_{2}$	(d) Can be used in cutting and welding.

Answer:

(i) → (d)
(ii) → (c)
(iii) → (b)
(iv) → (a)

Question:65

Match the terms in Column I with the relevant item in Column II.

Column I	Column II
(i) Electrolysis of waterproduces	(a)atomic reactor
(ii) Lithium aluminium hydride is used as	(b) polar molecule
(iii) Hydrogen chloride is a	(c) recombines on metal surface to generate high temperature
(iv)Heavy water is used as	(d) reducing agent
(v) Atomic hydrogen	(e) Hydrogen and oxygen

Answer:
(i) → (e) (ii) → (d) (iii) → (b) (iv) → (a) (v) → (c)

Question:66

Match the items in Column I with the relevant item in Column II.

Column I	Column II
(i) Hydrogen peroxide is used as	(a)zeolite
(ii)Used in Calgon method	(b) perhydrol
(iii) Permanent hardness of hard water is removed by	(c) Sodium hexametaphosphate
	(d) propellant

Answer:

i. → (b), (d)
ii. → (c)
iii. → (a), (c)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Assertion and Reason Type

Question:67

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the options given below each question.

Assertion (A): Permanent hardness of water is removed by treatment with washing soda.
Reason (R): Washing soda reacts with soluble magnesium and calcium sulphate to form insoluble carbonates.

(i) Statements A and R both are correct, and R is the correct explanation of A.
(ii) A is correct, but R is not correct.
(iii) A and R both are correct, but R is not the correct explanation of A.
(iv) A and R both are false.
Answer:

The answer is the option (i) Statements A and R both are correct, and R is the correct explanation of A.

Question:68

In the following questions, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the options given below each question.
Assertion (A): Some metals like platinum and palladium, can be used as storage media for Hydrogen.
Reason (R): Platinum and palladium can absorb large volumes of Hydrogen.
(i) Statements A and R both are correct, and R is the correct explanation of A.
(ii) A is correct, but R is not correct.
(iii) A and R both are correct, but R is not the correct explanation of A.
(iv) A and R both are false.

Answer:

The answer is the option (i) statements A and R both are correct, and R is the correct explanation of A.

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Long Answer Type

Question:69

Atomic Hydrogen combines with almost all elements, but molecular Hydrogen does not. Explain.
Answer:

Molecular Hydrogen is very stable, but atomic Hydrogen is very reactive, and the chemical behaviour of any molecule is determined by the bond dissociation enthalpy. In dihydrogen, the hydrogen bond dissociation enthalpy is very high, and therefore it only reacts with a certain number of elements.

Question:70

How can $D_2O$ be prepared from water? Mention the physical properties in which $D_2O$ differs from $H_2O$ . Give at least three reactions of $D_2O$ showing the exchange of Hydrogen with deuterium.

Answer:

Extended electrolysis of water can produce D2O; it differs from water as it has a high molecular mass.

		$H_2O$	$D_2O$
1.	Boiling Point	373K	373.4K
2.	Melting Point	273K	276.8K
3.	Molecular Mass	18.016	20.3

Reactions showing the exchange of Hydrogen with deuterium

$NaOH + D_{2}O \rightarrow NaOD + HOD$
$HCl + D_{2}O \rightarrow DCl + HOD$
$NH_{4}Cl + D_{2}O \rightarrow NH_{3}DCl + HOD$

Question:71

How will you concentrate $H_{2}O_{2}$ ? Show differences between structures of $H_{2}O_{2}$ and $H_{2}O$ by drawing their spatial structures. Also mention three important uses of $H_{2}O_{2}$ .
Answer:

To obtain hydrogen peroxide, you can use evaporation and barium peroxide by removing excess water. To make it more concentrated, use the process of distillation, and low pressure and then pure hydrogen peroxide can be obtained.

The spatial structures of $H_{2}O$ and $H_{2}O_{2}$ are as follows: -

Three important uses of $H_{2}O_{2}$ are as follows: -

Peroxide is used as a disinfectant and antiseptic in the market.
It is used to produce other chemicals in the industry and is also applied as a commercial bleaching product.
It finds great application in the textile industry.

Question:72

(i) Give a method for the manufacture of hydrogen peroxide and explain the reactions involved therein.
(ii) Illustrate oxidising, reducing and acidic properties of hydrogen peroxide with equations.
Answer:

Industrial preparation: $H_2O_2$ is prepared by the auto-oxidation of 2-alkylanthraquinols $H_2O (l) + H_2S (aq) \rightarrow$
$2Fe^{2+} + 2H^{+} + H_{2}O \rightarrow 2Fe^{3+} + 2 H_{2}O$
$PbS + H_{2}O_{2} \rightarrow PbSO_{4} + 4 H_{2}O$
(ii) (a) Reducing action in acidic medium
$2MnO^{-}_{4} + 6H^{+} + 5 H_{2}O_{2}\rightarrow 2Mn^{2+} + 8 H_{2}O + 5 O_{2}$
$HOCl + H_{2}O_{2} \rightarrow H_{3}O^{-} + Cl^{-} + O_{2}$
(b)Oxidising action in basic medium
$2Fe^{2+} + H_{2}O_{2} \rightarrow 2Fe^{3+} + 2OH^{-}$
(c)
$I_{2} + H_{2}O_{2} + OH�^{-} \rightarrow 2I^{-} + 2 H_{2}O + O_{2}$

Question:73

What mass of hydrogen peroxide will be present in 2 litres of a 5 molar solution? Calculate the mass of oxygen which will be liberated by the decomposition of 200 mL of this solution.

Answer:

We know that the molecular mass of $H_{2}O_2$ – 34
Mass of $H_{2}O_2$ present in 1 molar solution – 34 g
Therefore, the mass of $H_{2}O_2$ present in 2 litres of 1 molar solution – $2 \times 34 = 68g$
Hence, the mass of $H_{2}O_2$ that is present in 200ml solution of 1 molar $H_{2}O_2$ will be

$= \frac{34}{5}= 6.8g$
$2H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$
68g 32g
6.8g 3.2g

Question:74

A colourless liquid ‘A’ contains H and O elements only. It decomposes slowly on exposure to light. It is stabilised by mixing urea to store in the presence of light.
(i) Suggest possible structure of A.
(ii) Write chemical equations for its decomposition reaction in light.
Answer:

(i) A is Hydrogen Peroxide $(H_{2}O_{2})$ as it decomposes slowly on exposure to light and is stabilised by mixing urea to store in the presence of light.

Its possible structure is as follows: -

(ii) The chemical equations for its decomposition reaction in light are as follows-
$2H_{2}O_{2} (l)\rightarrow 2H_{2}O (l) + O_{2 }(g)$

Question:75

An ionic hydride of an alkali metal has significant covalent character and is almost unreactive towards oxygen and chlorine. This is used in the synthesis of other useful hydrides. Write the formula of this hydride. Write its reaction with $Al_{2}Cl_{6}$ .
Answer:

Since the ionic hydride of alkali metal has a significant covalent character, therefore, it is LiH. Since LiH is very stable, therefore, it is almost unreactive towards O₂?and Cl₂. It reacts with Al₂Cl₆? form lithium aluminum hydride.
$8 \mathrm{LiH}+\mathrm{Al}_{2} \mathrm{Cl}_{6} \rightarrow 2 \mathrm{LiAlH}_{4}+6 \mathrm{LiCl}$

Question:75

The ionic hydride of an alkali metal which has a significant covalent character and is almost unreactive towards oxygen and chlorine at moderate temperatures is, lithium hydride is unreactive. That is why it is used for the synthesis of other useful hydrides.

Its reaction to $Al_{2}Cl_{6}$ can be explained as follows: -

$8LiH + Al_{2}Cl_{6} \rightarrow 2LiAlH_{4} + 6LiCl$

Question:76

Sodium forms a crystalline ionic solid with dihydrogen. The solid is non-volatile and non- conducting in nature. It reacts violently with water to produce dihydrogen gas. Write the formula of this compound and its reaction with water. What will happen on electrolysis of the melt of this solid?
Answer:

Saline hydrides produce a dihydrogen gas during a violent reaction with water. The ionic solid crystal which is formed is non-conducting and non-volatile in nature and reacts aggressively with water for producing the dihydrogen gas.
The formula for this compound is NaH, its reaction with water is explained below: -
$NaH (s) + H_{2}O (aq) \rightarrow NaOH (aq) +H_{2} (g)$
Dihydrogen gas at the anode is liberated during electrolysis, which confirms the existence of H-ion

For students who wish to be able to reach out for queries and NCERT solutions from this chapter at any time so that they develop a strengthened understanding of the topic, there is a solution. Students can avail the option of NCERT Exemplar Class 11 Chemistry solutions chapter 9 PDF download for future studies.
Also read - NCERT Solutions for Class 11 Chemistry

Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

Class 11 Chemistry NCERT Exemplar solutions Chapter 9 include the following topics:

Position Of Hydrogen In The Periodic Table
Dihydrogen, H2
Occurrence
sotopes Of Hydrogen
Preparation Of Dihydrogen, H2
Laboratory Preparation Of Dihydrogen
Commercial Production Of Dihydrogen
Properties Of Dihydrogen
Physical Properties
Chemical Properties
Uses Of Dihydrogen
Hydrides
Ionic Or Saline Hydrides
Covalent Or Molecular Hydride
Metallic Or Non-stoichiometric (Or Interstitial ) Hydrides
Water Ex
Physical Properties Of Water Ex
Structure Of Water
Structure Of Ice
Chemical Properties Of Water
Hard And Soft Water
Temporary Hardness
Permanent Hardness
Hydrogen Peroxide (H202)
Preparation
Physical Properties
Structure
Chemical Properties
Storage
Uses
Heavy Water, D20

Dihydrogen as a Fuel

What The Students Will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

With the detailed and specific contents of Class 11 Chemistry NCERT exemplar solutions chapter 9 the students have the opportunity to have a detailed study of the element hydrogen. This chapter allows the students to extend their horizons and knowledge base about the chemical properties, structure, and physical properties of water which tend to further explain why hydrogen reacts in a specific manner. NCERT exemplar solutions for Class 11 Chemistry chapter 9 also discusses the various forms of existence of hydrogen such as Dihydrogen and its production and chemical and physical properties along with its preparation process. Another important concept discussed is the preparation, physical and chemical properties of hydrogen peroxide. Heavy water and the practical application of dihydrogen as a fuel have also been explored in the chapter.

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Chapter 1 Some Basic Concepts of Chemistry

Chapter 2 Structure of Atom

Chapter 3 Classification of Elements and Periodicity in Properties

Chapter 4 Chemical Bonding and Molecular Structure

Chapter 5 States of Matter

Chapter 6 Thermodynamics

Chapter 7 Equilibrium

Chapter 8 Redox Reactions

Chapter 10 The s-Block Elements

Chapter 11 The p-Block Elements

Chapter 12 Organic Chemistry – Some Basic Principles and Techniques

Chapter 13 Hydrocarbons

Chapter 14 Environmental Chemistry

Also, check

Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

The students who refer to NCERT Exemplar solutions for Class 11 Chemistry chapter 9 get to indulge in the in-depth study of hydrogen and all of its properties which include- occurrence, physical and chemical properties and its different forms of existence which include dihydrogen, hydrogen peroxide etc.
Another area of information discussed is around the position of hydrogen in the periodic table and how it affects and influences the chemical reactions of various elements. The students are informed about the practical applications of all the different forms of hydrogen.
The students will also be educated about the crucial facts about the physical and chemical properties, practical applications, storage and structure of the various forms and isotopes of hydrogen such as dihydrogen, hydrides, etc.

Also, check Chapter wise NCERT Solutions for Class 11 Chemistry

Chapter-1 - Some Basic Concepts of Chemistry

Chapter-2 - Structure of Atom

Chapter-3 - Classification of Elements and Periodicity in Properties

Chapter-4 - Chemical Bonding and Molecular Structure

Chapter-5 - States of Matter

Chapter-6 - Thermodynamics

Chapter-7 - Equilibrium

Chapter-8 - Redox Reaction

Chapter-9 - Hydrogen

Chapter-10 - The S-Block Elements

Chapter-11 - The P-Block Elements

Chapter-12 - Organic chemistry- some basic principles and techniques

Chapter-14 - Hydrocarbons

Chapter-15 - Environmental Chemistry

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NCERT Class 11 Chemistry Chapter 4 Notes Chemical Bonding and Molecular Structure - Download PDF

Hydrogen Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 9 Notes - Download PDF

The S-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 10 Notes - Download PDF

The P-Block Elements Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 11 Notes - Download PDF

Environmental Chemistry Class 11th Notes - Free NCERT Class 11 Chemistry Chapter 14 Notes - Download PDF

NCERT Class 11 Chemistry Chapter 3 Notes - Download PDF

NCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power

NCERT Solutions for Class 11 Chemistry Chapter 5 States of Matter

NCERT Class 11 Physics Chapter 5 Notes Laws of Motion - Download PDF

NCERT Class 11 Physics Chapter 3 Notes Motion in a straight Line - Download PDF

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NCERT Solutions for Miscellaneous Exercise Chapter 16 Class 11 - Probability

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NCERT Exemplar Class 11 Maths Solutions Chapter 10 Straight Lines

NCERT Exemplar Class 11 Maths Solutions Chapter 6 Linear Inequalities

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 1)

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NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: MCQ (Type 2)

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Short Answer Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Matching Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Assertion and Reason Type

NCERT Exemplar Class 11 Chemistry Solutions Chapter 9: Long Answer Type

Topics and Subtopics in NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

What The Students Will Learn From NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

NCERT Exemplar Class 11 Chemistry Solutions Chapter-Wise

Important Topics To Cover From NCERT Exemplar Class 11 Chemistry Solutions Chapter 9 Hydrogen

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