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Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution

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Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution

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Selina ICSE Solutions for Class 9 Chemistry Chapter 8 Atmospheric Pollution

Page No: 128

Question 1.
Define the following terms:
pollution
pollutant
air pollution
Solution:
Pollution may be defined as contamination of air, water or soil by undesirable amounts of materials or heat and is caused by the concentration of substances which have harmful effects.

Toxic and otherwise harmful substances which have an undesirable impact on different components of the environment and life forms are known as pollutants.

Air pollution means degradation of air quality due to concentration of harmful contaminants which affect human, plant and animal lives.

Question 2.
Name any four gaseous pollutants.
Solution:
Sulphur dioxide, hydrogen sulphide, carbon monoxide and hydrocarbons.

Question 3.
Name the compounds of sulphur that cause air pollution. Also state the harmful effects of sulphur compounds.
Solution:
Compounds of sulphur like sulphur dioxide, sulphur trioxide and hydrogen sulphide are pollutants.

Harmful effects of oxides of sulphur:

  • It causes headache, vomiting and even death due to respiratory failure.
  • It destroys vegetation and weakens building materials/constructions.
  • It mixes with smoke and fog to form smog, which is very harmful.

It is oxidised by atmospheric oxygen into sulphur trioxide (SO3) which combines with water to form sulphuricacid.
2SO2(g) + O2(g) →2SO3(g)
SO3(g) + H2O(l) → H2SO4(l)

Question 4.
State:
natural sources of air pollution
man-made sources of air pollution
Solution:
Volcanoes, decaying vegetation, forest fires and dust storms.
Automobiles, factories, industrial processes and decay of crop residue in rural areas.

Question 5.
a. How do oxides of nitrogen enter the atmosphere?
b. What are their harmful effects?
Solution:
Nitric oxide (NO) and nitrogen dioxide (NO2) enter the atmosphere in the following ways:

  1. On burning of fuels in furnaces, temperature increases. At high temperature, nitrogen and oxygen present in air combine to form oxides of nitrogen.
  2. When fuel burns in an internal combustion engine, oxides of nitrogen are produced, and they enter the atmosphere as exhaust gases from automobile engines.
  3. Nitric acid is formed by the reaction between atmospheric nitrogen and oxygen in the presence of electric discharge, which occurs during thunderstorms when there is lightning.
  4. Nitric oxide further reacts with atmospheric oxygen and ozone to form nitrogen dioxide.

Harmful effects of the oxide of nitrogen

  • Nitrogen dioxide is very harmful to plants and animals.
  • It causes irritation in the mucous membrane.
  • Large concentrations of NO2 may cause serious lung diseases.
  • Nitrogen dioxide causes serious injury to vegetation; it damages plant leaves.
  • In sunlight, nitrogen dioxide oxidises hydrocarbons to form photochemical smog. Photochemical smog causes eye irritation, asthma attacks and nasal and throat infections.

Question 6.
State the origin and health impact of smog.
Solution:
Smog is a dark, thick, dust and soot-laden fog pollutant which is a combination of oxides of nitrogen and sulphur and of partially oxidised hydrocarbons and their derivatives produced by industries and automobiles.
Smog is noxious and irritating. It reduces visibility, induces respiratory troubles and can cause death by suffocation.

Question 7.
What are the harmful effects of oxides of sulphur?
Solution:
Harmful effects of oxides of sulphur:

  1. It causes headache, vomiting and even death due to respiratory failure.
  2. It destroys vegetation and weakens building materials/constructions.
  3. It mixes with smoke and fog to form smog, which is very harmful.

It is oxidised by atmospheric oxygen into sulphur trioxide (SO3) which combines with water to form sulphuricacid.

2SO2(g) + O2(g) →2SO3(g)
SO3(g) + H2O(l) → H2SO4(l)

Question 8.
State the main sources and effects of carbon monoxide.
Solution:
Carbon monoxide is formed by incomplete combustion of fuels in homes, factories and automobiles.

Effects of carbon monoxide are as follows:

  • It is a highly poisonous gas.
  • It reduces the oxygen-carrying capacity of blood by an amount equivalent to the amount of haemoglobin converted to carboxyhaemoglobin.
    Haemoglobin + CO → Carboxyhaemoglobin
  • Because heart and brain are the two tissues most sensitive to oxygen depletion, they show the most serious effects of carbon monoxide exposure.
  • In high concentrations, carbon monoxide may kill by paralysing normal brain action.

Question 9.
Give the mechanism of the action of carbon monoxide.
Solution:

  • It is a highly poisonous gas.
  • When inhaled, it passes through the lungs directly into the blood stream. There it combines with haemoglobin, the substance which carries oxygen to body tissues. Because haemoglobin binds with carbon monoxide 200 times more strongly than oxygen, even low concentrations of carbon monoxide in air have magnified effects on the body.
  • It reduces the oxygen-carrying capacity of blood by an amount equivalent to the amount of haemoglobin converted to carboxyhaemoglobin.
    Haemoglobin + CO → Carboxyhaemoglobin

Question 10.
How can we control carbon monoxide poisoning?
Solution:
Carbon monoxide pollution can be controlled in the following ways:

  1. By switching over from internal combustion engines to electrically powered cars.
    Many pollution control devices are now installed in cars. Most of these devices help reduce pollution by burning gasoline completely. Complete combustion of gasoline produces only carbon dioxide and water vapour.
    2C8H18 + 5O2 → 16CO2 + 18H2O
  2. By using substitute fuels for gasoline: Natural gas in both compressed (CNG) and liquefied forms (LNG) is now increasingly being used as fuel. Alcohols are other feasible substitutes.
  3. By using catalytic convertors: Nitrogen oxide is reduced to nitrogen and oxygen in the presence of finely divided platinum or palladium as a catalyst.
    Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 1

Page No: 130

Question 1.
Why does rain water have pH less than 7?
Solution:
Carbon dioxide reacts with water to form weak carbonic acid which is slightly acidic having pH about 5.6.
Hence, the pH of rain water usually ranges between 5.6 and 3.5; at times, it can be as low as 2.

Question 2.
pH of acid rain is sometimes as low as 2. Explain.
Solution:Normal rain is only slightly acidic having pH about 5.6.
This is because carbon dioxide reacts with it to form weak carbonic acid.
CO2 + H2O → H2CO3
pH of acid rain usually ranges between 5.6 and 3.5; at times, it can be as low as 2.

Question 3.
Explain the formation of acid rain due to:
1. Oxides of sulphur
2. Oxides of nitrogen
Solution:

  • Sulphur is a non-metallic element found in coal and fuel oil. When these fuels are burned, sulphur combines with oxygen in air to form its gaseous oxides, sulphur dioxide (SO2) and sulphur trioxide (SO3).
    S + O2 → SO2
    2SO2 + O2 → 2SO3
  • Sulphur dioxide and sulphur trioxide react with water to form H2SO4 which is the main cause of acid rain.
    2SO2 + O2 + 2H2O → 2H2SO4
    SO3 + H2O → H2SO4
  • Nitric acid is formed by the combination of nitrogen and oxygen. Nitrogen and oxygen combine in the presence of thunder and lightning. Oxides of nitrogen are also produced by internal combustion engines.
    N2 + O2 → 2NO
  • Nitrogen oxide then gets oxidised in the atmosphere to nitrogen dioxide.
    2NO +O2 → 2NO2
  • Nitrogen dioxide combines with water to form a mixture of nitrous acid and nitric acid.
    2NO2 + H2O → HNO2 + HNO3

Question 4.
What are the causes of acid rain?
Solution:
The main causes of acid rain are the formation of mineral acids such as carbonic acid, nitric acid and sulphuric acid during rains.

Question 5.
Give the impact of acid rain:
1. on plants
2. on soil
3. on water bodies
Solution:
Acid rain causes loss of nutrients from plants, thus damaging their leaves.
It removes calcium and potassium both basic ingredients of soil, thus making it lose its fertility which ultimately damages forests.
Acid rain has serious ecological impacts as it affects water bodies too. The water of lakes and rivers is gradually becoming acidic due to acid rain which is affecting aquatic life.

Question 6.
How does a scrubber help in reducing the formation of acid rain?
Solution:
A scrubber can also be used to reduce the formation of acid rain. It is a device which absorbs gaseous pollutants. It is used for removing sulphur dioxide from a smoke stack, and usually consists of a fine spray of water and gas rising from the stack, which is passed through the scrubber where water absorbs sulphur dioxide.

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Page No: 131

Question 1.
What do you understand by Green House effect?
Solution:
Heating of the Earth and its environment due to solar radiation trapped by carbon dioxide and water vapour in the atmosphere is called greenhouse effect.

Question 2.
What are green house gases? How are they responsible for global warming?
Solution:
Gases which contribute to the greenhouse effect are called greenhouse gases. These gases are carbon dioxide, water vapour, oxides of nitrogen, methane, ozone and chlorofluorocarbons. Sunlight reaching Earth consists of three types of radiation-UV radiation, visible radiation and IR radiation. As sunlight passes through the atmosphere, most UV radiation is absorbed by ozone; 30% of IR radiation reaches the Earth’s surface, heating it up. As the Earth’s surface becomes hot, it starts emitting radiation with less energy than the incoming radiation and thus with longer wavelength. Some emitted IR radiation escapes from the Earth’s surface and some are absorbed by CO2, thus remaining on the Earth. Trapped radiation warms the Earth’s surface and lower layers of the atmosphere.

Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 3
Question 3.
State the sources and effects of the following gases:
1. Carbon dioxide
2. Methane
3. Water vapour
Solution:
Sources of carbon dioxide:

  • Burning of fossil fuels such as coal, natural gas and petroleum
  • Industrial processes such as manufacture of lime and those in fermentation units
  • Biological decay of plants
  • Respiration by animals, human beings and plants

Effects of carbon dioxide:

  • Greenhouse effect and global warming

Sources of methane:

  • Anaerobic decomposition of organic matter in soil, water and sediments
  • Incomplete combustion of fossil fuels

Effects of methane:

  • Greenhouse effect and global warming

Sources of water vapour:

  • Burning of hydrocarbons
  • Evaporation and transpiration

Effects of water vapour:

  • Greenhouse effect and global warming

Question 4.
State the ways of reducing the presence of green house gases.
Solution:
Ways of reducing the presence of greenhouse gases:

  • Minimise the use of automobiles: Depending on the situation, one can use a bicycle, the public transport system and car pools.
  • Plant more trees to increase green cover.
  • Avoid burning of dry leaves and wood.
  • Avoid smoking. It is illegal to smoke in public places and work places, because smoke is harmful not only for the one who is smoking but also for others sitting nearby.
  • Help people in understanding global warming; most people are unaware of it.

Question 5.
State the effects of green house gases on the atmosphere.
Solution:
Effects of global warming

  • Rise in sea level: Due to global warming, glaciers and polar ice caps have started to melt, and gradually this may lead to an increase in the sea level. This will in turn flood several coastal areas in countries such as India, Bangladesh, the Netherlands and the Maldives.
  • Global warming will cause more water to evaporate from water bodies, thus forming more water vapour. Because water vapour also contributes to the greenhouse effect, global warming will further increase.
  • Global warming can lead to changes in the rain pattern and thus shift in crop zones. For example, wheat-producing zones will shift from Russia and Canada to the less fertile polar regions.
  • Change in rain pattern due to global warming will also affect trees and plants in forests which are natural habitats of wild life. With destruction of forests, many species of wild life will also begin to die out.

Question 6.
State the role of a green house in growing plants.
Solution:
A greenhouse collects light and converts it to heat. It also stores thermal energy and helps moderate temperature and produces a controlled environment for plants to grow and thrive. It even offers protection from wind, rain, snow and other weather elements and protects fruits from invading pests and animals.

Question 7.
Our atmosphere acts as a green house. Explain.
Solution:
Our atmosphere contains greenhouse gases such as CO2, water vapour, O3, CH4, oxides of nitrogen and CFCs and allows the sunrays to come in. Sunlight reaching the Earth consists of three types of radiation-UV radiation, visible radiation and IR radiation. As sunlight passes through the atmosphere, most UV radiation is absorbed by ozone; 30% of IR radiation reaches the Earth’s surface, heating it up. As the Earth’s surface becomes hot, it starts emitting radiation with less energy than the incoming radiation and thus with longer wavelength. Some emitted IR radiation escapes from the Earth’s surface and some are absorbed by CO2, thus remaining on the Earth. Trapped radiation warms the Earth’s surface and lower layers of the atmosphere.

Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 3

Question 8.
How can we reduce global warming?
Solution:

Ways of reducing global warming:

  • Minimise the use of automobiles: Depending on the situation, one can use a bicycle, the public transport system and car pools.
  • Plant more trees to increase green cover.
  • Avoid burning of dry leaves and wood.
  • Avoid smoking. It is illegal to smoke in public places and work places, because smoke is harmful not only for the one who is smoking but also for others sitting nearby.
  • Help people in understanding global warming; most people are unaware of it.

Page No: 134

Question 1.
What is a pollutant?
Solution:
The toxic substances that have an undesirable impact on different components of the environment and are injurious to life and property are known as pollutants.

Question 2.
What is the effect of the following pollutants on living beings (one in each case)?
Fluorides
Smoke particles
Lead
Mercury compounds
Smog
Nitrogen oxide
Solution:
Effect of Pollutant on living beings:

  • Fluorides: Effects teeth and bones.
  • Smoke Particles: Cause asthma and lung diseases.
  • Lead: Damages the nervous and digestive systems and can cause cancer .
  • Mercury compounds: They cause disease like Minamata commonly found in fishermen.
  • Smog: It reduces visibility and induces respiratory troubles.
  • Nitrogen Oxide: Causes death of many plants.

Question 3.
What is air pollution? How does this pollution take place?
Solution:
Air pollution: Deterioration of air quality around us is called air pollution. It is defined as the presence of a contaminant in the atmosphere in a concentration large enough to injure human, plant and animal life.

Air pollution takes place due to the presence of gaseous pollutants like oxides of sulphur, hydrocarbons, smoke, oxides of carbon, oxides of nitrogen, dust, particulate pollutants like mist, spray and fume.

Question 4.
What are the components of clean, dry air?
Solution:
Components of air is:

Pure Air components By Volume percentage Concentration parts per million (ppm)
Nitrogen 78.09 780,900
Oxgyen 20.94 209,400
Inert Gases
Argon		 0.93 9300
Neon 18
Helium 5
Krypton 1
Xenon 1
Carbon-dioxide 0.03 315
Methane 1
Hydrogen 0.5
Natural pollutants
Oxides of nitrogen
Ozone

0.52
0.02

Question 5.
Name some particulate pollutants.
Solution:
Particulate Pollutant are dust, smoke, mist, spray and fume.

Question 6.
Why is cigarette-smoking harmful?
Solution:
Cigarette smoking is harmful not only for one who is smoking but also for sitting nearby and so one should avoid it. Tobacco smoke causes lung cancer and asthma.

Question 7.
What is smog? State its damaging effects.
Solution:
Smog: A smog is a pollutant which is a combination of oxides of sulphur and nitrogen, partially oxidized hydrocarbons and their derivatives produced by industries and automobiles from a dark, thick dust and soot laden fog known as smog.

Damaging Effect: Smog is noxious and irritating. It reduces visibility, induce respiratory troubles and can cause death by suffocation.
Photochemical Smog damages the tissues of certain plants and even decreases the yield of citrus fruits and grapes.

Question 8.
What do you understand by ppm?
Solution:
ppm means parts per million. That is, share in 10, 00000.

Question 9.
Describe the major air pollutants. How does carbon monoxide pollute our environment?
Solution:
Major air pollutants are: Large amounts of Carbon monoxide, Sulphur dioxide, H2S, Chlorine, HCl, Hydrocarbons and particulates. Particulate matter like sand, dust etc. Secondary pollutant like (PAN) peroxyaryl.

Question 10.
How do you propose to control:
a. carbon monoxide emission
b. SOx emission
Solution:
Control of: (i) Carbon monoxide, CO emission:
Emission of CO can be controlled by :

  • Switching over from internal combustion engine to electrically powered cars. (b) Using alcohols, CNG, LNG in place of gasoline.
  • By using Catalytic Mufflers or Convertors.
  • Using pollution control devices to burn gasoline completely.
  • Using lead free petrol.
  • By using catalytic convertors
    Nitrogen oxide is reduced to nitrogen and oxygen in the presence of finely divided platinum or palladium as catalyst.
    2NO → N2 + O2
    2NO2 → N2 + 2O2
  • Carbon monoxide changes to carbon dioxide in the presence of finely divided platinum as catalyst.
    CO → CO2 + H2O

(ii) (SOx) oxides of Sulphur emission: Oxides of sulphur (SO, SO) emission can be reduced by-
(a) Using coal or oil that has low sulphur content. (b) By using Scrubber, a device that absorbs gaseous pollutants.

Question 11.
Give the composition, causes and effects of acid rain.
Solution:
Acid rain

Factories in big cities release nitrogen dioxide and sulphur dioxide as their wastes. These gases dissolve in rainwater during rains and form nitrous acid and sulphurous acid. As the rain falls, these acids come down to the ground as an acid rain.

The normal rain is slightly acidic having a pH about 56 as carbon dioxide gas reacts with it to form a weak carbonic acid.
CO2 + H2O → H2CO3
(Carbonic acid)

The pH of acid rain ranges between 56 – 35 and in some cases pH can go even lower than 2.
The two forms of deposition of acid rains are:

  • Dry deposits-Particles containing sulphates and nitrates
  • Wet deposits-dew, rain, fog, smoke

Formation of acid rain

  • Acid rain refers to rain which has pH less than 5.6. It is mainly caused by atmospheric pollutants.
  • Natural sources: Bacterial decomposition, forest fires, volcanic eruptions.
  • Man made sources: Industries and smelting plants, automobile exhausts, power plants etc.
  • Oxides of nitrogen and sulphur interact with water vapour in presence of sunlight in the atmosphere to form nitric acid and sulphuric acid mist respectively. This mist remains as vapours at high temperatures and condenses at low temperatures. These acids mix with rain (snow or fog) and fall down on the Earth resulting in acid rain.

Causes of acid rain

  • The formation of mineral acids like carbonic acid, nitric acid and sulphuric acid is the main cause of acid rain.

Formation of Nitric acid and Nitrous acid

  • Nitrogen and oxygen (that is oxides of nitrogen) combines in the presence of thunder and lightning to form nitric acid.
  • They are also produced by internal combustion engines (automobile engines). This then gets oxidized in the atmosphere to nitrogen dioxide. Nitrogen dioxide combines with water to form a mixture of nitrous acid and nitric acid.

N2 + O2 → 2NO
(Nitrogen oxide)

2NO + O2 → 2NO2
(Nitrogen dioxide)

2NO2 + H2O → HNO2 + HNO3
(Nitrous acid) (Nitric acid)

Formation of Sulphuric acid and Sulphurous acid

Impurities in the coal: Coal used in power plants contains upto 4% sulphur. On combustion it forms pollutant sulphur dioxide (i.e, oxides of sulphur).
S + O2 → SO2
(Sulphur dioxide)

Sulphur dioxide reacts with water vapour to form sulphurous acid.
SO2 + H2O → H2SO3
(Sulphurous acid)

Sulphur dioxide can also be oxidized to sulphur trioxide.
2SO2 + O2 → 2SO3
(Sulphur trioxide)

Sulphur trioxide reacts with water vapour to form sulphuric acid.
SO3 + H2O H2SO4
(Sulphuric acid)

Impact of Acid rain

Changes the acidity of soil: The acids present in the acid rain like, nitric acid, nitrous acid and sulphuric acid, sulphurous acid increases the acidity of soil. It removes calcium and potassium minerals i.e., basic ingredients from the soil losing their fertility.

The hydrogen ions H+ which are added to the soil, when acid rain falls on the Earth interact chemically with existing soil minerals.
Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 4

Affects water bodies and marine organisms: The water of lakes and rivers is becoming acidic, which may no longer support aquatic life.

Material damage: It increases corrosion of metals, disintegrates paper and leather. Weakens building materials such as statues, marbles, sculptures, limestone, slate, mortar etc. These materials become pitted and weakened mechanically. The Taj Mahal in India faces this problem.
CaCO3 + H2SO4 → CaSO4 + CO2 + H2O
CaCO3 + 2HNO3 → Ca(NO3)2 + CO2 + H2O

Impact on living things: It damages forests. Acid rain gets absorbed by plants, animals directly or indirectly toxicity enters food chain affecting humans. They can affect a person’s breathing, at sufficiently high concentrations. Sulphur dioxide irritates the upper respiratory tract, which serve to expel soot particles and dust in the inhaled air. At even lower concentrations, it does still greater harm by injuring lung tissues.

Question 12.
Explain the effect of sulphur dioxide on the atmosphere.
Solution:
Harmful effects of oxides of sulphur:

  • It causes headache, vomiting and even death due to respiratory failure.
  • It destroys vegetation and weakens building materials/constructions.
  • It mixes with smoke and fog to form smog, which is very harmful.
  • It is oxidised by atmospheric oxygen into sulphur trioxide (SO3) which combines with water to form sulphuric acid (H2SO4). Sulphuric acid is the cause of acid rain.
    2SO2(g)  + O2(g) → 2SO3(g)
    SO3(g) + H2O →H2SO4(aq)

Question 13.
Explain the formation of ozone in the atmosphere.
Solution:
In the atmosphere ozone is formed by the action of ultraviolet rays of the sun on oxygen.
3O2 (g) → 2O3(g)

The high energy UV radiations break oxygen molecules into oxygen atoms.
O2 + Far UV → O + O
Oxygen molecule Oxygen atoms

Oxygen atom reacts with oxygen molecule to form ozone.
O + O2 → O3
Atom Molecule Ozone

The Net reaction is:
3O2 + Far UV → 2O3

Question 14.
What is the function of ozone in the atmosphere?
Solution:
It is formed by the action of ultraviolet rays of the Sun on oxygen.
O3 → O + O2

Ozone layer acts as a blanket in the atmosphere 16 km height above the Earths surface.
It absorbs harmful ultraviolet rays (UV radiations) coming from the Sun and prevents them to reach the surface of the Earth.

Ultraviolet rays have very harmful effects on living things. It causes skin cancer. It destroys many organic species which are necessary for life.

Thus it protects the life on earth from harmful effects of Ultra Violet Rays. Which can cause (a) Skin cancer (b) destroy many organic species necessary for life.

Question 15.
State the chemicals responsible for ozone layer destruction.
Solution:
Chemicals responsible for ozone destruction free radical chlorine (Cl) and nitrioxide (NO) are responsible for ozone depletion i.e. react with O3Free radical chlorine (Cl) is produced by UV rays from chlorofluoro carbons enter the atmosphere because of excessive use as solvents, Aerosol, Spray, Propellants, Refrigerants and blowing agents for plastic foams.

Chemicals responsible for the depletion of ozone layer

Fuel of planes: Burning of fuels of planes emits large quantity of nitric oxide and other gases in the atmosphere. Nitric oxide reacts with ozone and form nitrogen dioxide and nitrogen trioxide. This causes depletion of ozone.

NO(g) + O3(g) → NO2(g) + O2(g)
(Nitrogen dioxide)

NO2(g) + O3(g) → NO3(g) + O2(g)
(Nitrogen trioxide)

Excessive use of chlrofluro carbon: It is released by refrigerators and air conditioning systems.
It causes reduction of ozone layer that protects us from harmful ultraviolet rays (UV radiations) of the Sun.

The chlorofluro carbons are decomposed by the ultraviolet rays to highly reactive chlorine which is produced in the atomic form.
Selina Concise Chemistry Class 9 ICSE Solutions Atmospheric Pollution image - 5
The free radical [Cl] reacts with ozone to form chlorine monoxide.

Cl + O3 → ClO + O2
(Chlorine monoxide)

This causes depletion of ozone layer. Chlorine monoxide then reacts with atomic oxygen to produce more chlorine free radicals.

ClO + O → Cl + O2
(Free radical)
Again this free radical destroys ozone and the process continues giving rise to depletion of ozone layer.

Question 16.
Name any two:
natural sources of atmospheric pollution.
gases which are responsible for the formation of acid rain.
Solution:
Natural sources of atmospheric pollution:

  1. Decay of plants and animals
  2. Disintegration of rocks and soil
  3. SO2 and NO2 are gases responsible for acid rain

Question 17.
Explain the term ‘global warming’. State two ways by which global warming can be reduced.
Solution:
Global warming is the increase in temperature of Earth due to enhanced concentration of greenhouse gases (CFCs) in the atmosphere.

Two ways to reduce global warming:

  • Plant more trees to increase green cover
  • Minimise the use of automobiles

Question 18.
State two effects of ozone depletion.
Solution:
Effects of ozone depletion:

  1. UV rays of the Sun reach Earth and cause sun burn, premature ageing of the skin and skin cancer.
  2. UV radiation can also damage several parts of the eyes, including the lens, cornea, retina and conjunctiva.

Question 19.
What is the cause of acid rain? Give any two impacts of acid rain.
Solution:
Causes:

Sulphur and nitrogen oxides are emitted by burning fossil fuels. Such smoke and gases entering the atmosphere make a dilute soup of sulphuric and nitric acids. This falls on the land surface in the form of acid rain damaging the things on Earth.

Impacts:

  1. Acid rain accelerates the decay of building materials and paints, including buildings, statues and sculptures which are part of our nation’s culture and heritage.
  2. Acid rain causes respiratory problems in humans, especially for people suffering from asthma. It may cause throat irritation, dry cough and severe headache.

Question 20.
Explain the methods of preventing acid rain.
Solution:
Methods of preventing acid rain:

  1. By using coal or oil which has low sulphur content. This reduces the emission of oxides of sulphur and nitrogen responsible for acid rain.
  2. By using a scrubber, a device which absorbs gaseous pollutants.

Question 21.
State an advantage of CNG (Compressed Natural Gas).
Solution:
Using CNG causes less pollution. It does not contain lead, and it has low maintenance cost.

Question 22.
State how CFC break ozone layer.
Solution:
Depletion of O3 by CFC:

CFC is broken by UV rays of sunlight to produce [Cl] atom or free radical [Cl] which is highly reactive in the atomic form and it forms ClO(g) with O2.

CFCl3 → CFCl2 + Cl(atom)
Cl(g) + O3(g) → ClO(g) + O2(g)

This depletes ozone.

ClO further produces more[Cl] free radical and destroys more of O3,thereby resulting in ozone depletion.
ClO(g) + O(g)→ Cl(g) + O2(g)

Question 23.
Describe the methods of saving ozone layer.
Solution:
Methods to protect the ozone layer:

  1. Using alternative products such as HCFCs (hydrochlorofluorocarbons)
  2. Montreal Protocol, an international treaty, helps prevent ozone depletion.

Question 24.
Fill in the blanks:
The pollutants such as NO2, SO2 and SO3 dissolved in the moisture of air are the cause of ____________________.
Excessive release of carbon dioxide in the atmosphere is the cause of __________ effect which produces global warming.
Ozone layer prevents the harmful ________ radiation of the sun to reach the earth.
Decrease of the concentration of ozone in the stratosphere is the cause of formation of __________ holes.
Ozone depletion is mainly caused by the active __________ atoms generated from CFC in the presence of UV radiation.
Solution:

  1. acid rain
  2. greenhouse
  3. ultraviolet
  4. ozone
  5. chlorine

Question 25.
Select the correct answer:
a. Excessive release of carbon dioxide in the atmosphere is the cause of
i. Depletion of ozone
ii. formation of polar vartex
iii. global warming
iv. formation of smog
b. Inhalation of air polluted with carbon monoxide is dangerous because:
i. CO combines with O2 dissolved in blood.
ii. CO combines with haemoglobin of blood.
iii. CO removes water from the body and causes dehydration.
iv. CO causes coagulation of proteins in the body
c. Decrease of amount of ozone in stratosphere is called depletion of zone and it is caused by
i. UV radiations of sun
ii. Use of CFC compounds
iii. excessive use of detergents
iv. Use of polychlorinated biphenyls
Solution:

  1. global warming
  2. CO combines with haemoglobin of blood.
  3. Use of CFC compounds

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Selina ICSE Solutions for Class 9 Chemistry Chapter 9 Study of the First Element – hydrogen

Exercise 9

Solution 1.

(a) The position of hydrogen in the periodic table
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 1
Hydrogen is first element in the periodic table. It has an atomic number l and an atomic mass of 1.00794 amu, occupying group – IA. Its position is peculiar because it is grouped with metals although it is a non-metal properties. Hydrogen relate to Group IA as well as Group VII A.

(b) The properties of hydrogen resemble the properties of Group IA elements (Alkali metals), and some of it resembles the properties of Halogens (VIIA), so Hydrogen was put at the top of the periodic table so that the symmetry of the modern periodic table is not disturbed.

  1. All elements in Group- IA have one electron in outermost shell, so they havevalency one.
  2. These elements in Group-IA are good reducing agents.
  3. All elements of this group formsoxide which are highly basic and dissolves in water to form strong alkalis.
  4. They impartcolour to a flame.

Solution 2.

Similarity of hydrogen with alkali metals and halogens

Similarity of hydrogen with alkali metals [Group 1 (IA)] Similarity of hydrogen with halogens [Group 17 (VIIA)]
Electronic configuration Electronic configuration = 1. Thus, 1electro in the outermost valence shell.

Example:
H=1; Li=2, 1; Na=2,8,1; K=2,8,8,1

One electron less than the nearest noble gas.

Example:
H= 1 (He=2)
F= 2,7 (Ne=2,8)
Cl= 2,8,7 (Ar=2,8,8)
Ion formation Electropositive character exhibited.

H 1e– → H1+
Li 1e– → Li1+
Na 1e– → Na1+

 Electronegative character exhibited.

H + 1e– → H1-
F + 1e– → F1-
Cl + 1e– → Cl1-
Valency Electrovalency of one exhibited.

H1+ , Li1+ , Na1+

Electrovalency and covalencyexhibited.

Hydrogen:

forms NaH (electrovalent)forms CH4(covalent)

Chlorine:

forms NaCl (electrovalent)
forms CCl4 (covalent)
Reactions Strong affinity for non-metals (example: O, S, Cl)

Hydrogen: forms H2O; H2S; HCl

Sodium: forms Na2O; Na2S; NaCl

 __
Reducing agent Acts as a reducing agent.

Hydrogen:

CuO + H2 → Cu + H2O

Sodium:

CuO + 2Na → Cu + Na2O

 __
Atomicity __

Diatomic molecules are formed. (Two atoms linked by a single bond)

Hydrogen
H:H or H-H → H2 

Chlorine
Cl:Cl or Cl-Cl → Cl2

Resemblance with Halogens:

  1. Both exist in the form of diatomic molecules.
  2. Both show gaseous nature.
  3. Both have a valency of 1.
  4. Both are non-metals.
  5. Both lose electron to term anions.

Solution 3.

(a) Hydrogen is found in minute traces in the Earth’s crust and the Earth’s atmosphere. The atmosphere around the sun and stars is found to contain 1.1 % hydrogen.

(b) Henry Cavendish when prepared this gas from iron and dil. acids, he established its elementary nature and showed that when the gas burns in air, water is formed. It was on account of this property that Lavoisier in 1783 named it hydrogen (Greek word meaning water-former).

Solution 4.

(a) A monovalent metal
2Na + H2 → 2NaH
(Sodium hydride)

(b) A divalent metal
Ca + H2 → CaH2
(Calcium hydride)

Solution 5.

(a) Calcium: is not used in lab preparation of hydrogen because:

  1. The reaction and very violent and exothermic hence dangerous.
  2. The heat liberated ignites the hydrogen.
  3. Calcium is expensive.

(b) Iron: Iron reacts slowly at ordinary temperatures, hence requires heating. The hydrogen produced also contain impurities like sulphur dioxide and hydrogen sulphide. Hence, it is not used in lab preparation of hydrogen.

(c) Aluminium: It is not used in the lab preparation of hydrogen because oxides of this metal keep sticking to the surface of the metal. Thus the steam does not come in contact with metal and hence reaction stops. .

(d) Sodium: It is riot used in the lab preparation of hydrogen because the reaction is violent. The sodium melts into a globule and darts about freely on the surface of water hence the collection of hydrogen is difficult.

Solution 6.

Depending upon the nature of reaction taking place between metals and substances like air, water and acids, metals are arranged in a vertical series in order of their activity. Such a series is called activity series of metals.

The metals places near the top of the series are the most reactive, while those placed near the bottom are the least reactive.

When dilute hydrochloric acid or dilute sulphuric acid react with the metals above hydrogen in the activity series, they produce hydrogen. But the metals below hydrogen in the activity series do not.

Solution 7.

(a) Reactants: Nitrogen and hydrogen (Haber process)

Chemical equation:
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 2

Observation and conditions
Three volumes of hydrogen and one volume of nitrogen react at temperature 450oC-500oC at the pressure of 200-900 atm, in presence of a finely divided iron which acts as a catalyst, and promoter molybdenum.

(b) Reactants: Chlorine and hydrogen

Chemical equation:
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 3

Observation and conditions
Hydrogen and chlorine (in their equal volumes) react slowly in diffused sunlight but reacts explosively in direct sunlight. A spontaneous reaction takes place with the release of a large amount of energy.

(c) Reactants: Sulphur and hydrogen

Chemical equation:
H2 + S → H2S

Observation and conditions
Hydrogen gas when passed through molten sulphur, it reacts to give another gas, hydrogen sulphide.

(d) Reactants: Oxygen and hydrogen

Chemical equation:
2H2 + O2 → 2H2O

Observation and conditions
Hydrogen burns with a pop sound in oxygen. It burns with a pale blue flame forming water.

Solution 8.

(a) Among the given metals Zinc is most suitable.
(i) Copper: In case of copper, It is placed below hydrogen in the activity series. So it does not displace hydrogen from acid.
Cu + HCl → No reaction
(ii) In case of Mg; it is a very expensive metal.
(iii) In case of sodium, it reacts with explosively and violently.

(b) Among the given acids we prefer dilute sulphuric acid.
We reject concentrated sulphuric, dilute nitric and concentrated nitric acid because these are powerful oxidising agents and oxygen formed due to its decomposition oxidises the hydrogen to water.

(c) Modification: Collect the gas by downward displacement of water when all the air from the apparatus has been expelled. Drying Agent used is Calcium Chloride.

Solution 9.

(a) Iron reacts reversibly with steam. Hence the hydrogen formed is removed as it is released to prevent reduction of triferric tetraoxide.
Fe + 4H2O → Fe3O4 + 4H2 ?
(Steam)

(b)
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 4

Solution 10.

(a) The metal is magnesium
(b) Mg + H2O → MgO + H2

Solution 11.

(a) Substance A is CuO and substance B is Cu.

(b) Test for water
(i) It is neutral to litmus
(ii) It changes anhydrous copper sulphate into blue salt.

(c) When substance A i.e, CuO reacts with hydrogen, it removes oxygen and we get free metal i.e. Cu.

(d) CuO + H2 → Cu + H2O

(e) No, there is no reaction between substance B and dilute hydrochloric acid because copper does not displace hydrogen from acids.

(f) Cu + HCl → No reaction

Solution 12.

Magnesium lies above Hydrogen in reactivity series and can displace hydrogen from acid whereas, Mercury and silver lie below hydrogen in reactivity series and cannot displace hydrogen from acid and hence nothing happens.

Solution 13.

Soap bubbles containing hydrogen rapidly rise up in air as hydrogen is lighter than air.

Solution 14.

Bosch Process
Bosch process consists of following steps.

Step 1 :

Steam is passed over a hot coke (at 1000oC) in a special type of a furnaces called converters.
In this step carbon reacts with water to form carbon monoxide and hydrogen gas. This mixture is called water gas.Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 5

Step 2 :

In this step excess of steam is mixed with water gas and entire mixture is passed over heated ferric oxide and chromic oxide. Ferric oxide acts as catalyst and chromic oxide as promoter.
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 6

Step 3 :

In this step removal of carbon dioxide from reaction mixture takes place. The mixture of carbon dioxide and hydrogen is forced through cold water under pressure at 30 atmospheric pressure or through caustic potash solution which dissolve carbon dioxide leaving behind hydrogen gas.
2KOH + CO2 → K2CO3 + H2O

Step 4 :

In this last step, mixture is passed through ammonical solution of cuprous chloride solution so as to dissolve carbon monoxide. CO is removed by bubbling the gas through ammoniacal cuprous chloride solution. The moisture is removed by cooling the gas about 20°C when water vapoursfreeze. The pure and dry hydrogen gas is collected in steel cylinders.
CuCl + CO + 2H2O → CuCl.CO.2H2O

Solution 15.

(a) Cold water:

Sodium metal wrapped in small piece of wire gauze or Sodium amalgamated with mercury is used. This prevents sodium from darting about.
2Na + 2H2O → 2NaOH + H2 ↑

(b) Hot water:

Zn or Mg can be used.
Mg + H2O → MgO + H2 ↑
(boiling water)

(C) Steam:

Iron reacts with the steam and the reaction is reversible. Iron reacts with steam when it is red hot as hydrogen is blown out of contact with iron by the force of current of the steam.
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 7

Solution 16.

Metals

Mg, Al, Zn, Fe do not react with cold water. It reacts with boiling water liberating hydrogen gas but the reaction is very slow.
Mg, Al, Zn, Fe react with the hot steam in the heated state and form the corresponding oxide and hydrogen gas.
Iron reacts with the steam and the reaction is reversible.

Magnesium
Reaction of boiling water with steam is slow but Mg liberates Hydrogen rapidly with steam.

Mg + H2O → MgO + H2 ↑
(boiling water)

Aluminium

Gets coated with Al2O3 on rubbing with sand paper its oxide coating is removed and then it reacts with steam to produce hydrogen.

2Al + H2O → Al2O3 + 3H2 ↑
(steam)

Zinc

Zinc reacts with steam and produce zinc oxide and H2

Zn + H2O → ZnO + H2 ↑
(steam)

Iron

Iron produces H2 when red hot the reaction is reversible but as soon as hydrogen is produced it is blown out of contact with iron by the force of the current of steam.

Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 8
(steam)

Non-metals

Bosch Process (From Coke)
Bosch process consists of following steps.

Step 1 :

Steam is passed over a hot coke (at 1000oC) in a special type of a furnaces called converters.
In this step carbon reacts with water to form carbon monoxide and hydrogen gas. This mixture is called water gas.Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 9

Step 2 :

In this step excess of steam is mixed with water gas and entire mixture is passed over heated ferric oxide and chromic oxide. Ferric oxide acts as catalyst and chromic oxide as promoter.
Selina Concise Chemistry Class 9 ICSE Solutions Study of the First Element - hydrogen image - 10

Step 3 :

In this step removal of carbon dioxide from reaction mixture takes place. The mixture of carbon dioxide and hydrogen is forced through cold water under pressure at 30 atmospheric pressure or through caustic potash solution which dissolve carbon dioxide leaving behind hydrogen gas.

2KOH + CO2 → K2CO3 + H2O

Step 4 :

In this last step, mixture is passed through ammonical solution of cuprous chloride solution so as to dissolve carbon monoxide. CO is removed by bubbling the gas through ammoniacal cuprous chloride solution. The moisture is removed by cooling the gas about 20C when water vapoursfreeze. The pure and dry hydrogen gas is collected in steel cylinders.

CuCl + CO + 2H2O → CuCl.CO.2H2O

Solution 17.

(a) Zinc and iron lie above hydrogen in reactivity series and can displace hydrogen from acid.

Zn + H2SO4 → ZnSO4 + H2 ↑
(dilute)

Fe + H2SO4 → FeSO4 + H2 ↑
(dilute)

(b) Zn + 2HCl → ZnCl2 + H2 ↑
(dilute)

Fe + 2HCl → FeCl2 + H2 ↑
(dilute)

Copper lies below hydrogen. Thus, it cannot displace hydrogen from acids.

Solution 18.

Two alkalis which can displace hydrogen are NaOH and KOH.

Aluminium

2Al + 6NaOH → 2Na3AlO3 + 3 H2 ↑
(Sodium aluminate)

2Al + 2KOH + 2H2O → 2KAlO2 + 3 H2 ↑
(Potassium meta aluminate)

Zinc

Zn + 2NaOH → Na2ZnO2 + H2 ↑
(Sodium zincate)

Zn + 2KOH → K2ZnO2 + H2 ↑
(Potassium zincate)

Aluminium and Zinc have unique nature; They react with acids and can even react with hot concentrated alkalis to form hydrogen and a soluble salt. Salt s (oxides and hydroxide) of these metals are Amphoteric.

Question 19.
Complete and balance the following reactions.
(a) Na + H2O →_____________ +___________
(b) Ca + H2O →_____________ +___________
(c) Mg + H2O →_____________ +___________
(d) Zn + H2O →_____________ +___________
(c) Fe + H2O →_____________ +___________
(d) Zn + HCl →_____________ +___________
(e) Al + H2SO4 →_____________ +___________
(f) Fe + HCl →_____________ +___________
(g) Zn + NaOH →_____________ +___________
(h) Al + KOH + H2O→_____________ +___________

Solution:

(a) 2Na + 2H2O → 2NaOH + H2
(b) Ca + 2H2O → Ca(OH)2 + H2
(c) Mg + H2O → MgO + H2
(d) Zn + H2O → ZnO + H2
(e) 3Fe + 4H2O → Fe3O4 + 4H2
(f) Zn + 2HCl → ZnCl2 + H2
(g) 2Al + 3H2SO4  → Al2(SO4)3 + 3H2
(h) Fe + 2HCl → FeCl2 + H2
(i) Zn + 2NaOH → Na2ZnO2 + H2
(j) 2Al + 2KOH + 2H2O → 2KAlO2 + 3H2

Solution 20.

(a) Lead reacts with dilute sulphuric acid and hydrochloric acid and forms an insoluble coating of lead sulphate and lead chloride respectively. Hence, further reaction is prevented.

(b) When potassium and sodium react with dilute sulphuric acid, H2SO4 or dilute HCl, the reaction is highly explosive and practically not feasible.

Solution 21.

(a) Sodium hydroxide + zinc → hydrogen + Sodium zincate
(b) Calcium + water calcium → hydroxide + Hydrogen

Solution 22.

  1. Hydrogen gas is collected by the downward displacement of water. This is because-
    • It is virtually insoluble in water.
    • It forms an explosive mixture with air and therefore, cannot be collected by downward displacement of air, even though it is lighter than the air.
  2. A candle when brought near the mouth of the jar containing hydrogen gas starts burning but the candle is extinguished when pushed inside the jar because, Because hydrogen is a combustible gas but a non-supporter of combustion i.e. it burns itself but does not allow substances to burn in it.
  3. An oxy-hydrogen flame is used for welding and cutting metals because, when a mixture, of hydrogen and oxygen is burnt, temperatures as high as 2500°C is produced. This forms the basis for its use in oxy-hydrogen flame used in cutting and welding of flames.
  4. Apparatus for the laboratory preparation of hydrogen should be air tight and away from a naked flamebecause, mixture of hydrogen and air explodes violently when brought near a flame.

Question 23.
a. Helium is preferred to hydrogen for filling balloons because it is:
i. lighter than air
ii. almost as light as hydrogen
iii. non-combustible
iv. inflammable
b. Reacting with water, an active metal produces
i. oxygen
ii. nitric acid
iii. a base
iv. none of these
c. A metal oxide that is reduced by hydrogen is
i. Al2O3
ii. CuO
iii. CaO
iv. Na2O
d. Which of the following statements about hydrogen is incorrect?
i. It is an inflammable gas
ii. It is the lightest gas.
iii. It is not easily liquefied
iv. It is a strong oxidizing agent.
e. For the reaction PbO + H2→ Pb + H2O, which of the following statements is wrong?
i. H2 is the reducing agent.
ii. PbO is the oxidizing agent.
iii. PbO is oxidized to Pb.
iv. H2 is oxidized to H2O.
f. Which metal gives hydrogen with all of the following: water, acids, alkalis?
i. Fe
ii. Zn
iii. Mg
iv. Pb
g. Which of the following metals does not give hydrogen with acids?
i. Iron
ii. Copper
iii.Lead
iv. Zinc

Solution:
(a) (iii) non-combustible
(b) (iii) a base (c) (iii) CaO
(d) (iv) It is strong oxidising agent.
(e) (iii) PbO is oxidised to Pb
(f) (ii) Zn
(g) (ii) Copper

Question 11.
Choose terms from the options given in brackets to complete these sentences.
a. When CuO reacts with hydrogen,………………… is reduced and ……………….is oxidized to ………………… .
(CuO, H2,Cu,H2O)
b. Hydrogen is ………………… soluble in water.
(sparingly, highly, moderately)
c. Metals like …………….. , ……………… and ……………… give H2 with steam.
(iron, magnesium, aluminium, sodium , calcium)
d. Sodium ………………. reacts smoothly with cold water.
(metal, amalgam, in the molten state)
e. A metal …………….. hydrogen in the activity series gives hydrogen with …………… acid or …………… acid.
(above, below, dilute hydrochloric, concentrated hydrochloric, dilute sulphuric).

Solution 24.

(a) CuO, H2O
(b) sparingly
(c) amalgam
(d) iron, magnesium, aluminium
(e) above, dilute hydrochloric, dilute sulphuric.

Question 25.
Correct the following statements:

  1. Hydrogen is separated from CO by passing the mixture through caustic potash solution.
  2. All metals react with acids to give hydrogen.
  3. Hydrogen is dried by passing it through conc. H2SO4.
  4. Very dilute nitric acid reacts with iron to produce hydrogen.
  5. Conc. H2SO4 reacts with zinc to liberate hydrogen.

Solution:

(a) Hydrogen is used as a fuel in the form of coal gas, water gas and liquid hydrogen.
(b) All metals above hydrogen in reactivity series react with acids to give hydrogen.
(c) Metals like palladium or platinum or nickel absorb hydrogen at room temperature.
(d) The reaction of hydrogen with oxygen is explosive with pop sound.
(e) Concentrated sulphuric acid reacts with zinc to liberate sulphur dioxide.

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Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws

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Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws

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Selina ICSE Solutions for Class 9 Chemistry Chapter 7 Study of Gas Laws

Page No: 122

Question 1.
What do you understand by gas?
Solution:
The state of matter in which inter-particle attraction is weak and inter-particle space is so large that the particles become completely free to move randomly in the entire available space, is known as gas.

Question 2.
Give the assumptions of the kinetic molecular theory.
Solution:
The main assumption of the kinetic molecular theory of gases are as follows:

  1. All gases are made up of a large number of extremely small particles called molecules.
  2. There are large vacant spaces between the molecules of a gas so that actual volume of the molecules of a gas is negligible as compared to the total volume occupied by the gas.
  3. The molecules of a gas are always in a state of constant random motion in straight lines in all possible direction.
  4. There are negligible attractive forces between the molecules of a gas.
  5. There is no effect of gravity on the motion of the molecules of a gas.
  6. The average kinetic energy of the molecules of a gas is directly proportional to that of the Kelvin temperature of the gas.
  7. The molecules are perfectly elastic so that there is no net loss of energy during molecular collisions.
  8. The pressure of a gas is due to the bombardment of the molecules of a gas against the walls of a container.

Question 3.
During the practical session in the lab when hydrogen sulphide gas having offensive odour is prepared for some test, we can smell the gas even 50 metres away. Explain the phenomenon.
Solution:
In a laboratory, when hydrogen sulphide gas is prepared, it can be smelt even at 50 meters away. This is due to the phenomenon called Diffusion.

Diffusion is a process of intermixing of two substances kept in contact.

The inter-particle or inter molecular spaces in a gas are very large. When hydrogen sulphide gas is produced, its particle collides with air particles. Due to the collisions of particles, they start moving in all possible directions. As a result of which the two gases mix with each other forming a homogeneous mixture of a gas. Thus, the released gas can be smelt to a long distance.

Solution 4.

Pressure and volume relationship of gases-
Experiment: Take a 10 ml syringe fitted with a piston. Raise the latter to the 10 ml mark and wrap an adhesive tape over its nozzle. Fit the wrapped nozzle tightly into a hole, bored half way through a rubber stopper.

Observation: On placing some weight on the piston (to put pressure), the piston moves downward and reduces the volume of air. Gradually, put more weight. The piston moves further downward and the volume of the air is further reduced.

Now remove the weights one by one. You will notice that, on decreasing the pressure, the piston moves upward as such the volume of the air increases.

Conclusion:

  1. An increase in pressure at constant temperature causes a decrease in the volume of a gas; conversely, if the volume of a fixed mass of a gas at constant temperature is decreased, the pressure of the gas increases.
  2. A decrease in pressure at constant temperature causes a increase in the volume of a gas; conversely, if the volume of a fixed mass of a gas at constant temperature is increased, the pressure of the gas decreases.

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 1

Solution 5.

The molecular motion is directly proportional to the temperature.
As temperature increases, molecular motion increases because the molecule possesses certain kinetic energy. And as the temperature decreases, molecular motion also decreases. Thus, when temperature is zero, molecular motion stops or ceases.

Question 6.
State (i) the three variables for gas laws and (ii) SI units of these variables.
Solution:
The three variables for gas laws are:

  1. Volume, V
  2. Pressure, P
  3. Temperature, T

These three are called as the Standard variables. S.I. unit of volume is cubic meter (m3).
S.I. unit of pressure is Pascal (Pa).
S.I. unit of temperature is Kelvin (K) or degree Celsius (0C).

Question 7.
State Boyle’s Law.
Give its
i. Mathematical expression
ii. Graphical representation and
iii. Significance
Solution:
Boyle’s law: At constant temperature, the volume of a definite mass of any gas is inversely proportional to the pressure of the gas.                       Or
Temperature remaining constant, the product of the volume and pressure of the given mass of a dry gas is constant.

Mathematical representation:

According to Boyle’s Law,
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 2
Where K is the constant of proportionality if V’ and P’ are some other volume and pressure of the gas at the same temperature then,
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 3
temperature, a straight line passing through the origin is obtained.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 4
2. V vs P : Variation in volume (V) plotted against pressure (P) at a constant temperature, a hyperbolic curve in the first quadrant is obtained.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 5
3. PV vs P : Variation in PV plotted against pressure (P) at a constant temperature, a straight line parallel to X-axis is obtained.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 6

Significance of Boyles law:

According to Boyles law, on increasing pressure, volume decreases. The gas becomes denser. Thus, at constant temperature, the density of a gas is directly proportional to the pressure.

At higher altitude, atmospheric pressure is low so air is less dense. As a result, lesser oxygen is available for breathing. This is the reason that the mountaineers have to carry oxygen cylinders with them.

Question 8.
Explanation of Boyle’s Law on the basis of kinetic theory of matter.
Solution:
According to kinetic theory of matter, the number of particles present in a given mass and the average kinetic energy is constant.

If the volume of given mass of a gas is reduced to half of its original volume. The same number of particles will have half space to move.

As a result, the number of molecules striking the unit area of the walls of the container at given time will get doubled of the pressure will also get doubled.

Alternatively, if the volume of a given mass of a gas is doubled at constant temperature, same number of molecules will have double space to move. Thus, number of molecule striking the unit area of the walls of container at a given time will become one half of original value. Thus, pressure will also get reduced to half of original pressure. Hence, it is seen that if pressure increases, volume of a gas decreases at constant temperature and this is Boyle’s law.

Question 9.
The molecular theory states that the pressure exerted by a gas in a closed vessel results from the gas molecules striking against the walls of the vessel. How will the pressure change if
The temperature is doubled keeping the volume constant
The volume is made half of its original value keeping the T constant
Solution:
(a) Pressure will be doubled.
(b) Pressure remains the same.

Question 10.
a. State Charles’s law.
b. Give its
i. Graphical representation
ii. Mathematical expression and
iii. Significance
Solution:

Charless Law

At constant pressure, the volume of a given mass of a dry gas increases or decreases by 1/273 of its original volume at 00C for each degree centigrade rise or fall in temperature.

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 7
For Temperature = Conversion from Celsius to Kelvin
1 K = 0C + 273
For example,
20oC = 20 + 273 = 293 K

Graphical representation of Charles law

T vs V: The relationship between the volume and the temperature of a gas can be plotted on a graph, A straight line is obtained.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 8

Significance of Charles’ Law: Since the volume of a given mass of gas is directly proportional to its temperature, hence the density decreases with temperature. This is the reason that:

(a) Hot air is filled in the balloons used for meteorological purposes. (b) Cable wires contract in winters and expand in summers.

Question 11.
Explanation Of Charles’ Law on the basis of kinetic theory of matter is as follows:
Solution:
According to kinetic theory of matter, the average kinetic energy of the gas molecules is directly proportional to the absolute temperature. Thus, when the temperature of a gas is increased, the molecules would move faster and the molecules will strike the unit area of the walls of the container more frequently and vigorously. If the pressure is kept constant, the volume increases proportionately. Hence, at constant pressure, the volume of a given mass of a gas is directly proportional to the temperature (Charles’ law).

Question 12.
Define absolute zero and absolute scale of temperature. Write the relationship between °C and K.
Solution:
Absolute zero

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 9
Absolute or Kelvin scale of temperature

The temperature scale with its zero at – 273oC and each degree equal to one degree on the Celsius scale is called Kelvin or the absolute scale of temperature.

Conversion of temperature from Celsius scale to Kelvin scale and vice versa

The value on the Celsius scale can be converted into Kelvin scale by adding 273 to it.
For example,
20oC = 20 + 273 = 293 K

Question 13.
(a) What is the need for the Kelvin scale of temperature?
(b) What is the boiling point of water on the Kelvin scale? Convert it into centigrade scale.
Solution:
(a) The behaviour of gases shows that it is not possible to have temperature below 273.15C. This act has led to the formulation of another scale known as Kelvin scale. The real advantage of the Kelvin scale is that it makes the application and the use of gas laws simple. Even more significantly, all values on the Kelvin scale are positive.

(b) The boiling point of water on the Kelvin scale is 373 K.
Now, K = C + 273 and C = K – 273
Kelvin scale can be converted to degree Celsius by subtracting 273 from it. So, boiling point of water on centigrade sale is : 373 K – 273 = 100C

Question 14.
(a) Define STP or NTP.
(b) Why is it necessary to compare gases at STP?
Solution:
(a) Standard or Normal Temperature and Pressure (S.T.P. or N.T.P.)

The pressure of the atmosphere which is equal to 76 cm or 760 mm of mercury is referred to as S.T.P. or N.T.P. The full form for S.T.P. is Standard Temperature and Pressure or Normal Standard temperature and pressure denotes 0oC or 273K.

Value: The standard values chosen are 0oC or 273 K for temperature and 1 atmospheric unit (atm) or 760 mm of mercury for pressure.
The standard values chosen are 0oC or 273 K for temperature and 1 atmospheric unit (atm) or 760 mm of mercury for pressure.
Standard temperature = 0oC = 273 K
Standard pressure = 760 mm Hg
= 76 cm of Hg
= 1 atmospheric pressure (atm)

(b) Because the volume of a given mass of dry enclosed gas depends upon the pressure of the gas and temperature of the gas in Kelvin so to express the volume of the gases we compare these to S.T.P.

Solution 15.

(a) (i) C = oC (ii) K = 273K
(b) (i) 1 atm (ii) 760 mm Hg (iii) 76 cm Hg. (iv) 1, torr = 133.32 Pascal

Question 16.
What is the relationship between the Celsius and Kelvin scales of temperature?
Convert (i) 273°C to Kelvin and (ii) 293 K to °C.
Solution:
Temperature on Kelvin scale (K) = 273 + Temperature on Celsius scale
Or K = 273 + oC

(i) 273C in Kelvin
t oC = t K – 273
273oC = t K – 273
T K = 273 + 273 = 546 K
273oC = 546 K

(ii) 293 K in oC
t oC = 293 – 273
t oC = 20oC
293 K = 20oC

Question 17.
State the laws which are represented by the following graphs.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 43
Solution:

(a) Charles’s Law
(b) Boyles Law

Question 18.
Give reasons for the following:
(a) All temperature in the absolute (Kelvin) scale are in positive figures.
(b) Gases have lower density compared to solids or liquids.
(c) Gases exert pressure in all directions.
(d) It is necessary to specify the pressure and temperature of a gas while stating its volume.
(f) Inflating a balloon seems to violate Boyle’s law.
(g) Mountaineers carry oxygen cylinders with them.
(h) Gas fills the vessel completely in which it is kept.
Solution:
(a) The real advantage of the Kelvin scale is that it makes the application and use of gas laws simple. Even more significantly, all values on the scale are positive. Thus, removing the problem of negative (-) values on the Celsius scale.

(b) The mass of a gas per unit volume is very small due to the large intermolecular spaces between the molecules. Therefore, gases have low density. Whereas in solids and liquids, the mass is higher and intermolecular spaces are negligible.

(c) At a given temperature, the number of molecules of a gas striking against the walls of the container per unit time per unit area is the same. Thus, gases exert the same pressure in all directions.

(d) Since the volume of a gas changes remarkably with change in temperature and pressure, it becomes necessary to choose standard value of temperature pressure.

(e) According to Boyle’s Law, the volume of a given mass of a day gas is inversely proportional to its pressure at constant temperature.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 10
When a balloon is inflated, the pressure inside the balloon decreases and according to Boyle’s Law, the volume of the gas should increase. But this does not happen. On inflation of a balloon along with reduction of pressure of air inside balloon, the volume of air also decreases. And this violates Boyle’s law.

(f) Atmospheric pressure is very low at high altitudes, volume of air increases thus air becomes less dense. Because volume is inversely proportional to density. Hence, lesser volume of oxygen is available for breathing. Thus, mountaineers have to carry oxygen cylinders with them.

(g) In gas as inter-particle attraction is weak and inter-particle space is so large that the particles become completely free to move randomly in the entire available space and takes the shape of the vessel in which it is kept.

Question 19.
How did Charles’s law lead to the concept of absolute scale of temperature?
Solution:
The temperature scale with its zero at -273oC and where each degree is equal to degree on the Celsius scale is called the absolute scale of temperature.

The temperature -273oC is called the absolute zero. Theoretically, this is the lowest temperature that can never be reached. At this temperature all molecular motion ceases.

The temperature – 273oC is called absolute zero.
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 11

Question 20.
What is meant by aqueous tension? How is the pressure exerted by a gas corrected to account for aqueous tension?
Solution:
Gases like nitrogen, hydrogen are collected over water as shown in the figure. When the gas is collected over water. The gas is moist and contains water vapour. The total pressure exerted by this moist gas is equal to the sum of the partial pressures of the dry gas and the pressure exerted by water vapour: The partial pressure of water vapour is also known as Aqueous tension.Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 12
Ptotal = Pgas + Pwater vapour
Pgas = Ptotal – Pwater vapour
Actual Pressure of gas = Total pressure – Aqueous tension

Question 21.
State the following:

  1. Volume of a gas at 0 Kelvin
  2. Absolute temperature of a gas at 7°C
  3. Gas equation
  4. Ice point in absolute temperature
  5. STP conditions

Solution:
(a) The volume of gas is zero.
(b) The absolute temperature is 7 + 273 = 280 K
(c) The gas equation is-
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 13
(d) Ice point = 0 + 273 = 273 K
(e) Standard Temperature is taken as 273 K or °C
Standard pressure is taken as 1 atmosphere (atm) or 760 mm Hg.

Question 22.
Choose the correct answer:
a. The graph of PV vs P for a gas is
i. Parabolic
ii. Hyperbolic
iii. A straight line parallel to the X-axis
iv. A straight line passing through the origin
b. The absolute temperature value that corresponds to 27°C is
i. 200 K
ii. 300 K
iii. 400 K
iv. 246 K
c. Volume-temperature relationship is given by
i. Boyle
ii. Gay-Lussac
iii. Dalton
iv. Charles
d. If pressure is doubled for a fixed mass of a gas, its volume will become
i. 4 times
ii. ½ times
iii. 2 times
iv. No change
Solution:
(a) (iii) Straight line paralled to X- axis.
(b) (ii) 27C = 27 + 273 = 300 K
(c) (iv) Charles
(d) (ii) 1/2 times

Question 23.
Match the following:
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 42
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 14

Question 24.
Correct the following statements:
Volume of a gas is inversely proportional to its pressure at constant temperature.
Volume of a fixed mass of a gas is directly proportional to its temperature, pressure remaining constant.
0°C is equal to zero Kelvin.
Standard temperature is 25°C.
Boiling point of water is 273 K
Solution:
(a) Volume of a gas is directly proportional to the pressure at constant temperature.
(b) Volume of a fixed mass of a gas is inversely proportional to the temperature, the pressure remaining constant.
(c) -273C is equal to zero Kelvin.
(d) Standard temperature is 0°C
(e) The boiling point of water is -373 K.

Question 25
The average kinetic energy of the molecules of a gas is proportional to the ………….
The temperature on the Kelvin scale at which molecular motion completely ceases is called……………
If temperature is reduced to half, ………….. would also reduce to half.
The melting point of ice is …………. Kelvin
Solution:
(a) Absolute temperature
(b) Absolute zero
(c) Volume
(d) 273

Page No: 123 Numericals

Question Num 1
What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to 200 dm3 temperature remaining constant.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 51

Question Num 2.
2 litres of a gas is enclosed in a vessel at a pressure of 760 mmHg. If temperature remains constant, calculate pressure when volume changes to 4 dm3.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 50

Question Num 3.
At constant temperature, the effect of change of pressure on volume of a gas was as given below:
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 49
Plot the following graphs
1. P vs V
2. P vs 1/V
3. PV vs P
Interpret each graph in terms of a law.
Assuming that the pressure values given above are correct, find the correct measurement of the volume.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 44

 i. P vs. V:
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 45

At constant temperature, P is inversely proportional to V. Thus, the plot of V versus P will be a rectangular hyperbola.Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 46

According to Boyle’s law, at constant temperature, pressure of a fixed amount of gas varies inversely to its volume. The graph of pressure verses 1/V shows a positive slope.

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 47.

According to Boyle’s law, the product of pressure and volume is constant at constant temperature. The graph of PV versus P is constant which indicates that the given gas obeys Boyle’s law.

b. The correct measurements of the volume are given below:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 48

Question Num 4.
800 cm3 of gas is collected at 650 mm pressure. At what pressure would the volume of the gas reduce by 40% of its original volume, temperature remaining constant?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 15

Question Num 5.
A cylinder of 20 litres capacity contains a gas at 100 atmospheric pressure. How many flasks of 200 cm3capacity can be filled from it at 1 atmosphere pressure, temperature remaining constant?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 16

Question Num 6.
A steel cylinder of internal volume 20 litres is filled with hydrogen at 29 atmospheric pressure. If hydrogen is used to fill a balloon at 1.25 atmospheric pressure at the same temperature, what volume will the gas occupy?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 17

Question Num 7.
561 dm3 of a gas at STP is filled in a 748 dm3 container. If temperature is constant, calculate the percentage change in pressure required.
Solution:
Initial volume = V1 = 561 dm3
Final volume = V2 = 748 dm3
Difference in volume = 748 – 561 = 187 dm3
As the temperature is constant,
Decrease in pressure percentage =

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 53

Question Num 8.
88 cm3 of nitrogen is at a pressure of 770 mm mercury. If the pressure is raised to 880 mmHg, find by how much the volume will diminish, temperature remaining constant.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 18

Question Num 9.
A gas at 240 K is heated to 127°C. Find the percentage change in the volume of the gas (pressure remaining constant).
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 54

Question Num 10.
Certain amount of a gas occupies a volume of 0.4 litre at 17°C. To what temperature should it be heated so that its volume gets (a) doubled, (b) reduced to half, pressure remaining constant?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 19

Question Num 11
A gas occupies 3 litres at 0°C. What volume will it occupy at -20°C, pressure remaining constant?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 20

Question Num 12.
A gas occupies 500 cm3 at normal temperature. At what temperature will the volume of the gas be reduced by 20% of its original volume, pressure being constant?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 21

Question Num 13.
Calculate the final volume of a gas ‘X’ if the original pressure of the gas at STP is doubled and its temperature is increased three times.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 22

Question Num 14.
A sample of carbon dioxide occupies 30 cm3 at 15°C and 740 mm pressure. Find its volume at STP.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 23

Question Num 15.
50 cm3 of hydrogen is collected over water at 17°C and 750 mmHg pressure. Calculate the volume of a dry gas at STP. The water vapour pressure at 17°C is 14 mmHg.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 25

Page No: 124

Question Num 16.
At 0°C and 760 mmHg pressure, a gas occupies a volume of 100 cm3. Kelvin temperature of the gas is increased by one-fifth and the pressure is increased one and a half times. Calculate the final volume of the gas.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 26

Question Num 17.
It is found that on heating a gas its volume increases by 50% and its pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was heated.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 27

Question Num 18.
A certain mass of a gas occupies 2 litres at 27°C and 100 Pa. Find the temperature when volume and pressure become half of their initial values.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 28

Question Num 19
2500 cm3 of hydrogen is taken at STP. The pressure of this gas is further increased by two and a half times (temperature remaining constant). What volume will hydrogen occupy now?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 29

Question Num 20.
Taking the volume of hydrogen as calculated in Q.19, what change must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 (pressure remaining constant)?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 30

Question Num 21.
A given amount of gas A is confined in a chamber of constant volume. When the chamber is immersed in a bath of melting ice, the pressure of the gas is 100 cmHg.

  1. What is the temperature when the pressure is 10cmHg?
  2. What will be the pressure when the chamberis brought to 100°C

Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 31
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 32

Question Num 22.
A gas is to be filled from a tank of capacity 10,000 litres into cylinders each having capacity of 10 litres. The condition of the gas in the tank is as follows:
a Pressure inside the tank is 800 mmHg.
b. Temperature inside the tank is -3°C.
When the cylinder is filled, the pressure gauge reads 400 mmHg and the temperature is 0°C. Find the number of cylinders required to fill the gas.

Solution 22.

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 33

Question Num 23.
Calculate the volume occupied by 2 g of hydrogen at 27°C and 4 atmosphere pressure if at STP it occupies 22.4 litres.
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 34

Question Num 24.
What temperature would be necessary to double the volume of a gas initially at STP if the pressure is decreased to 50%?
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 35

Question Num 25.
Which will have greater volume when the following gases are compared at STP:
1.2/N2 at 25°C and 748 mmHg
1.25/O2 at STP
Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 36
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 37
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 38

Question Num 26.
Calculate the volume of dry air at STP that occupies 28 cm3 at 14°C and 750 mmHg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mmHg.

Solution:

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 39

Question Num 27
An LPG cylinder can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at 27°C. Because of a sudden fire in the building, the temperature rises. At what temperature will the cylinder explode?

Solution:
P = 14.9 atm
V = 28 cm3
T = ?
P1 = 12 atm
V = V1
]T1 = 300 K
Using gas equation,
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 56

Question Num 28.
22.4 litres of a gas weighs 70 g at STP. Calculate the weight of the gas if it occupies a volume of 20 litres at 27°C and 700 mmHg of pressure.
Solution:

 

Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 40
Selina Concise Chemistry Class 9 ICSE Solutions Study of Gas Laws image - 41

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Selina ICSE Solutions for Class 9 Chemistry Chapter 4 The Language of Chemistry

Page No: 8

Question 1.
What is a symbol? What information does it convey?
Solution:
A symbol is the short form which stands for the atom of a specific element or the abbreviations used for the names of elements.

  1. It represents a specific element.
  2. It represents one atom of an element.
  3. A symbol represents how many atoms are present in its one gram (gm) atom.
  4. It represents the number of times an atom is heavier than one atomic mass unit (amu) taken as a standard.
Question 2
Why is the symbol S for sulphur, but Na for sodium and Si for silicon?
Solution:
In most cases, the first letter of the name of the element is taken as the symbol for that element and written in capitals (e.g. for sulphur, we use the symbol S). In cases where the first letter has already been adopted, we use a symbol derived from the Latin name (e.g. for sodium/Natrium, we use the symbol Na). In some cases, we use the initial letter in capital together with a small letter from its name (e.g. for silicon, we use the symbol Si).
Question 3.
Write the full form of IUPAC. Name the elements represented by the following symbols:
Au, Pb, Sn, Hg
Solution:
The full form of IUPAC is International Union of Pure and Applied Chemistry.
Names of the elements:
Au – Gold
Pb – Lead
Sn – Tin
Hg – Mercury
Question 4.
If the symbol for Cobalt, Co, were written as CO, what would be wrong with it?
Solution:
Co stands for Cobalt. If we write CO, then it would mean that it is a compound containing two non-metal ions, i.e. carbon and oxygen, which forms carbon monoxide gas.

Question 5(d).
2H2
Solution:
(a) H stands for one atom of hydrogen.
(b) H2 stands for one molecule of hydrogen.
(c) 2H stands for two atoms of hydrogen.

Question 6.
What is meant by atomicity? Name the diatomic element.
Solution:
The number of atoms of an element that join together to form a molecule of that element is known as its atomicity. Diatomic molecules: H2, O2, N2, Cl2
Question 7(a).
Explain the terms ‘valency‘ and ‘variable valency‘.
Solution:
  1. Valency of Na is +1 because it can lose one electron.
  2. Valency of O is -2 because it can accept two electrons.

Variable valency: It is the combining capacity of an element in which the metal loses more electrons from a shell next to a valence shell in addition to electrons of the valence shell.

Question 7(b).
How are the elements with variable valency named? Explain with an example.
Solution:
If an element exhibits two different positive valencies, then
  1. for the lower valency, use the suffix -OUS at the end of the name of the metal
  2. for the higher valency, use the suffix -IC at the end of the name of the metal.

Example:

Element Lower valency Higher valency
Ferrum (Iron) Ferrous (Fe2+) Ferric (Fe3+)

Question 8.
Give the formula and valency of:

  1. aluminate ………………… .
  2. chromate ………….…….. .
  3. aluminium ………………. .
  4. cupric  ………………… .

Solution:

Name Formula Valency
a. Aluminate AlO2 -2
b. Chromate CrO4 -2
c. Aluminium Al +3
d. Cupric Cu +2

Question 9.b
What is the significance of formula?
Solution:
Chemical formula: The chemical formula of a substance (element or compound) is a symbolic representation of the actual number of atoms present in one molecule of that substance.
It also indicates the fixed proportion by weight in which atoms combine.
Rules:
(i) The positive and negative radicals are written side by side (+ve first) with their charge as a superscript on the right side.
(ii) Charges are then interchanged and written as a subscript.
(iii) The final formula is written without the sign of charge, e.g. Hg2O

  1. Hg1+O2-
  2. Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 1
  3. Hg2O

Question 10(a).
What do you understand by the following terms?
Acid radical
Solution:
Acid radical: The electronegative or negatively charged radical is called an acid radical.
Examples: Cl–, O2-

Question 10(b).
What do you understand by the following terms? Basic radical
Solution:
Basic radical: The electropositive or positively charged radical is called a basic radical.
Examples: K+, Na+

Question 11.

Match the following:

Compound Formula
(a) Boric acid  i. NaOH
(b) Phosphoric acid  ii. SiO2
(c) Nitrous acid  iii. Na2CO3
(d) Nitric acid  iv. KOH
(e) Sulphurous acid  v. CaCO3
(f) Sulphuric acid  vi. NaHCO3
(g) Hydrochloric acid  vii. H2S
(h) Silica (sand)  viii. H2O
(i) Caustic soda (sodium hydroxide)  ix. PH3
(j) Caustic potash (potassium hydroxide)  x. CH4
(k) Washing soda(sodium carbonate)  xi. NH3
(l) Baking soda(sodium bicarbonate)  xii. HCl
(m) Lime stone.(calcium carbonate)  xiii. H2SO3
(n) Water  xiv. HNO3
(o) Hydrogen sulphide  xv. HNO2
(p) Ammonia  xvi. H3BO3
(q) Phosphine  xvii. H3PO4
(r) Methane  xviii. H2SO4

Solution:

Compound Formula (Ans)
(a) Boric acid xvi. H3BO3
(b) Phosphoric acid xvii. H3PO4
(c) Nitrous acid xv. HNO2
(d) Nitric acid xiv. HNO3
(e) Sulphurous acid xiii. H2SO3
(f) Sulphuric acid xviii. H2SO4
(g) (a) Hydrochloric acid xii. HCl
(h) Silica (sand) ii. SiO2
(i) Caustic soda
(sodium hydroxide)		 i. NaOH
(j) Caustic potash (potassium hydroxide) iv. KOH
 (k) Washing soda
(sodium carbonate)		 iii. Na2CO3

 (l) Baking soda
(sodium bicarbonate)

 vi. NaHCO3
 (m) Lime stone
(calcium carbonate)		 v. CaCO3
(n) Water viii. H2O
(o) Hydrogen sulphide vii. H2S
(p) Ammonia xi. NH3
(q) Phosphine ix. PH3
(r) Methane x. CH4

Question 12.
Select the basic and acidic radicals in the following compounds.

  1. MgSO4
  2. (NH4)2SO4
  3. Al2(SO4)3
  4. ZnCO3
  5. Mg(OH)2

Solution:

Acidic radical Basic radical
MgSO4 SO4– Mg+
(NH4)2SO4 SO4– NH4+
Al2(SO4)3 SO4– Al3+
ZnCO3 CO3– Zn2+
Mg(OH)2 OH– Mg2+
Question 13.
Write chemical formula of the sulphate of Aluminium, Ammonium and Zinc.
Solution:
Valencies of aluminium, ammonium and zinc are 3, 1 and 2, respectively.
The valency of sulphate is 2.
Hence, chemical formulae of the sulphates of aluminium, ammonium and zinc are Al2(SO4)3, (NH4)2SO4 and ZnSO4.
Question 14.
The valency of an element A is 3 and that of element B is 2. Write the formula of the compound formed by the combination of A and B
Solution:
Formula of the compound = A2B3

Question 15.
Write the chemical names of the following compounds:

  1. Ca3(PO4)2
  2. K2CO3
  3. K2MnO4
  4. Mn3(BO3)2
  5. Mg(HCO3)2
  6. Na4Fe(CN)6
  7. Ba(ClO3)2
  8. Ag2SO3
  9. (CH3COO)2Pb
  10. Na2SiO3

Solution:
Chemical names of compounds:

  1. Ca3(PO4)2 – Calcium phosphate
  2. K2CO3 – Potassium carbonate
  3. K2MnO4 – Potassium manganate
  4. Mn3(BO3)2 – Manganese (II) borate
  5. Mg(HCO3)2 – Magnesium hydrogen carbonate
  6. Na4Fe(CN)6 – Sodium ferrocyanide
  7. Ba(ClO3)2 – Barium chlorate
  8. Ag2SO3 – Silver sulphite
  9. (CH3COO)2Pb – Lead acetate
  10. Na2SiO3 – Sodium silicate

Question 16
Write the basic radicals and acidic radicals of the following and then write the chemical formulae of these compounds.

  1. Barium sulphate
  2. Bismuth nitrate
  3. Calcium bromide
  4. Ferrous sulphide
  5. Chromium sulphate
  6. Calcium silicate
  7. Potassium ferrocyanide
  8. Stannic oxide
  9. Magnesium phosphate
  10. Sodium zincate
  11. Stannic phosphate
  12. Sodium thiosulphate
  13. Potassium manganate
  14. Nickel bisulphate

Solution:

Compounds Acidic
radical		 Basic
radical		 Chemical
formulae
Barium sulphate SO42- Ba2+ BaSO4
Bismuth nitrate NO3– Bi3+ Bi(NO3)3
Calcium bromide Br– Ca2+ CaBr2
Ferrous sulphide S2- Fe2+ FeS
Chromium sulphate SO42- Cr3+ Cr2(SO4)3
Calcium silicate SiO42- Ca2+ Ca2SiO4
Potassium ferrocyanide [Fe(CN)6]4- K1+ K4[Fe(CN)6]
Stannic oxide O2- Sn2+ SnO2
Magnesium phosphate (PO4)3- Mg2+ Mg3(PO4)2
Sodium zincate ZnO2- Na1+ Na2ZnO2
 Stannic phosphate (PO4)3- Sn4+ Sn3(PO4)4
 Sodium thiosulphate (S2O3)2- Na1+ Na2S2O3
Potassium manganate MnO42- K1+ K2MnO4
Nickel bisulphate HSO41- Ni3+ Ni(HSO4)3

Question 16.
Give the names of the following compounds.

  1. NaClO
  2. NaClO2
  3. NaClO3
  4. NaClO4

Solution:

  1. NaClO – Sodium hypochlorite
  2. NaClO2 – Sodium chlorite
  3. NaClO3 – Sodium chlorate
  4. NaClO4 – Sodium perchlorate

Question 18(a).
Complete the following statements by selecting the correct option :
The formula of a compound represents
i. an atom
ii. a particle
iii. a molecule
iv. a combination
Solution:
iii. The formula of a compound represents a molecule.

Question 18(b).
Complete the following statements by selecting the correct option :
The correct formula of aluminium oxide is
i. AlO3
ii. AlO2
 iii. Al2O3
Solution:
iii. The correct formula of aluminium oxide is Al2O3.

Question 18(c).
Complete the following statements by selecting the correct option :
The valency of nitrogen in nitrogen dioxide (NO2) is
i. one
ii. two
iii. three
iv. four
Solution:
iv. The valency of nitrogen in nitrogen dioxide (NO2) is four.

Page No: 13

Question 1.
Balance the following equations:
  1. Fe + H2O → Fe3O4 + H2
  2. Ca + N2 → Ca3N2
  3. Zn + KOH → K2ZnO2 + H2
  4. Fe2O3 + CO → Fe + CO2
  5. PbO + NH3 → Pb + H2O + N2
  6. Pb3O4 → PbO + O2
  7. PbS + O2 → PbO + SO2
  8. S + H2SO4 → SO2 + H2O
  9. S + HNO3 → H2SO4 + NO2 + H2O
  10. MnO2 + HCl → MnCl2 + H2O + Cl2
  11. C + H2SO4 → CO2 + H2O + SO2
  12. KOH + Cl2 → KCl + KClO + H2O
  13. NO2 +H2O → HNO2 + HNO3
  14.  Pb3O4 + HCl → PbCl2 + H2O + Cl2
  15.  H2O + Cl2 → HCl + O2
  16.  NaHCO3 → Na2CO3 + H2O + CO2
  17. HNO3 + H2S → NO2 + H2O + S
  18. P + HNO3 → NO2 + H2O + H3PO4
  19. Zn + HNO3  → Zn(NO3)2 + H2O + NO2

Solution:
Balanced chemical equations:

  1. 3Fe + 4H2O → Fe3O4 + 4H2
  2. 3Ca + N2 → Ca3N2
  3. Zn + 2KOH → K2ZnO2 + H2
  4. Fe2O3 + 3CO → 2Fe + 3CO2
  5. 3PbO + 2NH3 → 3Pb + 3H2O + N2
  6. 2Pb3O4 → 6PbO + O2
  7. 2PbS + 3O2 → 2PbO + 2SO2
  8. S + 2H2SO4 → 3SO2 + 2H2O
  9. S + 6HNO3 → H2SO4 + 6NO2 + 2H2O
  10. MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
  11. C + 2H2SO4 → CO2 + H2O + SO2
  12. 2KOH + Cl2 → KCl + KClO + H2O
  13. 2NO2 + H2O → HNO2 + HNO3
  14. Pb3O4 + 8HCl → 3PbCl2 + 4H2O + Cl2
  15. 2H2O + 2Cl2 → 4HCl + O2
  16. 2NaHCO3 → Na2CO3 + H2O + CO2
  17. 2HNO3 + H2S → 2NO2 + 2H2O + S
  18. P + 5HNO3 → 5NO2 + H2O + H3PO4
  19. Zn + 4HNO3  → Zn(NO3)2 + 2H2O + 2NO2

Page No: 17

Question 1.
Fill in the blanks:

  1. Dalton used symbol _____ for oxygen _____ for hydrogen.
  2. Symbol represents _____ atom(s) of an element.
  3. Symbolic expression for a molecule is called _____.  .
  4. Sodium chloride has two radicals. Sodium is a _____ radical while chloride is _____ radical.
  5. Valency of carbon in CH4 is _____ , in C2H6 _____, in C2H4 ___ and in C2H2 is ____.
  6. Valency of Iron in FeCl2 is _____  and in FeCl3 it is ____ .
  7. Formula of iron (ill) carbonate is _____ .

Solution:

  1. Dalton used symbol [O] for oxygen,[H] for hydrogen.
  2. Symbol represents gram atom(s) of an element.
  3. Symbolic expression for a molecule is called molecular formula.
  4. Sodium chloride has two radicals. Sodium is a basic radical, while chloride is an acid radical.
  5. Valency of carbon in CH4 is 4, in C2H64, in C2H44 and in C2H2 is 4.
  6. Valency of iron in FeCl2 is 2 and in FeCl3 it is 3.
  7. Formula of iron (III) carbonate is Fe2[CO3]3.

Question 2.
Complete the following table.

 

Acid Radicals

 

Basic Radicals

 Chloride Nitrate Sulphate Carbonate Hydroxide Phosphate
Magnesium MgCl2 Mg(NO3)2 MgSO4 MgCO3 Mg(OH)2  Mg3(PO4)2
Sodium
Zinc
Silver
Ammonium
Calcium
Iron (II)
Potassium

Solution:

Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 3

Question 3.
Sodium chloride reacts with silver nitrate to produce silver chloride and sodium nitrate

  1. Write the equation.
  2. Check whether it is balanced, if not balance it.
  3. Find the weights of reactants and products.
  4. State the law which this equation satisfies.

Solution:
(a) NaCl+ AgNO3 → NaNO3 + AgCl↓
(b) It is a balanced equation.
(c) Weights of reactants:NaCl – 58.44, AgNO3 – 169.87
Weights of products: NaNO3 – 84.99, AgCl – 143.32
NaCl + AgNO3 → NaNO3  + AgCl
(23+35.5) + (108+14+48) → (23+14+48) + (108+35.5)
58.5 + 170 → 85 + 143.5
228.5 g → 228.5 g
(d) Law of conservation of mass: Matter is neither created nor destroyed in the course of a chemical reaction.

Question 4(a).
What information does the following chemical equation convey? Zn + H2SO4 → ZnSO4+ H2
Solution:
(a) This equation conveys the following information:
  1. The actual result of a chemical change.
  2. Substances take part in a reaction, and substances are formed as a result of the reaction.
  3. Here, one molecule of zinc and one molecule of sulphuric acid react to give one molecule of zinc sulphate and one molecule of hydrogen.
  4. Composition of respective molecules, i.e. one molecule of sulphuric acid contains two atoms of hydrogen, one atom of sulphur and four atoms of oxygen.
  5. Relative molecular masses of different substances, i.e. molecular mass of
    Zn = 65
    H2SO4 = (2+32+64) = 98
    ZnSO4 = (65+32+64) = 161
    H2 = 2
  6. 22.4 litres of hydrogen are formed at STP.
Question 4(b).
What information do the following chemical equations convey? Mg + 2HCl → MgCl2+ H2
Solution:
(b) This equation conveys the following information:
  1. Magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen gas.
  2. 24 g of magnesium reacts with 2(1 + 35.5) = 73 g of hydrochloric acid to produce (24 + 71), i.e. 95 g of magnesium chloride.
  3. Hydrogen produced at STP is 22.4 litres.
Question 5(a).
What are polyatomic ions? Give two examples.
Solution:
(a) A poly-atomic ion is a charged ion composed of two or more atoms covalently bounded that can be carbonate (CO32-) and sulphate (SO42-)
Question 5(b).
Name the fundamental law that is involved in every equation.
Solution:
(b) The fundamental laws which are involved in every equation are:
  1. A chemical equation consists of formulae of reactants connected by plus sign (+) and arrow (→) followed by the formulae of products connected by plus sign (+).
  2. The sign of an arrow (→) is to read ‘to form’. It also shows the direction in which reaction is predominant.
Question 6(a).
What is the valency of : fluorine in CaF2
Solution:
(a) Valency of fluorine in CaF2 is -1.
Question 6(b).
What is the valency of :
sulphur in SF6
Solution:
(b) Valency of sulphur in SF6 is -6.

Question 6(c).

What is the valency of :

phosphorus in PH3
Solution:
(c) Valency of phosphorus in PH3 is +3.

Question 6(d).
What is the valency of : carbon in CH4
Solution:
(d) Valency of carbon in CH4 is +4.
Question 6(e).
What is the valency of :
nitrogen in the following compounds:
(i) N2O3 (ii) N2O5 (iii) NO2 (iv) NO
Solution:
(e) Valency of nitrogen in the given compounds:
  1. N2O3  = N is +3
  2. N2O5 = N is +5
  3. NO2 = N is +4
  4. NO = N is +2
Question 7.
Why should an equation be balanced? Explain with the help of a simple equation.
Solution:
According to law of conservation of mass, “matter can neither be created nor be destroyed in a chemical reaction”. This is possible only, if total number of atoms on the reactants side is equals to total number of atoms on products side. Thus, a chemical reaction should be always balanced.
Let us consider an example,
Fe + H2O → Fe3O4 + H2
In this equation number of atoms on both sides is not the same, the equation is not balanced.
The balanced form of this equation is given by,
3Fe + 4H2O  → Fe3O4 + 4H2
Question 8(a).
Write the balanced chemical equations of the following reactions. sodium hydroxide + sulphuric acid → sodium sulphate + water
Solution:
2NaOH + H2SO4 → Na2SO4 + 2H2O

Question 8(b).
Write the balanced chemical equations of the following reactions. potassium bicarbonate + sulphuric acid → potassium sulphate + carbon dioxide + water
Solution:
2KHCO3 + H2SO4 → K2SO4 + 2CO2 + 2H2O

Question 8(c).
Write the balanced chemical equations of the following reactions. iron + sulphuric acid → ferrous sulphate + hydrogen.
Solution:
Fe + H2SO4 → FeSO4 + H2

Question 8(d).
Write the balanced chemical equations of the following reactions. chlorine + sulphur dioxide + water →  sulphuric acid + hydrogen chloride
Solution:
Cl2 + SO2 + 2H2O → H2SO4 + 2HCl

Question 8(e).
Write the balanced chemical equations of the following reactions. silver nitrate → silver + nitrogen dioxide + oxygen”
Solution:
2AgNO3 → 2Ag + 2NO2 + O2

Question 8(f).
Write the balanced chemical equations of the following reactions.
copper + nitric acid → copper nitrate + nitric oxide + water
Solution:
3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O

Question 8(g).
Write the balanced chemical equations of the following reactions.
ammonia + oxygen → nitric oxide + water
Solution:
  Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 9

Question 8(h).
Write the balanced chemical equations of the following reactions.
barium chloride + sulphuric acid → barium sulphate + hydrochloric acid
Solution:
BaCl2 + H2SO4 → BaSO4 + 2HCl

Question 8(i).
Write the balanced chemical equations of the following reactions.
zinc sulphide + oxygen → zinc oxide + sulphur dioxide
Solution:
2ZnS + 3O2 → 2ZnO + 2SO2

Question 8(j).
Write the balanced chemical equations of the following reactions.
aluminium carbide + water → aluminium hydroxide + methane
Solution:
Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

Question 8(k).
Write the balanced chemical equations of the following reactions.
iron pyrites(FeS2) + oxygen → ferric oxide + sulphur dioxide
Solution:
 4FeS2 + 11O2 → 2Fe2O3 + 8SO2

Question 8(l).
Write the balanced chemical equations of the following reactions.
potassium permanganate + hydrochloric acid → potassium chloride + manganese chloride + chlorine + water
Solution:
2KMnO4 + HCl → 2KCl + 2MnCl2 + 5Cl2 + 8H2O

Question 8(m).
Write the balanced chemical equations of the following reactions.
aluminium sulphate + sodium hydroxide → sodium sulphate + sodium meta aluminate + water.
Solution:
Al2(SO4)3 + 8NaOH → 3Na2SO4 + 2NaAlO2 + 4H2O

Question 8(n).
Write the balanced chemical equations of the following reactions.
aluminium + sodium hydroxide + water → sodium meta aluminate + hydrogen
Solution:
2Al + 2NaOH + 2H2O → 2NaAlO2 + 3H2

Question 8(o).
Write the balanced chemical equations of the following reactions.
potassium dichromate + sulphuric acid → potassium sulphate + chromium sulphate + water + oxygen.
Solution:
2K2Cr2O7 + 8H2SO4 → 2K2SO4 + 2Cr2(SO4)3 + 8H2O + 3O2

Question 8(p).
Write the balanced chemical equations of the following reactions.
potassium dichromate + hydrochloric acid → Potassium chloride + chromium chloride + water + chlorine
Solution:
K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 7H2O + 3Cl2

Question 8(q).
Write the balanced chemical equations of the following reactions.
sulphur + nitric acid → sulphuric acid + nitrogen dioxide + water.
Solution:
S + HNO3 → H2SO4 + NO2 + H2O

Question 8(r).
Write the balanced chemical equations of the following reactions.
sodium chloride + manganese dioxide + sulphuric acid → sodium hydrogen sulphate  + manganese sulphate + water + chlorine.
Solution:
2NaCl + MnO2 + 3H2SO4 → 2NaHSO4 + MnSO4 + 2H2O + Cl2

Question 9(a).
Define atomic mass unit.
Solution:
Atomic mass unit (amu) is equal to one-twelfth the mass of an atom of carbon-12 (atomic mass of carbon taken as 12).

Question 9(b)(ii)
Calculate the molecular mass of the following:
(NH4)2CO3
Given atomic mass of Cu = 63·5, H = 1, O= 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of (NH4)2CO3
= (2 × 14) + (8 × 1) + 12 + (3 × 16)
= 28 + 8 + 12 + 48
= 96

Question 9(b)(iii)
Calculate the molecular mass of the following:
(NH2)2CO
Given atomic mass of Cu = 63·5, H = 1, O= 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of (NH2)2CO
= (14 × 2) + (4 × 1) + 12 + 16
= 28 + 4 + 12 + 16
= 60

Question 9(b)(iv)
Calculate the molecular mass of the following:
Mg3N2
Given atomic mass of Cu = 63·5, H = 1, O = 16, C = 12, N = 14, Mg = 24, S = 32
Solution:
Molecular mass of Mg3N2
= (3 × 24) + (2 × 14)
= 72 + 28
= 100

Question 10(a).
Choose the correct answer from the options given below.
Modern atomic symbols are based on the method proposed by
i. Bohr
ii. Dalton
iii. Berzelius
iv. Alchemist
Solution:
iii. Berzelius

Question 10(b).
Choose the correct answer from the options given below.
The number of carbon atoms in a hydrogen carbonate radical is
i. One
ii. Two
iii. Three
iv. Four
Solution:
One

Question 10(c).
Choose the correct answer from the options given below.
The formula of iron (III) sulphate is
i. Fe3SO4
ii. Fe(SO4)3
iii. Fe2(SO4)3
iv. FeSO4
Solution:
iii. Fe2(SO4)3

Question 10(d).
Choose the correct answer from the options given below.
In water, the hydrogen-to-oxygen mass ratio is
i. 1: 8
ii. 1: 16
iii. 1: 32
iv. 1: 64
Solution:
i. 1:8

Question 10(e).
Choose the correct answer from the options given below.
The formula of sodium carbonate is Na2CO3 and that of calcium hydrogen carbonate is
i. CaHCO3
ii. Ca(HCO3)2
iii. Ca2HCO3
iv. Ca(HCO3)3
Solution:
i. Ca(HCO3)2

Solution 11.
(a) A molecular formula represent The Molecule of an element or of a Compound.
(b) The molecular formula of water (H2O) represents 18 parts by mass of water.
(c) A balanced equation obeys the law of conservation of mass wherever unbalanced equation does not obey this law.
(d) CO and Co represent carbon-monoxide and cobalt respectively.

Solution 12.

  1. Relative molecular mass of CHCl3
    = 12 + 1 + (3 × 35.5)
    = 12 + 1 + 106.5
    = 119.5
  2. Relative molecular mass of (NH4)2 Cr2O7
    = (14 × 2) + (1× 8) + (52 × 2) + (16 × 7)
    = 28 + 8 + 104 + 112
    = 252
  3. Relative molecular mass of CuSO4· 5H2O
    = 63.5 + 32 + (16 × 4) + 5(2 + 16)
    = 159.5 + 90
    = 249.5
  4. Relative molecular mass of (NH4)2SO4
    = (2 × 14) + (8 × 1) + 32 + (4 × 16)
    = 28 + 8 + 32 + 64
    = 132
  5. Relative molecular mass of CH3COONa
    = (12 × 2) + (1× 3) + (16 × 2) + 23
    = 24 + 3 + 32 + 23
    = 82
  6. Potassium chlorate (KClO3)
    = 39.1+ 35.5 + (16 × 3)
    = 39.1+ 35.5 + 48
    = 122.6
  7. Ammonium chloroplatinate (NH4)2PtCl6
    = (14 × 2) + (1 × 8) + 195.08 + (35.5 × 6)
    = 28 + 8 + 195.08 + 213
    = 444.08

Solution 13.

Compound Empirical formula

(a) Benzene (C6H6)

 CH

CompoundEmpirical formula
(b) Glucose (C6H12O6)

 CH2O

CompoundEmpirical formula
(c) Acetylene (C2H2)

 CH

CompoundEmpirical formula
(d) Acetic acid (CH3COOH)

 CH2O

Solution 14.
Relative molecular mass of MgSO4·7H2O
=24 + 32 + (16 × 4) + 7(2 + 16)
=24 + 32 + 64 + 126
=246
26 g of Epsom salt contains 126 g of water of crystallisation.
Hence, 100 g of Epsom salt contains
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 5
The % of H2O in MgSO4·7H2O = 51.2

Solution 15.
(a)  Relative molecular mass of Ca(H2PO4)2
= 40.07 + (1 × 4) + (30.9 × 2) + (16 × 8)
= 40.07 + 4 + 61.8 + 128
= 233.87
233.87 g Ca(H2PO4)2 contains 61.8 g P
So, 100 g Ca(H2PO4)2 contains
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 6
The % of P in Ca(H2PO4)2 is 26.42%.
(b) Relative molecular mass of Ca3(PO4)2
= (40.07 × 3) + (30.9 × 2) + (16 × 8)
= 120.21 + 61.8 + 128
= 310.01
310.01 g Ca3(PO4)2 contains 61.8 g P
So, 100 g Ca(H2PO4)2 contains
(IMAGE)
The % of P in Ca(H2PO4)2 is 19.93%.

Solution 16.
Relative molecular mass of KClO3
= 39.09 + 35.5 + (3 × 16)
= 122.59 g
Selina Concise Chemistry Class 9 ICSE Solutions The Language of Chemistry image - 7
The percentages of K, Cl and O in KClO3 are 31.9%, 28.9% and 39.1%, respectively.

Solution 17.
Relative molecular mass of urea is

Element No. of atoms Atomic mass Total
N 2 14 28
C 1 12 12
H 4 1 4
O 1 16 16

[12 + 16 + 28 + 4] = 60
Hence, relative molecular mass of urea = 60
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Selina ICSE Solutions for Class 9 Chemistry Chapter 8 The Periodic Table

Page No. 79

Question 1.
What is the need for classification of elements?
Solution:
It is impossible for a chemist to study all the elements and their compounds. Hence, classification is a must.
Following are the reasons for the classification of elements:

  1. To study elements better
  2. To correlate the properties of the elements with some fundamental properties which are characteristic of all the elements
  3. To reveal relationships between elements

Question 2.
What was the basis of the earliest attempts made for classification and grouping of elements?
Solution:
The first classification of elements was into 2 groups-metals and non-metals.

Question 3(a).
A, B and C are the elements of a Dobereiner’s triad. If the atomic mass of A is 7 and that of C is 39, what should be the atomic
mass of B?
Solution:

(a)
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 1

Question 3(a).
Why was Dobereiner’s triad discarded?
Solution:
(b) Döbereiner failed to arrange all the known elements in the form of triads.
In the triad of fluorine (19), chlorine (35.5) and bromine (80), it is observed that the mean of the atomic masses of fluorine and bromine is ½(19 + 80) = 49.5, not 35.5.

Question 4.
Explain ‘Newland’s Law of Octaves.’ Why was the law discarded?
Solution:
Elements when arranged in the increasing order of their atomic weights are similar to the eighth and the first note of the musical scale. For example, the eighth element from lithium is sodium. Similarly, the eighth element from sodium is potassium. Thus, lithium and sodium provide any specific place for hydrogen.

  1. This classification did not work with heavier elements.
  2. Newland adjusted two elements Cobalt (Co) and Nickel (Ni) in the same slot.
  3. Fe, which resembles Co and Ni in properties, has been placed far away.

Question 5.
Did Dobereiners triads also exist in the columns of Newland’s Octaves? Compare and find out.
Solution:
Yes, Döbereiner’s triads also exist in the columns of Newland’s octaves. For example, the second column of Newlands classification has the elements Lithium (Li), Sodium (Na) and Potassium (K), which constitute a Döbereiner’s triad.

Question 6(a).
Lithium, sodium and potassium elements were put in one group on the basis of their similar properties. What are those similar properties?
Solution:
(a) Elements of lithium, sodium and potassium have the following similar properties:

  1. All these have one electron in the outermost shell.
  2. They form unipositive ions.
  3. They are good reducing agents.
  4. They are soft metals.
  5. They impart colour to the flame.
  6. Common name of the group is alkali metals [Group 1A].

Question 6(b).
The elements calcium, strontium and barium were put in one group or family on the basis of their similar properties.
What were those similar properties?
Solution:
(b)

  1. All of them are metals.
  2. Oxide of each of them is alkaline in nature.
  3. Each has valency 2.

Question 7(a).
What was Mendeleev’s basis for classification of elements?
Solution:
(a) Mendeleev’s basis for periodic classification:

  1. Similarities in the chemical properties of elements.
  2. Increasing order of atomic weights of elements.

Question 7(b).
Mendeleev’s contributions to the concept of periodic table laid the foundation for the Modern Periodic Table. Give reasons.
Solution:
(b) Mendeleev laid the foundation for the modern periodic table by showing periodicity of the properties of the elements by arranging the elements (63) then known into 8 groups, by leaving gaps for undiscovered elements and predicting their properties. He made separate groups for metals and non-metals. He also created periods in which the element gradually changes from metallic to non-metallic character. He was also able to show that the element in the same sub-group had the same valency.

Question 8.
State Mendeleev’s periodic law.
Solution:
Mendeleev’s periodic law: The physical and chemical properties of all the elements are a periodic function of their atomic masses.

Question 9(a).
Use Mendeleev’s Periodic Table to predict the formula of hydrides of carbon and silicon.
Solution:
(a) C is in Group 4. So, the hydride will be CH4 (Methane).
Si is in Group 4. So, the hydride will be SiH4 (Silane).

Question 9(b).
Use Mendeleev’s Periodic Table to predict the formula of oxides of potassium, aluminium and barium.
Solution:
(b)
K is in Group 1. So, the oxide will be K2O (Potassium oxide).
Al is in Group 3. So, the oxide will be Al2O3 (Aluminium oxide).
Ba is in Group 2. So, the oxide will be BaO (Barium oxide).

Question 10.
Which group of elements was missing from Mendeleev’s original periodic table?
Solution:
Anomalous pairs of elements were missing from Mendeleev’s periodic table.

Question 11.
State the merits of Mendeleev’s classification of elements.
Solution:
Merits of Mendeleev’s classification of elements:

  1. Grouping of elements
  2. Gaps for undiscovered elements: Mendeleev left some gaps in his periodic table for subsequent inclusion of elements not known at that time.
  3. He predicted the properties of the then unknown elements on the basis of properties of elements lying adjacent to the vacant slots (eka-aluminium and eka-silicon).

Question 12.
Why did Mendeleev’s leave some gaps in his periodic table os elements? Explain your answer with an example.
Solution:
He left gaps in the table for the undiscovered elements. He discovered the properties of such elements with the help of neighboring elements.
He discovered eka-silicon with atomic mass of 72 which was later named Germanium with atomic mass 72.6.

Question 13.
The atomic number of an element is more important to the chemist than its relative atomic mass. Why?
Solution:
Henry Moseley found that when cathode rays struck anodes of different metals, the wavelength of these metals was found to decrease in a regular manner of changing the metal of anode in the order of its position in the periodic table. By this, he concluded that the number of positive charges present in the nucleus due to protons (atomic number) is the most fundamental property of the element.

So, Henry Moseley found that the atomic number is a better fundamental property of an element compared to its atomic mass. This lead to the modern periodic law.

This law gave explanations for anomalies in Mendeleev’s classification of elements such as

  1. Position of isotopes with the same atomic number can be put in one place in the same group.
  2. Position of argon and potassium: Potassium with higher atomic number should come later, and argon with lower atomic number should come first.

Question 14.
Consider the following elements: Be, Li, Na, Ca, K. Name the elements of (a) same group (b) same period.
Solution:

Element At. No. Electronic distribution
Be 4 2, 2
Li 3 2, 1
Na 11 2, 8, 1
Ca 20 2, 8, 8
K 19 2, 8, 8, 2
  1. Same IA group (Li, Na, K) and IIA group (Be, Ca)
  2. In the second period (Be, Li) and in the fourth period (K, Ca)

Question 15(a).
Name an element whose properties were predicted on the basis of its position in Mendeleev’s periodic table.
Solution:
a. Eka-silicon

Question 15(b).
Name two elements whose atomic weights were corrected on the basis of their positions in Mendeleev’s periodic table.}
Solution:
b. Gold and Platinum

Question 15(c).
How many elements were known at the time of Mendeleev’s classification of elements?
Solution:
c. Only 63 elements were discovered at the time of Mendeleev’s classification of elements.

Page No. 86

Question 1(a).
State the modern periodic law.
Solution:
a. Modern periodic law: The physical and chemical properties of all elements are a periodic function of their atomic numbers.

Question 1(b).
How many periods and groups are there in the modern periodic table?
Solution:
b. Eighteen groups and seven periods

Question 2.
What is the main characteristic of the last elements in the periods of a periodic table? What is the general name of such elements?
Solution:
Last elements of each period have their outermost shell complete, i.e. 2 or 8 electrons.
The general name is inert gases or noble gases.

Question 3(a).
What is meant in the periodic table by a group?
Solution:
a. Vertical columns in a periodic table which have the same number of valence electrons and similar chemical properties are called a group.

Question 3(b).
What is meant in the periodic table by a period?
Solution:
b. In a periodic table, elements are arranged in the order of increasing atomic numbers in horizontal rows called periods.

Question 4.
From the standpoint of atomic structure, what determines which elements will be the first and which the last in a period of the periodic table?
Solution:
Atomic number determines which element will be the first and which will be the last in a period of the periodic table.

Question 5(a).
What are the following groups known as?

  1. Group 1
  2. Group 17
  3. Group 18

Solution:

(a)

  1. Group 1 is known as the alkali metals.
  2. Group 17 is known as the halogens.
  3. Group 18 is known as the transition elements.

Question 5(b).
Name two elements of each group.
Solution:
(b)

  1. Group 1: Lithium (Li), Sodium (Na)
  2. Group 17: Chlorine (Cl), Iodine (I)
  3. Group 18: Helium (He), Neon (Ne)

Question 6(a).
What is the number of elements in the 1st period?
Solution:
(a) There are two elements in the first period.

Question 6(b).
What is the number of elements in the 3rd period, of the modern periodic table?
Solution:
(b) There are eight elements in the third period.

Question 7(a).
How does number of (i) valence electrons (ii) valency; vary on moving from left to right in the second period of a periodic table?
Solution:

(a)

  1. The valence electrons in the same shell (outermost shell) increase progressively by one across the period. The first element hydrogen has one valence electron and helium has two valence electrons.
  2. On moving from left to right in a period, valency increases from 1 to 4, then falls to one and ultimately to zero in the last group.

Question 7(b).
How does number of (i) valence electrons (ii) valency; vary on moving from left to right in the third period of a periodic table?
Solution:

(b)

  1. Valence electrons in the same shell (outermost shell) increase progressively by one across the period. The first element sodium has one valence electron and magnesium has two valence electrons.
  2. On moving from left to right in the third period, valency increases from 1 to 7 and ultimately to zero in the last group.

Question 8.
How do atomic structures (electron arrangements) change in a period with increase in atomic numbers moving left to right?
Solution:
The size of atoms decreases when moving from left to right in a period. Thus, in a particular period, the alkali metal atoms are the largest and the halogen atoms are the smallest.

Question 9(a).
This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 2
Which of these is:

    1. a noble gas?
    2. a halogen?
    3. an alkali metal?
    4. an element with valency 4?

Solution:

(a)

  1. H and P are noble gases.
  2. G and O are halogens.
  3. A and I are alkali metals.
  4. D and L have valency 4.

Question 9(b).
This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 2
If A combines with F, what would be the formula of the resulting compound?
Solution:
(b) Li2O. A stands for lithium and F stands for oxygen. The valence of lithium is +1 and the valence of O is -2, i.e. A2F.

Question 9(c).
This question refers to elements of the periodic table with atomic numbers from 3 to 18. In the table below, some elements are shown by letters, even though the letters are not the usual symbols of the elements.
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 2
Solution:
(c) G has atomic number 9; therefore, its electronic arrangement is 2, 7.

Question 10.
Sodium and aluminium have atomic numbers 11 and 13, respectively. They are separated by one element in the periodic table, and have valencies 1 and 3 respectively. Chlorine and potassium are also separated by one element in the periodic table (their atomic numbers being 17 and 19, respectively) and yet both have valency 1. Explain.
Solution:
Na and Al have the capacity to donate an electron due to which the valency is positive, whereas Cl and K can only gain or lose one electron due to which their valency is -1 and +1, respectively. This is the only difference between these two.

Question 11.
Helium is an unreactive gas and neon is a gas of extremely low reactivity. What, if anything, do their atoms have in common.
Solution:
These elements have a full outermost subshell, which results in high stability. They only react with other elements in extreme circumstances.

Question 12(a).
In which part of a group would you separately expect the elements to have the greatest metallic character?
Solution:
a. The greatest metallic character can be expected at the bottom of the group.

Question 12(b).
In which part of a group would you separately expect the elements to have the largest atomic size?
Solution:
b. The largest atomic size can be expected at the lower part of the group.

Question 13.
What happens to the number of valence electrons in atoms of elements as we go down a group of the periodic table?
Solution:
The number of valence electrons remains the same as we go down a group.

Question 14(a).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 4
State which are metals, non-metals and noble gas in this table.
Solution:
(a) Metals: A and B; Non-metals: C; Noble gases: D and E

Question 14(b).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 5
State which is most reactive (i) metal (ii) non-metal
Solution:
(b) Most reactive

  1. Metals: Alkali metals (Group I); Caesium
  2. Non-metals: Halogens (Group 17); Fluorine

Question 14(c).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 6
Which type of ion will be formed by element A, B and C.
Solution:
(c)

Element A will form a positive ion 1+ (cation).
Element B will form a positive ion 2+ (cation).
Element C will form a negative ion 1– (anion).

Question 14(d).
The position of elements A, B, C, D and E in the periodic table are shown below:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 7
Which is larger in size (i) D or E (ii) B or C
Solution:
(d)

  1. E
  2. B

Question 15.
Write the electronic configuration of element 17T35.

  1. What is the group number of T?
  2. What is the period number of T?
  3. How many valence electrons are there in an atom of T?
  4. What is the valency of T?
  5. Is it a metal or non-metal?
  6. State number of protons and neutrons in T.

Solution:
K L M
Electronic configuration = 2, 8, 7

a. VIIA
b. Third period
c. Seven
d. Valency of T = -1
e. Non-metal
f. Protons = 17, Neutrons = 18

Page No. 91

Question 1.
Element P has atomic number 19. To which group and period, does P belong? Is it a metal or a non-metal? Why?
Solution:
Atomic number of P = 19
Electronic configuration = 2, 8, 8, 1
Group number of the element = 1
A Period number of the element = 4
P is a metal.

Question 2.
An element belongs to the third period and Group IIIA (13) of the periodic table. State:
a. the number of valence electrons,
b. the valency,
c. if it is a metal or non-metal?
d. the name of the element.
Solution:
a. 3
b. +3
c. Metal
d. Aluminium

Question 3.
Name and state the following with reference to the elements of the first three periods of the periodic table.

  1. Noble gas with duplet arrangement of electrons.
  2. Metalloid in Period 3.
  3. Valency of elements in Group 14 and 15.
  4. Noble gas having electronic configuration: 2, 8, 8.
  5. Group whose elements have zero valency.
  6. A covalent compound formed by an element in period 2 and ahalogen.
  7. Non-metallic element present in Period 3 of Groups 15 and 16
  8. An electrovalent compound formed by an alkaline earth metal and a halogen.
  9. Bridge elements of Period 3 of Group 1,2 and 3.
  10. Alkali metal in period 3 that dissolves in water giving a strong alkali.
  11. Typical elements of Groups 14 and 15.
  12. Alkaline earth metal in period 3.

Solution:

(a) Helium
(b) Silicon
(c) 4, 3
(d) Argon
(e) Noble gases
(f) Carbon tetrachloride (CCl4)
(g) Silicon, Phosphorus
(h) Sodium chloride (Na+Cl–)
(i) Li and Mg; Be and Al; B and Si
(j) Sodium
(k) Typical elements of Period 2 belonging to group 14 and 15 are carbon and nitrogen.
Typical elements of’ Period 3 belonging to group 14 to 15 are silicon and phosphorus.
(l) Beryllium

Question 4.
Match column A with column B
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 8
Solution:

Column A Column B
(a) Elements short of 1 electron in octet (v) Halogens
(b) Highly reactive metals (iii) Alkali metals
(c) Unreactive elements (ii) Noble gases
(d) Elements of groups 3 to 12 (i) Transition elements
(e) Radioactive elements (vi) Actinides
(f) Elements with 2 electrons in outermost orbit (iv) Alkaline earth metals

Question 5.
Complete the table:
Selina Concise Chemistry Class 9 ICSE Solutions The Periodic Table image - 9
Solution:

Atomic number Element Electronic configuration Select element of the same group
11 Sodium 2, 8, 1 K
15 Phosphorus 2, 8, 5 N
16 Sulphur 2, 8, 6 O
9 Fluorine 2, 7 Cl

Question 6.

  1. Write down the word that will correctly complete the following sentences:
  2. Relative atomic mass of a light element up to calcium is approximately_____ its atomic number.
  3. The horizontal rows in a periodic table are called _____ .
  4. Going across a period left to right, atomic size _____ .
  5. Moving left to right in the second period, number of valence electrons _____ .
  6. Moving down in the second group, number of valence electrons___________.

Solution:

a. Relative atomic mass of a light element up to calcium is approximately 20 its atomic number.
b. The horizontal rows in a periodic table are called periods.
c. Going across a period left to right, atomic size increases.
d. Moving left to right in the second period, number of valence electrons increases from 1 to 8.
e. Moving down in the second group, number of valence electrons remain same.

Question 7(a).
Name the alkali metals, How many electron(s) they have in their outermost orbit.
Solution:
Name of the alkali metals: Lithium, sodium, potassium, rubidium, cesium and francium
Electrons in the outermost orbit: 1

Question 7(b).
Take any one alkali metal and write its reaction with (i)oxygen (ii)water (iii)acid.
Solution:

  1. Reaction of alkali metal with oxygen – React rapidly with oxygen
    4Na + O2 → 2Na2O
  2. Reaction of alkali metal with water – React with water violently and produce hydrogen
    2M + 2H2O → 2MOH + H2
  3. Reaction of alkali metal with acid – React violently with dil. HCl and dil. H2SO4 to produce hydrogen
    2M + 2HCl → 2MCl + H2

Page No. 92

Question 8(a).
Name the method by which alkali metals can be extracted.
Solution:
a. Alkali metals can be extracted by the electrolysis of their molten salts.

Question 8(b).
What is the colour of the flame of sodium and potassium?
Solution:
b. The colour of the flame of sodium is golden yellow, and the colour of the flame of potassium is pale violet.

Question 9(a)
Name the first three alkaline earth metals.
Solution:
a. The first three alkaline earth metals are Beryllium, Magnesium and Calcium.

Question 9(b)
Write the reactions of first three alkaline earth metals with dilute hydrochloric acid.
Solution:
b. Reactions of the first three alkaline earth metals with dilute hydrochloric acid:
Be + 2HCl → BeCl2 + H2
Mg + 2HCl → MgCl2 + H2
Ca + 2HCl → CaCl2 + H2

Question 10(a)
How do alkaline earth metals occur in nature?
Solution:
a. Alkaline earth metals occur in nature in the combined state and not in the free state as they are very reactive.

Question 10(b)
Write the electronic configuration of the first two alkaline earth metals.
Solution:
b. Electronic configuration of the first two alkaline earth metals:
4Be: 1s22s2
12Mg: 1s22s22p63s2

Solution 11.
a. Group 17 elements are called halogens. The name halogens is from Greek halo (sea salt) and gens (producing, forming) and thus means ‘sea salt former’.

b. Group 17 elements or halogens:

  1. Reactivity: Halogens are the most reactive non-metals, their reactivity decreases down the group. Fluorine is the most reactive halogen and iodine is the least reactive halogen.
  2. Colour: Fluorine is a pale yellow gas, chlorine is a greenish yellow gas, bromine is a reddish brown liquid and iodine is a violet solid.
  3. Physical state: Gaseous

Solution 12.
(a)

  1. Group 17 elements react with metals to form metal halides which are neutral in nature.
  2. Group 17 elements react with non-metals to form acidic compounds such as hydrogen halides.

(b) Group 17 elements are highly reactive because of their closeness to the noble or stable gas configuration. They can easily achieve a noble gas electron structure.

Solution 13.
a. All the noble or inert gases have 8 electrons in their valence shell except helium which has two electrons in its valence shell.
b. Xenon or krypton from Group 18 can form compounds.

Solution 14.

  1. Helium gas is used in airships and balloons.
  2. Neon gas is used in neon lights. The brightly coloured advertising light works when an electric discharge is passed in a tube containing a little of a noble gas such as neon.

Question 9.
An element A has 2 electrons in its fourth shell. State:

  1. its atomic number
  2. its electronic configuration
  3. its valency
  4. position in the periodic table
  5. is it a metal or non metal
  6. is it an oxidising or reducing agent

Solution 15.

a. Ca
b. 1s22s22p63s23p64s2
c. 2
d. Group 2 Period 4
e. Metal
f. Reducing agent

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Selina ICSE Solutions for Class 9 Chemistry – Atomic Structure

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Selina ICSE Solutions for Class 9 Chemistry – Atomic Structure

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Selina ICSE Solutions for Class 9 Chemistry Chapter 7 Atomic Structure

Exercise 7

Solution 1.

The latest research on atom has proved that most of the postulates of Dalton’s atomic theory contradict. But Dalton was right that atoms take part in chemical reactions. Comparison of Dalton’s atomic theory with Modem atomic theory.

Dalton’s atomic theory Modern atomic theory
1. Atoms are indivisible particles. 1. Atoms are divisible into sub-atomic particles like protons, neutrons and electrons.
2. Atoms can neither be created nor destroyed. 2. Atoms can be created and destroyed by nuclear fusion and fission.
3. The atoms of an element are alike in all respect and differ from atoms of other elements. 3. The atoms of an element may not be alike in all respects, as it is seen in the case of isotopes. Isotopes which are atoms of the same element having the same atomic number but different mass numbers.

Solution 2.

(a) Inert elements: The elements which have complete outer most shell i.e. 2 or electrons. They ordinarily do not enter into any reaction.

(b) These exist as monoatoms because molecules of these elements contain only one atom.

(c) Valence electrons: The number of electrons present in the outermost shell or valence shell is known as valence electrons.

Solution 3.

The three isotopes of hydrogen differ only due to their mass number which is respectively 1,2 and 3 and named protium, deuterium and tritium.
Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 1

Solution 4.

Atomic number Name with valency
4 –
15 A solid non-metal of valency 3.
8 A gas of valency 2.
19 A metal of valency 1.
14 A non-metal of valency 4.

Solution 5.

Atom Atomic number Atomic mass No. of Protons No. of Electrons No. of Neutrons Electronic configuration
(a) Sodium 11 23 11 11 12 2, 8, 1
(b) Chlorine 17 35 17 17 18 2, 8, 7
(c) Oxygen 8 16 8 8 8 2, 6
(d) Carbon 6 12 6 6 6 2, 4

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 2

Solution 6.

The significance of the number of protons found in the atoms in each of different element is fixed its place in periodic table.

Solution 7.

Atomic numbers of –

X Y Z
6 9 12
(2,4) (2,7) (2,8,2)

(a) Y (2,7) forms Anion.
(b) Z (2, 8, 2) forms Cation.
(c) X (2,4) has four electrons in the valence shell.

Solution 8.

(a) X+1
(b) Oxidising agent, because it has ability to donate electron.

Solution 9.

(a) Mass number:The atomic mass number is defined as the sum of the number of protons and neutrons contained in the nucleus of an atom of that element. It is denoted by the symbol A.

(b) Ion: An atom or molecule that carries a positive or negative charge because of loss or gain of electrons.

(c) Cation: It is positively charged ion that is formed when an atom loses one or more electrons e.g. Na+, Hg2+, Ca2+ etc.

(d) Atom: It is defined as the smallest unit of matter which takes part in a chemical reaction.

(e) Element: It is a substance which cannot be split up into two or simpler substances by usual chemical methods of applying heat, light or electric energy. e.g. Hydrogen, Oxygen, Chlorine etc.

(f) Orbit: It is defined as a circular path around the nucleus in which electrons of the atom revolve.

Solution 10.

Atomic number = 2 Mass number = 4

Solution 11.

(a) (i) Atom E contains 7 protons.
(ii) Atom E has an electronic configuration 2, 7.

(b) Atom C Stands for 73Li
Atom D stands for 816O
Compound formula = Li2O

(c) Metals are: A and C Non-metals are: E, D, E

Solution 12.

No of electrons in M Shell = 2
Number of electrons in K and L shell will be 2, 8 and respectively.
Therefore, Electronic configuration will be: 2, 8, 2.
Atomic number = 2 + 8 + 2 = 12, Since, atomic number = Number of Protons
No of Protons = 12

Solution 13.

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 3
(a) (ii) Electronic configuration = 2, 8, 2
(b) Mass numbers are different of two isotopes of magnesium because of different number of neutron, that is, 12 and 14 respectively.

Solution 14.

Nucleons: Particles which constitute nucleus are called nucleons. Proton and neutrons are the nucleons.
At. weight of phosphorus = 31 Atomic number = 15
Total number of nucleons = 31 (No. of P + No. of N) No. of neutrons = 31 – 15 = 16
Electronic configuration
No. of electrons = 15 = 2,8,5
Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 4

Solution 15.

(a) Atoms of the same elements differing in the number of neutrons in their nuclei are known as isotopes. Thus, isotopes of an element have the same atomic number but different atomic mass number.
The fundamental particles is Neutrons which differs.

Uses of isotopes:

  1. Some isotopes are Redioactive; isotopes of cobalt are used for treating cancer and other diseases.
  2. An isotopes of uranium 235U is used as a fuel in nuclear reactor.

(b)

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 5

Solution 16.

In chemical reactions only electrons take part. The chemical properties depend upon the electronic configuration.

The isotopes of element 1735Cl and 1737Cl have same atomic number and hence, the same configuration. So they have same chemical properties. These differ only in physical contents and weights because neutrons contribute to the mass of an atom.

1735Cl and 1737Cl have different number of neutrons 18 and 20 respectively.

Solution 17.

The atomic masses of the isotopes of chlorine are 35 and 37. However in any given sample of chlorine gas, the isotopes occur in approximate 3 : 1, 75% of Cl35, and 25% of Cl37. Therefore, the relative atomic mass or atomic weight of chlorine is 35.5.

Fractional atomic weight of chlorine

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 6

Solution 18.

(a) Atomic number: The number of protons present in the nucleus of an atom is the atomic number of that atom. It is represented by the symbol Z.
Atomic number (Z) = Number of protons (p)

(b)

No. of Protons No. of Electrons No. of Neutrons Atomic number Mass number
35 Cl
17		 17 17 18 17 35
37 Cl
17		 17 17 20 17 37

(c) Electronic configuration of chlorine is 2, 8, 7.

Solution 19.

(a) Hydrogen
(b) Element of zero group i.e. He (Helium)
(c) Calcium (2, 8, 8, 2). Therefore, 2 electrons in valance shell. Hence valency is 2.
(d) Chlorine atoms: 1735Cl and 1737Cl
(e) K shell.

Solution 20.

(a) Physical properties depend on the Atomic mass and isotopes have different mass number. (A) i.e. they have different number of neutrons. So, isotopes have different physical properties.

(b) Argon does not react as, Argon has completely filled outer-most orbit. The atomic number of argon is 18. Therefore, electronic configuration is 2, 8, 8. There are 8 electrons in the outermost or valance shell. Therefore, argon does not react.

(c) Actual Atomic Mass is greater than mass number (P + N) since mass number is a whole number approximation of atomic mass unit. In fact Neutrons are slightly heavier than protons and atom includes the existence of over 200 sub-atomic particles.

(d) 1735Cl and 1737Cl are isotopes of chlorine element which differ in number of neutrons. Whereas chemical properties are determined by electronic configuration of an atom. Isotopes of an element are chemically alike.

Solution 21.

Element A

Atomic number = 7
Electronic configuration: = 2, 5
Valency of element A = 8-5 = (3–)

Element B

Electronic configuration 2,8,8 Valency of element B = Zero Element C
Number of Electrons 13 Electronic configuration: 2,8,3
Valancy of element C = 3+

Element D

Protons = 18 = electrons Electronic configuration: 2,8, 8
Valency of element D = Zero

Element E

Electronic configuration = 2, 8, 8, 1
Valency of element E = 1+
[ii] C and E are metals [iii] A is a Non-metal
[iv] B and D are inert gases. (A, C and E are not inert gases.)

Solution 22.

(a) C. Atomic nucleus
(b) A. 6
(c) C. 2,8,8,1

Solution 23.

Elements tend to combine with one another to attain the stable electronic configuration of the nearest noble gas/inert gas (Duplet or Octet)

(a) Sodium Chloride

The electronic configuration of Sodium is 2, 8, Thus, Sodium atom tends to lose one electron to attain the stable electronic configuration of neon gas 2, 8.

The electronic configuration of Chlorine is 2, 8, 7. Thus, Chlorine atom needs one electron to complete its octet and achieve the stable electronic configuration of Argon 2, 8, 8.

When sodium and chlorine atoms approach each other, the sodium atom, Na, loses an electron to form sodium ion, The cation, Na+, carries a single positive charge.
Na → Na+ + e–

The electron lost by sodium atom is transferred to the chlorine atom forming chloride ion, Cl–.
Cl + e– → Cl–

The sodium and chloride ions thus, achieve stable electronic configurations.
Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 7

(b) Hydrogen H2

Hydrogen atom has one electron in its valence shell. Hydrogen atom needs one more electron to complete its duplet and attain the stable electronic configuration of Helium.

In case of hydrogen H2 molecule, each of the two hydrogen H atoms contributes one electron so as to have one shared pair of electrons. Both the hydrogen atoms attain stable duplet structure resulting in the formation of a single covalent bond [H – H] between them.

Both the hydrogen atoms have equal attraction for the electrons. Thus, the shared pair of electrons remains equidistant from both the atoms.

Selina ICSE Solutions for Class 9 Chemistry - Atomic Structure image - 8

Solution 24.

Element Symbol Atomic Number Mass Number Number of Neutrons Number of Electrons Number of Protons
Li 3 6 4 3 3
Cl 17 35 18 17 17
Na 11 23 12 12 11
Al 13 27 14 13 13
S 15 32 16 15 15

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Selina ICSE Solutions for Class 9 Chemistry – Elements, Compounds and Mixtures

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Selina ICSE Solutions for Class 9 Chemistry – Elements, Compounds and Mixtures

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Selina ICSE Solutions for Class 9 Chemistry Chapter 3 Elements, Compounds and Mixtures

Exercise 3(A)

Solution 1.

An element is a pure substance composed of only one kind of atom. Example: C, H, O, Na, Ca, N etc.

Characteristics of an element:

  1. An element is made up of only one kind of atoms.
  2. An element is pure and homogeneous substance.
  3. An element has fixed melting and boiling points.
  4. An atom is the smallest particle of an element which takes part in a chemical reaction.
  5. An element may chemically react with another elements or compounds.
  6. An element can occur in solid, liquid or gaseous state.
  7. The molecules are made up of one or more atoms of the same or different elements.

Solution 2.
Selina ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures image - 1

Solution 3.

Two elements which show exception to the properties of:

Metals :-

  1. Mercury (Hg) is liquid at room temperature.
  2. Tungsten (W) is a poor conductor of electricity.

Non-metals:-

  1. Iodine is lustrous.
  2. Carbon is ductile.

Solution 4.

(a) Molecule: A molecule is the smallest particle of a pure substance (element or compound), and it has all the properties of that substance. It is composed of atoms. It is capable of existing in a free state.
Example: O2, H2, Cl2 are molecules.

(b) Atomicity: Atomicity is the number of atoms present in a molecule of an element.

(c) Compound: A compound is a pure substance composed of two or more elements combined chemically in a fixed proportion by mass. The properties of compounds are different from the properties of their constituent elements. Example: H2O, CO2 etc.

Solution 5.

(a) A diatomic element Nitrogen (N2)
(b) A tetratomic element Phosphorus (P)
(c) Monoatornic element Helium (He)
(d) Lustrous non-metal Iodine
(e) Liquid non-metal Bromine (Br2)
(f) A gas filled in electric bulbs Argon (Ar)
(g) A liquid metal Mercury (Hg)
(h) A non metal conductor of electricity Graphite
(i) A metal non malleable and non ductile Zinc (Zn)
(j) A lustrous non metal Graphite

Solution 6.

(i) Sodium chloride is obtained when sodium chemically combines with chlorine in ratio of 23:355 by weight.
(ii) When molten sodium chloride is subjected to electrolysis, the ratio by weight of sodium and chlorine librated at electrodes is 2:3.

Solution 7.

Type Substances Reason
Element Chlorine, Sulphur They cannot be split up into any simpler substance.
Compound Carbon dioxide It can be produces by chemical analysis of two or more simpler substances with different properties.
Mixture Honey, milk, sea water, gun powder, apple juice, brine, syrup and bronze These are produced by mere mixing of two or more substances in any proportions by weight.

Solution 8.

(a) This is because molecules have all the properites of that substance and is capable of existing in a free state, molecules are composed of atoms.

(b)

Element Compund
1. It is a pure substance which cannot be converted into simpler substances by any physical or chemical means. 1. It is a pure substance made up two or more elements combined chemically in a fixed ratio.
2. It is made up of only one kind of atoms. 2. It is made up of two or more different kinds of atoms.
3. The molecules are made up of one or more atoms. 3. The molecules are made up of two or more atoms.

Solution 9.

It is true that the elements can form different compounds.
Example: Hydrogen and oxygen combine to give two different compounds, water (H2O) and hydrogen peroxide (H2O2) under different conditions.

Solution 10.

Characteristics of a compound

  1. A compound is made up of one or more atoms of the same or different elements.
  2. It has a homogeneous composition.
  3. In a compound the elements are present in a fixed ratio by mass.
  4. The properties of a compound are different from those of its Constituent elements.

Solution 11.

The properties of compounds are different from the properties of their constituent elements. Example: H2O, FeS, C12H22O11

  1. H2O: Water is a liquid, while constituent elements, Hydrogen and Oxygen are gases.
  2. FeS: Iron sulphide is a black substance, not attracted by a magnet and insoluble in carbon disulphide. While constituent elements, Iron is grey colored, attracted by a magnet. Sulphur is a yellow colored, soluble in carbon disulphide.
  3. C12H22O11: Sugar is a crystalline solid, sweet to taste and soluble in water. But, its constituent elements, Carbon, is black insoluble solid. Hydrogen and Oxygen are invisible and odorless gases.

Solution 12.

A compound is a pure substance composed of two or more elements combined chemically in fixed proportions by mass. The properties of a compound are different from the properties of their constituent elements.

  1. H2O: Water is a liquid, while constituent elements, Hydrogen and Oxygen are gases.
  2. FeS: Iron sulphide is a black substance, not attracted by a magnet and insoluble in carbon disulphide. While constituent elements, Iron is grey colored, attracted by a magnet. Sulphur is a yellow colored, soluble in carbon disulphide.
  3. C12H22O11: Sugar is a crystalline solid, sweet to taste and soluble in water. But, its constituent elements, Carbon, is black insoluble solid. Hydrogen and Oxygen are invisible and odorless gases.

Solution 13.

A mixture cannot be represented by a chemical formula because constituents present in a mixture are in any ratio and they are not chemically united.

Solution 14.

(a) Air
(b) Concrete
(c) Milk

Solution 15.

Elements Compounds Mixtures
Mercury Sugar, Distilled water, Alcohol, Nitre, Washing soda, Rust, Marble Air, Milk, Wax, Sea-water, Paint, Brass, Bread, Soap, Tap water

Solution 16.

On adding sulphuric acid to water we will get a Homogeneous Mixture (true solution).

This mixture will have different densities and boiling points depending upon the amounts of acid and water. The properties of acid and water will remain same even after mixing.

Solution 17.

Iron and sulphur when mixed, forms a mixture. It can be identified as follows:

  • A grayish yellow mixture will be produced.
  • All the individual properties of iron andsulphur will be shown separately in a mixture. Iron particles will be attracted by magnet. Sulphur will dissolve in carbon disulphide. Iron and sulphur when heated forms a compound.

It can be identified as follows:

  • A grey dark solid will be produced.
  • The compound formed is homogeneous.
  • It is neither attracted by a magnet nor it is soluble in carbon disulphide.

Solution 18.

Mixture Compound
1. It is obtained by the physical combination of either elements, or compounds, or both. 1. It is obtained by the chemical combination of more than one element.
2. The composition of elements present in a mixture is not fixed. 2. The composition of elements present in a compound is fixed.
3. It shows the properties of all its constituent elements. 3. The properties of a compound are different from those of its elements.
4. Its constituents can be separated using physical methods. 4. Its constituents can be separated by using only chemical and electrochemical methods.
5. The mixtures can be homogeneous or heterogeneous. 5. A compound is always homogeneous in nature.

Solution 19.

No. Types of mixture Example Nature
1. Two solids Bronze (Zn, Cu, Sn) Homogeneous
2. A solid in liquid Sugar in water, Salt in water, Iodine in alcohol

Sugar in oil

 Homogeneous

Heterogeneous
3. Liquids Oil in water, Kerosene in water,

Acetone + water

Heterogeneous

Homogeneous
4. Liquid and gas Moisture in air Homogeneous

Solution 20.

Homogeneous mixtures Heterogeneous mixture
1. A mixture is said to be homogeneous if its constituents are uniformly distributed and are not physically distinct. 1. A mixture is said to be heterogeneous if its constituents are not uniformly distributed and are physically distinct.
2. They have the same composition and properties throughout their mass. 2. They have the different composition and properties in different parts of their mass.
3. For example: sugar solution, salt in water etc. 3. For example: Mixture of sand in water, Mixture of oil in water

Solution 21.

(a)

  1. The properties of ammonia are different from those of its components i.e nitrogen and hydrogen.
  2. When ammonia is formed from nitrogen and hydrogen, energy is given out.
  3. In ammonia, N and H are present in fixed ratio of 14 : 3 by mass.

(b) Air is considered as mixture not a compound because:

  1. The composition of air is different at different places and different altitudes.
  2. The constituents of air can be easily separated by physical means.
  3. The properties of air vary according to the properties of oxygen and nitrogen.

(c) Tap water does not have a fixed melting point. It may have various dissolved as well as undissolved impurities. Also it does not have a fixed composition.

Solution 22.

A pure substance is a homogeneous material with a definite, invariable chemical composition, and definite, invariable physical and chemical properties.

Sugar is a pure substance as all sample of sugar consists of same chemical composition. i.e. carbon, hydrogen and oxygen, C12H22O11.

Solution 23.

If dilute hydrochloric acid is added to the compound formed of iron and sulphur, Hydrogen sulphide gas is produced. (With a rotten egg smell).

Solution 24.

Selina ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures image - 2
Selina ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures image - 8

Solution 25.

When a mixture of powdered iron and sulphur is heated in a test tube, the powder starts melting and a pungent smell of a gas is given out. A dark grey solid is produced and this is called iron sulphide.

Solution 26.

Solute: A substance which gets dissolved in a solvent is called as solute.
Solvent: A substance in which solute gets dissolved in it is called as solvent.
Solution: A homogeneous mixture of two or more substances which are chemically non reacting, whose composition can be varied within certain limits is called a solution.
Solution = Solute + Solvent

Solution 27.

A homogeneous solution in which the size of the particle is about 10-10 m is called a true solution.

Properties of true solution:

  1. It is a homogeneous mixture.
  2. The solute particles are very small, about 10-10 m.
  3. It is clear and transparent.
  4. It does not scatter light:
  5. The components cannot be separated by filtration or by any physical means.
  6. The solute particles in a solution do not settle down.
  7. The particles are invisible.
  8. The particles diffuse rapidly

Solution 28.

Properties of suspension are:

  1. Suspension is a heterogeneous mixture.
  2. The solute particles in a suspension have a size of greater than 10-7 m.
  3. It is opaque in nature.
  4. The particles are visible to naked eye or under a simple microscope.
  5. The particles do not pass even through ordinary filter paper.
  6. The particles of a suspension do not diffuse.
  7. The particles in a suspension settle down on standing.
  8. It scatters light to some extent.

Solution 29.

A homogeneous looking heterogeneous mixture in which particles having a size between 10-10 m and 10-7 m dispersed in a continuous medium is called as colloid.

Sr. no. Dispersion medium Dispersed phase Common name of the system Examples
1. Solid Gas Solid foam Pumice stone, foam rubber etc.
2. Liquid Gas Foam or froth Soap lather, shaving cream foam, lemonade froth, etc.
3. Liquid Liquid Emulsion Milk, emulsified oils, medicines etc.
4. Solid Liquid Gel Cheese, Butter, Jams, Jellies, Boot polish etc.
5. Gas Liquid Aerosol of liquids Fog, mist, cloud, Liquid sprays etc.
6. Solid Solid Solid sol Glasses, Gems, Pigmented plastics etc.
7. Liquid Solid Sol Sulphur sol, Gold sol, Ferric hydroxide sol, Starch solution, Muddy water etc.
8. Gas Solid Aerosol of solids Smoke, dust etc.

 Solution 30.

The movement of colloidal particles towards a particular electrode under the influence of an electric field is called electrophoresis.

It is shown by only those solutions, the particles of which carry a charge, such as colloid, particles. Thus this phenomenon of electrophoresis can be used to find the nature of charge carried by colloidal particles in a colloidal system.

For example, If the colloidal particles carry positive charge, they move. Towards negatively charge electrode (cathode) when subjected to an electric field. If they carry negative charge, they move towards positively charged electrode (Anode).

Solution 31.

Sr no. Property True solution Suspension Colloids
1 Particle size Less than 10-10 m Greater than 10-7m Between 10-10 to 10-7 m
2 Filtrability Pass easily through ordinary filter paper as well as animal membranes. Do not pass even through ordinary filter filter paper animal membranes. Pass easily through ordinary paper but not through animal membranes.
3 Visibility of particles Invisible Visible to naked eye or under simple microscope. Visible under Ultra microscope.

Solution 32.

A suspension is a heterogeneous mixture in which very fine particles, about 10-7 m, of a solid are dispersed in any medium like a liquid or a gas.

Dispersed substance:

  1. Cannot pass through a filter paper nor a semi-permeable membrane.
  2. It is visible to the naked eye.
  3. They settle down after sometime.

Examples:

Chalk in water, Sand in water, Coagulated matter.

Solution 33.

Tyndall effect can be defined as, the scattering of beam of light by colloidal particles present in a colloidal solution.

Tyndall effect can be observed when a fine beam of light passes through a small hole in the dark room. This happens due to the scattering of light by the particles of dust or smoke present in the air.Selina ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures image - 3
Experiment- In two Separate jars, take (FeSO4) Ferrous sulphate solution (green colour) in A and milk solution diluted with water in B.

Focus a torch on these jars in dark place. Torch light will not be visible in Ferrous sulphate solution and will not show Tyndall Effect as FeSO4 solution is actually true solution whereas light will be visible in B jar containing milk solution which is colloidal or Emulsion.

Examples of Tyndall effect from our daily life :-

  1. Tyndall Effect can be observed when sun light passes through the Canopy of a dense forest. In the forest, mist contains tiny droplets of water, which act as colloidal particles dispersed in air.
  2. When light enters a room through d small slit, dust particles scatter the light.
  3. Tyndall effect is observed when colloidal solution is distinguished from a true solution.

Solution 34.

The effect is called as the Tyndall effect.
Tyndall effect is shown by colloidal solution.

Properties of colloidal solution:

  1. The particles of colloidal solution do not settle under gravity. They can be made to settle down by centrifugation.
  2. It is a heterogeneous mixture.

Solution 35.

(a) Liquid in water – e.g. Alcohol solution, ammonia solution.
(b) Non-aqueous solution – e.g. solution of iodine in alcohol- This is called Tincture iodine. (c) Solid in non-aqueous solvent – e.g. solution of stearicacid in ethanol.

Solution 36.

Common names of colloids:

  1. Solid -gas is Solid foam
  2. Solid -Liquid is Gel
  3. Liquid – Solid is Sol

Solution 37.

Solution Mixture Compound
1. The solute is not present in any fixed proportion but its composition is uniform. The composition is not uniform. The component may be present in any proportion.

The constituents of a compound are combined in a definite proportion.
2. The solute can be recovered by evaporating the solvent, i.e. by physical means. The components can be separated by ordinary physical means. The components can be separated by chemical means only

Solution 38.

The distributed substance in the solution is called as dispersed phase.
The medium in which distributed substance is dispersed is referred to as dispersion medium.

Solution 39.

Blood – colloidal solution Sugar solution – true solution Salt solution – true solution
Starch solution – colloidal solution Ink – colloidal solution

Solution 40.

Starch solution, milk, soap solution, blue vitriol will scatter light as these are colloidal solutions.

Exercise 3(B)

Solution 1.

The methods used to separate solid-liquid mixture are:

  1. Evaporation
  2. Distillation
  3. Filtration
  4. Sedimentation and decantation
  5. Centrifugation /churning
  6. Chromatography

Solution 2.

(a) Sand and water are separated by Filtration. Since sand is insoluble in water and forms Heterogeneous mixture. Water dissolves sodium chloride. The solution is than filtered with the help of a filter paper. Sand gets collected as a residue on the filter paper. And the dissolved sodium chloride in water can later be evaporated.

(b) Salt from an aqueous salt solution – Salt can be separated from aqueous salt solution by the process of evaporation. For this the aqueous salt solution (i.e. salt in water) is taken in a beaker and heated over flame. The liquid i.e. water will get evaporated leaving behind salt.

(c) Pure water from salt water- Pure water can be separated from salt water by the process of distillation. Distillation is the process of converting a liquid into vapour by heating and the subsequent condensation of the vapour back into a liquid. Water evaporates and recondenses in pure form and it is collected in a receiver. The salt residue remains in the distillation flask.
Selina ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures image - 4
(d) Tea leaves from prepared tea – Tea leaves can be separated from prepared tea by the process of filtration. For this purpose a sieve or mesh made of steel or nylon is taken through which prepared tea is passed, thereby leading to retaining of tea leaves on the mesh.

(e) Cream from milk – Cream can be separated from milk by the process of centrifugation. The mixture is taken in centrifuge tubes and placed in holders. The holders are then rotated rapidly. After sometime, the rotation is stopped and the centrifuge tubes are taken out. It is observed that separation of mixture takes place on the bases of density. i.e. the more denser components settles at the bottom and the other becomes the supernatant which can be separated. In case of mixture of cream and milk. Cream comes in the upper layer and is recovered from milk.

(f) Sugar from sugar solution – It is separated by crystallisation. Heat the solution to obtain a saturated solution. Filter the solution while hot and allowed to cool. Crystals of sugar will be formed which are collected.

(g) Dye from black ink – Dye from black ink is separated by the process of chromatography. In this the constituents of a mixture are separated by their absorption over an appropriate absorbing material.

Solution 3.

(a) Carbon – titrachloride and water form immiscible mixture and can be separated by Separating Funnel. Since, they are immiscible liquids and forms two distinct layers.

Carbon tetrachloride is heavier liquid with the density of 1.59 g/cm3. Water is lighter with the density of 1g/cm3. Thus, the denser liquid carbon tetra chloride forms a lower level in the funnel which can be separated out first. The remaining water can be collected in the separate beaker.
Selina ICSE Solutions for Class 9 Chemistry - Elements, Compounds and Mixtures image - 5
(b) Lead chloride and silver chloride can be separated by solvent extraction. For this the given mixture is dissolved in water. Silver chloride dissolves in water but lead chloride does not dissolve. The solution is filtered through a filter medium, lead chloride will be retained on the filter medium and the filtrate containing water and silver chloride is collected. Then the filtrate is evaporated to get silver chloride.

Solution 4.

The method of separation depends on both the type of mixture and the physical properties of its constituents. These are :
(i) The physical state of the constituents.
(ii) The differences in the physical properties of the constituents such as :
(a) boiling point, (b) melting point, (c) density, (d) magnetic properties, (e) ability to sublime, (f) volatility, (g) solubility in various solvents .

Solution 5.

It is the process of dissolving one of the components in a particular liquid when one of the components is soluble in water or in some other solvents and the other component is not.

For example, a mixture of iron filings and sulphur is separated by using carb disulphide as solvent A mixture of sodium chloride and chalk can also be separated by using water as the solvent for sodium chloride.

Solution 6.

(a) Mercury, silver
(c) Magnesium, sodium carbonate (NO2CO3)
(b) Calcium, lead
(d) Sodium

Solution 7.

(a) Sublimation: Sublimation is the change of state of matter from solid to gaseous state without passing through the liquid state.

For example, To separate the mixture of ammonium chloride and sodium chloride, sublimation is used. On heating ammonium chloride will change into vapour which will condense into a solid in the neck of the inverted funnel which can be scrapped off. Sodium chloride remains in the evaporating dish.

(b)

Evaporation Sublimation
1. Evaporation is change of state matter from solid state to gaseous state.

2. Evaporation takes place below its boiling point by supply of heat.

1. Sublimation is the change of state of matter from solid state to gaseous state.

2. Sublimation takes place below its melting point by supply of heat.

Solution 8.

(a) Distillation is the process of converting a liquid into vapour (by heating) and the subsequent condensation of the vapour back into a liquid.
(b) Fractional distillation is a process which involves distillation and collection of fractions or different liquids boiling at different temperatures.
(c) Centrifugation is the method of separating solids from a liquid, where the mixture is homogeneous.

Solution 9.

(a) Two miscible liquids can be separated by-

  1. Distillation or by
  2. Fractional Distillation depending on difference in their boiling points.

(b) Yes, mixture of chloroform (B. P. 61C) and carbon-tetrachloride (B. P. 77C) can be separated by Fractional Distillation as difference in their B. P. is less than 30C. Chloroform has a lower boiling point 61oC, so, it distils out first. It is collected in a receiver, leaving behind carbon tetrachloride (having boiling point 77oC).
Modification: A fractionating G-Column is fitted over distilling flask.

(c) Homogeneous Mixtures of solid and liquid are separated by Centrifugation.

Uses of Centrifugation

  1. For separating cream from milk and butter from curd in dairies.
  2. In washing machines to squeeze out water from the wet clothes.

Solution 10.

(a) Dissolve the mixture of carbon and sulphur in carbon disulphide. Sulphur gets dissolve in carbon disulphide. The undissolved solid (carbon) is removed by filtration.

The filtrate is evaporated to dryness in order to recover the soluble solid i.e. sulphur powder.

(b) (i) Separation of potassium chloride: Prepare the suspension of mixture in water. The potassium chloride dissolves, but neither carbon nor sulphur, Filter the suspension and collect clear filtrate of potassium chloride in a separate beaker. Evaporate the clear filtrate on low heat. The water evaporates leaving behind white potassium chloride.

(ii) Separation of sulphur: Dissolve the residue in carbon disulphide. Sulphur dissolves in carbon disulphide leaving behind carbon. Filter the suspension and collect clear filtrate. Evaporate the filtrate in shade. The carbon disulphide evaporates leaving behind sulphur.

(iii) Removal and purification of carbon: Wash the residue of carbon on filter paper with carbon disulphide, so as to remove any sulphur. Dry the residue in shade. Carbon disulphide evaporates leaving behind carbon.

Solution 11.

(a) Chromatography
(b) Sedimentation
(c) Evaporation
(d) Fractional distillation

Solution 12.

By Sublimation method

Ammonium chloride and potassium chloride are placed in an evaporating dish and covered with an inverted funnel. On heating ammonium chloride will change into vapour which will condense into a solid in the neck of the inverted funnel which can be scrapped off. Sodium chloride remains in the evaporating dish.

Solution 13.

When caustic soda is added to an aqueous solution of copper sulphate, a blue precipitate of Cu(OH)2 , is obtained. Cu(OH)2 will be separated from mixture by filtration.

Solution 14.

Three commercial materials obtained from fractional distillation of petroleum are as follows:

  1. Natural gases
  2. Kerosene oil
  3. Lubricating oil

Use of Natural gases:

  1. It is used as fuel.

Use of Kerosene oil:

  1. It is used as fuel in homes.
  2. It is used for lighting the lamps on roads and home.

Use of lubricating oil:

  1. It is used to lubricate different parts of machines and vechic1es.
  2. It is used to prepare black polish etc.

Solution 15.

(a) By Sublimation: Heat the impure iodine in a china dish over a low flame. When violet fumes of iodine starts coming out. Place a cold inverted funnel over the china dish. The violet fumes condense on the cooler sides of funnel to form tiny crystals of pure iodine.

(b) By Magnetic Separation: Roll a strong horse shoe magnet on the mixture. The iron filings cling to the magnet, leaving behind c

Solution 16.

(a) Alcohol from a mixture of alcohol and water can be separated through fractional distillation. Alcohol liberates first because it has a lower boiling point than water.

(b) The components which are soluble in water but their solubilities are different.
For example, (i) Sodium nitrate and sodium chloride. (ii) Potassium chlorate and potassium chloride.

Solution 17.

(a) Nitre and common salt:

Take a beaker and prepare a saturated solution of the mixture in boiling water. On cooling, nitre will crystallize out while common salt will remain in solution. The process is repeated two or three times to separate the components completely.

(b) Ammonia and hydrogen:

The mixture of gases is bubbled through a Woulfe’s bottle containing the solvent i.e. water in which Ammonia dissolves. The insoluble gas component i.e. hydrogen passes out through the delivery tube and is collected in gas jar. The soluble gas component is obtained from its aqueous solution by boiling the solution.

(c) Powdered Chalk and Sugar:

Dissolve given mixture in water. Sugar dissolves but chalk does not as it is insolube chalk. Filter the mixture. Chalk is residue and is dried in the folds of filter paper. Filtrate is crystallized to get crystals of sugar.

(d) Sand, table salt:

Dissolve mixture in water. Salt dissolves but sand do not. Filter the solution to collect sand on filter paper. Then the filtrate is evaporated to get salt.

Iron filings and naphthalene:

By magnetic separation. Iron fillings get cling to the magnet and are separated. Remaining will be naphthalene.

Solution 18.

(a) Ingredients of gun-powder are – (nitre,sulphurand charcoal):

Dissolve mixture in water. Nitre dissolves but not sulphur and charcoal. Filter the mixture and collect the clear filtrate of nitre. Heat filtrate to crystallisation point. On cooling crystals of nitre separate out. Filter the crystals and dry them.

Dissolve the sulphur and charcoal in Carbon disulphide. Sulphur dissolves but charcoal does not. Filter the solution. Allow the filtrate to evaporate in shade. Carbon disulphide evaporates leaving behind sulphur. Charcoal is dried on filter paper.

(b) Sulphur, Sand and common salt:

Dissolve mixture in water. Common salt dissolves but sulphur and sand are insoluble in water. Filter the solution. sulphur and sand will get collected on the filter paper. Evaporate the solution to get common salt Collect the residue (Sand and Sulphur powder)and dissolve it in carbon disulphide, CS2. Sulphur dissolves in carbon disulphide but sand is insoluble in carbon disulphide. Filter the solution and evaporate it to get sulphur. Sand is dried on filter paper.

(c) Carbon dioxide and Carbon Monoxide:

When mixture is passed through a long tube having a number of porous partitions, Carbon monoxide will diffuse more rapidly as compared to Carbon dioxide. Thus if there is sufficient partitions, in the end, Carbon dioxide comes out.

(d) Water and Sugar:

Take a beaker half filled with water and dissolve as much sugar as you can in it, with constant stirring. Now heat the solution and go on adding sugar till it stops dissolving. Filter solution while it is hot. Allow the solution to cool. The crystals of pure sugar settle down at the bottom of the beaker.

(e) Sand and Iodine:

The mixture is placed in a china dish and an inverted dry funnel is placed over it, with its stem closed with cotton wool. It is then gently heated at a low flame. Iodine sublimes on the cooler side of funnel in the form of fine powder crystals. The residue left behind in the china dish is sand.

Solution 19.

Remove Iron using a Magnetic Separation:

Dissolve mixture in water. Sodium Chloride dissolves but not sulphur and charcoal. Filter the solution and allow the filtrate to evaporate to get Sodium Chloride. Now dissolve Sulphur and Charcoal in carbon disulphide solution, CS2. Sulphur dissolves but not charcoal. Filter the solution. Allow the solution to evaporate to get sulphur. Dry, charcoal in folds of filter papers.

Solution 20.

(a) solid-solid mixture:

  1. Magnetic separation: Example- Iron + Sulphur
  2. Solvent extraction: Example- Carbon + Sulphur
  3. Sublimation: Example- NH4Cl + NaCl

(b) Solid-liquid mixture:

  1. Evaporation: Example- Salt water solution
  2. Distillation: Example- Impure water
  3. Filtration: Example- Chalk + Water

(c) liquid-liquid mixture:

  1. Separating funnel: Example- Water + Carbon tetra chloride
  2. Fractional distillation: Example- Benzene + Toluene
  3. Distillation: Example- Acetone + Water

Solution 21.

The mixture is separated by exploiting the following facts in the order given below:

  1. Saw dust by gravity
  2. Ammonium chloride is soluble in water.
  3. Iodine sublime on heating
  4. Sand is insoluble in water

Solution 22.

(a) Dissolve given mixture in hot water lead chloride dissolves but not lead sulphate.
Filter the mixture. Residue is lead sulphate. Dry it in the folds of filter paper.

(b) Dissolve given mixture in water, Na2CO3, dissolves but not ZnCO3. Filter it ZnCO3 is residue and is dried in the folds of filter paper.

(c) Separation of given mixture can be done by fractional distillation. Boiling point of Benzene is 80°C and is less as compared with toluene having boiling point 111°C. So, Benzene distills over first.

(d) PbCl2 from a mixture of PbCl2 and AgCl.
Dissolve given mixture in water. AgCl dissolves, but not PbCl2. Filter the solution.
Evaporate the filtrate to get PbCl2.

Solution 23.

(a) Solid to gaseous state.
(b) Iodine

Solution 24.

Dissolve given mixture in dilute nitric acid. Copper oxide dissolves, but not charcoal. Filter the mixture, residue is charcoal. Dry it in the folds of filter paper. Now heat the filtrate to get copper oxide.

Solution 25.

To separate nitrogen from the air:

  1. Air is passed through filters or an electronic precipitator to remove dust. (ii) Then air is repeatedly compressed by increasing the pressure.
  2. Air is then cooled by decreasing the temperature.
  3. Now air becomes liquefied.
  4. Liquid air containing carbon dioxide, oxygen, nitrogen and inert gases is subjected to fractional evaporation.
  5. Carbon dioxide separates out as solid ice at -78°C.
  6. Liquid containing has a lower boiling point of -196°C, so it distills out first. Oxygen with boiling point -183°C is left behind.

Solution 26.

The two main components (oxygen and nitrogen) are separated from the air as follows:

  1. Air is passed through filters or an electronic precipitator to remove dust.
  2. Then air is repeatedly compressed by increasing the pressure.
  3. Air is then cooled by decreasing the temperature.
  4. Now air becomes liquefied.
  5. Liquid air containing carbon dioxide, oxygen, nitrogen and inert gases is subjected to fractional evaporation.
  6. Carbon dioxide separates out as solid ice at -78°C.

Solution 27.

(a) Chromatogeaphy: The process of separation of different dissolved constituents or mixture by absorbing them over an appropriate adsorbent material is called chromatography.

(b) The principle on which this technique is based is the difference in the adsorption different substances on the surface of a solid medium.

(c) Chromatograms are the distinct coloured rings or zones which are formed due to separation of mixture containing coloured substances by the process of chromatography.

(d) Advantages of chromatography:

  1. It can be carried out by very small amount of material.
  2. The substances under investigation do not get washed in chromatographic separation.

Solution 28.

(a) Centrifugation is used in diagnostic laboratories for testing blood/urine samples. It is also used for separation of cream from milk and butter from curd in dairies.
(b) Chromatography is used for purification of number of industrial products.

Solution 29.

The methods employed to separate two liquids are:

  1. Distillation and fractional distillation
  2. By using separating funnel.

Solution 30.

The apparatus used to separate oil and water is separating funnel. In this the two immiscible liquids are separated on the basis of their density differences. Thus, the heavier liquid will settle at bottom and the lighter one will form upper laye

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Solution 31.

(a) Magnetic separation
(b) Solvent extraction
(c) Fractional crystallization
(d) Chromatography
(e) Boiling
(f) Separating funnel
(g) Distillation
(h) Evaporation
(i) Solvent extraction

Solution 32.

Take a rectangular sheet of filter paper approximately 10 em x 8em. Mark a horizontal line on the sheet about 2 cm from the lower edge and mark a cross (x) near the middle of the line. Prepare a mixture by mixing red and blue inks nearly-in equal proportions. With the help of dropper put about 2 to 3 drops of ink mixture on the cross (x) marked on the line. Allow it to dry. Fix the paper to a cork in a tall jar so that its lower end rests at the bottom of the jar. Pour a mixture of water and alcohol into the jar.

(1 : 1) so that about a 2 cm of filter paper dips in water-alcohol mixture. Leave it for about one hour. Solvent rises slowly by capillary action and reaches the cross marked on line and then goes on rising higher. Two coloured lines are seen at different heights. When the difference in heights is appreciable, remove the paper from jar and let it dry. In this way we can separate coloured constituents present in a mixture of ink.
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Solution 33.

(a) sublimation
(b) filtration
(c) immiscible, separating funnel
(d) sublimation
(e) methylated sprit

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Selina ICSE Solutions for Class 9 Chemistry – Matter and its Composition

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Selina ICSE Solutions for Class 9 Chemistry – Matter and its Composition

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Selina ICSE Solutions for Class 9 Chemistry Chapter 1 Matter and its Composition

Exercise 1

Solution 1.

(a) Melting point: The constant temperature at which a solid changes into liquid state by absorbing heat energy is called melting point.

(b) Boiling point: It is the temperature at which a liquid changes into vapour under atmospheric pressure.

(c) Evaporation: The slow passing of molecules of a liquid into gaseous state at a temperature below its boiling point.

(d) Freezing: It is a process in which a liquid changes into solid state by giving out heat energy.

Solution 2.

Boiling point of a liquid can be raised by increasing the atmospheric pressure.

Solution 3.

On heating, solid wax melts into liquid wax, which on further heating, is converted into wax vapours. These changes can be seen in a burning candle. The candle is made up of a solid wax. When we light a candle, the wax near its wick melts. The molten wax rises up the wick and is converted into wax vapour. The wax vapour mixes with oxygen in the air and burns. In a lighted candle, you can see the solid and the liquid states of wax. The vapour of wax can be seen rising from the wick for some time after the candle is put out.
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Solution 4.

(a) Sublimation: The process by which a solid directly change to its vapour state (or gaseous state) without passing through liquid state and vice versa is called sublimation.

(b) Liquefaction: It is a process of change of state of a substance from gaseous state to liquid state at a particular temperature. It is also known as condensation.

(c) Melting: It is process of changing from solid state to a liquid state at a particular temperature.

(d) Boiling: The process by which a liquid rapidly changes into a gaseous state, by absorbing the heat energy is called boiling.

Solution 5.

(a)

An atom A molecule
Atom is a smallest particle of an element. Molecule is a group of two or more atoms combined together so it is bigger.
Atom consists of nucleus (containing protons and neutrons) and electrons. Molecule consists of combination of two or more like or different atoms chemically bound together e.g. H2, HCl, NaCl etc.
Atom can neither be seen through naked eye nor through magnifying microscope. Molecule is not visible to naked eye, while can be seen through highly magnifying microscope.
Atom cannot be further divided. Molecule can further be divided to give individual atoms.
Atoms may or may not have independent existence. Molecules are capable of having independent existence. For example, atom of oxygen (O) has no independent existence while its molecule exists as O2in nature.

(b)

Boiling Evaporation
Boiling is the process in which liquid gets converted into gaseous state. Evaporation is a process in which the liquid gets converted into its gaseous form at any temperature below its boiling point.
Boiling occurs at the entire mass of the liquid. That is, it is a bulk phenomenon. Evaporation occurs on the surface of the liquid. That is, it is a surface phenomenon.
Boiling occurs rapidly. Evaporation is a slow process.
Boiling occurs at a specific temperature. Evaporation occurs at any temperature.

(c)

Melting Boiling
The process of changing from solid state to a liquid state at a particulartemperature is called melting or fusion. The process of change of liquid to vapour form all parts of the liquid at a particular temperature is called boiling.
Melting refers to the phenomenon when a solid transforms into a liquid. Boiling refers to the phenomenon when liquid transform into a gas.
Example: Melting of ice Example: Boiling of water

(d)

Gas Vapour
A substance exists as a gas at the room temperature and atmospheric pressure. A substance is a solid or liquid under ordinary condition but it is gaseousunder specific conditions.
It is present at ordinary conditions of temperature. Its temperature is lower than the boiling point of its liquid state.
e.g. – Nitrogen, oxygen. e.g. – Iodine, Camphor

Solution 6.

(a) Water boils of 100oC under 1 atmosphere pressure.
(b) At high altitude water boils below 100oC.
(c) A liquid evaporates below its boiling point.
(d) When a substance is heated kinetic energy of the particles increases.
(e) Solids have the negligible inter-particle space.
(f) Gases have the negligible inter-particle forces.

Solution 7.

(a) Increase in atmospheric pressure
(b) Sulphur
(c) Inter-conversion of state of matter

Solution 8.

(a) Sublimation
(b) Melting
(c) Evaporation
(d) Vaporisation

Solution 9.

(a) Increase in temperature favours Evaporation. When evaporation occurs, remaining liquid becomes cooler. The particles of the liquid absorb heat energy from surroundings to regain energy lost during evaporation which makes the surroundings cold.

(b) Earthen pot has pores which help in evaporation. Some of the water continuously seeps out from these pores. This water absorbs heat of vaporization from the remaining water and gets evaporate. Thus, the remaining water loses heat and gets cooled.

(c) This happens because, when the petrol changes from liquid state to the vapour state, is absorbs heat energy from the palm. The palm thus loses heat and gets cooled.

(d) In humid weather wet clothes take longer time to dry up due to the slow evaporation of water from their surface.

(e) Evaporation is a surface phenomenon. With increase in surface area, evaporation increases. Hot tea in Saucer cools faster than in a cup and hence we can sip faster.

Solution 10.

(a) Naphthalene balls become smaller day by day as they have very weak force of attraction operating between their particles, which break away from other particles from the surface of solid without heating.

(b) In gases the particles are far apart and there is enough space available for compression. Hence, gases can be compressed easily.

(c) Heat energy supplied increases the rate of vibration of the particles and decreases. The inter-particle attraction.

(d) Light has no mass and it does it occupy space. Thus, it is not considered as matter.

(e) According to ‘Law of Conservation of Mass’, “Mass can neither be created nor destroyed in a chemical reaction.” However, it may change from one form to other.

Solution 11.

In summers, we perspire more. Cotton being a good absorber of water helps in absorbing the sweat and exposes it to the atmosphere for evaporation. When sweat evaporates from our body, it takes heat from our body. The heat energy equal to the latent heat of vaporisation is absorbed from the body leaving the body cool.

Solution 12.

Balloon get heat from sun and on heating, the vibration of particles increases and the inter-particle force of attraction between them gets reduced, therefore, balloon bursts.

Solution 13.

Law of conservation of mass: It states that mass can neither be created nor destroyed in a chemical reaction. During any change, physical or chemical, matter is neither created nor destroyed. However it may change from one form to another.

Experimental Verification of Law of Conservation of Mass

Requirements: H-shaped tube called Landolt’s tube, Sodium chloride solution, silver nitrate solution, etc.

Procedure: A specially designed H-shaped tube is taken. Sodium chloride solution is taken in one limb ofthe tube and silver nitrate solution in the other limb as shown in figure. Both the limbs are now sealed and weighed. Now the tubes is averted so that the solutionscan mix up together and react chemically. The reaction takes place and a white precipitate of silver chloride is obtained.

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The tube is weighed again. The mass of the tube is found to be exactly the same as the mass obtained before inverting the tube. Thus, this experiment clearly verifies the law of conservation of mass
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Solution 14.

Law of conservation of mass is applied to a burning candle. A candle is made of solid wax. When it is lighted, wax near its wick melts and changes into to liquid form. The molten wax rises up the wick and is converted into wax vapours. The wax vapours their mix with oxygen in the air.

Thus, in burning of candle the matter is neither created nor destroyed but one form is changed into the other form.

Solution 15.

The reaction is:
Selina ICSE Solutions for Class 9 Chemistry - Matter and its Composition image - 4
Total mass of reactants = (6 g + 5.3 g) = 11.3 g
Total mass of products = (8.2 + 2.2 + 0.9) g = 11.3 g
As the total mass of reactants is equal to the total mass of products. Hence, the reaction follows Law of conservation of mass.

Solution 16.

The reaction will be as follows:
Methane + Oxygen → Carbon dioxide + Water

According to law of conservation of mass,
Total mass of reactants = Total mass of products
Mass of methane + mass of oxygen = Mass of carbon dioxide + Mass of Water
Mass of methane  +  32 g  =  22  +  18 g
Mass of methane = (40 – 32) = 8 g
8g of methane is required.

Solution 17.

Word equation for the reaction is:
Sodium + Chlorine → Sodium Chloride

According to law of conservation of mass,
Total mass of reactants = Total mass of products
Mass of sodium + Mass of chlorine = Mass of sodium chloride
23 g+ Mass of chlorine = 58.5 g
Mass of chlorine = 58.5 – 23 = 35.5 g
35.5 g of chlorine is needed.

Solution 18.

Word equation for the reaction is:
Magnesium + Oxygen → Magnesium oxide

According to law of conservation of mass,
Total mass of reactants = Total mass of products
Mass of magnesium + Mass of oxide = Mass of magnesium oxide
4.8g           +          3.2g     = Mass of magnesium oxide
Mass of magnesium oxide = 8 g

Solution 19.

(a) (iv) No fixed shape and size highly compressible.
(b) (i) The solid starts melting.
(c) (i) evolved
(d) (iv) decreases with increasing pressure.
(e) (iii) Iodine

Solution 20.

(a) does not, melting, boiling
(b) liquid, solid
(c) temperature, temperature
(d) gaseous, sublimation
(e) high and negligible

Solution 21.

(a) freezing
(b) less
(c) sublimation
(d) remains constant

Solution 22.

Column A Column B
(a) Constituent of matter Molecules
(b) No compressibility Solid
(c) Maximum expansion Gas
(d) Conversion of a gas into liquid Condensation

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