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PREFACE Chemistry is one of the oldest academic discipline and its roots lie in man’s fascination towards study of structure,
composition and properties of matter and the reactions by which matter converts from one form to the other.
NEET: Chemistry (Vol. I) not only adds great value towards a progressive society but also contributes greatly to other
branches of science like biology, physics, geology, astronomy, biotechnology etc. Thus chemistry serves to be the backbone
of all lifesciences.
Target’s “NEET: Chemistry (Vol. I)” has been compiled according to the notified syllabus for NEET-UG & ISEET, which
in turn has been framed after reviewing various state syllabi as well as the ones prepared by CBSE, NCERT and COBSE.
In the National-Eligibility-cum-Entrance Test (NEET), 25% weightage is given to Chemistry, as there are 45 questions
based on Chemistry, out of the total 180 questions.
Target’s “NEET: Chemistry (Vol. I)” comprises of a comprehensive coverage of theoretical concepts & Multiple Choice
Questions. In the development of each chapter we have ensured the inclusion of shortcuts & unique points represented as a
‘note’ for the benefit of students.
The flow of content & MCQs have been planned keeping in mind the weightage given to a topic as per the NEET-UG &
ISEET exam.
MCQs in each chapter comprise of mixture of questions based on theory & numericals and their level of difficulty is at par with
that of various competitive examinations like CBSE, AIIMS, CPMT, PMT, JIPMER, IIT, AIEEE, & the likes.
This edition of “NEET: Chemistry (Vol. I)” has been conceptualized with a complete focus on the kind of assistance
students would require to answer tricky questions, which would give them an edge required to score in this highly
competitive exam.
Lastly, we are grateful to the publishers of this book for their persistent efforts, commitment to quality & their unending
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Yours faithfully
Authors
No.
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All the best to all Aspirants!
Topic Name
Some Basic Concepts of Chemistry
Structure of Atom
Classification of Elements and Periodicity in Properties
Chemical Bonding and Molecular Structure
States of Matter: Gases and Liquids
Thermodynamics
Equilibrium
Redox Reactions
Hydrogen
s-Block Elements (Alkali and Alkaline earth metals)
Some p-Block Elements
Organic Chemistry – Some Basic Principles and Techniques
Alkanes
Alkenes
Alkynes
Aromatic hydrocarbons
Environmental Chemistry
Page No.
1
48
100
132
209
252
307
384
424
460
512
565
655
693
727
749
786
Chemistry (Vol. I)
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01
1.0
Some basic concepts of chemistry
General Introduction – Importance and Scope of Chemistry
¾
Scientists and their contribution:
Scientists
Joseph Louis Gay-Lussac
Amedeo Avogadro
Antoine Lavoisier
Joseph Proust
John Dalton
Contribution
Gay Lussac’s law of combining volumes of gases
Avogadro’s law
Law of conservation of mass
Law of definite composition
Law of multiple proportions
¾
Chemistry:
“Chemistry is defined as the study of composition, structure and properties of matter and the
reactions by which one form of matter may be converted into another form.”
¾
There are five important branches of chemistry such as:
CHEMISTRY
Physical
Chemistry
It deals with the
structure of matter, the
energy changes and
the theories, laws and
principles that explain
the transformation of
matter from one form
to another.
Inorganic
Chemistry
It deals with the
chemistry
of
elements other
than carbon and
their compounds.
Organic
Chemistry
It deals with
the chemistry
of carbon and
carbon
compounds.
Analytical
Chemistry
It deals with the
separation, extraction,
identification
and
quantitative
determination of the
composition
of
different substances.
Bio-Chemistry
It deals with the
substances which
are constituents of
living organisms.
Note:
Apart from the above, there are several other branches of chemistry as:
i.
Medicinal (pharmaceutical) Chemistry: It deals with the application of chemical research
techniques to the synthesis of pharmaceuticals.
ii.
Environmental Chemistry: It deals with the study of chemistry associated with soil, air and water
and also of the impact of human activities on the natural system.
iii. Green Chemistry: It deals with processes and products that eliminate or reduce the use or release of
hazardous substances.
iv. Food Chemistry: It deals with the chemical processes associated with all forms of food stuffs.
v.
Agrochemistry: It deals with the application of chemistry for agricultural production and food
processing.
vi. Geo Chemistry: It deals with the study of chemical composition and chemical processes associated
with Earth and the other plants.
vii. Astrochemistry: It deals with the study of the compositions and reactions of the chemical elements
and molecules found in the space and the interactions between this matter and radiation.
viii. Photochemistry: It deals with the interactions between light and matter.
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ix.
Electrochemistry: It deals with the study of chemical reactions in a solution and electron transfer
particularly within electrolytic solution.
x.
Solid State Chemistry: It deals with the structures, properties and chemical processes that occur in
the solid phase.
xi. Polymer Chemistry: It deals with the examination of structure and properties of macromolecules and
the study of new ways to synthesize those molecules.
xii. Nuclear Chemistry: It deals with the study of radioactivity, nuclear processes and nuclear properties.
xiii. Nano Chemistry: It deals with the production and reactions of nanoparticles and their compounds.
2
¾
Importance and scope of Chemistry:
Chemistry plays a very important role in our everyday lives. Our daily needs of food, clothing,
shelter, potable water, medicines, etc., are in one or the other manner connected with chemical
compounds, processes and principles. There is no aspect of life that is not associated with chemistry.
In fact, Chemistry is the single branch of science which profoundly influences the existence of human
beings, plants, animals as well as their habitat. Thus, mankind owes much to chemistry because it has
improved the quality of life.
¾
Some Applications of Chemistry:
i.
Chemistry in medicines and health care:
The chemical substances used for treatment of diseases by destroying the disease causing
agents (antigens) without causing harm to the host tissues are called drugs or medicines.
Some of the medicinal compounds are mentioned below:
a.
NAME OF THE
COMPOUND
Antipyretics
b.
Analgesics
c.
Tranquillizers
d.
Antiseptics
e.
Disinfectant
f.
Antimicrobials
g.
Antibiotics
h.
Antacids
i.
Antihistamines
FUNCTION/MEDICINAL
PROPERTY
Used to lower the temperature of the
body in high fever.
Used to relieve pain without causing
impairment of consciousness. These are
of two types:
i.
Narcotic drugs: These are sleep
inducing.
ii. Non-narcotic drugs: These do not
induce sleep.
Used for the treatment of stress, fatigue,
mild and severe mental diseases.
EXAMPLES
Aspirin (acetylsalicyclic acid),
paracetamol, phenacetin
Naproxen, Ibuprofen
Morphine, Codeine
Noradrenaline,
Iproniazid,
Phenelzine
(antidepressant
drugs)
Used to either kill or prevent the growth Furacin, Soframycin
of micro-organisms. Not harmful and
can be applied on living tissues.
Used to kill micro-organisms, but are Chlorine, Dettol, Bithional,
harmful to mankind and cannot be Iodine, Boric acid, Iodoform,
applied on living tissues.
Hydrogen peroxide
Used to cure infections caused by Salvarsan, prontosil
micro-organisms.
Produced by microbes and are used to Penicillin,
Ampicillin,
inhibit the growth of microbes.
Streptomycin, Neomycin
Used to neutralize excess acid in the Baking soda (NaHCO3) in
gastric juices and give relief from acid water,
omeprazole,
indigestion, acidity and gastric ulcers.
lansoprazole
Used to diminish or abolish the main Brompheniramine,
Terfenadine
actions of histamine released in the (Diametapp),
Dimithendine
body, thus prevent the allergic (Seldane),
(foristal)
reactions.
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j.
Anaesthetics
k.
Antifertility
drugs
ii.
Used to produce general or local General
anaesthetics:
insensibility to pain and other Chloroform, Diethyl ethers,
sensations.
Vinyl ethers.
Local anaesthetics:
Cocaine, Novocaine.
Used to control pregnancy.
Ethynylestradiol (Novestrol),
Mifepristone
Chemistry in food:
Many chemicals are added to food for their preservation and enhancing their appeal. These are
called food additives. While antioxidants, preservatives, fat emulsifiers, flour improvers are
added to increase the shelf life of the stored food, some additives like dyes, flavours and
sweetening agents are added to improve their cosmetic value. Some of these additives are
mentioned below:
a.
b.
c.
d.
iii.
NAME OF THE
FOOD
ADDITIVE
Food preservatives
FUNCTION
EXAMPLES
These are added to the food materials to
prevent their spoilage and to retain their
nutritive value for long periods.
These are used to enhance the taste of
food stuffs.
Butylated hydroxyanisole,
(BHA), Butylated hydroxy
toluene (BHT)
Taste enhancers
Monosodium
glutamate
(MSG) – commonly called
as aginomoto.
Artificial
These give sweetening effect to the food Aspartame (methylester)
sweetening agents and enhance its odour and flavour.
Alitame, sucralose.
Edible
food These are used to give an attractive Natural dye like carotene.
colours (dyes)
appeal to the food stuffs.
Other applications of chemistry:
a.
Fuel: These are used for transportation and power generation. Petroleum is a rich source
of organic compounds. It is fractionally distilled to obtain various fractions like gasoline,
kerosene, diesel and aviation fuel.
b.
Dyes: These impart colour to the textiles. A dye should have a suitable colour and
capacity to fix to the fibre.
Natural dyes include Indigo, alizarin etc.
Synthetic dyes include Azo dyes, pthalocyanin dyes, etc.
c.
Detergents: These are the substances which possess cleansing properties.
E.g. Sodium alkyl sulphates:
(C12H25 – OSO3Na: Sodium lauryl sulphate),
long chain alkyl benzene sulphonates:
C12H25
SO3Na
Sodium p-dodecyl benzene sulphonate.
¾
Matter and its constituents:
i.
The entire universe is made up of matter.
ii.
Matter is anything which has mass and occupies space.
iii. It exists in three physical forms viz., solid, liquid and gas.
a.
Solids have definite shape and definite volume.
b.
Liquids have definite volume but no definite shape.
c.
Gases have neither definite shape nor definite volume.
iv. These three states are interconvertible by changing the conditions of temperature and pressure.
heat
heat
ZZZ
X Liquid YZZZ
ZZZ
X Gas
Solid YZZZ
cool
cool
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Note:
Greek philosopher Democritus had suggested that matter is composed of extremely small atomio.
¾
Classification of matter:
Matter
Physical classification
Chemical classification
Solids
Liquids
Eg. NaCl
Gases
Eg. H2O
Eg. CH4
Mixtures
It comprises of two or more
substances (components) present in
any ratio in which the constituent
substances retain their separate
identities. Eg. Air, Tea, Brass (an
alloy of copper and zinc) etc
Homogeneous
It comprises of a single phase
in which components are
completely mixed with each
other and its composition is
uniform throughout.
Eg. Mixture of salt and
water.
Physical methods
Pure substance
It comprises of a single type
of particle present in a fixed
ratio in which all the
constituent particles are
same in their chemical
nature. Eg. Water, glucose,
sodium chloride etc
Heterogeneous
It comprises of two or more
phases present in the mixture
and its composition is not
uniform throughout.
Eg. Phenol-water
system,
silver
chloride-water
system etc.
Compounds
Pure substances which are Chemical methods
made up of two or more
components.
Eg. Water, ammonia, etc.
Inorganic
Eg. AlCl3
¾
4
Elements
Pure substances which
are made up of only one
component.
Eg. Ag, Au, Cu, etc.
Organic
Eg. CH3CHO
Unit and its need:
Definition:
“The arbitrarily decided and universally accepted standards used in the measurement of physical
quantities are called units.”
Some Basic Concepts of Chemistry
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Need:
i.
For calculation of experimental data.
ii.
For measurement of physical quantities such as mass, pressure, volume, length, time,
temperature, density, etc.
iii. Any measured property is expressed as a number along with an appropriate unit associated with
the property as only the number does not give any idea of the property.
¾
Various system in which units are expressed:
i.
Units are expressed in various systems like CGS (centimeter for length, gram for mass and
second for time), FPS (foot, pound, second) and MKS (meter, kilogram, second) systems etc.
ii.
In 1960, the general conference of weights and measures, proposed a revised metric system,
called International System of units i.e., SI system, abbreviated from its French name
Systeme Internationale d′ Units.
Note:
NASA’s Mars climate orbiter, the first weather satellite for Mars, was destroyed by heat. The failure
of the mission was due to confusion in estimating the distance between Earth and Mars in miles and
kilometers.
¾
Seven fundamental SI units:
No.
1.
2.
3.
4.
5.
6.
7.
¾
Fundamental quantity
Length
Mass
Time
Temperature
Amount of substance
Electric current
Luminous intensity
SI unit
Meter
Kilogram
Second
Kelvin
Mole
Ampere
Candela
Symbol
m
kg
s
K
mol
A
cd
Derived units
The units of all physical quantities can be derived from the seven fundamental SI units. These units
are known as derived units.
The table given below shows some common derived units.
No.
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
x.
xi.
Physical quantity
Area
Volume
Density
Velocity
Acceleration
Force
Pressure
Electric charge
Electric potential or
Potential difference
Energy (work or heat)
Concentration
xii.
Heat capacity
xiii.
Electrochemical equivalent
ix.
Relationship with fundamental unit
Length squared
Length cubed
Mass per unit volume
Distance travelled in unit time
Velocity change per unit time
Mass × acceleration
Force per unit area
Current × time
Energy per unit charge
Force × distance travelled
Mole per cubic metre
Cp = dH/dT
Cv = dE/dT
Z = E/F
Symbol
m2
m3
kg m−3
ms−1
ms−2
kg m s−2 (newton, N)
kg m −1 s−2
A s (coulomb, C)
kg m2s−2A−1 (J A−1 s−1
or volt V or J C−1)
kg m2 s−2 (J s−1)
mol m−3
kg m2 s−2 K−1 mol−1
(J K−1 mol−1)
kg C−1 (kg/coulomb)
Note: 1 Litre = 1 dm3
Some Basic Concepts of Chemistry
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Some common SI prefixes used for expressing big and small numbers:
Prefix
Tera−
Giga−
Mega−
myria−
kilo−
hecto−
deka−
deci−
centi−
milli−
micro−
nano−
pico−
femto−
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Symbol
T
G
M
my
k
h
da
d
c
m
µ
n
p
f
Magnitude
1012
109
106
104
103
102
10
10−1
10−2
10−3
10−6
10−9
10−12
10−15
Meaning (multiply by)
1 000 000 000 000
1 000 000 000
1 000 000
1 000 0 (this is now obsolete)
1 000
100
10
0.1
0.01
0.001
0.000 001
0.000 000 001
0. 000 000 000 001
0.000 000 000 000 001
Laws of chemical combination
¾
Chemical combination:
“The process in which the elements combine with each other chemically, to form compounds, is
called as chemical combination.”
¾
Laws of Chemical Combination:
One of the most important aspects of the subject of chemistry is the study of chemical reactions.
These chemical reactions take place according to certain laws called as “Laws of chemical
combination.”
i.
Law of conservation of mass:
The law was first stated by Russian scientist Lomonosove in the year 1765. Later in 1774,
French scientist, Antoine Lavoisier also stated the same law independently.
Statement:
It states that, “Mass is neither created nor destroyed during chemical combination of matter.”
Explanation:
a.
According to Lavoisier, total masses of the reactants before the reaction are found to be
same as that of total masses of the products formed after the reaction.
b.
Eg. AgNO3 + NaCl ⎯→ AgCl + NaNO3
1.70g
0.555g
1.435g
0.82g
ii.
Law of definite composition or constant proportions:
This law was first stated by French chemist Joseph Proust in (1799).
Statement:
It states that, “Any pure compound always contains the same elements in a definite proportion
by weight irrespective of its source or method of preparation.”
Explanation:
a.
In support of this law, it was experimentally proved that a naturally occurring pure
sample of copper carbonate contains 51.35 % copper by weight, 38.91 % carbon by
weight and 9.74 % oxygen by weight.
b.
Further, a pure sample of copper carbonate was synthesized in laboratory and it was
found that the percentage by weight of copper, carbon and oxygen were exactly identical
to that of the naturally occurring sample of copper carbonate.
c.
French scientist Berthollet opposed Proust’s law of definite proportion by giving
examples of the substances containing different proportions of elements.
d.
However, Berthollet’s objections were ruled out as the experimental work of analysis
mentioned by Berthollet was found to be based on impure samples or incomplete reactions.
Some Basic Concepts of Chemistry
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iii.
iv.
Law of multiple proportions:
This law was proposed by British scientist John Dalton in 1808.
Statement:
It states that, “If two elements, combine chemically with each other forming two or more
compounds with different compositions by weight, then the masses of the two interacting
elements in the two compounds are in the ratio of small whole numbers.”
Explanation:
a.
Hydrogen and oxygen combine to form two compounds H2O (water) and H2O2
(Hydrogen peroxide).
H2O :
2 parts of Hydrogen, 16 parts of Oxygen
2 parts of Hydrogen, 32 parts of Oxygen.
H2O2 :
b.
The masses of oxygen which combine with same mass of hydrogen in these two
compounds bear a simple ratio 1 : 2.
Law of reciprocal proportions:
This law was given by Richter in 1794.
Statement:
It states that, “When two different elements combine separately with the same weight of a third
element, the ratio in which they do so will be the same or some simple multiple of the ratio in
which they combine with each other.”
Explanation:
a.
Definite mass of an element A combines with two other elements B and C to form two
compounds.
b.
If B and C also combine to form a compound, their combining masses are in same
proportion or bear a simple ratio to the masses of B and C which combine with a constant
mass of A.
Eg.
H
NaH
(B) Na
v.
(A)
NaCl
HCl
Cl (C)
Hydrogen combines with sodium and chlorine to form compounds NaH and HCl
respectively.
NaH :
23 parts of sodium,
1 part of Hydrogen
HCl :
35.5 parts of chlorine,
1 part of Hydrogen
Sodium and chlorine also combine to form NaCl in which 23 parts of sodium and 35.5
parts of chlorine are present. These are the same parts which combine with one part of
hydrogen in NaH and HCl respectively.
Gay-Lussac’s law of combining volumes of gases:
This law was enunciated by Gay-Lussac in 1808.
Statement:
It states that,“When gases react together to produce gaseous products, the volumes of reactants
and products bear a simple whole number ratio with each other, provided volumes are
measured at same temperature and pressure.”
Explanation:
a.
Under similar conditions of temperature and pressure, 1 volume of hydrogen reacts with
1 volume of chlorine to give 2 volumes of hydrogen chloride.
H2
+ Cl2 ⎯→ 2HCl
1 volume
b.
c.
1 volume
2 volumes
Thus, the volume ratio of hydrogen: chlorine: hydrogen chloride is 1 : 1 : 2.
This is a simple whole number ratio and is also in agreement with their molar ratios when
they are involved in the reaction.
7
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Note:
i.
Gay-Lussac’s law of combining volumes is applicable only to gaseous reactions and not to
reactions involving solids and liquids.
ii.
The volumes of gases in the chemical reaction are not additive, though it appears to be additive.
However in case of hydrogen-oxygen reaction, 2 volumes of hydrogen and 1 volume of oxygen
equal to 3 volumes of reactants get converted into 2 volumes of product steam.
iii. Similarly, in case of formation of ammonia, 1 volume of nitrogen reacts with 3 volumes of
hydrogen equal to 4 volumes of reactants get converted into 2 volumes of product ammonia.
1.2
Dalton’s atomic theory
¾
Dalton’s atomic theory:
John Dalton, an English school teacher, proposed the atomic theory in the year 1808.
According to him, “Atom is the smallest indivisible particle of a substance.”
Postulates/Assumptions:
Dalton made the following assumptions in his theory:
i.
All matters are made up of tiny, indestructible, indivisible unit particles called atoms.
ii.
Atoms are the smallest particles of an element and molecules are the smallest particles of a
compound.
iii. All atoms of the same element have same size, shape, mass and all other properties.
iv. Atoms of different elements have different properties.
v.
Compounds are formed when atoms of different elements combine.
vi. The atoms in a compound unite in small whole number ratios like 1:1, 1:2, 1:3, 2:1, 2:3, etc.
vii. A chemical reaction involves only the separation, combination or rearrangement of integer
number of atoms.
viii. During a chemical reaction, atoms are neither created nor destroyed.
Note:
The number of atom present in a molecule of a substance is called Atomicity.
¾
∴
Avogadro’s Law:
Avogadro, in the year 1811, combined Gay -Lussac’s law and Dalton’s theory to propose Avogadro’s
law.
Statement :
It states that, “Equal volumes of all gases, under identical conditions of temperature and pressure,
contain equal number of molecules.”
OR
“At constant pressure and temperature, volume of a gas is directly proportional to the number of
molecules.”
V ∝ number of molecules (P, T constant)
Explanation:
i.
If equal volumes of three gases i.e. Hydrogen (H2), Oxygen (O2) and Chlorine (Cl2) are taken in
different flasks of the same capacity under similar conditions of temperature and pressure, all
the flasks are found to have the same number of molecules.
ii.
However, these molecules may differ in size and mass.
1L of H2 gas
at N.T.P
1L of O2 gas
at N.T.P
1L of Cl2 gas
at N.T.P
Illustration of Avogadro’s hypothesis.
8
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Eg.
a.
b.
c.
d.
e.
f.
g.
Hydrogen + Chlorine ⎯→ Hydrogen chloride
[1 vol ]
[1 vol ]
[2 vol]
Applying Avogadro’s hypothesis, assuming that 1 volume contains n molecules, it
follows that
Hydrogen + Chlorine ⎯→ Hydrogen chloride
2n molecules
n molecules n molecules
Dividing throughout by 2n, we get
1
1
molecule +
molecule ⎯→ 1 molecule
2
2
This means that 1 molecule of hydrogen chloride contains 1 / 2 molecule of hydrogen and
1 / 2 molecule of chlorine.
Now, 1 / 2 molecule of hydrogen can exist because one molecule of hydrogen contains 2
atoms of hydrogen and therefore 1/ 2 molecule of hydrogen contains one atom of
hydrogen.
Similarly, 1/2 molecule of chlorine contains an atom of chlorine because chlorine is also
a diatomic molecule.
Thus, one molecule of hydrogen chloride is formed from one atom of hydrogen and one
atom of chlorine.
Note:
This generalization is in agreement with Dalton’s atomic theory.
1.3
Concept of Elements, Atoms and Molecules
¾
Elements:
i.
“An element is defined as a substance which cannot be separated into simpler substances by
any chemical process.”
Eg. Platinum, Nickel, Cobalt, etc.
ii.
A pure substance is made up of only one kind of atoms having the same atomic number.
iii. The smallest particle of an element is the atom.
iv. Elements can be divided into two groups namely
a.
naturally occurring and
b.
artificially synthesized one
v.
There are about 118 elements out of which about 92 elements are naturally occurring and
nearly 26 elements are synthesized in laboratory.
vi. The most abundant element in the earth’s crust is oxygen.
vii. Artificially synthesized elements have a very short life as they breakup into more stable lighter
elements.
viii. Most of the elements are solids, while eleven of them are gases and only two are liquids.
ix. The two liquids are Mercury and Bromine.
x.
Elements can be broadly divided into four categories.
Classification of elements:
Elements
Metals
NonMetals
Metalloids
Noble
Gases
Characteristics
Generally solid, hard, malleable, ductile, high tensile
strength, lustre and good conductors of heat and electricity.
Generally non-lustrous, brittle, poor conductors of heat and
electricity.
Elements that have properties which lie in between those of
metals and non- metals.
Group of six elements that do not combine with other
elements and tend to exist by themselves.
They are chemically inactive. Hence, are also known as inert
elements.
Some Basic Concepts of Chemistry
Example
Copper, iron, zinc,
etc.
Sulphur, phosphorus,
nitrogen,etc.
Arsenic, tin, bismuth,
etc.
Neon, helium, argon,
etc.
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¾
Chemical symbols:
i.
An abbreviated form in which the name of an element is represented is called as a symbol.
ii.
Chemists represent elements by symbols of one or two letters.
iii. The first letter of the symbol is always capital and the second letter if present, is always small.
iv. The symbols of most of the elements are derived from the English names of the elements.
v.
In some cases, Latin names of the elements are used to derive the symbols.
Eg. Aluminium − Al , Einsteinium − Es, Gold − Au (Aurum)
¾
Compounds:
i.
“Compounds are defined as substances of definite compositions which can be decomposed into
two or more substances by a simple chemical process.”
Eg. Methane, ammonia, urea, etc.
ii.
The properties of all the substances and elements obtained on decomposition of the compounds
are completely different.
Eg. Carbon is combustible and oxygen supports combustion, but carbon dioxide is used as a
fire extinguisher.
¾
Atoms:
i.
“The smallest indivisible particle of an element is called atom.”
ii.
Every atom of an element has a definite mass of the order of 10−26 kg and has a spherical shape
of radius of the order of 10−15 m.
iii. The smallest atom of an element is that of hydrogen with mass 1.667 × 10−26 kg.
iv.
Atoms may or may not exist freely.
v.
Atoms of almost all the elements can react with one another to form compounds.
Note:
Name
Monoatomic
Diatomic
Polyatomic
10
Number of atoms in a molecule
Only one atom
Two atoms
More than two atoms
Example
noble gases, some metals, carbon, silicon, etc.
Hydrogen (H2), Oxygen (O2), Nitrogen (N2), etc.
Phosphorus (P4), Sulphur (S8), etc.
¾
Molecules:
i.
“A molecule is an aggregate of two or more atoms of definite composition which are held
together by chemical bonds.”
OR
“The smallest particle of a substance (element or compound) which is capable of independent
existence is called a molecule.”
ii.
It has all the properties of the original compound.
iii. It cannot be divided into its constituent atoms by simple methods.
iv. Only under drastic conditions, a molecule can be decomposed into its constituent atoms.
v.
The properties of the constituent atoms of a compound and the molecule of compound are
completely different.
¾
Phlogiston theory:
i.
The phlogiston theory was proposed by Ernst Stahl (1660 - 1734).
ii.
Phlogiston was described as a substance in a combustible material which is given off when the
material burns.
iii. This theory persisted for about 100 years and was a centre of much controversy.
iv. Antoine Lavoisier proved that the flammable air produced by Cavandish was a new gas and
named it as hydrogen gas.
v.
During the end of the eighteenth century, much work was done with gases, especially by
Joseph Black, Henry Cavendish, Joseph Priestley and Carl Scheele.
vi. Priestley was a very conservative scientist. Even after his discovery of oxygen, he still believed
in phlogiston theory.
Some Basic Concepts of Chemistry
TARGET Publications
1.4
Chemistry (Vol. I)
Atomic and molecular masses
¾
Atomic mass:
i.
In 1961, the International Union of Chemists selected a new unit for expressing the atomic
masses.
ii.
They accepted the stable isotope of carbon (12C) with mass number of 12 as the standard for
comparing the atomic and molecular masses of elements and compounds.
iii. “Atomic mass is the average relative mass of an atom of an element as compared to the mass of
an atom of carbon (C12) taken as 12”.
Mass of an atom
Atomic mass =
1
mass of an atom of C12
12 th
Note:
i.
1 a.m.u = 1.66056 × 10−24 g , where a.m.u stands for atomic mass unit.
1.6736 × 10−24 g
= 1.00780 a.m.u = 1.0080 a.m.u
ii.
Mass of hydrogen atom =
1.66056 × 10−24 g
iii.
iv
v.
.
¾
Mass of oxygen-16 (16O) = 15.995 a.m.u
Recently the unit of atomic mass ‘a.m.u.’ is been replaced by ‘u’ known as unified mass.
“Gram atomic mass is the quantity of an element whose mass in grams is numerically equal to
its atomic mass”.
OR
“Atomic mass of an element expressed in grams is the gram atomic mass or it is also called
gram atom”.
Eg. The atomic mass of oxygen = 16 a.m.u
Therefore, gram atomic mass of oxygen = 16g.
Average atomic mass:
i.
Majority of elements occur in nature as mixtures of several isotopes.
ii.
Isotopes are the different atoms of same elements possessing different atomic masses but same
atomic number.
iii. The average relative mass depends upon the isotopic composition of that particular element.
iv. The best way to define the atomic mass of the elements is to determine the atomic mass of each
isotope separately and then combine them in the ratio of their proportion of occurrence. This is
called average atomic mass.
v.
Each element has a number of isotopes with different isotopic masses.
vi. While calculating the atomic mass of an element, a weighed average of the isotopic masses of
the isotopes of the element is taken, considering the relative quantity of isotopes.
vii. Thus, it is the average mass of an atom of the element which is used in calculating the atomic
mass weight of the element.
Eg.
a.
Chlorine has two isotopes, 85Cl and 37Cl, present in 75 % and 25 % proportion
respectively. Hence, the atomic mass of chlorine is the weighed average of these two
isotopic masses.
i.e., (35.0 × 0.75) + (37.0 × 0.25) = 35.5.
b.
Aston’s mass spectrometer proved that neon exists in nature in the form of a mixture of
three isotopes,
1.
Neon-20 with atomic mass 19.9924 u with natural abundance 90.92 %
2.
Neon-22 with atomic mass 21.9914 u with natural abundance 8.82 %
3.
Neon-21 with atomic mass 20.9940 u with natural abundance 0.26 %
∴
Average atomic mass of Ne,
11
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Chemistry (Vol. I)
TARGET Publications
(Atomicmassof 20 Ne × % of 20 Ne) + (Atomicmassof 22 Ne × % of 22 Ne)
+ (Atomicmassof 21 Ne× % of 21 Ne)
100
= [(19.9924 u × 90.92) + (21.9914 u × 8.82) + (20.994 u × 0.26)]/100
= 20.1713 u
=
¾
1.5
Molecular mass:
“Molecular mass of a substance is defined as the ratio of mass of one molecule of a substance to
1
of the mass of one atom of 12C.”
th
12
It is also the algebraic sum of atomic masses of constituent atoms present in the molecule.
Characteristics of molecular mass:
i.
Like atomic mass, molecular mass is expressed as a.m.u.
ii.
It is the number that indicates comparative mass of a molecule of a compound with respect to
1
of the mass of one atom of 12C.
th
12
iii. Gram molecular mass is the molecular mass expressed in grams.
iv. 1 gram molecular mass is also known as 1 gram molecule.
Eg.
Molecular mass of CO2 = 44 a.m.u
Therefore gram molecular mass of CO2 = 44 g
Mole concept and molar mass
¾
Mole Concept:
i.
A mole is defined as the amount of substance that contains the same number of entities (atoms,
molecules, ions or other particles), as present in 12 g (or 0.012 kg) of the 12C isotope.
ii.
The quantity of a substance equal to its atomic mass or molecular mass in grams is referred as
1 mole of a substance.
iii. Avogadro Number (NA):
“The number of atoms, molecules, ions, or electrons etc. present in 1 mole of a substance is
found to be equal to 6.023 × 1023, which is called Avogadro Number (NA).”
Thus, NA = 6.023 × 1023 molecules or ions or electrons per mol.
Eg.
1 mole of hydrogen atoms = 6.023 × 1023 hydrogen molecules.
1 mole of sodium ions = 6.023 × 1023 sodium ions.
1 mole of electrons = 6.023 × 1023 electrons.
iv. The volume of 1 mole of any pure gas at standard temperature and pressure is always constant
and is equal to 22.414 L or 0.022414 m3. This value is called as Avogadro’s molar volume or
molar gas volume at STP.
Eg. 1 mole of chlorine gas = 22.4 L or 0.0224 m3
¾
Molar Mass:
i.
The mass of one mole of a substance is called its molar mass (M)
ii.
The units of molar mass are g mol−1 or kg mol−1.
iii. The molar mass is equal to atomic mass or molecular mass expressed in grams, depending upon
whether the substance contains atoms or molecules.
Note:
Mole Triangle:
The relationship between the mass of a gas with number of moles, volume of a gas at STP and the
number of molecules is given by the mole triangle.
12
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
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Number of
fundamental
particles
Multiplied by
Avogadro’s number
Mass of
substance
Divided by
molecular mass
Divided by
Avogadro’s number
Number of
moles
Multiplied by
molecular mass
Multiplied by
22.4 dm3
Divided by
22.4 dm3
Volume
occupied by
gas at STP
in dm3
Mole Triangle
Note:
a.
b.
c.
d.
e.
f.
g.
h.
i.
j.
1.6
Mass of an element
Atomic mass
Atomic mass
Mass of one atom =
6.023×1023
One mole of atoms =
One mole of molecules = 6.023 × 1023 molecules
= Gram molecular mass of the substance
Molecular mass
Mass of one molecule =
6.023×1023
Mass of the compound
Moles of a compound =
Molecular mass
Mass of thesubstance
Number of moles (n) =
Molar mass of the substance
Number of molecules = n × Avogadro’s number
Volume of gas at S.T.P = n × 22.414 L
Volume occupied by 1 mole of a gas at N.T.P = 22.4L
Molecular mass = Vapour density × 2
Percentage composition and empirical and molecular formula
¾
Percentage composition:
i.
Percentage composition of a compound is the relative mass of each of the constituent element
in 100 parts of it.
ii.
Percentage composition can be calculated as mass percentage.
iii. Mass percentage gives the mass of each element expressed as the percentage of the total mass.
iv.
Mass percentage of an element =
Mass of the element in1mole of compound
× 100
Molar mass of the compound
Eg.
Calculate the mass percentage composition of glucose:
The formula of glucose = C6H12O6
Molar mass of glucose = (6 × 12) + (12 × 1) + (6 × 16) = 180
The formula of glucose shows that there are 6−C atoms, 12−H atoms and 6−O atoms.
Mass of 6−C atoms = (6 × 12) = 72
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13
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⎛ 72 ⎞
Mass % of C = ⎜
⎟ × 100 = 40.0 %
⎝ 180 ⎠
Mass of 12−H atoms = 12 × 1 = 12
⎛ 12 ⎞
Mass % of H = ⎜
⎟ × 100 = 6.67 %
⎝ 180 ⎠
Mass of 6−O atoms = 6 × 16 = 96
⎛ 96 ⎞
Mass % of O = ⎜
⎟ × 100 = 53.33 %
⎝ 180 ⎠
¾
Chemical formula:
i.
Elements and compounds are represented by symbols and formulae respectively. A chemical
formula gives the representation of a molecule of a substance in terms of symbols of various
elements present in it.
Eg.
Ammonia is represented by the formula NH3, carbon dioxide by CO2, copper sulphate by
CuSO4, etc. The determination of a formula of the substance involves the chemical analysis of
a.
The constituent elements present.
b.
The relative amount of elements of each type present in a given mass of the compound.
ii.
The chemical formula may be of two types:
a.
Empirical formula:
“The empirical formula of a compound is defined as a chemical formula indicating the
relative number of constituent atoms in a molecule in the simplest ratio.”
Eg. Molecular formula of benzene = C6H6
∴
Empirical formula = CH
b.
Molecular formula:
“The formula which gives the actual number of each kind of atoms in one molecule of the
compound is called the molecular formula of the compound.”
It is an integral multiple of empirical formula.
Eg. Molecular formula of benzene = C6H6.
Thus, it has six atoms of carbon and six atoms of hydrogen.
iii. Empirical and molecular formula of some molecules are given below:
Compound
Hydrogen peroxide
Benzene
Glucose
Sucrose
Naphthalene
iv.
Empirical Formula
HO
CH
CH2O
C12H22O11
C5H4
Molecular Formula
H2O2
C6H6
C6H12O6
C12H22O11
C10H8
Molecular formula and empirical formula are related as:
Molecular Formula = n × Empirical formula
where ‘n’ is a simple whole number and may have values 1, 2, 3 …..
n =
Molecular mass
Empiricalformula mass
Eg.
The molecular mass of benzene is 78. The empirical formula of benzene is CH and therefore,
its Empirical formula mass is 13.
Thus,
Molecular mass
78
n =
=
= 6
13
Empiricalformula mass
Therefore, molecular formula of benzene = 6 × (CH) = C6H6
14
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¾
The various steps involved in determining the empirical formula are:
Step 1:
Divide the percentage of each element by its atomic mass. This gives the moles of atoms of various
elements in the molecule of the compound.
Moles of atoms =
Percentage of an element
Atomic mass of the element
Step 2:
Divide the result obtained in the above step by the smallest value among them to get the simplest ratio
of various atoms.
Step 3:
Make the values obtained above to the nearest whole number by multiplying if necessary by a
suitable integer. This gives the simplest whole number ratio.
Step 4:
Write the symbols of the various elements side by side and insert the numerical value at the right
hand lower corner of each symbol. The formula thus obtained represents the empirical formula of the
compound.
¾
Steps for determination of the Molecular Formula of a compound:
Step 1:
Determine the empirical formula as described above.
Step 2:
Calculate the empirical formula mass by adding the atomic masses of the atoms in the empirical
formula.
Step 3:
Determine the molecular mass of the compound.
Molecular mass can be determined by the following formulae:
i.
Molecular mass = Vapour Density × 2
ii.
Molecular mass of an acid = Equivalent mass × basicity of the acid
iii. Molecular mass of a base = Equivalent mass × acidity of the base
iv. Molecular mass = Equivalent mass × no. of e− gained or lost.
Step 4:
Determine the value of ‘n’ as, n =
Molecular mass
Empirical formula mass
Change ‘n’ to the nearest whole number.
Step 5:
Multiply empirical formula by ‘n’ to get the molecular formula.
Molecular formula = n × Empirical formula
Eg.
A compound with molar mass 159 was found to contain 39.62% copper and 20.13% sulphur. Suggest
molecular formula for the compound (Cu = 63, S = 32 and O = 16)
Solution:
% copper + % sulphur = 39.62 + 20.13 = 59.75
This is less than 100%. Hence compound contains adequate quantity of oxygen so that total
percentage of elements is 100%.
Hence, % of oxygen = 100 − 59.75 = 40.25%
15
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
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Element
% of Element
At. Mass
Atomic ratio
39.62
= 0.63
63
Cu
39.62
63
S
20.13
32
20.13
= 0.63
32
O
40.25
16
40.25
= 2.51
16
Simplest ratio
0.63
=1
0.63
0.63
=1
0.63
2.51
=4
0.63
Hence, empirical formula is CuSO4 and empirical formula mass = 63 + 32 + 16 × 4 = 159
Molecular mass = empirical formula mass
Therefore, molecular formula = empirical formula = CuSO4
Note:
Molecular mass is the sum of the atomic masses of all the the atoms as given in the molecular formula of
the substance.
1.7
Chemical reactions, stoichiometry and calculations based on stoichiometry
¾
Chemical Reactions:
i.
“A chemical reaction is a process in which a single substance or many substances interact with
each other to produce one or more substances.”
ii.
Chemical reactions are represented in terms of chemical equations.
iii. Chemical equation is a statement of a chemical reaction in terms of the symbols and formulae
of the species involved in the reaction.
iv. The chemical equation may be defined as, “the brief representation of a chemical change in
terms of symbols and formulae of substances involved in it”.
v.
The chemical reaction when written in the form of chemical equation is always in balanced
form and the masses are always conserved.
vi. The substances which react with each other to bring about the chemical changes are known as
reactants.
vii. Whereas, the substances which are formed as a result of the chemical change are known as
products.
Eg.
AgNO3 + NaCl ⎯→ AgCl + NaNO3
Reactants
¾
16
Products
Stoichiometry:
Stoichiometry means quantitative relationship among the reactants and the products in a reaction.
1N2 (g) + 3H2 (g) ⎯→ 2NH3 (g)
1, 3, and 2 are coefficients of N2, H2 and NH3 respectively. These coefficients of reactants and
products in the balanced chemical reaction are called stoichiometric coefficients.
The stoichiometric calculations involve the following steps:
i.
Write the correct formula of the reacting substances and products. Care must be taken to satisfy
valencies of the atoms of the compound.
ii.
For writing the balanced chemical equation, following three steps must be followed:
Step 1:
Write the names of the reactants with ‘+’ sign separating the reacting substances on the left
hand side. Then draw an arrow from left to right and to the right side of the arrow, write the
names of all the products with ‘+’ sign separating them. Thus, in case of a reaction involving
burning of methane in oxygen, producing carbon dioxide and water, the reaction is written as,
methane + oxygen ⎯→ carbon dioxide + water
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
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Step 2:
Rewrite the chemical equation in terms of chemical formula of each substance as shown.
CH4 (g) + O2 (g) ⎯→ CO2 (g) + H2O(l)
Step 3:
Balance the mass of the chemical reaction by selecting the proper whole number coefficients
for each reactant and product as shown.
CH4(g) + 2O2 (g) ⎯→ CO2 (g) + 2H2O(l)
This is the balanced chemical equation.
Note:
Some Common Compounds:
No.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
¾
Compound
Phosphoric acid
Sodium phosphate
Ferric phosphate
Aluminium phosphate
Copper phosphate
Ferrous phosphate
Hydrogen chloride
Potassium chloride
Sodium chloride
Cuprous chloride
Formula
H3PO4
Na3PO4
FePO4
AlPO4
Cu3(PO4)2
Fe3(PO4)2
HCl
KCl
NaCl
Cu2Cl2
No.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
Compound
Cupric chloride
Ferrous chloride
Ferric chloride
Stannous chloride
Stannic chloride
Hydrogen sulphate
Sodium sulphate
Copper sulphate
Ferrous sulphate
Ferric sulphate
Formula
CuCl2
FeCl2
FeCl3
SnCl2
SnCl4
H2SO4
Na2SO4
CuSO4
FeSO4
Fe2(SO4)3
Mass relationship:
i.
A balanced chemical reaction may be used to establish the weight relationships of reactants and
products.
ii.
This is based on the law of conservation of mass, which states that, total mass of reactants is
always equal to total mass of the products.
Atomic masses: (Na = 23, Cl = 35.5)
2Na(s)
+ Cl2(g)
⎯→ 2 NaCl(s)
(2 atoms)
[2 × 23]
(1 molecule)
[1 × 35.5 × 2]
⎯→
⎯→
(2 molecules)
[2 (23 + 35.5)]
[46 g ]
[71g]
⎯→
[117g]
117g
⎯→
117g
¾
Limiting reactants:
“It is the reactant that reacts completely but limits further progress of the reaction.”
¾
Excess reactant:
“It is the reactant which is taken in excess than the limiting reactant.”
+
O2(g) ⎯→ 2H2O(l)
Eg. 2H2(g)
3g
2g
xg
Limiting
Excess
reactant
reactant
Some Basic Concepts of Chemistry
Water
17
Chemistry (Vol. I)
¾
TARGET Publications
Calculations based on stoichiometry:
Solving of stoichiometric problems is very important. It requires grasp and application of mole
concept, balancing of chemical equations and care in the conversion of units.
The problems based upon chemical equations may be classified as:
i.
Mole to mole relationships:
In these problems, the moles of one of the reactants/products is to be calculated if that of other
reactants/products are given.
ii.
Mass-mass relationships:
In these problems, the mass of one of the reactants/products is to be calculated if that of the
other reactants/products are given.
iii.
Mass-volume relationship:
In these problems, mass or volume of one of the reactants or products is calculated from the
mass or volume of other substances.
iv.
Volume-volume relationship:
In these problems, the volume of one of the reactants/products is given and that of the other is
to be calculated.
The main steps for solving such problems are:
a.
Write down the balanced chemical equation.
b.
Write down the moles or gram atomic or gram molecular masses of the substances whose
quantities are given or have to be calculated. In case, there are two or more atoms or
molecules of a substance, multiply the mole or gram atomic mass or molecular mass by
the number of atoms or molecules.
c.
Write down the actual quantities of the substances given. For the substances whose
masses or volumes have to be calculated, write the symbol of interrogation (?).
d.
Calculate the result by a unitary method.
Eg.
Calculate, how many moles of methane are required to produce 22g of CO2(g) after combustion.
(Atomic masses, C = 12u, H = 1u, O = 16u)
Solution:
The balanced combustion chemical reaction is,
CH4(g) + 2O2(g) ⎯→ CO2(g) + 2H2O(g)
1 mol
1 mol
44g
?
22 g
44 g of CO2(g) are produced from 1 mole of CH4
22 g of CO2(g) will be produced by burning
1 × 22
moles of CH4.
44
= 0.5 mole of CH4.
18
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
Multiple Choice Questions
1.0
1.
2.
3.
4.
General Introduction – Importance and
Scope of Chemistry
The branch of chemistry which deals with
carbon compounds is called _____ chemistry.
(A) Organic
(B) Inorganic
(C) Carbon
(D) Bio
The branch of chemistry which deals with the
separation, identification and quantitative
determination of the composition of different
substances is called _____ chemistry.
(A) Organic
(B) Inorganic
(C) Analytical
(D) Bio
_____ chemistry deals
elements other than
compounds.
(A) Organic
(C) Inorganic
with the chemistry of
carbon and of their
(B)
(D)
The branch of chemistry that deals with the
structure of matter, the energy changes and the
theories, laws and principles that explain the
transformation of matter from one form to
another is called _____ chemistry.
(A) Inorganic
(B) Organic
(C) Analytical
(D) Physical
_____ chemistry is the chemistry of the
substances consisting of living organisms.
(A) Organic
(B) Physical
(C) Inorganic
(D) Bio
6.
Which of the following branch of chemistry
deals with the application of chemical research
techniques
to
the
synthesis
of
Pharmaceuticals ?
(A) Nano chemistry
(B) Polymer chemistry
(C) Medicinal chemistry
(D) Green chemistry
Nuclear chemistry deals with
(A) Chemical processes associated with
food stuffs.
(B) Structure, properties and chemical
processes occuring in the solid phase.
(C) Production
and
reactions
of
nanoparticles.
(D) Study
of
radioactivity,
nuclear,
processes and nuclear properties.
Some Basic Concepts of Chemistry
Which of the following drugs is used as a
tranquilizer?
(A) Naproxen
(B) Furacin
(C) Noradrenaline
(D) Ampicillin
9.
Aspirin is used as an
(A) Antipyretic
(C) Antihistamine
(B)
(D)
Antibiotic
Tranquilizer
10.
Aspartame is used as a food additive to
(A) Impart attractive colour to the food stuff.
(B) give sweetening effect to the food and
enhance its odour and flavour.
(C) To prevent food spoilage
(D) To retain the food nutritive value for a
longer time.
11.
Indigo is used as a/an
(A) Natural dye
(C) Artificial dye
(B)
(D)
Detergent
Fuel
12.
Greek philosopher _____ had suggested that
matter is composed of extremely small
atomio.
(A) Dalton
(B) Aristotle
(C) Ptolemy
(D) Democritus
13.
Which of the following is CORRECT?
(A) 1 L = 1 dm3
(B) 1 L = 10 dm3
3
(C) 10 L = 1 dm
(D) 1 L = 1 m3
14.
The prefix femto is used for expressing
(A) 109
(B) 10−12
(C) 10−15
(D) 105
15.
A ______ is a simple combination of two or
more substances in which the constituent
substances retain their separate identities.
(A) compound
(B) mixture
(C) alloy
(D) amalgam
16.
If two or more phases are present in a mixture
then it is called a _____ mixture.
(A) heterogeneous
(B) homogeneous
(C) homologous
(D) heterologous
17.
Mixture of all gases constitute _____ system.
(A) homogeneous
(B) heterogeneous
(C) homologous
(D) heterologous
18.
Mixture of liquids constitute _____ system.
(A) homogeneous
(B) heterogeneous
(C) either (A) or (B)
(D) neither (A) nor (B)
19
Physical
Bio
5.
7.
8.
Chemistry (Vol. I)
19.
Phenol−water system is a/an
(A) element.
(B) compound.
(C) homogeneous system.
(D) heterogeneous system.
20.
Which out of the following is NOT a
homogeneous mixture?
(A) Air
(B) Solution of salt in water
(C) Solution of sugar in water
(D) Smoke
21.
Which one of the following is NOT a mixture?
(A) Iodized table salt
(B) Gasoline
(C) Liquefied Petroleum Gas (LPG)
(D) Distilled water
1.1
Laws of chemical combination
22.
Law
by
(A)
(B)
(C)
(D)
23.
24.
25
26.
20
TARGET Publications
27.
Which of the following is the best example of
law of conservation of mass?
[NCERT 1975]
(A) 12 g of carbon combines with 32 g of
oxygen to form 44 g of CO2
(B) When 12 g of carbon is heated in a
vacuum there is no change in mass.
(C) A sample of air increases in volume
when heated at constant pressure but its
mass remains unaltered.
(D) The weight of a piece of platinum is the
same before and after heating in air.
28.
‘n’ g of substance X reacts with ‘m’ g of
substance Y to form ‘p’ g of substance R and
‘q’ g of substance S. This reaction can be
represented as, X + Y = R + S. The relation
which can be established in the amounts of the
reactants and the products will be
(A) n − m = p − q
(B) n + m = p + q
(C) n = m
(D) p = q
29.
If law of conservation of mass was to hold
true, then 20.8 g of BaCl2 on reaction with
9.8 g of H2SO4 will produce 7.3 g of HCl and
BaSO4 equal to
(A) 11.65 g
(B) 23.3 g
(C) 25.5 g
(D) 30.6 g
30.
The law of definite composition was proposed
by
(A) Lomonosove
(B) Antoine Lavoisier
(C) Joseph Proust
(D) Dalton
31
If water samples are taken from sea, rivers,
clouds, lake or snow, they will be found to
contain H2 and O2 in the fixed ratio of 1 : 8.
This indicates the law of
(A) Multiple proportion
(B) Definite proportion
(C) Reciprocal proportion
(D) None of these
32.
The percentage of copper and oxygen in
samples of
CuO obtained by different
methods were found to be the same. This
illustrates the law of
[AMU 1982, 92]
(A) Constant proportion
(B) Conservation of mass
(C) Multiple proportion
(D) Reciprocal proportion
of conservation of mass was first stated
Lomonosove
Antoine Lavoisier
Joseph Proust
Dalton
After a chemical reaction, the total mass of
[MP PMT 1989]
reactants and products
(A) Is always increased
(B) Is always decreased
(C) Is not changed
(D) Is always less or more
The sum of the masses of reactants and
products is equal in any physical or chemical
reaction. This is in accordance with
(A) Law of multiple proportion
(B) Law of definite composition
(C) Law of conservation of mass
(D) Law of reciprocal proportion
The law of conservation of mass holds good
for all of the following except
(A) All chemical reactions.
(B) Nuclear reactions.
(C) Endothermic reactions.
(D) Exothermic reactions.
1.5 g of hydrocarbon on combustion in excess
of oxygen produces 4.4 g of CO2 and 2.7 g of
H2O, The data illustrates
(A) Law of conservation of mass
(B) Law of multiple proportion
(C) Law of constant composition
(D) Law of reciprocal proportion
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
33.
Irrespective of the source, pure sample of
water always yields 88.89 % mass of oxygen
and 11.11 % mass of hydrogen. This is
explained by the law of [Kerala CEE 2002]
(A) Conservation of mass
(B) Constant composition
(C) Multiple proportions
(D) Constant volume
40.
Different proportions of oxygen in the various
oxides of nitrogen proves the law of
[MP PMT 1985]
(A) Equivalent proportion
(B) Multiple proportion
(C) Constant proportion
(D) Conservation of mass
34.
A sample of pure carbon dioxide, irrespective
of its source contains 27.27 % carbon and
2.73 % oxygen. The data supports
[AIIMS 1992]
(A) Law of constant composition
(B) Law of conservation of mass
(C) Law of reciprocal proportion
(D) Law of multiple proportion
41.
Hydrogen and oxygen combine to form H2O2
and H2O containing 5.93 % and 11.2 %
Hydrogen respectively. The data illustrates
(A) Law of conservation of mass
(B) Law of constant proportion
(C) Law of reciprocal proportion
(D) Law of multiple proportion
35.
Zinc sulphate contains 22.65 % of zinc and
43.9 % of water of crystallization. If the law
of constant proportions is true, then the weight
of zinc required to produce 20 g of the crystals
will be
(A) 45.3 g
(B) 4.53 g
(C) 0.453 g
(D) 453 g
42.
Which one of the following pairs of
compounds illustrates the law of multiple
proportion?
[EAMCET 1989]
(A) H2O, Na2O
(B) MgO, Na2O
(C) Na2O, BaO
(D) SnCl2, SnCl4
36.
A sample of calcium carbonate (CaCO3) has
the following percentage composition :
Ca = 40 %; C = 12 %; O = 48 %
If the law of constant proportions is true, then
the weight of calcium in 4 g of a sample of
calcium carbonate from another source will be
(A) 0.016 g
(B) 0.16 g
(C) 1.6 g
(D) 16 g
43.
Which of the following pairs of substances
illustrates the law of multiple proportions ?
[CPMT 1972, 78]
(A) CO and CO2
(B) H2O and D2O
(C) NaCl and NaBr
(D) MgO and Mg(OH)2
37.
The law of definite proportion is NOT
applicable to nitrogen oxide because
[EAMCET 1981]
(A) Nitrogen atomic weight is not constant
(B) Nitrogen molecular weight is variable
(C) Nitrogen equivalent weight is variable
(D) Oxygen atomic weight is variable
44.
38.
The law of multiple proportions was given by
(A) Proust
(B) Dalton
(C) Avogadro
(D) Lavoisier
Two samples of lead oxide were separately
reduced to metallic lead by heating in a
current of hydrogen. The weight of lead from
one oxide was half the weight of lead obtained
from the other oxide. The data illustrates
[AMU 1983]
(A) Law of reciprocal proportions
(B) Law of constant proportions
(C) Law of multiple proportions
(D) Law of equivalent proportions
39.
In SO2 and SO3, the ratio of the masses of
oxygen which combine with a fixed mass of
sulphur is 2 : 3. This is an example of the law
of
(A) Constant proportion
(B) Multiple proportion
(C) Reciprocal proportion
(D) Gay Lussac
45.
1.0 g of an oxide of A contains 0.5 g of A.
4.0 g of another oxide of A contains 1.6 g of
A. The data indicates the law of
(A) Reciprocal proportion
(B) Constant proportion
(C) Conservation of energy
(D) Multiple proportion
21
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
46.
47.
48.
49.
50.
22
In compound A, 1.00 g of nitrogen unites with
0.57 g of oxygen. In compound B, 2.00 g of
nitrogen combines with 2.24 g of oxygen. In
compound C, 3.00 g of nitrogen combines
with 5.11 g of oxygen. These results obey the
[CPMT 1971]
(A) Law of constant proportion
(B) Law of multiple proportion
(C) Law of reciprocal proportion
(D) Dalton’s law of partial pressure
Two elements, X (Atomic mass 16) and Y
(Atomic mass 14) combine to form
compounds A, B and C. The ratio of different
masses of Y which combines with fixed mass
of X in A, B and C is 1:3:5. If 32 parts by
mass of X combines with 84 parts by mass of
Y in B, then in C, 16 parts by mass of X will
combine with
(A) 14 parts by mass of Y
(B) 42 parts by mass of Y
(C) 70 parts by mass of Y
(D) 82 parts by mass of Y
Two elements X and Y have atomic masses of
14 and 16 respectively. They form a series of
compounds A, B, C, D and E in which for the
same amount of element; X, Y is present in
the ratio 1 : 2 : 3 : 4 : 5. If the compound A
has 28 parts by mass of X and 16 parts by
mass of Y, then the compound of C will have
28 parts by mass of X and
[NCERT 1971]
(A) 32 parts by mass of Y
(B) 48 parts by mass of Y
(C) 64 parts by mass of Y
(D) 80 parts by mass of Y
One part of an element A combines with two
parts of an element B. Six parts of the element
C combines with four parts of element B. If A
and C combine together the ratio of their
weights will be governed by
(A) Law of definite proportion
(B) Law of multiple proportion
(C) Law of reciprocal proportion
(D) Law of conservation of mass
2 g of hydrogen combines with 16 g of oxygen
to form water and with 6 g of carbon to form
methane. In carbon dioxide, 12 g of carbon is
combined with 32 g of oxygen. These figures
illustrate the law of
(A) Multiple proportion
(B) Constant proportion
TARGET Publications
(C)
(D)
Reciprocal proportion
Conservation of mass
51.
6 g of carbon combines with 32 g of sulphur to
form CS2. 12 g of C also combines with 32 g
of oxygen to form carbondioxide. 10 g of
sulphur combines with 10 g of oxygen to form
sulphur dioxide. Which law is illustrated by
them?
(A) Law of multiple proportion
(B) Law of constant composition
(C) Law of Reciprocal proportion
(D) Gay Lussac’s law
52.
______ law of combining volumes states that
“Whenever gases combine, they do so in
simple ratio by volumes”.
(A) Avogadro’s
(B) Gay Lussac’s
(C) Dalton’s
(D) Thomson’s
53.
1 L of N2 combines with 3 L of H2 to form 2 L
of NH3 under the same conditions. This
illustrates the
(A) Law of constant composition
(B) Law of multiple proportions
(C) Law of reciprocal proportions
(D) Gay-Lussac’s law of gaseous volumes
54.
The balancing of chemical equation is based
upon_____
(A) Law of combining volumes
(B) Law of multiple proportions
(C) Law of constant volume
(D) Law of definite proportions
55.
Which of the following reactions has the ratio
of volumes of reacting gases and the product
as 1 : 2 : 2 ?
(A) 2CO(g) + O2(g) ⎯→ 2CO2(g)
(B) O2(g) + 2H2(g) ⎯→ 2H2O(g)
(C) H2(g) + F2(g) ⎯→ 2HF(g)
(D) N2(g) + 3H2(g) ⎯→ 2NH3(g)
56.
How many litres of ammonia will be formed
when 2 L of N2 and 2 L of H2 are allowed to
react?
(A) 0.665
(B) 1.0
(C) 4.00
(D) 1.33
57.
What volume of ammonia would be formed
when 0.36 dm3 of nitrogen reacts with
sufficient amount of hydrogen? (all volumes
are measured under same conditions of
temperature and pressure)
(A) 0.36 dm3
(B) 0.72 dm3
(C) 0.18 dm3
(D) 0.12 dm3
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
58.
The volume of oxygen required for the
complete combustion of 0.25 cm3 of CH4 at
S.T.P is
(A) 0.25 cm3
(B) 0.5 cm3
(D) 1 cm3
(C) 0.75 cm3
1.2
Dalton’s atomic theory
59.
Who proposed the atomic theory?
(A) Democritus
(B) Newton
(C) Thompson
(D) Dalton
60.
According to Dalton’s atomic theory, an atom
can
(A) be created.
(B) be destroyed.
(C) neither be created nor destroyed.
(D) be created and destroyed.
1.3
Concepts
Molecules
of
Elements,
Atoms
and
68.
Substances which CANNOT be decomposed
into two different substances by chemical
process are called
(A) Elements
(B) Molecules
(C) Compounds
(D) All of these
69.
_____ elements exist naturally.
(A) 92
(B) 35
(C) 118
(D) 105
70.
The most abundant element in the earth’s crust
is
(A) Iron
(B) Aluminium
(C) Oxygen
(D) Nitrogen
61.
Dalton assumed that _____ are the smallest
particles of a compound.
(A) atoms
(B) molecules
(C) ions
(D) elements
71.
Which of the following is indivisible by
chemical methods?
(A) atom
(B) molecule
(C) compound
(D) mixture
62.
On the basis of his assumptions, Dalton
explained
(A) Law of conservation of mass
(B) Law of multiple proportion
(C) Both (A) and (B)
(D) Neither (A) nor (B)
72.
Atom is the smallest particle of a/an
(A) compound
(B) substance
(C) mixture
(D) element
73.
Atoms have a mass of the order
(B) 10−15 kg
(A) 10−26 kg
−26
(D) 10−15 g
(C) 10 g
74.
A/An _____ is an aggregate of two or more
atoms of definite composition which are held
together by chemical bonds.
(A) ion
(B) molecule
(C) compound
(D) mixture
75.
Atoms have a radius of the order
(A) 10−26 m
(B) 10−15 µm
(D) 10−15 m
(C) 10−15 mm
76.
The number of atoms in 5.4 g of NH3 is
approximately
(B) 2 × 1023
(A) 8 × 1023
(C) 4 × 1023
(D) 6 × 1023
77.
The number of atoms of oxygen present in
11.2 L of ozone at N.T.P. are
(B) 6.02 × 1023
(A) 3.01 × 1022
23
(D) 1.20 × 1019
(C) 9.03 × 10
78.
_____ is a substance of definite composition
which can be decomposed into two or more
substances by a simple chemical process.
(A) Compound
(B) Element
(C) Solution
(D) Mixture
63.
The number of atoms present in a molecule of
a substance is called ________
(A) Atomicity
(B) Volume
(C) Density
(D) Mass
64.
Atomicity of mercury vapour is_____
(A) 1
(B) 2
(C) 3
(D) 4
65.
Atomicity of ammonium phosphate molecule
is_____
(A) 5
(B) 10
(C) 15
(D) 20
66.
67.
Avogadro’s law distinguishes between
(A) cations and anions.
(B) atoms and molecules.
(C) atoms and ions.
(D) molecules and ions.
Two containers of the same size are filled
separately with H2 gas and CO2 gas. Both the
containers under the same temperature and
pressure will contain the same
(A) number of atoms
(B) mass of gas
(C) number of molecules
(D) number of electrons
Some Basic Concepts of Chemistry
23
Chemistry (Vol. I)
TARGET Publications
79.
The phlogiston theory was suggested for
(A) neutralisation reaction.
(B) oxidation reaction.
(C) reduction reaction.
(D) combustion reaction.
80.
Antoine Lavoisier proved that the flammable
air produced by Cavendish was a new gas and
named it as
(A) Oxygen
(B) Hydrogen
(C) Methane
(D) Nitrogen
1.4
Atomic and Molecular masses
81.
The modern atomic mass scale is based on
(A) 12C
(B) 16O
1
(C) H
(D) 13C
82.
Recently the unit of atomic mass amu is
replaced by
(A) u
(B) mol
(C) g
(D) kg
83.
1 amu is equal to
1
(A)
of C − 12
12
(C) 1 g of H2
84.
(B)
(D)
1
of O − 16
14
1.66 × 10−23 kg
The number of atoms in 6 amu of He is
(A) 18
(B) 18 × 6.022 × 1023
(C) 54
(D) 54 × 6.023 × 1023
85.
The element whose atom has mass of
10.86 × 10−26 kg is
(A) Boron
(B) Calcium
(C) Silver
(D) Zinc
86.
An atom of an element weighs 1.792 × 10–22 g,
atomic mass of the element is
(A) 108
(B) 17.92
(C) 1.192
(D) 64
87.
88.
24
The number of gram atoms of oxygen present
in 0.3 gram mole of (COOH)2.2H2O is
(A) 0.6
(B) 1.8
(C) 1.2
(D) 3.6
The sulphate of a metal M contains 9.87 % of
M. This sulphate is isomorphous with
ZnSO4.7H2O. The atomic mass of M is
[IIT 1991]
(A) 40.3 u
(B) 36.3 u
(C) 24.3 u
(D) 11.3 u
89.
_____ mass of a substance is defined as the
ratio of mass of one molecule of a substance
th
1
of mass of one 12C atom.
to
12
(A) Chemical
(B) Molecular
(C) Molar
(D) Gram molar
90.
The molecular mass of carbon dioxide is 44.
What is the unit of molecular mass?
(A) g
(B) mol
(C) a.m.u
(D) mol g−1
91.
Vapour density of a gas is 22. What is its
molecular mass?
[AFMC 2000]
(A) 33
(B) 22
(C) 44
(D) 11
92.
The vapour density of gas A is four times that
of B. If molecular mass of B is M, then
molecular mass of A is
(A) M
(B) 4M
(C) 3M
(D) 2M
93.
Boron has two stable isotopes, 10B (19 %) and
11
B (81 %). The atomic mass that should
appear for boron in the periodic table is
[CBSE PMT 1990]
(A) 10.8 amu
(B) 10.2 amu
(C) 11.2 amu
(D) 10.0 amu
1.5
Mole concept and molar mass
94.
Avogadro’s number is
(A) number of atoms in one gram of
element.
(B) number of millilitres which one mole of
a gaseous substance occupies at N.T.P.
(C) number of molecules present in a gram
molecular mass of a substance.
(D) All of these.
95.
Avogadro number is the number of particles
present in
(A) 1 molecule
(B) 1 atom
(C) 1 kg
(D) 1 mole
96.
NA = _________ atoms mol−1.
(A) 6.021 × 1021
(B) 6.024 × 1024
15
(C) 6.051 × 10
(D) 6.023 × 1023
97.
One _____ is the collection of 6.023 × 1023
atoms /molecules/ions.
(A) kg
(B) g
(C) mole
(D) cm
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
98.
The number of molecules in 22.4 cm3 of
nitrogen gas at STP is
(A) 6.023 × 1020
(B) 6.023 × 1023
20
(D) 22.4 × 1023
(C) 22.4 × 10
99.
Number of molecules in 0.4 g of He is
(B) 6.023 × 1022
(A) 6.023 × 1023
(C) 3.011 × 1023
(D) 3.011 × 1022
100. If NA is the Avogadro’s number then number
of valence electrons in 4.2 g of nitride ions
N3− is
(B) 4.2 NA
(A) 2.4 NA
(C) 1.6 NA
(D) 3.2 NA
101. 11.2 cm3 of hydrogen gas
contains____
(A) 0.0005 mol
(B) 0.01 mol
(C) 0.029 mol
(D) 3.011 × 1023 molecules
at
STP,
102. The number of molecules present in 0.032 mg
of methane is
(A) 12.046 × 1017
(B) 1.2044 × 1017
(D) 2 × 10−6
(C) 12.044 × 107
103. What is the mass of 0.5 mole of ozone
molecule?
(A) 8 g
(B) 16 g
(C) 24 g
(D) 48 g
104. At STP, 2 g of helium gas (molar mass = 4)
occupies a volume of
(A) 22.4 dm3
(B) 11.2 dm3
3
(C) 5.6 dm
(D) 2 dm3
105. The number of molecules in 16 g of oxygen is
(A) 6.023 × 1023
(B) 3.011 × 1023
(C) 3.011 × 1022
(D) 1.5 × 1023
106. The number of sulphur atoms present in
0.2 moles of S8 molecules is
(A) 4.82 × 1023
(B) 9.63 × 1022
(C) 9.63 × 1023
(D) 1.20 × 1023
107. 19.7 kg of gold was recovered from a
smuggler. How many atoms of gold were
recovered? (Au = 197)
[Pb. CET 1985]
(A) 100
(B) 6.023 × 1023
(C) 6.023 × 1024
(D) 6.023 × 1025
Some Basic Concepts of Chemistry
108. How many moles of electrons weigh one
kilogram?
(A) 6.023 × 1023
1
(B)
× 1031
9.108
6.023
× 1054
(C)
9.108
1
(D)
× 108
9.108 × 6.023
109. The number of molecule at NTP in 1 mL of an
ideal gas will be
(A) 6 × 1023
(B) 2.69 × 1019
23
(C) 2.69 × 10
(D) 2.69 × 1034
110. 4.4 g of an unknown gas occupies 2.24 L of
volume under NTP conditions. The gas may
be
[MP PMT 1995]
(B) CO
(A) CO2
(C) O2
(D) SO2
111. One mole of CO2 contains
(A) 6.023 × 1023 atoms of C
(B) 6.023 × 1023 atoms of O
(C) 18.1 × 1023 molecules of CO2
(D) 3 g atoms of CO2
112. How many grams are contained in 1 g atom of
Na?
(A) 13 g
(B) 23 g
1
(C) 1 g
(D)
g
23
113. One mole of oxygen weighs______.
(A) 8 g
(B) 32 g
(C) 1 g
(D) 64 g
114. 1 mol of CH4 contains
(A) 6.02 × 1023 atoms of H
(B) 4 g atom of Hydrogen
(C) 1.81 × 1023 molecules of CH4
(D) 3.0 g of carbon
115. The mass of carbon present in 0.5 mole of
K4[Fe(CN)6] is
(A) 1.8 g
(B) 18 g
(C) 3.6 g
(D) 36 g
116. How many molecules are present in one gram
[AIIMS 1982]
of hydrogen?
(A) 6.023 × 1023
(B) 3.012 × 1023
(C) 2.512 × 1023
(D) 1.512 × 1023
117. The number of moles of sodium oxide in
620 g is
[BHU 1992]
(A) 1 mole
(B) 10 moles
(C) 18 moles
(D) 100 moles
25
Chemistry (Vol. I)
118. How many atoms are contained in one mole of
sucrose (C12H22O11)?
[Pb. PMT 2002]
23
(A) 45 × 6.023 × 10 atoms/mole
(B) 5 × 6.623 × 1023 atoms/mole
(C) 5 × 6.023 × 1023 atoms/mole
(D) 40 × 6.023 × 1023 atoms/mole
119. One mole of P4 molecule contains
(A) 1 molecule
(B) 4 molecules
1
(C)
× 6.022 × 1023 atoms
4
(D) 24.092 × 1023 atoms
120. Total
(A)
(B)
(C)
(D)
number of atoms in 44 g of CO2 is
6.023 × 1023
6.023 × 1024
1.807 × 1024
18.06 × 1022
121. The mass of 1 atom of hydrogen is
(A) 1 g
(B) 0.5 g
(C) 1.66 × 10−24 g
(D) 3.2 × 10−24 g
122. The mass of 2.01 × 1023 molecules of CO is
(A) 9.3 g
(B) 7.2 g
(C) 1.2 g
(D) 3 g
123. How many moles of Helium gas occupy
22.4 L at 0 °C at 1 atm pressure?
(A) 0.11
(B) 0.90
(C) 1.0
(D) 1.11
TARGET Publications
128. 2 moles of H2 at NTP occupy a volume of
(A) 11.2 litre
(B) 44.8 litre
(C) 2 litre
(D) 22.4 litre
129. If the density of water is 1 g/cm3, then the
volume occupied by one molecule of water is
approximately
[Pb. PMT 2004]
(A) 18 cm3
(B) 22400 cm3
(C) 6.02 × 10−23 cm3 (D) 3.0 × 10−23 cm3
130. The number of molecules in 8.96 L of a gas at
0°C and 1 atmospheric pressure is
approximately
[BHU 1993]
23
(A) 6.023 × 10
(B) 12.04 × 1023
23
(D) 24.09 × 1022
(C) 18.06 × 10
131. Number of g of oxygen in 32.2 g of
Na2SO4.10H2O is
[Haryana PMT 2000]
(A) 20.8
(B) 22.4
(C) 2.24
(D) 2.08
132. Number of moles of water in 488 gm of
BaCl2.2H2O are (Ba = 137)
(A) 2 moles
(B) 4 moles
(C) 3 moles
(D) 5 moles
133. The mass of 1 × 1022 molecules of
CuSO4.5H2O is
[CBSE PMT 1999; MH CET 2003]
(A) 41.51 g
(B) 415.1 g
(C) 4.151 g
(D) 4151 g
124. The mass of a molecule of water is
[Bihar CEE 1995]
(A) 3 × 10–26 kg
(B) 3 × 10–25 kg
(D) 2.5 × 10–26 kg
(C) 1.5 × 10–26 kg
134. Mass of H2O in 1000 kg CuSO4.5H2O is
(Cu = 63.5)
(A) 3.607 kg
(B) 36.07 kg
(C) 360.7 kg
(D) 3607 kg
125. The mass of 1 molecule of N2 is
(A) 3.24 × 10−22 g
(B) 3.45 × 10−25 g
(C) 4.65 × 10−23 g
(D) 4.56 × 1022 g
135. The number of molecules in 16 g of methane is
(A) 3.0 × 1023
(B) 6.023 × 1023
16
16
(C)
× 1023
(D)
× 1023
6.02
3.0
126. The mass of a molecule of the compound
C60H122 is
(A) 1.4 × 10−21 g
(B) 1.09 × 10−21 g
(C) 5.025 × 1023 g
(D) 16.023 × 1023 g
127. The number of moles of oxygen in 1 L of air
containing 21 % oxygen by volume, in
standard conditions, is
[CBSE PMT 1995; Pb. PMT 2004]
(A) 0.186 mol
(B) 0.21 mol
(C) 2.10 mol
(D) 0.0093 mol
26
136. The number of water molecules in 1 litre of
water is
[EAMCET 1990]
(A) 18
(B) 18 × 1000
(C) NA
(D) 55.55 NA
137. The numbers of moles of BaCO3 which
contain 1.5 moles of oxygen atoms is
[EAMCET 1991]
(A) 0.5
(B) 1
(C) 3
(D) 6.02 × 1023
138. 1.24 g of P is present in 2.2 g of
(A) P4S3
(B) P2S2
(C) PS2
(D) P2S4
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
139. 2 g of oxygen contains number of atoms equal
to that in
[BHU 1992]
(A) 0.5 g of hydrogen (B) 4 g of sulphur
(C) 7 g of nitrogen (D) 2.3 g of sodium
140. If 1021 molecules are removed from 200 mg of
CO2, then the number of moles of CO2 left are
[IIT 1983]
(B) 28.8 × 10−3
(A) 2.89 × 10−3
(D) 1.68 × 10−2
(C) 0.288 × 10−3
141. Mole triangle is the relationship between the
mass of a gas, the number of moles, the
volume at S.T.P. and the
(A) number of electrons.
(B) number of molecules.
(C) pressure at S.T.P.
(D) temperature at S.T.P.
1.6
Percentage composition and empirical and
molecular formula
142. _____ of a compound is the chemical formula
indicating the relative number of atoms in the
simplest ratio.
(A) Empirical formula
(B) Molecular formula
(C) Empirical mass
(D) Molecular mass
143. _____ indicates the actual number
constituent atoms in a molecule.
(A) Empirical formula
(B) Molecular formula
(C) Empirical mass
(D) Molecular mass
of
144. The mass percentage of each constituent
element present in 100 g of compound is
called its
(A) Molecular composition
(B) Atomic composition
(C) Percentage composition
(D) Mass composition
145. If two compounds have the same empirical
formula but different molecular formulae, they
must have
[MP PMT 1986]
(A) Different percentage composition
(B) Different molecular mass
(C) Same viscosity
(D) Same vapour density
146. The percentage of oxygen in NaOH is
[CPMT 1979]
(A) 40
(B) 60
(C) 8
(D) 10
Some Basic Concepts of Chemistry
147. Percentage of nitrogen in urea is about
(A) 46 %
(B) 85 %
(C) 18 %
(D) 28 %
148. The percentage composition of carbon in urea,
[CO(NH2)2] is
(A) 40 %
(B) 50 %
(C) 20 %
(D) 80 %
149. What is the % of H2O in Fe(CNS)3.3H2O?
(A) 45
(B) 30
(C) 19
(D) 25
150. The percentage of P2O5 in diammonium
hydrogen phosphate (NH4)2HPO4 is
[CPMT 1992]
(A) 23.48
(B) 46.96
(C) 53.79
(D) 71.00
151. A 400 mg iron capsule contains 100 mg of
The
ferrous
fumarate
(CHCOO)2Fe.
percentage of iron present in it is
approximately
(A) 33 %
(B) 25 %
(C) 14 %
(D) 8 %
152. Which pair of species have same percentage
of carbon?
(A) CH3COOH and C6H12O6
(B) CH3COOH and C2H5OH
(C) HCOOCH3 and C12H22O11
(D) C6H12O6 and C12H22O11
153. Empirical formula of glucose is
(B) C6H11O6
(A) C6H12O6
(C) CHO
(D) CH2O
154. The empirical formula of C2H2 is _____.
(A) C2H4
(B) CH
(D) all of these
(C) CH4
155. A compound (80 g) on analysis gave C = 24 g,
H = 4 g, O = 32 g. Its empirical formula is
[CPMT 1981]
(A) C2H2O2
(B) C2H2O
(C) CH2O2
(D) CH2O
156. Which of the following has same molecular
formula and empirical formula?
(A) CO2
(B) C6H12O6
(C) C2H4
(D) all of these
157. The molecular mass of an organic compound
is 78. Its empirical formula is CH. The
molecular formula is
(B) C2H2
(A) C2H4
(C) C6H6
(D) C4H4
27
Chemistry (Vol. I)
158. The empirical formula of an acid is CH2O2,
the probable molecular formula of acid may be
[AFMC 2000]
(A) CH2O
(B) CH2O2
(C) C2H4O2
(D) C3H6O4
159. The empirical formula of a compound is
CH2O. If 0.0835 mole of the compound
contains 1.0 g of hydrogen, then the molecular
formula of the compound is
(A) C6H12O6
(B) C5H10O5
(C) C4H8O8
(D) C3H6O3
160. On analysis, a certain compound was found to
have 254 g of iodine (At. mass 127) and 80 g
oxygen (At. mass 16). What is the molecular
formula of the compound?
(A) IO
(B) I2O
(D) I2O5
(C) I5O3
161. The molecular formula of the compound with
molecular mass 159, containing 39.62 % Cu
and 20.13 % S is
(A) Cu2S
(B) CuS
(D) CuSO4
(C) CuSO3
162. A compound made of two elements A and B is
found to contain 25 % A (At. mass 12.5) and
75 % B
(At. mass 37.5). The simplest
formula of the compound is
(A) AB
(B) AB2
(D) A3B
(C) AB3
163. Two elements X (At. mass 75) and Y (At.
mass 16) combine to give a compound having
75.8 % X. The formula of the compound is
(A) XY
(B) XY2
(C) X2Y2
(D) X2Y3
164. An oxide of a metal (M) contains 40 % by
mass of oxygen. Metal (M) has atomic mass
of 24. The empirical formula of the oxide is
(A) M2O
(B) MO
(D) M2O4
(C) M2O3
165. Two oxides of metal contain 27.6 % and 30 %
oxygen respectively. If the formula of first
oxide is M3O4 then formula of second oxide is
(A) MO
(B) M2O
(C) M2O3
(D) MO2
28
TARGET Publications
166. 14 g of element X combines with 16 g of
oxygen. On the basis of this information,
which of the following is a CORRECT
statement?
(A) The element X could have an atomic
mass of 7 and its oxide formula is XO
(B) The element X could have an atomic
mass of 14 and its oxide formula is X2O
(C) The element X could have an atomic
mass of 7 and its oxide formula is X2O
(D) The element X could have an atomic
mass of 14 and its oxide formula is XO2
1.7
Chemical reactions, stoichiometry
calculations based on stoichiometry
and
167. ________ is the quantitative relationship
between the reactants and products in a
balanced chemical equation.
(A) Stoichiometry
(B) Complexometry
(C) Chemistry
(D) Reactions
168. The starting material which takes part in
chemical reaction is called
(A) product.
(B) reactant.
(C) catalyst.
(D) starter.
169. _____ reactant is the reactant that reacts
completely but limits further progress of the
reaction.
(A) Oxidizing
(B) Reducing
(C) Limiting
(D) Excess
170. 3 g of H2 reacts with 29 g of O2 to yield water.
Which is the limiting reactant?
(A) H2
(B) O2
(D) none of these
(C) H2O
171. _____ reactant is the reactant which is taken in
excess than the limiting reactant.
(A) Oxidizing
(B) Reducing
(C) Limiting
(D) Excess
172. A _____ chemical reaction may be used to
establish the weight relationships of reactants
and products.
(A) thermal
(B) molecular
(C) balanced
(D) molar
173. The set of numerical coefficient that balances
the equation
K2CrO4 + HCl ⎯→ K2Cr2O7 + KCl + H2O is
[Kerala CEE 2001]
(A) 1, 1, 2, 2, 1
(B) 2, 2, 1, 1, 1
(C) 2, 1, 1, 2, 1
(D) 2, 2, 1, 2, 1
Some Basic Concepts of Chemistry
TARGET Publications
Chemistry (Vol. I)
174. One mole of calcium phosphide on reaction
with excess of water gives
[IIT 1999]
(A) One mole of phosphine
(B) Two mole of phosphoric acid
(C) Two moles of phosphine
(D) One mole of phosphorus pentoxide
184. Hydrogen reacts with nitrogen to form
ammonia as: N2(g) + 3H2(g) ⎯→ 2NH3(g)
The amount of ammonia that would be
produced if 200 g of H2 reacts with N2 is
(A) 1032.2 g
(B) 11332 g
(C) 1133.3 g
(D) 8692.6 g
175. For the reaction : A + 2B ⎯→ C
5 moles of A and 8 moles of B will produce
(A) 5 moles of C
(B) 4 moles of C
(C) 8 moles of C
(D) 13 moles of C
185. What volume of Hydrogen gas, at 273 K and 1
atm pressure will be consumed in obtaining
21.6 g of elemental boron (At. mass = 10.8)
from the reduction of boron trichloride by
Hydrogen ?
[AIEEE 2003]
(A) 22.4 L
(B) 89.6 L
(C) 67.2 L
(D) 44.8 L
176. The moles of O2 required for reacting with
6.8 g of ammonia in the following reaction
(..... NH3 +..... O2 ⎯→ ..... NO + ..... H2O) is
(A) 5
(B) 2.5
(C) 1
(D) 0.5
177. One mole of potassium dichromate completely
oxidises ______ number of moles of ferrous
sulphate in acidic medium.
(A) 1
(B) 3
(C) 5
(D) 6
178. 27 g of Al (At. mass = 27) will react with
oxygen equal to
(A) 24 g
(B) 8 g
(C) 40 g
(D) 10 g
179. 1.2 g of Mg (At. mass 24) will produce MgO
equal to
(A) 0.05 mol
(B) 0.03 mol
(C) 0.01 mol
(D) 0.02 mol
180. If 0.5 mol of BaCl2 is mixed with 0.2 mol of
Na3PO4, the maximum number of moles of
Ba3(PO4)2 that can be formed is
(A) 0.7
(B) 0.5
(C) 0.3
(D) 0.1
186. 12 g of Mg (At. mass 24) will react
[MNR 1985]
completely with acid to give
(A) One mole of H2
1
(B)
mole of H2
2
2
(C)
mole of O2
3
1
1
mol of H2 and
mol of O2
(D) Both
2
2
187. H2 evolved at STP on complete reaction of 27
g of Aluminium with excess of aqueous NaOH
would be
[CPMT 1991]
(A) 22.4 litres
(B) 44.8 litres
(C) 67.2 litres
(D) 33.6 litres
188. What mass of CaO will be obtained by heating
3 mole of CaCO3? [At. mass of Ca = 40]
(A) 150 g
(B) 168 g
(C) 16.8 g
(D) 15 g
181. If one mole of ethanol (C2H5OH) completely
burns to form carbon dioxide and water, the
mass of carbon dioxide formed is about
(A) 22 gm
(B) 45 gm
(C) 66 gm
(D) 88 gm
189. How much of NaOH is required to neutralise
1500 cm3 of 0.1N HCl? (Na = 23)
[KCET 2001]
(A) 40 g
(B) 4 g
(C) 6 g
(D) 60 g
182. Complete combustion of 0.858 g of compound
X gives 2.63 g of CO2 and 1.28 g of H2O. The
lowest molecular mass X can have is
(A) 43 g
(B) 86 g
(C) 129 g
(D) 172 g
190. How many grams of caustic potash is required
to completely neutralize 12.6 g HNO3?
(A) 22.4 g of KOH
(B) 1.01 g of KOH
(C) 6.02 g of KOH
(D) 11.2 g of KOH
183. 1.12 mL of a gas is produced at STP by the
action of 4.12 mg of alcohol (ROH) with
methyl magnesium iodide. The molecular
mass of alcohol is [Roorkee 1992; IIT 1993]
(A) 16.0
(B) 41.2
(C) 82.4
(D) 156.0
Some Basic Concepts of Chemistry
191. The mass of CaCO3 produced when carbon
dioxide is passed in excess through 500 mL of
0.5 M Ca(OH)2 will be
(A) 10 g
(B) 20 g
(C) 50 g
(D) 25 g
29
Chemistry (Vol. I)
192. What mass of calcium chloride in grams
would be enough to produce 14.35 g of AgCl?
(At. Mass of Ca = 40, Ag = 108)
(A) 5.50 g
(B) 8.295 g
(C) 16.59 g
(D) 11.19 g
193. The amount of sulphur required to produce
100 moles of H2SO4 is
(A) 3.2 × 103 g
(B) 32.65 g
(C) 32 g
(D) 3.2 g
194. What should be the weight of 50 % HCl which
reacts with 100 g of limestone ?
(A) 50 % pure
(B) 25 % pure
(C) 10 % pure
(D) 8 % pure
195. The conversion of oxygen to ozone occurs to
the extent of 15 % only. The mass of ozone
that can be prepared from 67.2 L of oxygen at
STP will be
(A) 144 g
(B) 96 g
(C) 640 g
(D) 64 g
Miscellaneous
196. An element forms two oxides containing
respectively 53.33 and 36.36 percent of
oxygen. These figures illustrate the law of
(A) Conservation of mass
(B) Constant proportion
(C) Reciprocal proportion
(D) Multiple proportion
197. One sample of atmospheric air is found to
have 0.03 % of carbon dioxide and another
sample 0.04 %, this is an evidence that
(A) The law of constant composition is not
always true
(B) The law of multiple proportions is true
(C) Air is a compound
(D) Air is a mixture
198. Among the following pairs of compounds, the
one that illustrates the law of multiple
proportions is
(A) NH3 and NCl3
(B) H2S and SO2
(C) CuO and Cu2O (D) CS2 and FeSO4
199. An example of a chemical change is
(A) the melting of an ice cube
(B) the boiling of gasoline
(C) the frying of an egg
(D) all of these
200. ________ is the sum of the atomic masses of
all the atoms as given in the molecular
formula of the substance.
(A) Molecular mass (B) Empirical mass
(C) Percentage mass (D) Equivalent mass
30
TARGET Publications
201. Molecular mass = vapour density × 2, is valid
for
(A) metals.
(B) non-metals.
(C) solids.
(D) gases.
202. One mole of H2O corresponds to
(A) 22.4 litres at 1 atm and 25 °C
(B) 6.02 × 1023 atoms of hydrogen and
6.02 × 1023 atoms of oxygen
(C) 18 g
(D) 1 g
203 4.48 litres of methane at N.T.P. corresponds to
(A) 1.2 × 1022 molecules of methane
(B) 0.5 mole of methane
(C) 3.2 g of methane
(D) 0.1 mole of methane
204. 6.023 × 1023 electrons make an electrical
charge of one
(A) electron volt
(B) avogadro
(C) coulomb
(D) faraday
205. The mass of a substance that displaces
22.4 litre air at NTP is
(A) Molecular mass (B) Empirical mass
(C) Equivalent mass (D) All of these
206. The number of atoms present in 0.1 mole of
P4 are
(A) 2.4 × 1023 atoms
(B) 6.02 × 1022 atoms
(C) same as in 0.2 mole of S8
(D) same as in 3.1 g of phosphorus
207. Number of moles of water in 1 L of water with
density 1 g/cc are
(A) 55.56
(B) 45.56
(C) 56.55
(D) 46.55
208. Which one of the following pairs of gases
contain the same number of molecules?
(A) 16 g of O2 and 14 g of N2
(B) 8 g of O2 and 22 g of CO2
(C) 28 g of N2 and 22 g of CO2
(D) 32 g of O2 and 32 g of N2
209. 54 grams of aluminium (atomic mass = 27)
will react with how many grams of oxygen ?
(A) 16 g
(B) 48 g
(C) 40 g
(D) 15 g
210. The largest number of molecules is in
[BHU 1997]
(A) 34 g of water
(B) 28 g of CO2
(C) 46 g of CH3OH
(D) 54 g of N2O5
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
211. Which of the following has least mass?
(A) 2 g atom of nitrogen
(B) 3 × 1023 atoms of C
(C) 1 mole of S
(D) 7.0 g of Ag
218. 14 g of nitrogen represents
(A) 6.02 × 1023 N2 molecules
(B) 22.4 L of N2 at N.T.P.
(C) 11.2 L of N2 at N.T.P.
(D) 28 gm of nitrogen.
212. Which of the following contains maximum
number of atoms?
[JIPMER 2000]
21
(A) 6.023 × 10 molecules of CO2
(B) 22.4 L of CO2 at S.T.P.
(C) 0.44 g of CO2
(D) None of these
219. In the reaction,
213. 4.0 g of caustic soda (molar mass 40) contains
same number of sodium ions as are present in
(A) 10.6 g of Na2CO3 (molar mass 106)
(B) 58.5 g of NaCl (Formula mass 58.5)
(C) 100 mL of 0.5 M Na2SO4 (Formula
mass 142)
(D) 1 mol of NaNO3 (molar mass 85)
214. Four containers of 2L capacity contain
dinitrogen as described below. Which one
contains maximum number of molecules
under similar conditions ?
(A) 2.5 g of N2 molecules
(B) 4 g of N atoms
(C) 40 g of N atoms
(D) 84 g of dinitrogen
215. Which of the following contains the largest
number of atoms?
(A) 11 g of CO2
(B) 4 g of H2
(C) 5 g of NH3
(D) 8 g of SO2
216. 4.4 g of CO2 and 2.24 litre of H2 at STP are
mixed in a container. The total number of
molecules present in the container will be
(A) 6.023 × 1023
(B) 1.2046 × 1023
(C) 2 moles
(D) 6.023 × 1024
217. Number of moles of KMnO4 required to
oxidize one mole of Fe(C2O4) in acidic
medium is
[Haryana CEE 1996]
(A) 0.6
(B) 0.167
(C) 0.2
(D) 0.4
Some Basic Concepts of Chemistry
+
−
+ 6Cl(aq)
+ 3H2(g),
2Al(s) + 6HCl(aq) ⎯→ 2Al3(aq)
(A)
(B)
(C)
(D)
6 L HCl(aq) is consumed for every 3L
H2(g) produced.
33.6 L H2(g) is produced regardless of
temperature and pressure for every mole
of Al that reacts.
67.2 L H2(g) at STP is produced for every
mole of Al that reacts.
11.2 L H2(g) at STP is produced for every
mole of HCl(aq) consumed.
220. Number of water molecules in a drop of water,
if 1 mL of water has 20 drops and A is
Avogadro number, is
(A) 0.5 A / 18
(B) 0.05 A
(C) 0.5 A
(D) 0.05 A / 18
221. M is the molecular mass of KMnO4. The
equivalent mass of KMnO4 when it is
converted into K2MnO4 is
M
(A) M
(B)
3
M
M
(D)
(C)
5
7
222. Volume of a gas at N.T.P. is 1.12 × 10−7 cc.
Calculate the number of molecules in it.
(A) 3.01 × 1020
(B) 3.01 × 1012
(C) 3.01 × 1023
(D) 3.01 × 1024
223. Under similar conditions, oxygen and nitrogen
are taken in the same mass. The ratio of their
volume will be_______.
(A) 7 : 8
(B) 3 : 5
(C) 6 : 5
(D) 9 : 2
31
Chemistry (Vol. I)
TARGET Publications
9 Answers to Multiple Choice Questions
1. (A)
11. (A)
21. (D)
31. (B)
41. (D)
51. (C)
61. (B)
71. (A)
81. (A)
91. (C)
101. (A)
111. (A)
121. (C)
131. (B)
141. (B)
151. (D)
161. (D)
171. (D)
181. (D)
191. (D)
201. (D)
211. (B)
221. (A)
2. (C)
12. (D)
22. (A)
32. (A)
42. (D)
52. (B)
62. (B)
72. (D)
82. (A)
92. (B)
102. (A)
112. (B)
122. (A)
132. (B)
142. (A)
152. (A)
162. (A)
172. (C)
182. (A)
192. (A)
202. (C)
212. (B)
222. (B)
3. (C)
13. (A)
23. (C)
33. (B)
43. (A)
53. (D)
63. (A)
73. (A)
83. (A)
93. (A)
103. (C)
113. (B)
123. (C)
133. (C)
143. (B)
153. (D)
163. (B)
173. (D)
183. (C)
193. (A)
203. (C)
213. (C)
223. (A)
4. (D)
14. (C)
24. (C)
34. (A)
44. (C)
54. (C)
64. (A)
74. (B)
84. (A)
94. (C)
104. (B)
114. (B)
124. (A)
134. (C)
144. (C)
154. (B)
164. (B)
174. (C)
184. (C)
194. (A)
204. (D)
214. (D)
5. (D)
15. (B)
25. (B)
35. (B)
45. (D)
55. (B)
65. (D)
75. (D)
85. (D)
95. (D)
105. (B)
115. (D)
125. (C)
135. (B)
145. (B)
155. (D)
165. (C)
175. (B)
185. (C)
195. (A)
205. (A)
215. (B)
6. (C)
16. (A)
26. (A)
36. (C)
46. (B)
56. (C)
66. (B)
76. (B)
86. (A)
96. (D)
106. (C)
116. (B)
126. (A)
136. (D)
146. (A)
156. (A)
166. (C)
176. (D)
186. (B)
196. (D)
206. (A)
216. (B)
7. (D)
17. (A)
27. (A)
37. (C)
47. (C)
57. (B)
67. (C)
77. (C)
87. (B)
97. (C)
107. (D)
117. (B)
127. (D)
137. (A)
147. (A)
157. (C)
167. (A)
177. (D)
187. (D)
197. (D)
207. (A)
217. (A)
8. (C)
18. (C)
28. (B)
38. (B)
48. (B)
58. (B)
68. (A)
78. (A)
88. (C)
98. (B)
108. (D)
118. (A)
128. (B)
138. (A)
148. (C)
158. (B)
168. (B)
178. (A)
188. (B)
198. (C)
208. (A)
218. (C)
9. (A)
19. (D)
29. (B)
39. (B)
49. (C)
59. (D)
69. (A)
79. (D)
89. (B)
99. (B)
109. (B)
119. (D)
129. (D)
139. (B)
149. (C)
159. (A)
169. (C)
179. (A)
189. (C)
199. (C)
209. (B)
219. (D)
10. (B)
20. (D)
30. (C)
40. (B)
50. (C)
60. (C)
70. (C)
80. (B)
90. (C)
100. (A)
110. (A)
120. (C)
130. (D)
140. (A)
150. (C)
160. (D)
170. (A)
180. (D)
190. (D)
200. (A)
210. (A)
220. (D)
" Hints to Multiple Choice Questions
29.
∴
∴
BaCl2 + H2SO4 ⎯→ HCl + BaSO4
20.8 + 9.8 = 7.3 + x
x = 23.3 g
35.
100 g zinc sulphate crystals are obtained from 22.65 g of zinc.
22.65
1 g of zinc sulphate crystal will be obtained from =
g of zinc.
100
∴
∴
20 g of zinc sulphate crystals will be obtained from =
36.
100 g of CaCO3 is obtained from 40 g of Ca
40
1 g of CaCO3 will be obtained from
g of Ca
100
∴
22.65
× 20 = 4.53 g of zinc.
100
40
× 4 = 1.6 g
100
∴
weight of Ca in 4 g of a sample of calcium carbonate from another source will be =
47.
∴
In compound B, 32 parts of X react with 84 parts of Y.
In compound B, 16 parts of X react with 42 parts of Y.
In compound C, 16 parts of X react with x parts of Y.
The ratio of masses of Y which combines with fixed mass of X in compounds B and C is 3:5
B
C
∴
32
x=
42
x
3
5
42 × 5
= 70
3
Some Basic Concepts of Chemistry
TARGET Publications
Chemistry (Vol. I)
48.
A
16
C
x
x = 16 × 3
x = 48
57.
N2 +
3 H2 ⎯→ 2 NH3
(1 vol.) (3 vol.)
(2 vol.)
1
: 3
: 2
x
0.36 : 3 × 0.36 :
2 × 1.08
x=
= 0.72 dm3
3
58.
∴
CH4 + 2O2 ⎯→ CO2 + 2H2O
(1 vol.) (2 vol.)
(1 vol.) (2 vol.)
1 cm3 of CH4 requires 2 cm3 of O2
0.25 cm3 of CH4 requires 0.5 cm3 of O2
64.
In vapour state, atomicity of mercury is 1.
65.
Ammonium phosphate [(NH4)3PO4] contains 3N, 12H, 1P and 4O, atoms, hence total number of atoms are
20, hence atomicity is 20.
76.
Molecular mass of NH3 = 17
Amount of NH3 = 5.4 g
5.4
Number of moles =
= 0.32 mole
17
1 mole of NH3 = 6.023 × 1023 atoms
6.023 × 1023 × 0.32
0.32 mole of NH3 =
atoms
1
≈ 2 × 1023 atoms
∴
∴
∴
77.
∴
84.
∴
∴
∴
85.
∴
1
3
22.4 L of ozone ≡ 6.023 × 1023 atoms
6.023 × 1023 × 11.2
11.2 L of ozone ≡
22.4
= 3.011 × 1023 atoms
Hence, number of oxygen atoms in ozone (O3)
= 3 × 3.011 × 1023
= 9.03 × 1023 atoms
1
amu
3
x atom of He ≡ 6 amu
x
1×6=
3
x = 6 × 3 = 18 atoms
1 atom of He ≡
Atomic mass of the given element
= 6.023 × 1023 × 10.86 × 10–26 kg
= 65.4 × 10–3 kg
= 65.4 g
The element whose atom has mass of 10.86 × 10–26 kg is Zinc.
Some Basic Concepts of Chemistry
33
Chemistry (Vol. I)
86.
Atomic mass of an element
= 1.792 × 10−22 × 6.023 × 1023
≈ 108
87.
∵ 1mole (COOH)2.2H2O has 96 g of oxygen
∴
0.3 mole (COOH)2.2H2O has 96 × 0.3 = 28.8 g
28.8
Number of gram atoms of oxygen =
16
= 1.8
∴
88.
∴
∴
TARGET Publications
As the given sulphate is isomorphous with ZnSO4.7H2O, its formula would be MSO4.7H2O; Let m be the
atomic mass of M, molecular mass of MSO4.7H2O
= m + 32 + 64 + 126 = m + 222
m
Hence % of M =
× 100
m + 222
= 9.87 (given)
100 m = 9.87 m + 222 × 9.87
90.13 m = 222 × 9.87
222 × 9.87
m=
= 24.3 u
90.13
91.
Molecular mass = Vapour density × 2
= 22 × 2
= 44
92.
∵
Molecular mass = Vapour density × 2
M
Vapour density of A is 4 ×
= 2M
2
(Vapour density of A is four times that of B)
∴
Molecular mass of A = 2M × 2 = 4M
93.
Contribution of 10B = 10.0 × 0.19
= 1.9 amu
….(i)
11
Contribution of B = 11.0 × 0.81
= 8.91 amu
….(ii)
Adding (i) and (ii) = 1.9 + 8.91 = 10.81 amu
Thus, the average atomic mass of boron is 10.81 amu.
99.
Molecular mass of He = 4
Amount of He = 0.4 g
∴
Number of moles of He =
∵
0.4
= 0.1 mole
4
1 mole of He = 6.023 × 1023 molecules
∴
0.1 mole of He =
0.1 × 6.023 × 1023
1
= 6.023 × 1022 molecules
100. 14 g N3– ions have 8NA valence electrons
∴
4.2 g of N3– ions have valence electrons
8N A × 4.2
=
14
= 2.4NA
34
Some Basic Concepts of Chemistry
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101. 1 cm3 = 0.001 L
∴
11.2 cm3 = 0.001 × 11.2 = 0.0112 L
Number of moles in 11.2 cm3 of H2 is =
0.0112
= 0.0005 mol
22.4
102. Molecular Mass of CH4 = 12 + 4 = 16 g mol−1
Amount of CH4 = 0.032 mg = 3.2 × 10−5g
3.2 ×10−5
= 2 × 10−6 mol
16
1 mole of CH4 contains 6.023 × 1023 molecules of CH4
2 × 10−6 moles of CH4 contains = 2 × 10−6 × 6.023 × 1023 = 12.046 × 1017 mol
Number of moles of CH4 =
∴
103. Ozone has molecular structure as O3
∵
1 mole of ozone = 48 g
∴
0.5 mole of ozone =
0.5 × 48
= 24 g
1
104. 4 g of Helium gas = 22.4 dm3 of volume
22.4 × 2
∴
2 g of Helium gas =
= 11.2 dm3
4
105. Molecular mass of oxygen = 32 g/mol
Amount of oxygen = 16 g
16
∴
Number of moles of oxygen =
= 0.5 mole
32
∵
1 mole of oxygen = 6.023 × 1023 molecules
∴
0.5 moles of oxygen =
6.023 × 1023 × 0.5
= 3.011 × 1023 molecules
1
106. Number of S atoms = 6.023 × 1023 × 0.2 × 8
≈ 9.63 × 1023
107. Gram atomic mass of Au = 197
19700
= 100
197
Number of Au atoms = 100 × 6.023 × 1023 = 6.023 × 1025
108. 1 mole ≡ 6.023 × 1023 electrons
Mass of 1 electron is 9.108 × 10−31 kg
∴
6.023 × 1023 electrons mass is 6.023 × 1023 × 9.108 × 10−31 kg
1
Now, one kg of electron contains =
mole
−31
9.108 × 10 × 6.023 × 1023
1
× 108 mole
=
9.108 × 6.023
Number of moles in 19700 g =
109. ∵ 22400 mL at NTP has 6.023 × 1023 molecule
∴
1 mL at NTP has =
6.023 × 1023
= 0.0002688 × 1023 = 2.69 × 1019 mol
22400
Some Basic Concepts of Chemistry
35
Chemistry (Vol. I)
TARGET Publications
110. ∵ 2.24 L of gas has mass = 4.4 g
∴
4.4
× 22.4 = 44 g
2.24
So, given gas is CO2 because CO2 has molecular mass = 44 g
22.4 L of gas has mass =
114. 1 mole of CH4 contains 4 mole of hydrogen atom i.e., 4 g atom of hydrogen.
115. 1 mole of K4[Fe(CN)6] ≡ 12 × 6 g of carbon
0.5 × 12 × 6
∴
0.5 mole of K4[Fe(CN)6] ≡
g ≡ 36 g of carbon
1
116. ∵ 2 g of hydrogen = 6.023 × 1023 molecules
∴
1 g of hydrogen =
6.023 × 1023
= 3.012 × 1023 molecule.
2
117. Sodium oxide = Na2O
Molecular mass = 46 +16 = 62
∵
62 g of Na2O = 1 mole
∴
620 g of Na2O = 10 moles
118. 1 mole of sucrose contains 6.023 × 1023 molecules.
∵
1 molecule of sucrose has 45 atoms
∴
6.023 × 1023 molecule of sucrose has 45 × 6.023 × 1023 atoms/mole
119. 1 mole of P ≡ 6.023 × 1023 molecules
∵
1 molecule of P4 ≡ 4 atoms
∴
6.023 × 1023 molecules ≡ 4 × 6.023 × 1023 ≡ 24.092 × 1023 atoms
120. 1 mole of CO2 contains 6.023 × 1023 molecules
∵
1 molecule of CO2 ≡ 3 atoms
∴
6.023 × 1023 molecules of CO2 ≡ 3 × 6.023 × 1023 atoms ≡ 1.807 × 1024 atoms
121. Mass of 1 atom of hydrogen =
1
= 1.66 × 10−24 g
23
6.023×10
122. 6.023 × 1023 molecules ≡ 28 g of CO
28 × 2.01 × 1023
∴
2.01 × 1023 molecules ≡
g of CO ≡ 9.3 g of CO
6.023 × 1023
123. 4 g of He ≡ 22.4 L of He at 0 °C and at 1 atm pressure
Weight in g of He
4
∴
Number of moles =
=
= 1 mole
Molecular mass of He
4
124. 6.023 × 1023 molecules = 18 g of water
18 × 1
∴
1 molecule of water =
= 3 × 10–23g = 3×10–26 kg
6.023 × 1023
125. 6.023 × 1023 molecules of N2 ≡ 28 g of N2
28 × 1
∴
1 molecule of N2 ≡
≡ 4.65 × 10–23 g of N2
23
6.023 × 10
36
Some Basic Concepts of Chemistry
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126. Mass of Molecule =
Molar mass in gram
6.023 ×10 23
= Molar mass in amu ×1.66 × 10−24 g
= 842 × 1.66 × 10−24 (∵ Molar mass of C60H122 = 842)
= 1.4 × 10−21
127. 1 L of air = 210 cc of O2
22400 cc = 1 mole
∵
∴
210 cc =
1
× 210 = 0.0093 mol
22400
128. 1 mole of H2 ≡ 22.4 L
∴
2 mole of H2 ≡ 44.8 L
M
M
or M = V; 18 g = 18 mL
129. d = ; 1 =
V
V
∵
6.023 × 1023 molecule of water has volume = 18 cc or cm3
∴
1 molecule of water has volume =
18
= 2.988 × 10–23 ≈ 3× 10–23 cm3
23
6.023 × 10
130. ∵ 22.4 L of a gas at STP has number of molecules = 6.023 × 1023
∴
8.96 L of a gas at STP has number of molecules =
6.023 × 1023 × 8.96
= 2.409 × 1023 = 24.09 × 1022 mol
22.4
131. Na2SO4.10H2O = 2 × 23 + 32 + 4×16 + 10 × 18 = 46 + 32 + 64 + 180 = 322 g
322g Na2SO4.10H2O contains 224g of oxygen
∵
∴
32.2 g Na2SO4.10H2O contains
32.2 × 224
= 22.4 g of oxygen
322
132. Molecular mass of BaCl2.2H2O ≡ 137 + 35.5 × 2 + 2 × 18 ≡ 244 g
∵
244 g of BaCl2.2H2O ≡ 2 mole of water
∴
488 g of BaCl2.2H2O ≡
488 × 2
≡ 4 moles of water
244
133. Gram−molecular mass of CuSO4.5H2O = 250 g
Number of molecules in 250 g (one mole) of CuSO4.5H2O = 6.023 × 1023
Let the mass of 1 × 1022 molecule of CuSO4.5H2O = x g
x
1×1022
So,
=
or
23
250
6.023×10
250 ×1×1022
6.023×1023
x = 4.151 g
x=
134. Molecular mass of CuSO4.5H2O ≡ 249.5 g
∵
249.5 g of CuSO4.5H2O ≡ 90 g of H2O
∴
1000×103 g of CuSO4.5H2O ≡
106 × 90
g of H2O ≡ 360721 g of H2O ≡ 360.7 kg of H2O
249.5
135. 16 g of CH4 = 1mole = 6.023 × 1023 molecules
Some Basic Concepts of Chemistry
37
Chemistry (Vol. I)
TARGET Publications
∵
M
(d = density, M = mass, V = volume)
V
Since d = 1
So, M = V
18 g = 18 mL
18 mL or 18 g of water = NA molecules (NA = avogadro’s number)
∴
1000 mL =
136. d =
NA
× 1000 = 55.55 NA mol
18
137. ∵ 3 moles of oxygen is present in 1 mole of BaCO3
∴
1.5 moles of oxygen is present in BaCO3 =
1
1
× 1.5 =
= 0.5 mol
3
2
138. ∵ (31 × 4 = 124)g of P is present in 220g of P4S3
∴
1.24 g P is present in =
220
× 1.24 = 2.2 g of P4S3
124
2 1
= mole
16 8
4 1
= mole
Similarly, 4 g of sulphur =
32 8
139. 2 g of oxygen contains atom =
140. 200 mg of CO2 = 200 × 10–3 = 0.2 g
∵
44 g of CO2 = 6.023 × 1023 molecules
∴
6.023 × 1023
× 0.2 = 0.0274 × 1023 = 2.74 × 1021 molecules
44
Now 1021 molecules are removed,
So, remaining molecules = 2.74 × 1021 – 1021 = 1021(2.74 – 1) =1.74×1021 molecules
Now, 6.023 × 1023 molecules = 1mole
1 × 1.74 × 1021
= 0.289 × 10–2 = 2.89 × 10–3 moles
1.74 × 1021 molecules =
6.023 × 1023
0.2 g of CO2 =
146. ∵ 40 g of NaOH contains 16 g of oxygen
∴
100 g of NaOH contains =
16
× 100 = 40 % oxygen
40
147. Urea- (NH2 – CO – NH2)
∵
60 g of urea contains 28 g of nitrogen
28
× 100 = 46.66 %
60
≈ 46 %
148. 60g of urea contains 12g of carbon
12
× 100 = 20%
100g of urea contains
60
∴
100 g of urea contains
149. In Fe(CNS)3.3H2O
3 × 18
% of H2O =
× 100 = 19 %
284
38
Some Basic Concepts of Chemistry
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150. 2(NH4)2HPO4
≡ P2O5
2(36 + 1 + 31 + 64) = 62 + 80
264
= 142
(Molecular mass
(Molecular mass of
of the salt)
P2O5)
Molecular mass of P2 O5
142
% of P2O5 =
× 100 =
× 100 = 53.79 %
Molecular mass of salt
264
151. Molecular mass of (CHCOO)2Fe = 170
∴
∴
56
× 100 mg = 32.9 mg
170
32.9 mg of Fe is present in 400 mg of capsule
32.9
% of Fe in the capsule =
× 100 = 8.225 ≈ 8 %
400
Fe present in 100 mg of (CHCOO)2Fe =
152. Molecular mass of CH3COOH ≡ 60
∵
60 g of CH3COOH ≡ 12 × 2 g of carbon
12 × 2
× 100 ≡ 40 %
60
∴
Percentage of carbon ≡
∵
Similarly,
Molecular mass of C6H12O6 ≡ 180
180 g of C6H12O6 ≡ 12 × 6 g of carbon
∴
12 × 6
× 100 ≡ 40 %
180
Therefore, both the compounds have same percentage of carbon.
Percentage of carbon ≡
153. Molecular formula = n(Empirical formula)
= 6(CH2O)
= C6H12O6 (Glucose)
∴
Empirical formula = CH2O
155. C = 24 g, H = 4 g, O = 32 g
Hence number of moles is
24
4
32
= 2;
= 4;
=2
12
1
16
So, Molecular formula = C2H4O2
So, Empirical formula = CH2O
(Simplest formula)
156. CO2 has same molecular and empirical formula.
157. n =
Molecular mass
78
=
=6
Emperical formula mass 13
Molecular formula = n (Empirical formula)
= 6 (CH)= C6H6
158. Empirical formula of the acid is CH2O2
(Empirical formula)n = Molecular formula
n = whole number multiple i.e. 1,2,3,4......
If n = 1; then the molecular formula will be CH2O2
Some Basic Concepts of Chemistry
39
Chemistry (Vol. I)
TARGET Publications
159. ∵ 0.0835 mole of compound contains 1 g of hydrogen
∵
1
= 11.97 ≈ 12 g of hydrogen.
0.0835
12 g of hydrogen is present only in C6H12O6
∴
Molecular formula = C6H12O6
∴
1 mole of compound contains =
160.
∴
Element
Amount (g)
I
254
O
80
Atomic ratio
254
127 = 2
80
16 = 5
Simplest ratio
2
2=1×2=2
5
2 = 2.5 × 2 = 5
Molecular formula of compund is I2O5
161. As percentage of Cu = 39.62 %
S = 20.13 %
∴
Percentage of O = 100 – (39.62 + 20.13) = 40.25 %
Element
% of Element
At. Mass
∴
∵
∴
162.
Atomic ratio
39.62
= 0.63
63
Cu
39.62
63
S
20.13
32
20.13
= 0.63
32
O
40.25
16
40.25
= 2.51
16
Simplest ratio
0.63
=1
0.63
0.63
=1
0.63
2.51
=4
0.63
Empirical formula = CuSO4
159
Molecular mass
n=
=
=1
Empiricalformula mass 159
Molecular formula = (CuSO4) × n = (CuSO4) × 1 = CuSO4
Elements
% Composition
Atomic Mass
A
25
12.5
B
75
37.5
Atomic ratio
25
= 2
12.5
75
= 2
37.5
Simplest ratio
2
=1
2
2
=1
2
Hence, the simplest formula of the compound is AB (1 : 1).
163. % of X = 75.8
∴
% of Y = 100 – 75.8 = 24.2
Element
% Composition
X
75.8
Y
24.2
Atomic ratio
75.8
= 1.011
75
24.2
= 1.513
16
Simplest ratio
1.011
=1
1.011
1.513
= 1.5 ≈ 2
1.011
Thus, empirical formula of the compound is XY2
Hence, the Molecular formula of the compound = XY2
40
Some Basic Concepts of Chemistry
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164. % of oxygen ≡ 40 %
% of metal (M) ≡ 100 – 40 ≡ 60 %
∴
Element
% Composition
Oxygen
40
Metal (M)
60
Atomic ratio
40
16 = 2.5
Simplest ratio
2.5
2.5 = 1
60
24 = 2.5
2.5
2.5 = 1
Empirical formula of the oxide is MO (1 : 1)
165. Percentage of oxygen in M3O4 = 27.6 %
∴
Percentage of M in M3O4 = 100 – 27.6 = 72.4 %
Now, to find out the formula of second oxide, we have to calculate the atomic mass of metal (M).
Let the atomic mass of metal (M) be x
∴
Molecular mass of M3O4 = x + 16 × 4 = x + 64
x
∴
Percentage of M =
× 100 = 72.4 (calculated)
x + 64
x
∴
72.4 =
× 100
x + 64
∴
100x = 72.4x + 64 × 72.4
∴
27.6x = 64 × 72.4
64 × 72.4
∴
x=
= 168
27.6
As, number of metal (M) in M3O4 = 3
168
∴
Atomic mass of M =
= 56
3
Now, we know that percentage of oxygen in second oxide is 30 %
∴
Percentage of metal (M) = 100 – 30
= 70 %
∴
Formula of second oxide is given as follows:
∴
Elements
% Composition
M
70
O
30
Atomic ratio
70
= 1.25
56
30
= 1.875
16
Simplest ratio
1.25
= 1×2 = 2
1.25
1.875
=1.5×2=3
1.25
Simplest formula of the second oxide is M2O3
173. 2K2CrO4 + 2HCl ⎯→ K2Cr2O7 + 2KCl + H2O
174. Ca3P2 + 6H2O ⎯→ 2PH3 + 3Ca(OH)2
Calcium
Phosphide
Phosphine
175. A + 2B ⎯→ C
∵
2 mole of B ≡ 1 mole of C
∴
8 mole of B ≡
8 ×1
≡ 4 mole of C
2
Some Basic Concepts of Chemistry
41
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TARGET Publications
5
O2 ⎯→ 2NO + 3H2O
2
5
(2 × 17) g of NH3 ≡
mole of O2
2
6.8 × 5
6.8 g of NH3 ≡
mole of O2 ≡ 0.5 mole of O2
2 × 17 × 2
176. 2NH3 +
∵
∴
177. K2Cr2O7 + 6FeSO4 + 7H2SO4 ⎯→ 3Fe2(SO4)3 + Cr2(SO4)3 + 7H2O + K2SO4
∴
1 mole of K2Cr2O7 oxidizes 6 mole of FeSO4 completely.
∵
4Al + 3O2 ⎯→ 2Al2O3
(108 g) (96 g)
(204 g)
108 g Al reacts with 96 g of O2
∴
27 g Al will react with =
178.
96 × 27
= 24 g of O2
108
179. 24 g of Mg ≡ 1 mol of MgO
1.2 × 1
∴
1.2 g of Mg forms ≡
≡ 0.05 mol of MgO
24
180. 3BaCl2 + 2Na3PO4 ⎯→ Ba3(PO4)2 + 6NaCl
∵
2 mol of Na3PO4 ≡ 1 mol of Ba3(PO4)2
∴
0.2 mol of Na3PO4 ≡
0.2 × 1
≡ 0.1 mol
2
181. 1 mole of ethanol (C2H5OH) completely burns to form 2 mole of carbon dioxide and 3 mole of water
∵
1 mole of carbon dioxide ≡ 44 g
∴
2 mole of carbon dioxide ≡ 88 g
12 2.63
12 WCO2
×
182. % C =
×
× 100 =
× 100 = 83.6%
44 0.858
44
W
2 1.28
2 WH 2O
×
%H=
× 100 = 16.6 %
×
× 100 =
18 0.858
18
W
∴
∴
Element
% (A)
At.wt. (B)
C
H
83.6
16.6
12
1
Molecular formula is C3H7
C3H7 = (12 × 3) + (7 × 1)
= 36 + 7 = 43 g
183. ROH + CH3MgI ⎯→ CH4 + Mg
1mol
A
B
6.97
16.6
Ratio
Simplest ratio
1×3
2.38 × 3
≈3
≈7
OR
I
1mol = 22400mL
∴
42
1.12 mL is obtained from 4.12 mg
22400 mL will be obtained from
4.12
× 22400 = 82400 mg = 82.4 g
1.12
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Chemistry (Vol. I)
184. N2 + 3H2 ⎯→ 2NH3
(28 g) (6 g)
(34 g)
6 g of H2 produces 34 g of NH3
34 × 200
∴
200 g of H2 produces =
= 1133.3 g
6
3
H2 ⎯→ B + 3HCl;
2
21.6
(Number of moles of B) =
= 2 mole
10.8
3
B ≡ H2
2
3
For 1 mole of B ≡
mole of H2
2
2×3
For 2 mole of B ≡
≡ 3 mole of H2
2
V = 3 × 22.4 L = 67.2 L
185. BCl3 +
∴
∴
186. Mg+2 ≡ H2
Mg2+ + 2HCl ⎯→ MgCl2 + H2↑
∵
24 g of Mg = 1 mole of H2
∴
12 g of Mg =
12
1
=
mole of H2
24
2
3
H2
2
27 g
Hence, volume of hydrogen evolved is
3
× 22.4 = 33.6 L
2
∆
→ CaO + CO2
188. CaCO3 ⎯⎯
(94)
(56) (44)
1 mol of CaCO3 ≡ 56 g of CaO
3 × 56
∴
3 mole of CaCO3 ≡
≡ 168 g of CaO
1
187. H2O + Al + NaOH ⎯→ NaAlO2 +
189. N =
W(g) ×1000
V × Eq.wt.
W(g) × 1000
1500 × 40
0.1 × 1500 × 40
W=
1000
W=6g
0.1 =
∴
∴
190. HNO3 + KOH ⎯→ KNO3 + H2O
12.6
= 0.2 mole; HNO3 ≡ KOH
63
∵
0.2 mole of HNO3 ≡ 0.2 mole of KOH
∴
0.2 × Molecular mass (KOH) ≡ 0.2 × 56
≡ 11.2 g of KOH
Some Basic Concepts of Chemistry
43
Chemistry (Vol. I)
191. N =
TARGET Publications
W(g) × 1000
V × Molecular mass
0.5 × 500 × 100
= 25 g
W(g) =
1000
192. CaCl2 + Ag+ ⎯→ 2AgCl + Ca2+
(110) (108)
(2 × 143.5) (40)
∴
110 g of CaCl2 ≡ 2 × 143.5 g of AgCl
∴
For 14.35 g of AgCl,
110 × 14.35
≡ 5.5 g
Mass of CaCl2 required ≡
2 × 143.5
193. 1 mole of H2SO4 ≡ 32 g of S
100 × 32
∴
100 moles of H2SO4 ≡
≡ 3.2 × 103 g of S
1
194. 50 % HCl itself means 50 g HCl reacts with 100 g sample
50
∴
% Purity =
× 100 = 50 %
100
195. ∵ 22.4 L of O2 ≡ 48 g of O3
∴
67.2 L of O2 ≡
48 × 67.2
≡ 144 g of O3
22.4
203. 22.4 L ≡ 16 g of CH4
16 × 4.48
∴
4.48 L ≡
≡ 3.2 g of CH4
22.4
206. 1 mole of P4 = 6.023 × 1023 atoms
∴
0.1 mole of P4 = 6.023 × 1023 × 0.1 atoms = 6.023 × 1022 atoms
∴
0.1 mole of P4 = 6.023 × 1022 × 4 atoms = 2.4 × 1023 atoms
207. n =
Mass
1 kg
=
= 55.56 mol
−3
Molar mass 18×10 kg mol−1
16 1
=
32 2
14 1
=
14 g of N2 has number of moles =
28 2
Number of moles are same, so number of molecules are same.
208. 16 g of O2 has number of moles =
209.
4Al + 3O2 ⎯→ 2Al2O3
(108 g) (96 g)
(204 g)
∵
54 g Al will react with
210. (A)
∵
∴
44
96 × 54
= 48 g of O2
108
34 g of water :
18 g H2O = 6.023 × 1023 molecules
6.023 × 1023
× 34
18
= 11.38 × 1023 molecules
34 g H2O =
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
(B)
28 g of CO2 :
∵
44 g CO2 = 6.023 × 1023 molecules
∴
28 g CO2 =
(C)
46 g of CH3OH :
∵
32 g CH3OH = 6.023 × 1023 molecules
∴
46 g CH3OH =
(D)
54 g of N2O5 :
∵
108 g of N2O5 = 6.023 × 1023 molecules
∴
54 g of N2O5 =
211. (A)
(B)
∴
(C)
(D)
212. (A)
(B)
(C)
6.023 × 1023
× 28 = 3.83 × 1023 molecules
44
6.023 × 1023
× 46 = 8.658 × 1023 molecules
32
6.023 × 1023
× 54 = 3.01 × 1023 molecules.
108
2 g atom of nitrogen = 28 g
6.023 × 1023 atoms of C has mass = 12 g
12 × 3 × 1023
3 × 1023 atoms of C has mass =
=6g
6.023 × 1023
1 mole of S has mass = 32 g
7.0 g of Ag
So, lowest mass = 6 g of C
6.023 × 1021 molecules of CO2 :
Number of atoms = 3 × 6.023 × 1021 = 18.069 × 1021 atoms
22.4 L of CO2 :
Number of atoms = 6.023 × 1023 × 3 = 18.069 × 1023 atoms
0.44 g of CO2 :
Number of moles =
0.44
×6.023×1023moles
44
= 6.023 × 1021 moles
= 3×6.023× 1021 atoms
= 18.069 × 1021 atoms
213. ∵ 40 g of caustic soda (NaOH) contains 6.023 × 1023 ions of Na
∴
4 × 6.023 × 1023
= 6.023 × 1022 ions
40
Similarly, for 100 mL of 0.5 M Na2SO4
4 g of NaOH =
∵
1000 mL of 1M Na2SO4 = 2 × 6.023 × 1023 ions of Na
∴
100 mL of 0.5M Na2SO4 =
214. (A)
100 × 0.5 × 2 × 6.023 × 1023
= 6.023 × 1022 ions
1000 × 1
2.5 g of N2 molecules :
∵
28 g of N2 ≡ 6.023 × 1023 molecules
∴
2.5 g of N2 ≡
6.023 × 1023 × 2.5
≡ 5.4 × 1022 molecules
28
Some Basic Concepts of Chemistry
45
Chemistry (Vol. I)
TARGET Publications
(B). 4 g of N atoms :
∵
14 g atom of N ≡ 6.023 × 1023 molecules
(D)
6.023 × 1023 × 4
≡ 17.2 × 1022 molecules
14
40 g of N atoms :
14 g atom of N ≡ 6.023 × 1023 molecules
6.023 × 1023 × 40
≡ 172.1 × 1022 molecules
40 g of N atoms ≡
14
84 g of dinitrogen :
∵
28 g of dinitrogen ≡ 6.023×1023 molecules
∴
(C)
∴
∴
215. (A)
(B)
(C)
(D)
4 g atom of N ≡
6.023 × 1023 × 84
≡ 180.7 × 1022 molecules
28
Maximum number of molecules are present in 84 g of dinitrogen.
11 g of CO2 :
44 g of CO2 ≡ 6.023 × 1023 atoms
6.023×1023 ×11
11 g of CO2 =
= 1.505 × 1023 atoms
44
4 g of H2:
1.008 g of H2 ≡ 6.023 × 1023 atoms
6.023×1023 × 4
4 g of H2
=
= 23.900 × 1023 atoms
1.008
5 g of NH3 :
17 g of NH3 ≡ 6.023 × 1023 atoms
6.023×1023 × 5
5 g of NH3 =
= 1.77 × 1023 atoms
17
8 g of SO2 :
48 g of SO2 ≡ 6.023 × 1023 atoms
6.023×1023 × 8
= 1.0 × 1023 atoms
8 g of SO2 =
48
84 g of dinitrogen ≡
216. 44 g of CO2 ≡ 6.023 × 1023 molecules
6.023 × 1023 × 4.4
∴
4.4 g of CO2 ≡
≡ 0.6023 × 1023 molecules
44
Similarly,
22.4 L of H2 ≡ 6.023 × 1023 molecules
6.023 × 1023 × 2.24
≡ 0.6023 × 1023 molecules
∴
2.24 L of H2 ≡
22.4
∴
Total number of molecules present in the container is
= 0.6023 × 1023 + 0.6023 × 1023
= (0.6023 + 0.6023) × 1023
= 1.2046 × 1023 molecules
217. Equivalent of KMnO4 = Equivalent of Fe(C2O4)
x×5=1×3
x = 0.6 mol
46
Some Basic Concepts of Chemistry
Chemistry (Vol. I)
TARGET Publications
218. ∵
∴
28 g of N2 ≡ 22.4 L of N2
14 g of nitrogen ≡
22.4 × 14
≡ 11.2 L of N2
28
220. Number of molecules = n × A
Mass
n=
Molar Mass
∴
∴
∴
Mass of 1 mL of water = 1g
1
n=
18
1
×A
18
1
0.05A
Number of molecules in 1 drop of water =
×A=
mol
18 × 20
18
Number of molecules in 1 mL (20 drops) of water =
221. Equivalent mass =
∴
Molecular mass (M)
Number of e− gain or lost
When KMnO4 is converted in K2MnO4
transfer of 1 electron takes place.
M
Equivalent mass =
=M
1
222. n =
Number of molecules
NA
Also, n =
Volumeof gas at N.T.P.
22.4 L
∴
Number of molecules
Volumeof gasat N.T.P.
=
NA
22.4 L
∴
x
1.12 × 10−10 L
=
6.023 × 1023
22.4 L
∴
x=
1.12 × 10−10 × 6.023 × 1023
= 3.01 × 1012 mol
22.4
223. Molecular mass of nitrogen = 14
Molecular mass of oxygen = 16
∴
Ratio = 14 : 16 = 7 : 8
Some Basic Concepts of Chemistry
47