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Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 8 Elements of Group 1 and 2 Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 8 Elements of Group 1 and 2

Question 1.
Why is hydrogen studied separately even though it appears at the top of group 1?
Answer:
Even though hydrogen appears at the top of the group 1 containing alkali metals, it is studied separately because many of its properties differ from that of the alkali metals.

Question 2.
Give reason: Hydrogen (H₂) molecule is also referred to as dihydrogen.
Answer:

• The nucleus of a hydrogen atom consists of one positively charged proton i.e., nuclear charge of +1 and one extranuclear electron.
• As this electron is in direct influence of nuclear attraction, hydrogen has a little tendency to lose this electron.
• However, it can easily pair with the other electron forming a covalent bond.
• Therefore, it exists in diatomic form as H₂ molecule and hence, it is also referred to as dihydrogen.

Question 3.
Why does hydrogen occur in a diatomic form rather than in monoatomic form under normal
conditions?
Answer:

• Hydrogen atom has only one electron in its valence shell having electronic configuration 1s¹.
• It can acquire stable configuration of helium by sharing this electron with another hydrogen atom.
• Therefore, it shares its single electron with electron of the other H-atom to achieve stable inert gas configuration of He.
• Thus, hydrogen readily forms diatomic molecule and exists as H₂ rather than in monoatomic form.

Question 4.
Write a note on occurrence of hydrogen.
Answer:

• In the free state hydrogen exists as dihydrogen gas.
• Hydrogen is most abundant element in the universe and constitutes 70% of the total mass of the universe.
• Hydrogen is also the principal element in the solar system.
• On the earth, hydrogen is the tenth most abundant element on mass basis and the third most abundant element on atom basis.

Question 5.
State whether the following statements are TRUE or FALSE. Correct the false statement.
1. Electronic configuration of hydrogen is 1s¹.
ii. H⁺ ion formed by loss of the electron from hydrogen atom exists freely.
iii. H⁺ is nothing but a proton.
iv. Metastable metallic hydrogen was discovered at Harvard university, USA, in January 2017.
Answer:
i. True
ii. False
Hydrogen atom does not exist freely and is always associated with other molecules i.e., H₃O⁺.
iii. True
iv. True

Question 6.
Explain the laboratory methods for preparation of dihydrogen.
Answer:
Laboratory methods for preparation of dihydrogen:
i. By action of dilute HCl on zinc granules: Zinc granules on reaction with dilute hydrochloric acid (HCl) liberates hydrogen gas.

ii. By action of aqueous NaOH on zinc: Zinc on reaction with aqueous sodium hydroxide (NaOH) forms soluble sodium zincate and liberates hydrogen gas.

Question 7.
Describe the industrial method of preparation of dihydrogen by electrolysis of pure water.
Answer:
i. Pure water is a poor conductor of electricity. Therefore, a dilute aqueous solution of acid or alkali is used to prepare dihydrogen by electrolysis.
ii. For example, electrolysis of dilute aqueous solution of sulphuric acid yields two volumes of hydrogen at cathode and one volume of oxygen at anode.

Question 8.
How is pure dihydrogen (> 99.5% purity) gas obtained from barium hydroxide?
Answer:
Electrolysis of warm aqueous solution of barium hydroxide using nickel electrodes yields pure dihydrogen (> 99.5% purity).

Question 9.
Explain the terms:
i. Syngas
ii. Water-gas shift reaction.
Answer:
i. Syngas:

• Syngas is the mixture of CO and H₂. It is also called ‘water-gas’.
• It is used for the synthesis of CH₃OH and many hydrocarbons, hence, the name syngas or ‘synthesis gas’.
• Production of syngas is also the first stage of gasification of coal.

ii. Water-gas shift reaction:
The carbon monoxide in the water-gas is transformed into carbon dioxide by reacting with steam in presence of iron chromate as catalyst.

This reaction is called water-gas shift reaction.

Question 10.
Complete the following chemical reactions:

Answer:

Question 11.
Enlist physical properties of dihydrogen.
Answer:
Physical properties of dihydrogen:

• Dihydrogen is a colourless, tasteless and odourless gas.
• It bums with a pale blue flame.
• It is a nonpolar and water-insoluble gas.
• It is lighter than air.

Question 12.
What is the action of dihydrogen on the following?
i. Metals
ii. Dioxygen
Answer:
i. Action of dihydrogen on metals:
a. Dihydrogen combines with all the reactive metals including alkali metals, calcium, strontium and barium at high temperature to form metal hydrides.
b. For example: Dihydrogen combines with metallic sodium at high temperature to yield sodium hydride.

ii. Action of dihydrogen on dioxygen: Dihydrogen reacts with dioxygen in the presence of catalyst or by heating to form water. This reaction is highly exothermic.

Question 13.
Explain the effect of high bond dissociation energy of H-H bond on chemical reactivity of dihydrogen?
Answer:

• The bond dissociation energy of H-H bond is very high i.e, 436 kJ mol^-1. and thus, it does not react easily under normal conditions.
• However, at high temperature or in the presence of catalysts, hydrogen combines with many metals and non-metals to form corresponding hydrides and halides respectively.

Question 14.
What happens when dihydrogen reacts with halogens?
Answer:
i. Dihydrogen reacts with halogens (X₂) to give the corresponding hydrogen halides (HX).

ii. Dihydrogen reacts with fluorine to form hydrogen fluoride even at very low temperature (-250°C) in dark.

iii. However, the reaction with iodine requires a catalyst as the vigour of reaction of dihydrogen decreases with increasing atomic number of halogen.

Question 15.
Explain the reducing nature of hydrogen with chemical reactions.
Answer:
Dihydrogen reduces oxides and ions of some metals that are less reactive than iron, to the corresponding number of halogen metals at moderate temperature.
e.g.
i. CuO_(s) + H_2(g) → Cu_(s) + H₂O_(l)
ii. Fe₃O_4(s) + 4H_2(g) → 3Fe_(s) + 4H₂O_(s)
iii. Pd²⁺_(aq) + H_2(g) → Pd_(s) + 2H⁺_(aq)

Question 16.
What is hydrogenation?
Answer:
Hydrogenation is the reaction in which hydrogen gas reacts with unsaturated organic compounds in the presence of a catalyst to form hydrogenated (saturated) compounds.

Question 17.
How does dihydrogen react with various organic compounds to give useful, commercially important products?
Answer:
i. Hydrogenation of unsaturated organic compounds:
e.g. Hydrogenation of unsaturated organic compounds such as vegetable oil using nickel catalyst gives saturated organic compounds such as solid edible fats like vanaspati ghee.

ii. Hydroformylation of olefins and subsequent reduction of aldehyde to form alcohol:

Question 18.
Explain hydroformylation reaction of olefins using a suitable example.
Answer:
Hydroformylation of olefins gives aldehydes which on further reduction gives alcohols.
e.g. i. Hydroformylation of propene gives butyraldehyde.

ii. Butyraldehyde further undergoes reduction to give n-butyl alcohol.

Question 19.
What are the uses of dihydrogen?
Answer:
Dihydrogen is used in

• the production of ammonia.
• the formation of vanaspati ghee by catalytic hydrogenation of oils.
• rocket fuel (mixture of liquid hydrogen and liquid oxygen).
• the preparation of important organic compounds like methanol in bulk quantity.
  \(2 \mathrm{H}_{2(\mathrm{~g})}+\mathrm{CO}_{(\mathrm{g})} \stackrel{\text { Cobaltcatalyst }}{\longrightarrow} \mathrm{CH}_{3} \mathrm{OH}_{(l)}\)
• the preparation of hydrogen chloride (HCl) and metal hydrides.

Question 20.
Justify the placement of hydrogen in the group of alkali metals with the help of reaction with halogens.
Answer:
i. Hydrogen on reaction with halogens (X₂) give compounds with general formula HX.
e.g. H₂ + Cl₂ → 2HCl
ii. Similarly, alkali metals (M) on reaction with halogens (X₂) give compounds with general formula MX.
e.g. 2Na + Cl₂ → 2NaCl
iii. Thus, H₂ and alkali metals are monovalent elements and more electropositive than halogens. This similarity justifies the position of hydrogen in the group 1.

Question 21.
What do you mean by s-block elements? Where are they placed in the modern periodic table?
Answer:

• Elements of group 1 and group 2 in which the last electron enters into ‘ns’ subshell are s-block elements.
• The s-block elements are placed on the extreme left in the modem periodic table.

Question 22.
Name elements of group 1 and group 2.
Answer:

• Group 1 of the periodic table consists of the elements: hydrogen, lithium, sodium, potassium, rubidium, caesium and francium.
• Group 2 of the periodic table consists of elements: beryllium, magnesium, calcium, strontium, barium and radium.

Question 23.
What are alkali metals?
Answer:
The elements of group 1 except hydrogen are collectively called alkali metals.

Question 24.
What are alkaline earth metals?
Answer:
The elements of group 2 are collectively called alkaline earth metals because they occur as minerals in rocks.

Question 25.
Write a note on occurrence of group 1 and group 2 elements:
Answer:
i. Group 1 (alkali metals):

• Two elements of group 1 i.e., sodium and potassium are the sixth and seventh most abundant elements present in the earth’s crust.
• However, francium does not occur appreciably in nature because it is radioactive and has short half-life period.

ii. Group 2 (alkaline earth metals):

• The elements magnesium and calcium are found abundantly in earth’s crust.
• Radium is radioactive and is not easy to find.

Question 26.
Give reasons: s-block elements are never found in free state in nature.
Answer:

• s-Block elements contain group 1 and group 2 elements.
• The general outer electronic configuration of the group 1 elements is ns¹ and that of the group 2 elements is ns².
• The loosely held s-electrons in the valence shell of these elements can be easily removed to form metal ions.
• As a result, they are highly reactive in nature and always found in combined state.

Hence, s-block elements are never found in free state in nature.

Note: Electronic configurations of group 1 elements

Note: Electronic configurations of group 2 elements

Question 27.
Describe the physical properties of alkali and alkaline earth metals.
Answer:

• All the alkali and alkaline earth metals are silvery white in appearance.
• Due to their large atomic size they have low density.
• Both alkali and alkaline earth metals are soft, however, alkaline earth metals are harder than the alkali metals.
• Alkali metals are the most electropositive elements while alkaline earth metals are comparatively less electropositive than alkali metals.

Question 28.
Explain why do the group 1 and group 2 elements form diamagnetic and colourless compounds.
Answer:

• Unipositive ions of all the elements of group 1 have inert gas configuration and hence, they have no unpaired electron.
• Similarly, group 2 elements can lose their two valence shell electrons and form divalent ions that have inert gas configuration with no unpaired electrons.

Hence, due to the absence of unpaired electrons, compounds formed by group 1 and group 2 elements are diamagnetic and colourless.

Question 29.
Why do the properties of lithium and beryllium differ from the rest of the group 1 and group 2 elements?
Answer:
The properties of lithium and beryllium differ from the rest of the group 1 and group 2 elements due to their extremely small size and comparatively high electronegativity.

Question 30.
Complete the following table.

Group 1 elements	Group 2 elements
……………….	Alkaline earth metals
Outer electronic configuration: ……………….	Outer electronic configuration: ns2
Monovalent positive ions	……………….

Answer:

Group 1 elements	Group 2 elements
Alkali metals	Alkaline earth metals
Outer electronic configuration: ns1	Outer electronic configuration: ns2
Monovalent positive ions	Divalent positive ions

Question 31.
State the trends in the following properties of group 1 and group 2 elements down a group.
i. Atomic radii
ii. Ionic radii
iii. Ionization enthalpy
iv. Electronegativity
v. Standard reduction potential
Answer:

Sr. no.	Property	Down a group
i.	Atomic radii	Increases
ii.	Ionic radii	Increases
iii.	Ionization enthalpy	Decreases
iv.	Electronegativity	Decreases
V.	Standard reduction potential	Decreases

Question 32.
Give reasons: Potassium superoxide is used in breathing equipment used for mountaineers and in submarines and space.
Answer:
i. Potassium superoxide has ability to absorb carbon dioxide and give out oxygen at the same time.

ii. Due to this property of KO₂, it is used in breathing equipment used for mountaineers and in submarines and space.

Question 33.
What is the oxidation state of:
i. Na in Na₂O₂?
ii. K in KO₂?
Answer:
i. Oxidation state of Na in sodium peroxide (Na₂O₂):
Let x be the oxidation state of Na in Na₂O₂.
The net charge on peroxide ion \(\left(\mathrm{O}_{2}^{2-}\right)\) is -2.
Since any compound is electrically neutral, it has an overall charge as zero.
∴ 2x + (-2) = 0
∴ x = + 1
∴ Oxidation state of Na in Na₂O₂ is +1.

ii. Oxidation state of K in potassium dioxide/potassium superoxide (KO₂):
Let x be the oxidation state of K in KO₂
The net charge on superoxide ion \(\left(\mathrm{O}_{2}^{-}\right)\) is -1.
Since any compound is electrically neutral, it has an overall charge as zero.
∴ x + (-1) = 0
x = + 1
∴ Oxidation state of K in KO₂ is + 1.
[Note: Oxidation state of alkali metal is always +1.]

Question 34.
Magnesium strip slowly tarnishes on keeping in air but metallic calcium is readily attacked by air. Explain.
Answer:

• The reactivity of group 2 metals increases with increasing atomic radius and lowering of ionization enthalpy
  down the groups.
• Thus, calcium has lower ionization enthalpy. Therefore, calcium is more reactive than magnesium.
• Hence, Mg reacts slowly with air forming a thin film of oxide resulting into tarnishing, whereas Ca reacts readily at room temperature with oxygen and nitrogen in the air.

Question 35.
What happens when alkali metals react with hydrogen and halogens?
Answer:
i. Reaction with hydrogen: Alkali metals react with hydrogen at high temperature to form the Corresponding metal hydrides.

ii. Reaction with halogens: All the alkali metals react vigorously with halogens to produce their ionic halide salts.

[Note: As we move down the group, the reactivity of alkali metals towards hydrogen and halogens decreases in the following order: Li > Na > K > Rb > Cs.]

Question 36.
NaCl is an ionic compound but LiCl has some covalent character, explain.
Answer:

• Li⁺ ion has very small size and therefore, the charge density on Li⁺ is high.
• Due to this, it has high tendency to distort the electron cloud around the negatively charged chloride ion (Cl^–) which is larger in size.
• This results in partial covalent character of the LiCl bond.
• Na⁺ ion cannot distort the electron cloud of Cl^– due to the bigger size of Na⁺ compared to Li⁺.

Hence, NaCl is an ionic compound but LiCl has some covalent character.

Question 37.
Why is lithium iodide most covalent in nature among alkali halides?
Answer:

• Among the alkali metal ions, Li⁺ ion is the smallest cation while among halides, anion I^– has the largest size.
• Thus, electron cloud around I^– ion is easily distorted by Li⁺ ion leading to polarisation of anion and covalency.
• Also, the difference in electronegativities of Li and I is small.

Hence, lithium iodide is most covalent in nature among alkali halides.

Question 38.
Explain the reactivity of alkaline earth metals towards:
i. Water
ii. Hydrogen
iii. Halogens
Answer:
i. Reaction with water:
a. The elements of group 2 (alkaline earth metals) react with water to form metal hydroxide and evolve hydrogen gas.

b. Be does not react with water at all, Mg reacts with boiling water while Ca, Sr, Ba react vigorously even with cold water.

ii. Reaction with hydrogen: All alkaline earth metals except beryllium (Be), when heated with hydrogen form MH₂ type hydrides.

iii. Reaction with halogens: All the alkaline earth metals combine with halogens at high temperature to form their corresponding halides.

[Note: As we move down the group, the chemical reactivity of alkaline earth metals increases in the order Mg < Ca < Sr < Ba.]

Question 39.
Complete the following chemical equations.

Answer:

Question 40.
Describe the reducing nature of group 1 and group 2 elements.
Answer:
The reducing power of an element is measured in terms of standard electrode potential (E0) corresponding to the following transformation i.e, tendency to lose electron.
\(\mathrm{M}_{(\mathrm{s})} \longrightarrow \mathrm{M}_{(\mathrm{aq})}^{+}+\mathrm{e}^{-}\)
i. Reducing nature of group 1 elements:

• All the alkali metals have high negative values of E⁰ which indicates that they have strong reducing nature and hence, they can be used as strong reducing agents.
• Lithium is the most powerful and sodium is the least powerful reducing agent in the group.

ii. Reducing nature of group 2 elements:

• All the alkaline earth metals have high negative values of stanard reduction potential (E⁰) and are strong reducing agents.
• However, reducing power of alkaline earth metals is less than that of alkali metals.

Question 41.
Explain the nature of the solution formed by group 1 and group 2 metals in liquid ammonia.
Answer:
i. The alkali metals are soluble in liquid ammonia and thus, they dissolve in it giving deep blue solutions which are conducting in nature.
M + (x + y)NH₃ → [M(NH₃)_x]⁺ + [e(NH₃)_y]^–
ii. The blue colour of the solution is due to the ammoniated electron.
iii. These solutions are paramagnetic and on standing slowly liberate hydrogen resulting in the formation of amide.
\(\mathrm{M}_{(\mathrm{am})}^{+}+\mathrm{e}^{-}+\mathrm{NH}_{3(l)} \longrightarrow \mathrm{MNH}_{2(\mathrm{am})}+\frac{1}{2} \mathrm{H}_{2(\mathrm{~g})}\)
(where ‘am’ denotes solution in ammonia.)
iv. As a result, the blue colour of the solution changes to bronze and the solution becomes diamagnetic.
v. Similarly, the alkaline earth metals are also soluble in liquid ammonia which give deep blue-black coloured solutions.
M + (x + 2y) NH₃ → [M(NH₃)_x]²⁺ + 2[e(NH₃)_y]^–

Question 42.
Explain: Diagonal relationship in group 1 and group 2.
Answer:

• Elements belonging to the same group are expected to exhibit similarity and gradation in their properties.
• However, first alkali metal, lithium, and the first alkaline earth metal, beryllium, do not fulfil this expectation.
• Thus, lithium shows many differences when compared with the remaining alkali metals and shows similarity with magnesium, the second alkaline earth metal.
• Similarly, beryllium shows many differences with remaining alkaline earth metals and shows similarity with aluminium, the second element of the next main group i.e., group 13.
• The relative placement of these elements with similar properties in the periodic table is across a diagonal and thus, it is called diagonal relationship.

Question 43.
Explain diagonal relationship between lithium and magnesium with respect to:
i. Property of chlorides
ii. Thermal decomposition of their carbonates
Answer:
Both lithium and magnesium show similarities in various physical and chemical properties as follows:
i. Property of chlorides: Chlorides of lithium (LiCl) and magnesium (MgCl₂) are deliquescent as group 2 elements form deliquescent chlorides. These chlorides form corresponding hydrates (LiCl.2H₂O and MgCl₂.8H₂O) on crystallization from their aqueous solutions.

ii. Thermal decomposition of carbonates: Heating of lithium carbonate and magnesium carbonate results in their easy decomposition to form corresponding oxides and carbon dioxide (CO₂).

Question 44.
Mention the properties of lithium that differ from rest of the alkali group metals.
Answer:

• Reaction with nitrogen: Only lithium from alkali group metals reacts with nitrogen present in the air on heating, while rest of the members do not react with nitrogen.
• Thermal decomposition of carbonates: Alkali metal carbonates show no reaction on heating, while lithium carbonate decomposes on heating to form the corresponding oxide and liberate carbon dioxide gas.
• Property of chlorides: Lithium (LiCl) is deliquescent and forms corresponding hydrate (LiCl.2H₂O). Other alkali chlorides are not deliquescent and do not form hydrates.

Question 45.
Write a note on the diagonal relationship between Be and Al.
OR
What are the similarities between beryllium and aluminium?
Answer:
i. Beryllium is placed in the group 2 and period 2 of the modem periodic table. It resembles aluminium which is placed in group 13 and period 3.

ii. Due to nearly same charge to radius ratio of their ions, beryllium (\(\frac {2}{31}\) = 0.065) and aluminium (\(\frac {3}{53.55}\) = 0.056) exhibit diagonal relationship.
iii. Due to diagonal relationship, Be and Al show following similarities in their properties:
a. Nature of bonding: Both Be and Al have tendency to form covalent chlorides.

b. Lewis acids: BeCl₂ and AlCl₃ act as Lewis acids.
c. Solubility in organic solvents: BeCl₂ and AlCl₃ are soluble in organic solvents.
d. Nature of oxide: Both Be and Al form amphoteric oxides.

Question 46.
Explain the amphoteric nature of aluminium oxide with the help of reactions.
Answer:
Al₂O₃ (magnesium oxide) reacts with both acid (HCl) as well as base (NaOH) to form the corresponding products and therefore, it is amphoteric in nature.

Question 47.
Beryllium shows many differences with other alkaline earth metals. Discuss these differences with respect to chlorides and oxides.
Answer:
1. Properties of chlorides: Beryllium chloride is covalent whereas chlorides formed by other alkaline earth metals are ionic in nature. Beryllium chloride is a strong Lewis acid whereas chlorides formed by other alkaline earth metals are not Lewis acids. Beryllium chloride is soluble in organic solvents whereas chlorides formed by other alkaline earth metals are insoluble in organic solvents.

2. Properties of oxide: Beryllium oxide is amphoteric whereas oxides formed by other alkaline earth metals are basic in nature.

Question 48.
Complete the following chemical equations.

Answer:

Question 49.
Write the uses of
i. alkali metals
ii. alkaline earth metals
Answer:
i. Uses of alkali metals:

• Lithium metal is used in long-life batteries used in digital watches, calculators and computers.
• Liquid sodium is used for heat transfer in nuclear power stations.
• Potassium chloride is used as a fertilizer.
• Potassium is used in manufacturing potassium superoxide (KO₂) for oxygen generation. It is good absorbent of carbon dioxide.
• Caesium is used in photoelectric cells.

ii. Uses of alkaline earth metals:

• Beryllium is used as a moderator in nuclear reactors.
• Alloy of magnesium and aluminium is widely used as structural material and in aircrafts.
• Calcium ions are important ingredient in biological system, essential for healthy growth of bones and teeth.
• Barium sulphate is used in medicine as barium meal for intestinal X-ray.
• Radium is used in radiotherapy for cancer treatment.

Question 50.
State the importance of sodium and potassium in biological system.
Answer:

• Sodium ion is present as the largest supply in all extracellular fluids. These fluids provide medium for transporting nutrients to the cells.
• The concentration of sodium ion in extracellular fluid regulates the flow of water across the membrane.
• Sodium ions participate in the transmission of nerve signals.
• Potassium ions are the most abundant ions within the cells. They are required for maximum efficiency in the synthesis of proteins and also in oxidation of glucose.

Question 51.
How are the following ions of group 2 elements biologically important?
i. Mg²⁺
ii. Ca²⁺
Answer:
i. Magnesium ion (Mg²⁺)

• Mg²⁺ ions are important part of chlorophyll in green plants.
• They play an important role in the breakage of glucose and fat molecules, synthesis of proteins with enzymes and regulation of cholesterol level.

ii. Calcium ion (Ca²⁺)

• Ca²⁺ ions are important for bones and teeth in the form of apatite [Ca₃(PO₄)₂].
• They play an important role in blood clotting.
• Ca²⁺ ions are required for contraction and stretching of muscles.
• They are also required to maintain the regular beating of heart.

Question 52.
Explain Solvay process for manufacture of sodium carbonate.
Answer:
Sodium carbonate (Na₂CO₃) is commercially prepared by Solvay process. Preparation of sodium carbonate by Solvay process involves two stages.
i. In the first stage of Solvay process, carbon dioxide gas is bubbled through a concentrated solution of NaCl which is saturated with NH₃. This results in the formation of ammonium bicarbonate. Crystals of sodium bicarbonate separate as a result of the following reactions.


ii. Ammonium bicarbonate and sodium chloride undergoes double decomposition reaction to form sodium bicarbonate. As sodium bicarbonate has low solubility, it precipitates out in the form of crystals.
iii. In the second stage, the separated crystals of sodium bicarbonate are heated to obtain sodium carbonate (Na₂CO₃).

iv. NH₄Cl obtained in this process is treated with slaked lime, Ca(OH)₂, to recover NH₃ while CaCl₂ is obtained as a byproduct.

Question 53.
How is ammonia recovered in Solvay process? Name the important by-product obtained in the step?
Answer:
i. Ammonium chloride (NH₄Cl) is obtained during the Solvay process which is used for the preparation of Na₂CO₃. When NH₄Cl is treated with slaked lime, Ca(OH)₂, ammonia is recovered.

ii. Calcium chloride is obtained as an important by-product in this reaction.

Question 54.
Why potassium carbonate cannot be obtained by Solvay process?
Answer:
Potassium hydrogen carbonate (KHCO₃) is highly water soluble and cannot be precipitated out by reacting with potassium choride (KCl) and hence, potassium carbonate (K₂CO₃) cannot be obtained by Solvay process.

Question 55.
What is the action of heat on crystalline sodium carbonate (washing soda)?
Answer:
i. On heating washing soda (decahydrate of sodium carbonate) up to 373 K, it loses water molecules to form corresponding monohydrate.

ii. On heating above 373 K, monohydrate further loses water and changes into white anhydrous powder called soda ash.

Question 56.
Give reason: Aqueous solution of sodium carbonate is alkaline in nature.
Answer:
i. Sodium carbonate is hydrolysed by water as shown in the reaction given below.

ii. One of the products formed as a result of hydrolysis is NaOH which is a strong base.
Hence, aqueous solution of sodium carbonate is alkaline in nature due to formation of strong base (NaOH).

Question 57.
What are the uses of sodium carbonate?
Answer:
Uses of sodium carbonate:

• Due to its alkaline properties, sodium carbonate has an emulsifying effect on grease and dirt and hence, it is used as a cleaning material.
• It is used to make hard water soft as it precipitates out the soluble calcium and magnesium salts in hard water as carbonates.
  For example: Ca(HCO₃)_2(aq) + Na₂CO_3(aq) → CaCO_3(s) + 2NaHCO_3(aq)
• It is used for commercial production of soap and caustic soda.
• Sodium carbonate is used as an important laboratory reagent.

Question 58.
Describe the preparation of sodium hydroxide by Castner-Kellner process.
OR
Explain the electrolysis method for preparation of sodium hydroxide.
Answer:
i. Sodium hydroxide (caustic soda) is commercially obtained by the electrolysis of aqueous sodium chloride solution (brine) in Castner-Kellner cell (mercury cathode cell).
ii. In Castner-Kellner cell, mercury is used as cathode, carbon rod as anode and brine solution is used as electrolyte which is subjected to electrolysis.

iii. During electrolysis, the following reactions take place:
a. At cathode: Sodium ions get reduced to metallic sodium, which combines with mercury to form sodium amalgam (Na-Hg).
\(\mathrm{Na}^{+}+\mathrm{e}^{-} \stackrel{\mathrm{Hg}}{\longrightarrow} \mathrm{Na} \text {-amalgam }\)
b. At anode: Chloride ions are oxidized and thus, chlorine gas is evolved.
\(\mathrm{Cl}^{-} \longrightarrow \frac{1}{2} \mathrm{Cl}_{2}+\mathrm{e}^{-}\)

iv. Sodium amalgam is then treated with water to obtain sodium hydroxide and hydrogen gas.

Question 59.
Enlist the physical properties of sodium hydroxide.
Answer:
Physical properties of sodium hydroxide:

• Sodium hydroxide (NaOH) is a white deliquescent solid.
• It has a melting point of 591 K.
• It is highly water soluble and gives a strongly alkaline solution.
• The surface of sodium hydroxide solution absorbs atmospheric CO₂ to form Na₂CO₃.

Question 60.
Explain how sodium hydroxide is commercially important.
Answer:
Commercial uses of sodium hydroxide:

• Sodium hydroxide is used in purification of bauxite (the aluminium ore).
• It is used in commercial production of soap, paper, artificial silk and many chemicals.
• It is used for mercerising cotton fabrics.
• It is used in petroleum refining.
• It is also used as an important laboratory reagent.

Question 61.
Calcium carbonate occurs naturally in which forms?
Answer:
Calcium carbonate (CaCO₃) occur naturally in the form of chalk, limestone and marble.

Question 62.
Describe the various methods used for preparation of calcium carbonate.
Answer:
i. a. Calcium carbonate is prepared by passing carbon dioxide through solution of calcium hydroxide (slaked lime). This results in the formation of water insoluble solid calcium carbonate.

b. However, excess carbon dioxide transforms the precipitate of CaCO₃ into water-soluble calcium bicarbonate and therefore, it has to be avoided.

ii. Calcium carbonate can also be prepared by adding solution of calcium chloride to a solution of sodium carbonate. This results in the formation of calcium carbonate as precipitate.

Question 63.
Why controlled addition of CO₂ is essential during preparation of calcium carbonate from slaked lime?
Answer:
When excess of CO₂ is present, it leads to the formation of water-soluble calcium bicarbonate.

Hence, while preparing calcium carbonate from slaked lime, controlled addition of CO₂ is essential.

Question 64.
Mention some physical properties of calcium carbonate.
Answer:

• Calcium carbonate is soft, light, white powder.
• It is practically insoluble in water.

Question 65.
What happens when:
i. calcium carbonate is thermally decomposed?
ii. calcium carbonate reacts with dilute mineral acids?
Answer:
i. When calcium carbonate is heated to 1200 K, it decomposes into calcium oxide along with evolution of carbon dioxide gas.

ii. Calcium carbonate reacts with dilute mineral acids such as HCl and H₂SO₄ to give the corresponding calcium salt and liberate carbon dioxide gas.

Question 66.
Give the important uses of calcium carbonate.
Answer:

• Calcium carbonate in the form of marble is used as building material.
• It is used in the manufacture of quicklime (CaO) which is the major ingredient of cement.
• A mixture of CaCO₃ and MgCO₃ is used as flux in the extraction of metals from their ores.
• It is required for the manufacture of high-quality paper.
• It is an important ingredient in toothpaste, chewing gum, dietary supplements of calcium and filler in cosmetics.

Question 67.
Match the pairs.

	Column A		Column B
i.	Castner-Kellner cell	a.	Na2CO
ii.	Slaked lime	b.	CaCO3
iii.	Solvay process	c.	NaOH
iv.	Limestone	d.	Ca(OH)2

Ans:
i – c,
ii – d,
iii – a,
iv – b

Question 68.
How is hydrogen peroxide prepared by the action of cold dilute H₂SO₄ on
i. Hydrated barium peroxide?
ii. sodium peroxide (Merck process)?
Answer:
Preparation of hydrogen peroxide by the action of cold dilute H₂SO₄ on
i. hydrated barium peroxide: When hydrated barium peroxide is treated with ice-cold dilute sulphuric acid, the precipitate of barium sulphate is obtained. This precipitate is then filtered off to get hydrogen peroxide solution.

ii. Sodium peroxide (Merck process): When small quantity of sodium peroxide is added to ice-cold solution of dilute sulphuric acid with stirring, it gives hydrogen peroxide.

Question 69.
Explain how hydrogen peroxide can be obtained by electrolysis method.
Answer:
i. H₂O₂ can be manufactured by electrolysis of 50% H₂SO₄. In this method, 50% solution of H₂SO₄ is subjected to an electrolytic oxidation to form peroxydisuiphuric acid at anode.

ii. On hydrolysis, peroxy sulphuric yields hydrogen peroxide.

iii. This method can be used for the laboratory preparation of D₂O₂.

Question 70.
Describe the industrial method for preparation of hydrogen peroxide.
OR
How is hydrogen peroxide obtained from 2-ethylanthraquinol?
Answer:
i. Industrially hydrogen peroxide is prepared by air-oxidation of 2-ethylanthraquinol.

ii. 2-Ethylanthraquinone is reduced back to 2-ethylanthraquinol by catalytic hydrogenation.

Question 71.
What are the physical properties of hydrogen peroxide?
Answer:

• Pure H₂O₂ is a very pale blue coloured liquid.
• Its boiling point is 272.4 K.
• H₂O₂ is miscible in water and forms a hydrate (H₂O₂. H₂O).

Question 72.
How is strength of H₂O₂ solution expressed?
Answer:

• Strength of aqueous solution of H₂O₂ is expressed in ‘volume’ units i.e., volume strength.
• The commercially marketed 30% (by mass) solution of H₂O₂ has volume strength of 100 volume.
• It means that 1 mL of 30% solution of H₂O₂ will give 100 mL oxygen at STP.

Thus, Volume strength refers to the volume of oxygen (O₂) in litres at STP obtained by decomposition of 1 litre of the sample.

Question 73.
Write reactions depicting oxidising and reducing action of hydrogen peroxide in acidic medium.
Answer:
H₂O₂ acts as a mild oxidising as well as reducing agent.
i. Oxidising action of H₂O₂ in acidic medium.
\(2 \mathrm{Fe}_{(\mathrm{aq})}^{2+}+2 \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{H}_{2} \mathrm{O}_{2(\mathrm{aq})} \longrightarrow 2 \mathrm{Fe}_{(\mathrm{aq})}^{3+}+2 \mathrm{H}_{2} \mathrm{O}_{(l)}\)

ii. Reducing action of H₂O₂ in acidic medium.
\(2 \mathrm{MnO}_{4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{Mn}^{2+}+8 \mathrm{H}_{2} \mathrm{O}+5 \mathrm{O}_{2}\)

Question 74.
Enlist uses of hydrogen peroxide.
Answer:

• Hydrogen peroxide is used as mouthwash, germicide and mild antiseptic.
• It is used as a preservative for milk and wine.
• It is used as a bleaching agent for soft materials, due to its mild oxidising property.
• Due to its reducing property, is used as an antichlor to remove excess chlorine from fabrics which have been bleached by chlorine.
• Nowadays it is used in environmental chemistry for pollution control and restoration of aerobic condition of sewage water.

Question 75.
Describe preparation and properties of lithium aluminium hydride (LAH).
Answer:
i. Preparation: Lithium hydride when treated with aluminium chloride, gives lithium aluminium hydride, (LiAlH₄).

ii. Properties:

• Lithium aluminium hydride is a colourless solid.
• It reacts violently with water and even with atmospheric moisture.

Question 76.
How is lithium aluminium hydride (LAH) useful in organic synthesis?
Answer:
i. LAH is a source of hydride (H^–) and therefore, it is used as a reducing agent in organic synthesis.
For example:

ii. It is useful in the preparation of PH₃ (phosphine).
4PCl₃ + 3LiAlH₄ → 4PH₂ + AlCl₃ + LiCl

Question 77.
Complete the following reactions by mentioning the reagent/reaction conditions under which these reactions are carried out.

Answer:

Question 78.
Calculate % (by mass) of a H₂O₂ solution which is 45.4 volume.
Answer:
Given: 45.4 volume H₂O₂ solution
To find: % (by mass) of H₂O₂
Formula: Percentage (%) by mass = \(\frac{\text { Mass of solute }}{\text { Mass of solution }} \times 100\)
calculation: 45.4 volume H₂O₂ solution means 1 L of this solution will liberate 45.4 L of O₂ at STP.
Hydrogen peroxide (H₂O₂) decomposes as:

Ans: % (by mass) of H₂O₂ in 45.4 volume H₂O₂ solution is 13.6%.

Question 79.
Calculate the strength (g/L) of 20 volume solution of hydrogen peroxide.
Solution:
Given: 20 volume H₂O₂ solution
To find: Strength of H₂O₂ (g/L)
Formula: 20 volume H₂O₂ solution means that 1 L of this solution will liberate 20 L of oxygen at S.T.P. Let us calculate the amount of H₂O₂ (in grams) which gives 20 L of oxygen at S.T.P. This amount will be present in 1 L of 20 volume solution of H₂O₂.
Hydrogen peroxide (H₂O₂) decomposes as:

22.7 L of O₂ at S.T.P. is produced from H₂O₂ = 68 g
20 L of O₂ at S.T.P. is produced from H₂O₂ = \(\frac {68}{22.7}\) × 20 = 59.912g = 59.912 g/litre
Ans: Strength of H₂O₂ in 20 volume H₂O₂ solution is 59.912 g/L.

Question 80.
Calculate the volume strength of a 5% solution of hydrogen peroxide.
Solution:
Given: 5% solution of H₂O₂
To find: Volume strength of H₂O₂ solution
Calculation: 100 mL of solution contains 5 g of H₂O₂
1000 mL of solution will contain \(\frac {5}{100}\) × 1000 = 50 g of H₂O₂

68 g of H₂O₂ will give O₂ at S.T.P. = 22.7 L
50 g of H₂O₂ is present in 1000 mL of H₂O₂ or 1 L of H₂O₂
50 g of H₂O₂ will give O₂ at S.T.P. = \(\frac {22.7}{68}\) × 50 = 16.691 L
∴ 1 L of H₂O₂ gives O₂ at S.T.P. = 16.691 L
∴ Strength of H₂O₂ = 16.691 volume
Ans: The given 5% H₂O₂ solution is equivalent to 16.691 volume solution of hydrogen peroxide.

Question 81.
Naina is a school going kid. Every morning her mother makes her drink a glass of milk. When she asked her mother that why she has to drink a glass of milk daily, her mother told her that it is beneficial in maintaining healthy bones and teeth. Regular consumption of milk is recommended because it is a rich source of calcium.
i. In which form is calcium important for bones and teeth?
ii. Calcium belongs to which family in the modern periodic table?
iii. Write its electronic configuration.
iv. Calcium contain how many valence electrons?
v. Give any two-biological importance of calcium.
Answer:
i. Calcium is important for bones and teeth in the form of apatite [Ca(PO₄)₂].
ii. It belongs to the family of alkaline earth metals in the modem periodic table.
iii. Electronic configuration of 20Ca is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s².
iv. Calcium contains two valence electrons as it has two electrons in its outermost shell (4s).

v. Calcium ion (Ca²⁺)

• Ca²⁺ ions are important for bones and teeth in the fonn of apatite [Ca₃(PO₄)₂].
• They play an important role in blood clotting.
• Ca²⁺ ions are required for contraction and stretching of muscles.
• They are also required to maintain the regular beating of heart.

Multiple Choice Questions

1. Of all the elements present in the periodic table, ………… has the simplest atomic structure.
(A) lithium
(B) beryllium
(C) hydrogen
(D) helium
Answer:
(C) hydrogen

2. Electronic configuration of hydrogen is similar to that of ……………
(A) transition elements
(B) inert gases
(C) alkaline earth metals
(D) alkali metals
Answer:
(D) alkali metals

3. Isotopes are atoms of the same element having different ………….. number.
(A) neutron
(B) proton
(C) electron
(D) Both (A) and (B)
Answer:
(A) neutron

4. Tritium, \(\left({ }_{1}^{3} \mathrm{H}\right)\) ……………
(A) is an isotope of hydrogen
(B) contains one electron, one proton and two neutrons
(C) is a beta particle emitter
(D) all of these
Answer:
(D) all of these

5. In the electrolysis of acidified water using, ………….. is liberated at the anode.
(A) dihydrogen
(B) sulphate ions
(C) oxygen
(D) chloride ions
Answer:
(C) oxygen

6. Water gas is a mixture of ………….
(A) CO + H₂
(B) CO₂ + H₂
(C) O₂ + H₂
(D) CO + O₂
Answer:
(A) CO + H₂

7. During production of dihydrogen by water-gas shift reaction, which of the following is present as an impurity?
(A) Carbon monoxide
(B) Carbon dioxide
(C) Calcium carbonate
(D) Calcium oxide
Answer:
(B) Carbon dioxide

8. Solution is used to remove carbon dioxide present in the mixture along with dihydrogen.
(A) Sodium hydroxide
(B) Hydrochloric acid
(C) Magnesium chloride
(D) Sodium arsenite
Answer:
(D) Sodium arsenite

9. The reaction between dihydrogen and …………… is highly exothermic.
(A) halogens
(B) dioxygen
(C) dinitrogen
(D) metals
Answer:
(B) dioxygen

10. The elements of group 1 and group 2 belong to which block of the modem periodic table?
(A) d-block
(B) s-block
(C) p-block
(D) f-block
Answer:
(B) s-block

11. Which of the following is NOT an alkaline earth metal?
(A) Beryllium
(B) Barium
(C) Calcium
(D) Caesium
Answer:
(D) Caesium

12. The common oxidation state for alkali metals is …………….
(A) +2
(B) +1
(C) +3
(D) +4
Answer:
(B) +1

13. All alkaline earth metals have ………….. valence electrons in the outermost orbit.
(A) one
(B) two
(C) three
(D) four
Answer:
(B) two

14. Electronic configuration of potassium with respect to nearest noble gases is …………..
(A) [He]2s¹
(B) [Ne]3s¹
(C) [Ar]4s¹
(D) [Kr]5s¹
Answer:
(C) [Ar]4s¹

15. Which of the following is radioactive alkali metal?
(A) Rubidium
(B) Caesium
(C) Francium
(D) Beryllium
Answer:
(C) Francium

16. Which of the following element is rarest amongst s-block elements?
(A) Strontium
(B) Barium
(C) Radium
(D) Calcium
Answer:
(C) Radium

17. Which of the following is FALSE?
(A) Alkali metals readily loose electron to form monovalent M⁺ ions.
(B) In a group, from Li to Cs, atomic and ionic radii increase with atomic number.
(C) The monovalent ions of alkali metals are larger in size than the parent atoms.
(D) Ionization enthalpies decrease down the group from Li to Cs.
Answer:
(C) The monovalent ions of alkali metals are larger in size than the parent atoms.

18. The first ionization enthalpies of alkaline earth metals are …………. than those of the corresponding alkali metals.
(A) higher
(B) lower
(C) same
(D) none of these
Answer:
(A) higher

19. Which of the following alkaline earth metal does NOT react with water?
(A) Sr
(B) Mg
(C) Ca
(D) Be
Answer:
(D) Be

20. Oxides and hydroxides of alkaline earth metals except beryllium are ………….. in nature.
(A) acidic
(B) basic
(C) amphoteric
(D) neutral
Answer:
(B) basic

21. ………… is an excellent absorbent of carbon dioxide.
(A) KO₂
(B) KCl
(C) KOH
(D) KHCO₃
Answer:
(A) KO₂

22. Lithium shows diagonal relationship with ……………
(A) beryllium
(B) magnesium
(C) calcium
(D) boron
Answer:
(B) magnesium

23. The diagonal relationship between Li and Mg is due to the similarity in ……………
(A) ionic sizes
(B) electronegativity value
(C) polarizing power
(D) all of the above
Answer:
(D) all of the above

24. The alkali metal that reacts with nitrogen directly to form nitride is ……………
(A) Li
(B) Na
(C) K
(D) Rb
Answer:
(A) Li

25. In the Solvay process, the chief products are ……………
(A) CaCO₃ and Ca(HCO₃)₂
(B) Na₂CO₃ and NaHCO₃
(C) Na₂SO₄ and NaHSO₄
(D) CaCl₂ and Ca(NO₃)₂
Answer:
(B) Na₂CO₃ and NaHCO₃

26. In Castner-Kellner process for preparation of sodium hydroxide, ………….. is subjected to electrolysis.
(A) NaCl
(B) NaOH
(C) Na₂O
(D) Na2CO₃
Answer:
(A) NaCl

27. Which of the following method of preparation of H₂O₂ is known as Merck’s method?
(A) BaO₂.8H₂O(s) + H₂SO₄(aq) → BaSO₄(s) + H₂O₂(aq) + 8H₂O(l)
(B) Na₂O₂ + H₂SO₄ → Na₂SO₄ + H₂O₂
(C) BaO₂ + H₂O + CO₂ → BaCO₃↓ + H₂O₂
(D) 3BaO₂ + 2H₃PO₄ → Ba₃(PO₄)₂↓ + 3H₂O₂
Answer:
(B) Na₂O₂ + H₂SO₄ → Na₂SO₄ + H₂O₂

Balbharti Maharashtra State Board 11th Chemistry Important Questions Chapter 10 States of Matter Important Questions and Answers.

Maharashtra State Board 11th Chemistry Important Questions Chapter 10 States of Matter

Question 1.
What are the three distinct physical forms of a substance?
Answer:
The three distinct physical forms of a substance are solid, liquid and gas.

Question 2.
What are the different forms (physical states) in which water exists?
Answer:
Water exists in the three different forms solid ice, liquid water and gaseous vapours.

Question 3.
Give the differences between the three states of matter.
OR
State the properties of three states of matter.
Answer:

Question 4.
With suitable diagram, explain how three states of matter are interconvertible by exchange of heat.
Answer:

• On heating, solid changes to liquid, which on further heating changes to gas.
• On cooling, gas condenses to liquid, which on further cooling change to solid.

Question 5.
What are intermolecular forces? Explain the role of these forces in different states of matter.
Answer:

• Intermolecular forces are the attractive forces as well as repulsive forces present between the neighbouring molecules.
• The attractive force decreases with the increase in distance between the molecules.
• The intermolecular forces are strong in solids, less strong in liquids and very weak in gases. Thus, the three physical states of matter can be determined as per the strength of intermolecular forces.
• The physical properties of matter such as melting point, boiling point, vapor pressure, viscosity, evaporation, surface tension and solubility can be studied with respect to the strength of attractive intermolecular forces between the molecules.
• During the melting process, intermolecular forces are partially overcome, whereas they are overcome completely during the vapourization process.

Question 6.
Name the different types of intermolecular forces.
Answer:
The types of intermolecular forces are as follows:

1. Dipole-dipole interactions
2. Ion-dipole interactions
3. Dipole-induced dipole interactions
4. Hydrogen bonding
5. London dispersion forces

Question 7.
Write a short note on dipole moment.
Answer:
Dipole moment:
i. Dipole moment (µ) is the product of the magnitude of the charge (Q) and the distance between the centres of positive and negative charge (r).
ii. It is designated by a Greek letter (µ) (mu) and its unit is Debye (D).
iii. Dipole moment is a vector quantity and is depicted by a small arrow with tail in the positive centre and head pointing towards the negative centre.
iv. For example, HCl is a polar molecule.

The crossed arrow above the structure represents an electron density shift. Thus, polar molecules have permanent dipole moments.

Question 8.
Explain dipole-dipole interactions.
Answer:
Dipole-dipole interactions:
i. When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of another polar molecule. Interaction between such molecules is termed as dipole-dipole interaction.
ii. These forces are generally weak, with energies of the order of 3-4 kJ mol^-1 and are significant only when molecules are in close contact, i.e., in a solid or a liquid state.
iii. For example, C₄H₉Cl, (butyl chloride), CH₃ – O – CH₃ (dimethyl ether), ICl (iodine chloride, B.P. 27 °C), are dipolar liquids.
iv. The molecular orientations due to dipole-dipole interaction in ICl liquid is shown in the following figure:

v. More polar the substance, greater the strength of its dipole-dipole interactions.

Question 9.
Explain the effect of dipole moment on boiling point with an example.
Answer:
Higher the dipole moment, stronger are the intermolecular forces. Thus, higher is the boiling point.
e.g. Dipole moment of dimethyl ether (CH₃ – O – CH₃) is 1.3 D while that of ethane (CH₃ – CH₂ – CH₃) is 0.1 D. Since, dipole moment of dimethyl ether is higher than that of ethane, the boiling point of dimethyl ether is higher than that of ethane.
Note: Dipole moments and boiling points of some compounds:

Question 10.
Explain ion-dipole interactions.
Answer:
Ion-dipole interactions:
i. An ion-dipole force is the result of electrostatic interactions between an ion (cation or anion) and the partial charges on a polar molecule.
ii. The strength of this interaction depends on the charge and size of an ion. It also depends on the magnitude of dipole moment and size of the molecule.
iii. Ion-dipole forces are particularly important in aqueous solutions of ionic substances. When an ionic compound is dissolved in water, the ions get separated and surrounded by water molecules. This process is called hydration of ions.
iv. For example, Na⁺ ion (cation) – H₂O interaction is shown in the following figure:

v. The charge density on Na+ is more concentrated than the charge density on Cl^– because Na⁺ is smaller in size than Cl^–. This makes the interaction between (Na⁺) and negative end of the polar H₂O molecule stronger than the corresponding interaction between (Cl^–) and positive end of the polar H₂O molecule.
vi. More the charge on cation, stronger is the ion-dipole interaction. For example, Mg²⁺ ion has higher charge and smaller ionic radius (78 pm) than Na⁺ ion (98 pm), hence Mg²⁺ ion is surrounded (hydrated) more strongly with water molecules and exerts strong ion-dipole interaction.
Thus, the strength of interaction increases with increase in charge on cation and with decrease in ionic size or radius.
Therefore, ion-dipole forces increase in the order: Na⁺ < Mg²⁺ < Al³⁺.

Question 11.
Write a short note on dipole-induced dipole interactions.
Answer:
Dipole-induced dipole interactions:
i. When polar molecules (like H₂O, NH₃) and nonpolar molecules (like benzene) approach each other, the polar molecules induce dipole in the nonpolar molecules. Hence, ‘Temporary dipoles’ are formed by shifting of electron clouds in nonpolar molecules.
ii. For example, ammonia (NH₃) is polar and has permanent dipole moment while Benzene (C₆H₆) is nonpolar and has zero dipole moment. The force of attraction developed between the polar and nonpolar molecules is of the type dipole-induced dipole interaction. This is shown in the following figure:

Question 12.
Explain briefly London dispersion forces.
Answer:
London dispersion forces:

• In nonpolar molecules and inert gases, only dispersion forces exist.
• Dispersion forces are also called as London forces or van der Waals forces.
• It is the weakest intermolecular force that develops due to interaction between two nonpolar molecules.
• In general, all atoms and molecules experience London dispersion forces, which result from the motion of electrons.
• At any given instant of time, the electron distribution in an atom may be asymmetrical, giving the atom a short-lived dipole moment. This momentary dipole on one atom can affect the electron distribution in the neighbouring atoms and induce momentary dipoles in them. As a result, weak attractive force develops.
• For example, substances composed of molecules such as O₂, CO₂, N₂, halogens, methane gas, helium and other noble gases show van der Waals force of attraction.
• The strength of London forces increases with increase in molecular size, molecular mass and number of electrons present in an atom or molecule.

Question 13.
Give reason: Benzene has zero dipole moment and has no dipole-dipole forces yet it exists in liquid state.
Answer:

• Benzene (C₆H₆) is nonpolar molecule and has zero dipole moment.
• In benzene, only London forces exist due to momentary dipoles.
• The strength of London forces increases with increase in molecular size, molecular mass and number of electrons present in an atom or molecule.
• Hence, due to the presence of London forces, benzene exists in liquid state.

Question 14.
Explain the term polarizability.
Answer:
Polarizability:

• When two nonpolar molecules approach each other, attractive and repulsive forces between their electrons and the nuclei will lead to distortions in the size of electron cloud, a property referred to as polarizability.
• Polarizability is a measure of how easily an electron cloud of an atom is distorted by an applied electric field.
• It is the property of atom.
• The ability to form momentary dipoles, that means, the ability of the molecule to become polar by redistributing its electrons is known as polarizability of the atom or molecule.

Question 15.
Describe how London dispersion forces affect the boiling points of isomeric compounds like n-pentane and neopentane.
Answer:

• More the spread out of shapes, higher the dispersion forces present between the molecules.
• London dispersion forces are stronger in a long chain of atoms where molecules are not compact.
• For example, n-Pentane boils at 309.4 K, whereas neopentane boils at 282.7 K.
• Both the substances have the same molecular formula, C₅H₁₂, but n-pentane is longer and somewhat spread out, whereas neopentane is more spherical and compact.

Question 16.
Write a short note on hydrogen bonding.
Answer:
Hydrogen bonding:

• The electrostatic force of attraction between positively polarised hydrogen atom of one molecule and a highly electronegative atom (which may be negatively charged) of other molecule is called as hydrogen bond.
• Strong electronegative atoms that form hydrogen bonds are nitrogen, oxygen and fluorine.
• A hydrogen bond is a special type of dipole-dipole attraction.
• Hydrogen bonds are generally stronger than usual dipole-dipole and dispersion forces, and weaker than true covalent or ionic bonds.
• Hydrogen bond is denoted by (….) dotted line.
  e.g. Water (H₂O) and ammonia (NH₃) show hydrogen bonding.

Question 17.
Explain intramolecular and intermolecular hydrogen bond with suitable examples.
Answer:
i. Hydrogen bond which occurs within one single molecule represents intramolecular hydrogen bond.
e.g. H-bonding in ethylene glycol:

ii. A hydrogen bond present between two like or unlike molecules represents intermolecular hydrogen bond.
e.g. H-bonding in H-F:

Question 18.
How does hydrogen bonding influence boiling points of compounds?
Answer:

• Due to the presence of hydrogen bonding in the compounds, more energy is required to break the bonds.
• Therefore, boiling point is more in case of liquid molecules containing hydrogen bond.
• Hydrogen bonds can be quite strong with energies up to 40 kJ/mol.
• The boiling point generally increases with increase in molecular mass, but the hydrides of nitrogen (NH3), oxygen (H₂O) and fluorine (HF) have abnormally high boiling points due to the presence of hydrogen bonding between the molecules.

[Note: Due to presence of H-bond, viscosity’ of liquid increases. Hydrogen bonds play vital role in determining structure and properties of proteins and nucleic acids present in all living organisms.]

Question 19.
Observe the following figure and answer the questions:

i. What do the dotted lines represent?
ii. A water molecule can form how many H-bonds?
iii. Is this an example of intramolecular H-bonding?
Answer:
i. The dotted lines represent hydrogen bonds.
ii. A water molecule can form four H-bonds.
iii. No, it is an example of intermolecular H-bonding.

Question 20.
Explain the relation between intermolecular forces and thermal energy.
Answer:

• Thermal energy is the measure of kinetic energy of the particles of matter that arises due to movement of particles.
• It is directly proportional to the temperature; that means, thermal energy increases with increase in temperature and vice versa.
• Three states of matter are the consequence of a balance between the intermolecular forces of attraction and the thermal energy of the molecules.
• If the intermolecular forces are very weak, molecules do not come together to make liquid or solid unless thermal energy is decreased by lowering the temperature.
• When a substance is to be converted from its gaseous state to solid state, its thermal energy (or temperature) has to be reduced. At this stage, the intermolecular forces become more important than thermal energy of the particles.

Note: Comparison of intermolecular forces:

Question 21.
State true or false. Correct the false statement.
i. London dispersion force is the weakest intermolecular force that develops due to interaction between two nonpolar molecules.
ii. More the charge on cation, stronger is the ion-dipole interaction.
iii. A hydrogen bond is a special type of dipole-induced dipole attraction.
iv. Thermal energy is directly proportional to the temperature.
Answer:
i. True
ii. True
iii. False,
A hydrogen bond is a special type of dipole-dipole attraction.
iv. True

Question 22.
Name the major intermolecular forces between:
i. Cl₂ and CBr₄
ii. SiH₄ molecules
iii.He atoms in liquid He
iv. HCl molecules in liquid HCl
v. He and a polar molecule
vi. Water molecules
Answer:
i. London dispersion forces
ii. London dispersion forces
iii. London dispersion forces
iv. Dipole-dipole interactions
v. Dipole-induced dipole
vi. Hydrogen bonding

Question 23.
Why is the chemistry of atmospheric gases an important subject of study?
Answer:
The chemistry of atmospheric gases is an important subject of study as it involves air pollution. O₂ in air is essential for survival of aerobic life.

Question 24.
What are the measurable properties of gases?
OR
Explain the following measurable properties of gases in detail: Mass, volume, pressure, temperature and diffusion.
Answer:
Measurable properties of gases are as follows:
i. Mass:

• The mass (m) of a gas sample is measure of the quantity of matter it contains.
• It can be measured experimentally.
• The SI unit of mass is kilogram (kg).
  1 kg = 10³ g.
• The mass of a gas is related to the number of moles (n) by the expression:
  n = \(\frac{\text { mass in grams }}{\text { molar mass in grams }}=\frac{\mathrm{m}}{\mathrm{M}}\)

ii. Volume:

• Volume (V) of a sample of gas is the amount of space it occupies.
• It is expressed in terms of different units like Litres (L), millilitres (mL), cubic centimetre (cm³), cubic metre (m³) or decimetre cube (dm³).
• The SI Unit of volume is cubic metre (m³).
• Most commonly used unit to measure the volume of the gas is decimetre cube or litre.

iii. Pressure:

• Pressure (P) is defined as force per unit area.
  Pressure = \(\frac{\text { Force }}{\text { Area }}=\frac{\mathrm{f}}{\mathrm{a}}\)
• Pressure of gas is measured with ‘manometer’ and atmospheric pressure is measured by ‘barometer’.
• The SI unit of pressure is pascal (Pa) or Newton per metre square (N m^-2).

iv. Temperature:

• It is the property of an object that determines direction in which energy will flow when that object is in contact with another object.
• In scientific measurements, temperature (T) is measured either on the Celsius scale (°C) or the Kelvin scale (K).
• The SI unit of temperature of a gas is Kelvin (K).
• The Celsius and Kelvin scales are related by the expression: T(K) = t °C + 273.15

v. Density: Density (d) of a substance is the mass per unit volume.
d = \(\frac{\mathrm{m}}{\mathrm{V}}\)
∴ The SI unit of density is kg m^-3.
In the case of gases, relative density is measured with respect to hydrogen gas and is called vapour density.
∴ Vapour density = \(\frac{\text { Molar mass }}{2}\)

vi. Diffusion:
a. Diffusion is the process of mixing two or more gases to form a homogeneous mixture.
b. The volume of gas diffused per unit time is the rate of diffusion of that gas.

c. SI Unit for rate of diffusion is dm³ s^-1 or cm³ s^-1.

Question 25.
Convert:
i. 3.5 atm to mm Hg
ii. 1520 torr to atm
iii. 5 m³ to dm³
iv. 580 °c to Kelvin
Answer:
I. 3.5 atm to mm Hg:
1 atm = 760 mm Hg
∴ 3.5 atm = 3.5 × 760
= 2660 mm Hg

ii. 1520 torr to aim:
1 atm = 760 torr
∴ 1 torr = \(\frac {1}{760}\) atrn
∴ 1520 torr = \(\frac {1520}{760}\) = atm

iii. 5 m³ to dm³:
1 m³ = 10³ dm³
∴ 5m³ = 5 × dm³ = 5000 dm³

iv. 580 °C to Kelvin:
T(K) = t °C + 273.15
∴ T(K)= (580 °C) + 273.15 = 853.15 K

Question 26.
Name four measurable properties that are essential to study behaviour of gases.
Answer:

• Pressure
• Volume
• Temperature
• Number of moles

Question 27.
Explain Boyle’s law with the help of a diagram.
Answer:
Boyle’s law (Pressure-Volume relationship):
i. Statement: For a fixed mass (number of moles ‘n’) of a gas at constant temperature, the pressure (P) of the gas is inversely proportional to the volume (V) of the gas.
ii. Explanation:
The mathematical expression of Boyle’s law is:
P ∝ \(\frac{1}{\mathrm{~V}}\) (at constant T and n) k₁
∴ P = \(\frac{\mathrm{k}_{1}}{\mathrm{~V}}\) (where, k₁ is the proportionality constant)
On rearranging the above equation,
∴ PV = k₁ = constant
This implies that at constant temperature, product of pressure and volume of the fixed amount of a gas is constant.
Thus, when a fixed amount of a gas at constant temperature (T) occupying volume V₁ initially at pressure (P₁) undergoes expansion or compression, volume of the gas changes to V₂ and pressure to P₂.
According to Boyle’s law,
P₁V₁ = P₂V₂ = constant

iii. Schematic illustration of Boyle’s law:

Question 28.
Give the different graphical representations of Boyle’s law.
Answer:
i. Graph of pressure (P) versus volume (V) of a gas at constant temperature:
If the pressure (P) is plotted against volume (V) at constant temperature, a curve is obtained.

As the pressure increases, the volume decreases exponentially. The product of pressure and volume is always constant (PV = k). For a given mass of a gas, the value of k varies only with temperature.
[Note: Each curve is an isotherm as it is plotted at constant temperature, (iso = constant, therm = temperature).]

ii. Graph of PV versus pressure (P) of a gas constant temperature:
If the product of pressure and volume (PV) is plotted against pressure (P), a straight line is obtained parallel to x-axis (pressure axis).

iii. Graph of pressure (P) of a gas versus reciprocal of volume (1/V) at constant temperature:
If the pressure (P) of the gas is plotted against (1/V), a straight line is obtained passing through the origin.

[Note: At high pressure, deviation from Boyle’s law is observed in the behaviour of gases.]

Question 29.
Derive the relationship between density and pressure.
Answer:
Relationship between density and pressure:
With increase in pressure, gas molecules get closer and the density (d) of the gas increases. Hence, at constant temperature, pressure is directly proportional to the density of a fixed mass of gas.
From Boyle’s law,
PV = k₁ …….(1)
∴ V = \(\frac{\mathrm{k}_{1}}{\mathrm{P}}\)
But, d = \(\frac{\mathrm{m}}{\mathrm{V}}\)
On substituting, V from equation (2),
d = \(\frac{\mathrm{m}}{\mathrm{K}_{1}}\) × P
∴ d = k P …….(3)
where k = New constant
∴ d ∝ P
Above equation shows that at constant temperature, the pressure is directly proportional to the density of a fixed mass of the gas.

Question 30.
Write a short note on absolute temperature scale.
Answer:
Absolute temperature scale:

• Absolute temperature scale is related to Celsius temperature scale by the equation:
  T K = t °C + 273.15
• This also called thermodynamic scale of temperature.
• The units of this absolute temperature scale is called (K) in the honour of Lord Kelvin who determined the accurate value of absolute zero as -273.15 °C in the year 1854.

Question 31.
Explain Charles’ law with the help of a diagram.
Answer:
Charles’ law (Temperature-Volume relationship):
i. Statement: At constant pressure, the volume of a fixed mass of a gas is directly proportional to its temperature in Kelvin.

ii. Explanation:
For an increase of every degree of temperature, volume of the gas increases by \(\frac{1}{273.15}\) of its original value at 0 °C. This is expressed mathematically as follows:
V_t = V₀ + \(\frac{t}{273.15} V_{0}\) ………….(1)
Where V_t and V₀ are the volumes of the given mass of gas at the temperatures t °C and 0 °C. Rearranging the Eq. (1) gives

The equation (3) on rearrangement takes the following form:
\(\frac{\mathrm{V}_{\mathrm{t}}}{\mathrm{T}_{\mathrm{t}}}=\frac{\mathrm{V}_{0}}{\mathrm{~T}_{0}}\)
From this, a general equation can be written as follows:
\(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) …………(4)
\(\frac{\mathrm{V}}{\mathrm{T}}\) = K₂ = constant (at constant P and n)
∴ V = k₂T OR V ∝ T ……(5)
The equation (4) is the mathematical expression of Charles’ law.

iii. Schematic illustration of Charles’ law:

This shows that at constant pressure, gases expand on heating and contract on cooling.

Question 32.
Give the different graphical representation of Charles’ law.
Answer:
Graph of volume versus temperature at constant pressure:

• According to Charles’ law, the graph of volume of a gas (at given constant pressure, say P₁) versus its temperature in Celsius, is a straight line with a positive slope.
• On extending the line to zero volume, the line intercepts the temperature axis at -273.15 °C.
• At any other value of pressure, say P₂, a different straight line for the volume temperature plot is obtained, but we get the same zero-volume temperature intercept at -273.15 °C.
• The straight line of the volume versus temperature graph at constant pressure is called isobar.

[Note: Zero volume for a gas sample is a hypothetical state. In practice, all the gases get liquified at a temperature higher than -273.15 °C. This temperature is the lowest temperature that can be imagined but practically cannot be attained. It is the absolute zero temperature on the Kelvin scale (0 K).]

Question 33.
Write the statement for Gay-Lussac’s law.
Answer:
Statement for Gay-Lussac’s law:
At constant volume, pressure (P) of a fixed amount of a gas is directly proportional to its absolute temperature (T).

Question 34.
Give the mathematical expression for Gay-Lussac’s law.
Answer:
Gay-Lussac’s law (Pressure-Temperature relationship):
i. Statement: At constant volume, pressure (P) of a fixed amount of a gas is directly proportional to its absolute temperature (T).
ii. Explanation:
Gay-Lussac’s law can be mathematically expressed as:
P ∝ T
∴ P = k₃T
∴ \(\frac{\mathrm{P}}{\mathrm{T}}\) = Constant (at constant V and n)
Thus, according to Gay-Lussac’s law,
\(\frac{\mathrm{P}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2}}{\mathrm{~T}_{2}}\) = constant

Question 35.
Give the graphical representation of Gay-Lussac’s law.
Answer:
Graph of pressure versus temperature of a gas at constant volume:
When a graph is plotted between pressure (P) in atm and temperature (T) in kelvin, a straight line is obtained It is known as isochore.

Question 36.
Define the following terms:
i. Isotherm
ii. Isobar
iii. Isochore
Answer:
i. A graph of pressure (P) versus volume (V) at a constant temperature is known as isotherm.
ii. A graph of volume (V) versus absolute temperature (T) at a constant pressure is known as isobar.
iii. A graph of pressure (P) versus absolute temperature (T) at a constant volume is known as isochore.

Question 37.
State and explain Avogadro law.
Answer:
Avogadro law (Volume-Amount relationship):
i. Statement: Equal volumes of all gases at the same temperature and pressure contain equal number of molecules.
ii. Explanation:
V is directly proportional to n (number of moles) at constant ‘P’ and ‘T’.
V ∝ n
V = k₄ × n (where, k₄ is proportionality constant)
∴ \(\frac{\mathrm{V}}{\mathrm{n}}\) = constant (at constant T and P)
Note: Representation of Avogadro law

Question 38.
What is molar volume?
Answer:
The volume occupied by one mole of an ideal gas at STP is 22.414 L. This volume is known as molar volume.

Question 39.
Derive the relation between density of a gas and its molar mass.
Answer:
Relation between density of a gas and its molar mass:
According to Avogadro’s law, V ∝ n
Now, n = \(\frac{\mathrm{m}}{\mathrm{M}}\) (where, m is the mass of the gas and M is the molar mass of the gas)
∴ V ∝ \(\frac{\mathrm{m}}{\mathrm{M}}\)
M ∝ \(\frac{\mathrm{m}}{\mathrm{V}}\)
But, \(\frac{\mathrm{m}}{\mathrm{V}}\) = d (where, d is the density of the gas)
∴ d ∝ M
Thus, density of a gas is directly proportional to its molar mass.

Question 40.
The volume occupied by a given mass of a gas at 298 K is 25 mL at 1 atmosphere pressure.
Calculate the volume of the gas if pressure is increased to 1.25 atmosphere at constant temperature.
Solution:
Given: P₁ = Initial pressure = 1 atm, V₁ = Initial volume = 25 mL
P₂ = Final pressure = 1.25 atm
To find: V₂ = Final volume of the gas
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law, P₁V₁ = P₂V₂
Substituting the values of P₁, V₁ and P₂ in the above expression, we get
V₂ = \(\frac{P_{1} V_{1}}{P_{2}}=\frac{1 \times 25}{1.25}\) = 20 mL
Ans: The volume occupied by the gas is 20 mL.

Question 41.
The volume of a given mass of a gas is 0.6 dm3 at a pressure of 2 atm. Calculate the volume of the gas if its pressure is increased to 2.4 at the same temperature.
Solution:
Given: P₁ = Initial pressure = 2 atm
V₁ = Initial volume of given mass of the gas = 0.6 dm³
P₂ = Final pressure = 2.4 atm
To find: V₂ = Final volume of the gas
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
V₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{P}_{2}}=\frac{2 \times 0.6}{2.4}=0.5 \mathrm{dm}^{3}\)
Ans: The volume of the given gas is 0.5 dm³.

Question 42.
What will be the minimum pressure required to compress 500 dm³ of air at 5 bar to 200 dm³ at 25 °C.
Solution:
Given: P₁ = Initial pressure = 5 bar
V₁ = Initial volume = 500 dm³; V₂ = Final volume = 200 dm³
To find: P₂ = Final pressure
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
∴ P₂ = \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~V}_{2}}=\frac{5 \times 500}{200}\) = 12.5 bar
Ans: The minimum pressure required to compress 500 dm³ of air at 5 bar to 200 dm³ at 25 °C is 12.5 bar.

Question 43.
A balloon has certain volume at sea level. At what pressure (in kPa) will its volume be increased by 40% if the temperature is kept constant?
Solution:
Given: P₁ = Initial pressure = 101.325 kPa (∵ The pressure at sea level = 101.325 kPa)
V₁ = Initial volume at sea level = 100 dm³ (assumption)
V₂ = Final volume = (100 + 40) = 140 dm³
To find: P₂ = Final pressure
Formula: P₁V₁ = P₂V₂ (at constant n and T)
Calculation: According to Boyle’s law,
P₁V₁ = P₂V₂ (at constant n and T)
P₂ = \(\frac{P_{1} V_{1}}{V_{2}}\)
∴ P₂ = \(\frac{101.325 \times 100}{140}\) = 72.375 kPa
Ans: The pressure at which volume of the given balloon will be increased by 40% at a given temperature is 72.375 kPa.

Question 44.
At 300 K, a certain mass of a gas occupies 1 × 10^-4 dm³ volume. Calculate its volume at 450 K and at the same pressure.
Solution:
Given: T₁ = Initial temperature = 300 K, V₁ = Initial volume = 1 × 10^-4 dm³,
T₂ = Final temperature = 450 K
To find: V₂ = Final volume
Formula: \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (at constant n and P)
Calculation: According to Charles’ law, at constant pressure.

Ans: The volume of given gas becomes 1.5 × 10^-4 dm³ at the temperature of 450 K and same pressure.

Question 45.
A certain mass of a gas occupies a volume of 0.2 dm³ at the temperature, x K. Calculate the volume of the gas if its absolute temperature is doubled at same pressure.
Solution:
Given: V₁ = Initial volume = 0.2 dm³, T₁ = Initial temperature = x K
T₂ = Final temperature = 2 × x = 2x K
To find: V₂ = Final volume of the gas
\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The volume of given gas becomes 0.4 dm³ when the temperature is doubled.

Question 46.
The volume of a given mass of a gas at 0 °C is 0.2 dm³. Calculate its volume at 100 °C, if the pressure remains the same.
Solution:
Given: V₁ = Initial volume = 0.2 dm³, T₁ = Initial temperature = 0 °C = 273.15 K,
T₂ = Final temperature = 100 °C = 100 + 273.15 K = 373.15 K
To find: V₂ = Volume at 100 °C
Formula: \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\) (at constant n and P)
Calculation: According to Charles’ law,

Ans: The volume of gas at 100 °C is 0.273 dm³.

Question 47.
A glass container is sealed with a gas at 0.8 atm pressure and at 25 °C. The glass container sustains a pressure of 2 atm. Calculate the temperature to which the gas can be heated before bursting the container.
Solution:
Given: P₁ = Initial pressure = 0.8 atm, P₂ = Final pressure = 2 atm
T₁ = Initial temperature = 25 °C = 25 + 273.15 K = 298.15 K
To find: T₂= Final temperature

Ans: The temperature to which the gas can be heated before bursting the container is 472 °C.

Question 48.
An 8.0 L of sample at 0 °C and 5.6 atm of pressure contains 2.0 moles of a gas. If more 1.0 mole of gas at the same temperature and pressure is added, calculate the final volume.
Solution:
Given: V₁ = Initial volume = 8.0 L
n₁ = Initial mol = 2.0 mol
n₂ = Final mol = (2.0 + 1.0) = 0.3 mol
To find: V₂ = Final volume

Ans: The final volume is 12 L.

Question 49.
What is an ideal gas equation?
Answer:
Ideal gas equation is obtained by combining three gas laws, namely, Boyle’s law, Charles’ law and Avogadro law. Mathematically, it is given as:
PV = nRT
where,
P = Pressure of gas, V = Volume of gas, n = number of moles of gas,
R = Gas constant, T = Absolute temperature of gas

Question 50.
Derive the ideal gas equation.
Answer:
According to Boyle’s law,
V ∝ \(\frac{1}{\mathrm{P}}\) (at constant T and n) ………..(1)
According to Charles’ law,
V ∝ T (at constant P and n) …….(2)
According to Avogadro’s law,
V ∝ n (at constant P and T) …….(3)
Combining relations (1), (2) and (3), we get
V ∝ \(\frac{\mathrm{nT}}{\mathrm{P}}\)
Converting this proportionality into an equation by introducing a constant of proportionality (‘R’ known as gas constant), we get
∴ V = \(\frac{\mathrm{nRT}}{\mathrm{P}}\)
On rearranging the above equation, we get
PV = nRT
where,
P = Pressure of gas,
V = Volume of gas,
n = number of moles of gas,
R = Gas constant,
T = Absolute temperature of gas.
This is the ideal gas equation or equation of state.
[Note: In the ideal gas equation, R is called gas constant or universal gas constant, whose value is same for all the gases. In this equation, if three variables are known, fourth can be calculated. The equation describes the state of an ideal gas. Hence, it is also called as equation of state.]

Question 51.
Deduce values of gas constant ‘R’ in different units.
Answer:
i. R in SI Unit (in Joules): Value of R can be calculated by using the SI units of P, V and T. Pressure P is measured in N m^-2 or Pa, volume V in meter cube (m³) and temperature T in Kelvin (K).

ii. R in litre atmosphere: If pressure (P) is expressed in atmosphere (atm) and volume in litre (L) or decimeter cube (dm³) and Temperature in kelvin (K), (that is, old STP conditions), then value of R is,
R = \(\frac{1 \mathrm{~atm} \times 22.414 \mathrm{~L}}{1 \mathrm{~mol} \times 273.15 \mathrm{~K}}\)
∴ R = 0.0821 L atm K^-1 mol^-1
OR
R = 0.0821 dm³ atm K^-1 mol^-1

iii. R in calories: We know, 1 calorie = 4.184 Joules
∴ R= \(\frac{8.314}{4.184}\) = 1.987 ≅ 2 cal K^-1 mol^-1

Question 52.
Derive the following expression:
M = \(\frac{\text { mRT }}{\text { PV }}\)
Answer:
According to ideal gas equation,
PV = nRT
n = \(\frac{\mathrm{PV}}{\mathrm{RT}}\)
Now, for a known mass ‘m’ of gas having molar mass ‘M’, number of moles ‘n’ is given as:
n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
Therefore, \(\frac{m}{M}=\frac{P V}{R T}\)
On rearranging the equation, we get
M = \(\frac{\text { mRT }}{\text { PV }}\)

Question 53.
Derive the expression for combined gas law.
Answer:
The ideal gas equation is written as
PV = nRT …….(1)
On rearranging equation (1), we get,

The ideal gas equation used in this form is called combined gas law.

Question 54.
Derive the relation between density, molar mass and pressure.
Answer:
Relation between density, molar mass and pressure:
According to ideal gas equation,
PV = nRT …..(1)
On rearranging equation (1), we get
\(\frac{\mathrm{n}}{\mathrm{V}}=\frac{\mathrm{P}}{\mathrm{RT}}\) ……….(2)
Now, n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
On substituting the value of n, equation (2) becomes
\(\frac{\mathrm{m}}{\mathrm{MV}}=\frac{\mathrm{P}}{\mathrm{RT}}\)
\(\frac{\mathrm{d}}{\mathrm{M}}=\frac{\mathrm{P}}{\mathrm{RT}}\) ……….(3)
where d = \(\frac{\mathrm{m}}{\mathrm{V}}\) = density of the gas
On rearranging the equation, we get
M = \(\frac{\mathrm{dRT}}{\mathrm{P}}\) ……(4)
This equation can be used to calculate molar mass of a gas in terms of its density.

Question 55.
State Boyle’s law in terms of density.
Answer:
Boyle’s law in terms of density is stated as ‘At constant temperature, pressure of a given mass of gas is directly proportional to its density’.

Question 56.
Derive the expression that relates partial pressure with mole fraction of a gas.
Answer:
The partial pressures of individual gases can be written in terms of ideal gas equation as follows:


Thus, partial pressure of a gas is obtained by multiplying the total pressure of mixture by mole fraction of that gas.

Question 57.
What is water vapour?
Answer:
The ‘gas’ above the surface of liquid water is described as water vapour.

Question 58.
Write a short note on aqueous tension.
Answer:
Aqueous tension:
i. When the liquid water is placed into a container and air above is pumped away and the container is sealed, then the liquid water evaporates and only water vapour remains in the above space. After sealing, the vapour pressure increases initially, then slows down as some water molecules condense back to form liquid water. After a few minutes, the vapour pressure reaches a maximum value, which is called the saturated vapour pressure. The pressure exerted by saturated water vapour is called aqueous tension (P_aq).
ii. Aqueous tension increases with increase in temperature.

Question 59.
Explain how pressure of a dry gas can be calculated using aqueous tension.
Answer:
i. When a gas is collected over water in a closed container, it gets mixed with the saturated water vapour in that space. Therefore, the measured pressure corresponds to the pressure of the mixture of that gas and the saturated water vapour in that space.
ii. Pressure of pure and dry gas can be calculated by using the aqueous tension. It is obtained by subtracting the aqueous tension from the total pressure of the moist gas.
∴ P_{Dry gas} = P_Total – P_aq
i.e., P_{Dry gas} = P_Total – Aqueous Tension
Note: Aqueous tension of water (vapour pressure) as a function of temperature.

Question 60.
A sample of N₂ gas was placed in a flexible 9.0 L container at 300 K at a pressure of 1.5 atm. The gas was compressed to a volume of 3.0 L and heat was added until the temperature reached 600 K. What is the new pressure inside the container?
Solution:
Given: V₁ = Initial volume = 9.0 L, V₂ = Final volume = 3.0 L,
P₁ = Initial pressure = 1.5 atm
T₁ = Initial temperature = 300 K, T₂ = Final temperature = 600 K
To find: P₂ = Final pressure
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law,

Ans: The new pressure inside the container is 9 atm.

Question 61.
A gas at 772 mm Hg and at 35 °C occupies a volume of 6.851 L. Calculate its volume at STP.
Solution:
Given: V₁ = Initial volume = 6.851 L
P₁ = Initial pressure = 772 mm Hg, P₂ = Final pressure = 760 mm Hg
T₁ = Initial temperature = 35 °C = 35 + 273.15 K = 308.15 K
T₂ = Final temperature = 273.15 K
To find: V₂ = Final volume
Formula: \(\frac{\mathrm{P}_{1} \mathrm{~V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}}{\mathrm{~T}_{2}}\)
Calculation: According to combined gas law,

Ans: The volume of gas at STP is 6.169 L.

Question 62.
Find the temperature in °C at which volume and pressure of 1 mol of nitrogen gas becomes 10 dm³ and 2.46 atmosphere respectively.
Solution.
P = 2.46 atm, V = 10 dm³, n = 1 mol, R = 0.0821 dm³-atm K^-1 mol^-1
To find: Temperature (T)
Formula: PV = nRT
According to ideal gas equation,
PV = nRT
∴ T = \(\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{2.46 \times 10}{1 \times 0.0821}\)
T = 299.63 K
Temp, in °C = 299.63 – 273.15 = 26.48 °C
Ans: The temperature of the nitrogen gas under the given conditions is 26.48 °C.

Question 63.
Calculate the temperature of 5.0 mol of a gas occupying 5 dm³ at 3.32 bar.
(R = 0.083 bar dm³ K^-1 mol^-1)
Solution:
Given: n = number of moles = 5.0 mol, V = volume = 5 dm³
P = pressure = 3.32 bar, R = 0.083 bar dm³ K^-1 mol^-1
To find: Temperature (T)
Formula: PV = nRT
Calculation: According to ideal gas equation,
PV = nRT
∴ T = \(\frac{\mathrm{PV}}{\mathrm{nR}}=\frac{3.32 \times 5}{5.0 \times 0.083}\) = 40 K
Ans: The temperature of the gas is 40 K.

Question 64.
Calculate the number of moles of hydrogen gas present in a 0.5 dm³ sample of hydrogen gas at a pressure of 101.325 kPa at 27 °C.
Solution:
Given: V = 0.5 dm³ = 0.5 × 10^-3 m³, P = 101.325 kPa = 101.325 × 10³ Pa = 101.325 × 10³ Nm^-2
T = 27 °C = 27 + 273.15 K = 300.15 K, R = 8.314 J K^-1 mol^-1
To find: n = number of moles of gas
Formula: PV = nRT
Calculation: According to ideal gas equation,

Ans: The number of moles of hydrogen gas present in the given volume is 0.020 moles.

Question 65.
A mixture of 28 g N₂, 8 g He and 40 g Ne has 20 bar pressure. What is the partial pressure of each of these gases?
Solution:
Given: m_N2 = 28 g, m_He = 8 g, m_Ne = 40 g,
P_Total = 20 bar
To find: Partial pressure of each gas
Formula: P₁ = x₁ × P_Total
Calculation: Determine the number of moles (n) of each gas using the formula: n = \(\frac{\mathrm{m}}{\mathrm{M}}\)
Determine the mole fraction of each gas using the formula:

Ans: The partial pressure of nitrogen, helium and neon are 4 bar, 8 bar and 8 bar respectively.

Question 66.
What is an ideal gas?
Answer:
Ideal gas:

• The gases which obey’ ideal gas equation over a complete range of temperature and pressure are called ideal gases.
• For an ideal gas, the ratio of PV/RT = 1.
• In an ideal gas, there are no interactive forces between the molecules and the molecular volume is negligibly small compared to the volume occupied by the gas. The gas particles are considered as point particles.

Question 67.
What are real gases?
Answer:
Real gases:

• Gases, which do not obey ideal gas equation under all the conditions of temperature and pressure are called real gases.
• For real gases, the ratio of PV/RT will be either greater than 1 or less than 1.
• Real gases show deviation from ideal gas behaviour at higher pressures and lower temperatures.
• The intermolecular attractive forces are not negligible in real gases.
• In real gases, the actual volume of the molecules cannot be neglected as compared to the total volume of the container.

Question 68.
Explain the reason for deviations of gases from ideal behaviour.
Answer:
A deviation from the ideal behaviour is observed at high pressure and low temperature. It is due to two reasons.

• The intermolecular attractive forces are not negligible in real gases. These do not allow the molecules to collide the container wall with full impact. This results in decrease in the pressure.
• At high pressure, the molecules are very close to each other. The short range repulsive forces start operating and the molecules behave as small but hard spherical particles. The volume of the molecule is not negligible.
  Therefore, very less volume is available for molecular motion.
• At very low temperature, the molecular motion becomes slow and the molecules are attracted to each other due to the attractive force. Hence, the behaviour of the real gas deviates from the ideal gas behaviour.
• Deviation with respect to pressure can be studied by plotting pressure (P) vs volume (V) curve at a given temperature.
• From the graph, it is clear that at very high pressure, the measured volume is more than theoretically calculated volume assuming ideal behaviour. However, at low pressure, measured and theoretically calculated volumes approach each other.

Question 69.
What is compressibility factor (Z)?
Answer:
Compressibility factor (Z):
i. It is defined as the ratio of product PV and nRT.
Z = \(\frac{\mathrm{PV}}{\mathrm{nRT}}\)
ii. Deviation from ideal behaviour is measured in terms of compressibility factor.
iii. For ideal gases, Z = 1 under all conditions of temperature and pressure. Therefore, the graph of Z versus P will be a straight line parallel to pressure axis.
iv. For gases that deviate from ideal behaviour, value of Z deviates from unity.
Note: Variation of compressibility factor for some gases

Question 70.
Show that the compressibility factor can be represented as Z = \(\frac{V_{\text {real }}}{V_{\text {ideal }}}\)
Answer:
For real gas,

Thus, the compressibility factor (Z) is the ratio of actual molar volume of a gas to its molar volume if it behaved ideally, at that temperature and pressure.

Question 71.
Explain: Liquefaction of CO₂ with the help of pressure vs volume isotherm.
Answer:
Most gases behave like ideal gases at high temperature. For example, the PV curve of CO₂ gas at 50 °C is like the ideal Boyle’s law curve. As the temperature is lowered, the PV curve shows a deviation from the ideal Boyle’s law curve. At a particular value of low temperature, the gas gets liquified at certain increased value of pressure. For example, CO₂ gas liquifies at 38.98 °C and 73 atmosphere pressure. This is the highest temperature at which liquid CO₂ can exist. Above this temperature, liquid CO₂ cannot form even if very high pressure is applied. Other gases also show similar behaviour.

Question 72.
Define: Critical temperature, critical volume and critical pressure.
Answer:
i. The temperature above which a substance cannot be liquified by increasing pressure is called its critical temperature (T_c).
ii. The molar volume at critical temperature is called the critical volume (V_c).
iii. The pressure at the critical temperature is called the critical pressure (P_c).

Note: Critical constants for common gases

i. N₂ and O₂, have T_c values much below 0 °C and their P_c values are high. Consequently, liquefaction of O₂ and N₂ (and air) requires compression and cooling.
ii. The T_c value of CO₂ nearly equals the room temperature; however, its P_c value is very high. Therefore, CO₂ exists as gas under ordinary condition.

Question 73.
Water has T_c = 647.1 K and P_c = 220.6 bar. What do these values imply about the state of water under ordinary conditions?
Answer:
The T_c and P_c values of water are very high compared to the room temperature and common atmospheric pressure. As a result, water exists in liquid state under ordinary condition of temperature and pressure.

Question 74.
Gases possess characteristic critical temperature which depends upon the magnitude of intermolecular forces between the gas particles. Critical temperatures of ammonia and carbon dioxide are 405.5 K and 304.10 K respectively. Which of these gases will liquefy first when you start cooling from 500 K to their critical temperature?
Answer:
When cooling of ammonia and carbon dioxide gas is started from 500 K, then ammonia reaches its critical temperature first (i.e., 405.5 K.) and hence, it is also the first to get liquefied.
When the cooling is continued further, carbon dioxide gas is liquefied as it reaches its critical temperature (i.e., 304.10 K).

Question 75.
CO₂ has T_c = 38.98 °C and P_c = 73 atm. How many phases of CO₂ coexist at
i. 50 °C and 73 atm
ii. 20 °C and 50 atm.
Answer:
i. 50 °C and 73 atm represent a condition for CO₂ above its T_c. Therefore, under this condition CO₂ exists only as single phase.
ii. 20° C and 50 atm represent a condition for CO₂ below its T_c. Therefore, under this condition two phases of CO₂, namely, liquid and gas can coexist.

Question 76.
In which of the following cases, water will have the highest and the lowest boiling point?
i. Water is boiled in an open vessel.
ii. Water is boiled in a pressure cooker.
iii. Water is boiled in an evacuated vessel.
Answer:
Higher the pressure to which a liquid is exposed, higher will be its boiling point. The pressure to which water is exposed is maximum in the pressure cooker and minimum in the evacuated vessel. Therefore, boiling point of water is highest in a pressure cooker and lowest in an evacuated vessel.

Question 77.
Define: Liquid state
Answer:
Liquid state is the intermediate state between solid state and gaseous state.

Question 78.
Give reason: Liquid possesses properties such as fluidity, definite volume and ability to take shape of the bottom of the container in which it is placed.
Answer:
Molecules of liquid are held together by moderately strong intermolecular forces and can move about within the boundary of the liquid. As a result, liquid possesses properties such as fluidity, definite volume and ability to take shape of the bottom of the container in which it is placed.

Question 79.
Name some measurable properties of liquid.
Answer:

• Density
• Boiling point
• Freezing point
• Vapour pressure
• Surface tension
• Viscosity.

Question 80.
What are the factors affecting vapour pressure?
Answer:
Factors affecting vapour pressure:

• Nature of liquid: Liquids having relatively weak intermolecular forces possess high vapour pressure. Such liquids are called volatile liquids.
  e. g. Petrol evaporates quickly than motor oil. Hence, petrol has higher vapour pressure than motor oil.
• Temperature: When the liquid is gradually heated, its temperature rises and its vapour pressure increases.

Question 81.
Explain how temperature affects surface tension.
Answer:
Surface tension is a temperature dependent property. When attractive forces are large, surface tension is large. Surface tension decreases as the temperature increases. With increase in temperature, kinetic energy of molecules increases. So, intermolecular forces of attraction decrease, and thereby surface tension decreases.

Question 82.
Mention some applications of surface tension.
Answer:
Applications of surface tension:

• Cleaning action of soap and detergent is due to the lowering of interfacial tension between water and oily substances. Due to lower surface tension, the soap solution penetrates into the fibre, surrounds the oily substance and washes it away.
• Efficacy of toothpastes, mouthwashes and nasal drops is partly due to presence of substances having lower surface tension. This increases the efficiency of their penetrating action.

Question 83.
Give reason: Liquid droplets acquire spherical shape.
Answer:
For a given volume of a liquid, spherical shape always imparts minimum surface area thereby reducing the surface tension. Hence, liquid droplets acquire spherical shape.

Question 84.
Define: Coefficient of viscosity
Answer:
Coefficient of viscosity is defined as the degree to which a fluid resists flow under an applied force, measured by the tangential frictional force per unit area per unit velocity gradient when the flow is laminar.

Question 85.
Describe various factors affecting viscosity of a liquid.
Answer:
Factors affecting viscosity of a liquid:
i. Temperature: Viscosity is a temperature dependent property.
Viscosity ∝ \(\frac{1}{\text { Temperature }}\)
ii. Nature of liquid: Viscosity also depends on molecular size and shape. Larger molecules have more viscosity and spherical molecules offer the least resistance to flow and therefore are less viscous. Greater the viscosity, slower is the liquid flow.

Question 86.
Describe three daily life instances where viscosity plays an important role.
Answer:

• Lubricating oils are viscous liquids. Gradation of lubricating oils is done on the basis of viscosity. A good quality lubricating oil does not change its viscosity with increase or decrease in temperature.
• Increase blood viscosity than the normal value is taken as an indication of cardiovascular disease.
• Glass panes of old buildings are found to become thicker with time near the bottom. This indicates that glass is not a solid but a supercooled viscous liquid.

Question 87.
For an experiment, a scientist fills different gases in four flasks as shown below:

i. What is the ratio of the number of molecules of the gases in flask A to flask B?
ii. Calculate the pressure exerted by nitrogen gas in flask B if the temperature is doubled.
iii. If the scientist transfers the gas in flask D to another flask of 2.5 L at 1 atm pressure, what will be the temperature of the gas in the new flask?
Answer:
i. 1:1
ii. P ∝ T (when V and n are constants)
∴ If temperature is doubled, pressure also doubles.
∴ P = 2 atm
iii. \(\frac{\mathrm{V}_{1}}{\mathrm{~T}_{1}}=\frac{\mathrm{V}_{2}}{\mathrm{~T}_{2}}\) (when P and n are constants)
∴ \(\frac{1}{298}=\frac{2.5}{T_{2}}\)
∴ T₂ = 745 K

Question 88.
A balloon containing 0.6 mol of helium gas has a volume of 1.5 L.
i. Assuming that the temperature and pressure remains constant, what happens to the volume of the balloon if an additional 0.6 mol of helium is added?
ii. Assuming that the temperature and pressure remains constant, what happens to the volume of the balloon if 0.3 mol of helium is removed?
Answer:
i. The volume of the balloon increases.
ii. The volume of the balloon decreases.

Question 89.
In an experiment conducted to study the diffusion of gases using same experimental conditions, following data were recorded.
Gas A: 50 cm³ of gas A takes 7 minutes to diffuse from one container to the adjacent container.
Gas B: 50 cm³ of gas B takes 10 minutes to diffuse from one container to the adjacent container.
i. What is the rate of diffusion of gas A?
ii. What is the rate of diffusion of gas B?
iii. Which gas has higher molecular mass?
Answer:
i. Volume of gas A diffused = 50 cm³
Time required for diffusion = 7 minutes = 7 × 60 seconds

∴ The rate of diffusion of gas A is 0.12 cm³ s^-1.
ii. The rate of diffusion of gas B is 0.083 cm³ s^-1.
iii. Gas B has higher molecular mass.

Multiple Choice Questions

1. Which of the following is CORRECT for both gases and liquids?
(A) Indefinite volume
(B) Definite shape
(C) Indefinite shape
(D) Definite volume
Answer:
(D) Definite volume

2. The composition of …………. in air is about 78% by volume.
(A) CO₂
(B) O₂
(C) N₂
(D) Ar
Answer:
(C) N₂

3. Which of the following expression at constant pressure represents Charles’s law?
(A) V ∝ \(\frac{1}{\mathrm{~T}}\)
(B) V ∝ \(\frac{1}{\mathrm{~T}^{2}}\)
(C) V ∝ T
(D) V ∝ d
Answer:
(C) V ∝ T

4. The pressure of 2 mole of ideal gas at 546 K having volume 44.8 L is …………….
(A) 2 atm
(B) 3 atm
(C) 7 atm
(D) 1 atm 10²³
Answer:
(A) 2 atm

5. At 300 K, the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen at 4 bar. The molar mass of gaseous molecule is …………….
(A) 28 g mol^-1
(B) 56 g mol^-1
(C) 112 g mol^-1
(D) 224 g mol^-1
Answer:
(C) 112 g mol^-1

6. As the temperature increases, average kinetic energy of molecules increases. What would be the effect of increase of temperature on pressure provided the volume is constant?
(A) Increases
(B) Decreases
(C) Remains same
(D) Becomes half
Answer:
(A) Increases

7. Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is ………….. (Atomic wt. of Cl = 35.5 u)
(A) 1.46
(B) 0.46
(C) 1.64
(D) 0.64
Answer:
(B) 0.46

8. The number of moles of H₂ in 0.224 L of hydrogen gas at STP (273 K, 1 atm) assuming ideal gas behaviour is …………..
(A) 1
(B) 0.1
(C) 0.01
(D) 0.001
Answer:
(C) 0.01

9. The volume occupied by 11.5 g of carbon dioxide at STP is approximately equal to:
(A) 5.9 L
(B) 22.5 L
(C) 86 L
(D) 259 L
Answer:
(A) 5.9 L

10. Which of the following is CORRECT regarding a fixed amount of ideal gas?
(A) Doubling the temperature, doubles the volume, provided the pressure remains the same.
(B) Doubling the temperature, halves the volume, provided the pressure remains the same.
(C) Doubling the pressure, doubles the volume, provided the temperature remains the same.
(D) Doubling the volume, doubles the pressure, provided the temperature remains the same.
Answer:
(A) Doubling the temperature, doubles the volume, provided the pressure remains the same.

11. When one mole of an ideal gas is heated from 300 K to 360 K at constant pressure of 1 atm, its volume …………….
(A) increases from V to 6.0V
(B) increases from V to 3.6V
(C) increases from V to 1.2V
(D) increases from V to 1.6V
Answer:
(C) increases from V to 1.2V

12. The partial pressure of a gas is obtained by multiplying the total pressure of mixture by …………… of that gas.
(A) molar mass
(B) moles
(C) mass
(D) mole fraction
Answer:
(D) mole fraction

13. The highest temperature at which liquid CO₂ can exist is ……………..
(A) 18.98 °C
(B) 38.98 °C
(C) 50.0 °C
(D) 73.9 °C
Answer:
(B) 38.98 °C

14. The SI unit of surface tension is ……………..
(A) Pascal
(B) N s m^-2
(C) km^-2 s
(D) N m^-1
Answer:
(D) N m^-1