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Chapter –

1. WHAT IS CHEMISTRY?
“Chemistry is the branch of science that deals with the composition, properties and interaction
of matter”.
While going through this package, if you have a glance at your surroundings you would
observe various substances that have different forms and appearances. Some substances are gases,
some are liquids while the others are solids. Some of them are hard and shinning but others are soft
and dull. You would also observe that different substances behave differently: Iron rusts but gold
does not, copper conducts electricity but sulphur does not. The question now arises is how can these
and a vast number of other observations be explained? Chemistry tries to answer to all these queries
through experiments and formulation of an interpretation or hypothesis that explains the results.
The hypothesis, in turn, can be used to make more predictions and to suggest more experiments
until a consistent explanation or theory of known observation is finally arrived at.
It is important to keep in mind that scientific theories are not laws of nature. All they do is to
represent the best explanations of experimental results that we can come up with at the present
time. Some currently accepted theories will eventually be modified and others may be replaced
altogether if new experiments uncover results that present theories can’t explain.
Anything which has mass and occupies space is known as matter. Water, gold, rocks,
buildings, plants, animals and people are matter. Matter having specific use is termed as material.
For example cement, glass, wood, paper etc. are matter but are also termed as material on account of
their specific use. Solid, liquid and gaseous states represent the three different states of aggregation
and provide a basis of the physical classification of matter. Gaseous state of
matter at very high temperatures containing gaseous ions and free electrons is referred to as the
plasma state.
Description of matter merely in terms of aggregation does not suffice and it becomes
necessary to make use of another property to obtain a more useful classification of matter. One such
useful way studied by you at the secondary stage, is to classify matter into the following categories:
pure substances (elements, compounds) and mixtures. An element consists of only one kind of
atoms. Molecules are identifiable units of matter consisting of two atoms of the same element or of
different elements combined in a definite ratio. Mono, di or polyatomic molecules or extended
structures with the same kind of atoms constitute elements while in a compound two or more than
two different types of atoms are present. Thus a sample of pure hydrogen or oxygen gas is an
example of element whereas water is a compound. Another point of distinction between
an element and a compound is that when two or more elements combine to give a compound they
loose their individual chemical characteristics. Hence, hydrogen and oxygen which are gases at room
temperature lose their chemical identity when they form water whose properties are different from
its constituents. In a mixture, each of the constituents retains its characteristic properties. For
example sugar retains its characteristic sweetness even in its aqueous solution. Mixture may be
homogenous or heterogeneous. Heterogeneous mixtures are those in which the mixing is not
uniform and which therefore have regions of different compositions. Sand with sugar, water with
BASIC CONCEPT ON CHEMISTRY
1
MATTER
2
SOME BASIC CONCEPTS OF CHEMISTRY

gasoline and dust with air are all heterogeneous mixtures. Homogenous mixtures are those in which
the mixing is uniform and which, therefore, have a constant composition throughout. Air is a gaseous
mixture of primarily oxygen and nitrogen, sea water is a liquid mixture of primarily sodium chloride
dissolved in water and brass is a solid mixture of copper and zinc.
Matter
Mixture Pure substances
Homogeneous mixture
e.g.: sugar solution
Heterogeneous mixture
e.g.: sand with sugar, water with gasoline
Elements
e.g.: H2, O2
Compounds
e.g.: H2O
A substance is a distinct type of matter. All samples of a substance have the same properties
throughout. There are two kinds of substances, elements and compounds. Thus pure water is a
familiar example of a substance. All samples of pure water have the same melting and boiling point
whereas sea water is not a substance; it contains both salt and other dissolved substances and water.
3.1 THE ATOMS
The fact that all substances obeyed the laws of chemical combination by mass (explained
later) made the scientists to speculate about the ultimate particles of matter. The most famous of
these speculations is due to John Dalton.
DALTON’S ATOMIC THEORY
The main postulates of this atomic theory are
 Matter is discrete (i.e., discontinuous) and is made up of atoms. An atom is the smallest
(chemically) indivisible particle of an element, which can take part in a
chemical change.
 Atoms of the same element are identical in all respects, size, shape, structure etc. and
especially mass.
 Atoms of different elements have different properties and different masses.
 Atoms can neither be created nor destroyed. So a chemical reaction is nothing but a
rearrangement of atoms and the same number of atoms must be present before and after the
reaction.
 A compound is formed by the union of atoms of one element with atoms of another in a fixed
ratio of small whole numbers (1 : 1, 1 : 2, 2 : 3 etc).
All the postulates of Dalton’s atomic theory have been proved to be incorrect.
(i) Anatomis divisiblein thesense thatithas gotsubatomic particles.
(ii) The existence of isotopes for most elements shows that atoms of the same element need not
have the same mass. The atomic mass of an element is, in fact, a mean of the atomic masses of
the different isotopes of the element.
(iii)Part of atomic mass can be destroyed and an equivalent amount of energy is released during
nuclear fission.
(iv)Atoms combine in fixed integral ratios; however, there are instances where atoms combine
in nonintegral ratios. e.g., in zinc oxide, zinc and oxygen have not combined in exactly
an integral ratio. The atomic ratio of Zn : O = (1 + x) : 1, where x is a very small fraction.
Compounds of this kind are called nonstoichiometric compounds or Berthollide
THE CONCEPTS OF ATOMS AND MOLECULES
3

compounds as against compounds whose formulae are in accordance with atomic theory and
Proust’s law of definite proportions.
However, these aspects have not affected the basic philosophy of Dalton.
3.2 ATOMIC MASS
Dalton gave the idea of atomic mass in relative terms, that is, the average mass of one atom
relative to the average mass of the other. The relative atomic masses were referred to as the atomic
masses. It was found that 1.00 g of hydrogen combines with 8.0 g of oxygen to form water. In order
to find out atomic mass of oxygen (relative to hydrogen) it is required to know the relative numbers
of hydrogen and oxygen atoms in water. However during Dalton’s time it was not known how many
hydrogen and oxygen atoms are present in a water molecule. Now we know that a water molecule
has two hydrogen and one oxygen atom. So, atomic mass of oxygen on the hydrogen scale is given by
hydrogen
of
atom
one
of
mass
lative
Re
oxygen
of
atom
one
of
mass
lative
Re
= 16.
Thus, the atomic mass of oxygen relative to hydrogen is 16.
A scale based on oxygen, as it was considered more reactive forming a large number of
compounds, eventually replaced Dalton’s hydrogen based scale. Hence,
Atomic mass of an element =
oxygen
of
atom
1
of
mass
16
1
element
the
of
atom
1
of
mass

For a universally accepted atomic mass unit in 1961, Carbon12 isotope was selected as
standard. This scale depends on measurements of atomic mass by a mass spectrometer.
We can make accurate measurements of mass on this instrument by comparing mass of an atom
with the mass of a particular atom chosen as the standard and is arbitrarily assigned a mass of
exactly 12 atomic mass unit.
Atomic mass of an element =
12
carbon
of
atom
1
of
mass
12
1
element
the
of
atom
1
of
mass


One atomic mass unit (amu) is therefore, a mass unit equal to exactly one twelfth the mass of
a carbon12 atom. However, the new symbol ‘u’ (unified mass) is used now a days in place of amu.
Before moving forward, certain facts about atomic mass are summarized below:
 Atomic mass is not a mass but a number.
 Atomic mass is not absolute but relative to the mass of the standard reference
element (C12).
 Gram atomic mass is atomic mass expressed in grams, but it has a special significance with
reference to a mole (which is discussed later).
3.3 AVERAGE ATOMIC MASS
A particular element may consist of several isotopes with different atomic masses. For such
species the atomic mass calculated is the average relative atomic mass. The average relative atomic
mass depends upon the isotopic composition or fractional abundance i.e. fraction of the total number
of atoms that is composed of that particular isotope evaluated through mass spectrometer. Thus, the
average relative atomic mass of Neon, whose fractional abundance is known can be evaluated as
Average atomic mass =
100
abundance)
percentage
mass
molar
Isotopic
( 

Isotope Fractional Abundance
20Ne 0.9051
21Ne 0.0027
22Ne 0.0922

Average atomic mass of Ne = (20 × 0.9051) + (21 × 0.0027) + (22 × 0.0922) = 20.1871 u.
3.4 THE MOLECULE
Avogadro suggested that the fundamental chemical unit is not an atom but a molecule, which may
be a cluster of atoms held together in some manner causing them to exist as a unit. The term molecule
means the smallest particle of an element or a compound that can exist free and retain all its
properties.
Consider a molecule of sulphur dioxide. It has been established that it contains one atom of
sulphur and two atoms of oxygen. This molecule can be split up into atoms of sulphur and oxygen. So
the smallest particle of sulphur dioxide that can exist free and retain all its properties is the molecule
of sulphur dioxide. A compound molecule should contain at least 2 different atoms.
The term molecule is also applied to describe the smallest particle of an element which can
exist free. Thus a hydrogen molecule is proved to contain 2 atoms; when it is split up into atoms,
a change in properties is observed (You may know that nascent hydrogen which may be thought of
as atomic hydrogen is a more powerful reducing agent than ordinary hydrogen).
Molecules of elementary gases like hydrogen, oxygen, nitrogen, chlorine, etc., contain
2 atoms in a molecule; they are diatomic. Molecules of noble gases like helium, neon, argon, krypton
and xenon are monoatomic. Molecules of phosphorus contain 4 atoms (tetratomic) while those of
sulphur contain 8 atoms.
The number of atoms of an element in a molecule of the element is called its atomicity.
3.5 MOLECULAR MASS
It is the number of times a molecule is heavier than
12
1
th the mass of an atom of C12.
Molecular mass =
atom
12
C
one
of
mass
12
1
molecule
one
of
mass


 Molecular mass is not a mass but a number.
 Molecular mass is relative and not absolute.
 Molecular mass expressed in grams is called grammolecular mass.
 Molecular mass is calculated by adding all the atomic masses of all the atoms in
a molecule. Thus, the molecular mass of oxygen which contains 2 atoms in a molecule
would be (2  16) = 32. The molecular mass of carbon dioxide, which contains 1 atom of
carbon and 2 atoms of oxygen would be [12 + (2  16)] = 44. Molecular mass of sulphuric
acid, which contains 2 atoms of hydrogen, 1 atom of sulphur and 4 atoms of oxygen is
[(2  1) + (1  32) + (4  16)] = 98.
 Molecular mass is now called relative molecular mass.
Note: The formula mass of a substance is the sum of the atomic masses of all atoms in a
formula unit of the compound. Sodium chloride, NaCl is an ionic substance and we do not talk about
its molecular mass. Formula mass of NaCl = 58.5 u (for Na = 23.0 u and Cl = 35.5 u).

Key points
1. Dalton’s atomic theory considers the matter to be made up of atoms.
2. 1 amu is defined as th
12
1
of the mass of a C12
atom.
3. Average atomic mass =
100
abundance)
percentage
mass
atomic
Isotopic
( 

4. Molecule is the smallest particle of an element or a compound that can exist free and retain all its properties.
5. The number of atoms of an element in a molecule is called its atomicity.
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 which are known as laws of
chemical combination. These are as follows:
1. Law of conservation of mass.
2. Law of definite proportion.
3. Law of multiple proportion.
4. Gay Lussac’s law of gaseous volumes.
5. Avogadro’s law.
4.1 LAW OF CONSERVATION OF MASS
The great French chemist, Antoine Laurent Lavoisier established that when combustion is
carried out in a closed container ,the mass of the combustion products was exactly equal to the mass
of the consumed reactants. For example, when hydrogen gas burns and combines with oxygen to
yield water (H2O ), the mass of the water formed is equal to the mass of the hydrogen and oxygen
consumed. In other words, mass is neither created nor destroyed in chemical reaction.
Illustration 1
Question: When 4.2 g of NaHCO3 is added to a solution of acetic acid (CH3COOH) weighing 3 g,
it is observed that 2.2 g of CO2 is released into the atmosphere. The mass of
CH3COOH and H2O left behind is found to weigh 5g. Show that these observations
are in agreement with the law of conservation of mass, assuming the reactants are
completely consumed.
Solution: The reaction can be expressed by
CH3COOH + NaHCO3  CH3COONa + H2O + CO2 
Residue
Net weight of reactants = weight of CH3COOH + weight of NaHCO3
= 3 g + 4.2 g = 7.2 g
Net weight of products = weight of residue + weight of CO2
= 5 g + 2.2 g = 7.2 g
Thus, the net weight of reactants = weight of products.
Hence, the law of conservation of mass holds good.
4.2 LAW OF DEFINITE PROPORTION OR CONSTANT COMPOSITION
Further investigations in the decades following Lavoisier led the French chemist Joseph
Proust formulate a second fundamental chemical principle that we now call the law of definite
proportion, which states that a sample of pure compound, always consists of the same elements
combined together in the same proportions by mass. For example one molecule of ammonia NH3
LAWS OF CHEMICAL COMBINATION
4

always contains one atom of nitrogen and three atoms of hydrogen or 14.0 grams of nitrogen and 3.0
grams of hydrogen.
Illustration 2
Question: 6.488 g lead combines directly with 1.002 g of oxygen to form lead dioxide (PbO2).
Lead dioxide is also produced by heating lead nitrate and it was found that the
percentage of oxygen present in lead dioxide is 13.38 percent. Use these data to
illustrate the law of constant composition.
Solution: To verify the law of constant composition the percentage of oxygen or lead in the lead
dioxide obtained from both the experiments should be constant. So, moving stepwise,
lets evaluate the percentage of oxygen in experiment 1.
In experiment 1,
Weight of lead = 6.488 g
Weight of lead dioxide = mass of lead + mass of oxygen = 6.488 + 1.002 = 7.490 g (In
accordance with the law of conservation of mass)
 Percentage of oxygen in the dioxide thus formed
= 100
dioxide
lead
of
weight
oxygen
of
weight
 = 100
490
.
7
002
.
1
 =13.38%
Now, the percentage of oxygen present in lead dioxide obtained from experiment 2 is
13.38 % (given).
Since, the percentage composition of oxygen in both the samples of PbO2 is identical, the
above data illustrates the law of constant composition.
4.3 LAW OF MULTIPLE PROPORTIONS
This law was studied by Dalton and may be defined as when two elements combine to form
two or more chemical compounds, then the ratio of the mass of one of these elements, which
combine with a fixed mass of the other bears a simple ratio to one another.
For example, carbon combines with oxygen to form two compounds, namely, carbon dioxide and
carbon monoxide.
C +
2
1
O2  CO
12g 16g 28g
C + O2  CO2
12g 32g 44g
In carbon dioxide, 12 parts by weight of carbon combine with 32 parts by weight of oxygen
while in carbon monoxide, 12 parts by weight of carbon combine with 16 parts by weight of oxygen.
Therefore, the weights of oxygen which combine with a fixed weight of carbon (12 parts) in carbon
monoxide and carbon dioxide are 16 and 32 respectively. These weights of oxygen bear a simple
ratio of 16 : 32 or 1 : 2 to each other.
Illustration 3
Question: Two oxides of a certain metal were separately heated in a current of hydrogen
until constant weights were obtained. The water produced in each case was
carefully collected and weighed. 2 grams of each oxide gave respectively, 0.2517
gram and 0.4526 gram of water. Show that these results establish the law of
multiple proportion.
Solution: To verify the law of multiple proportion in each case. The weight of oxygen combining
with the fixed weight of the metal in the two different oxides should bear a simple ratio
to one another. So, to prove this, we move stepwise as follows:
Step 1. Calculate the weight of oxygen in each oxide.

Here, we are given :
Weight of each oxide = 2.0 g
Weight of water produced in case I = 0.2517 g
Weight of water produced in case II = 0.4526 g
18 g of H2O  16 g of oxygen
i.e. 18 g of water contain oxygen = 16 g
 0.2517 g of water contains oxygen = 2517
.
0
18
16
 g = 0.2237 g
and 0.4526 g of water contains oxygen = 4526
.
0
18
16
 g = 0.4023 g
Step 2. Calculate the weight of oxygen which would combine with 1 g of metal in each oxide.
In case I, Weight of metal oxide = 2g and weight of oxygen = 0.2237 g
 Weight of metal = 2 – 0.2237 = 1.7763 g
 Weight of oxygen which combines with 1.7763 g of metal = 0.2237 g
 Weight of oxygen which combines with 1 g of metal =
7763
.
1
2237
.
0
g = 0.1259 g
In case II, Weight of metal oxide = 2g and weight of oxygen = 0.4023 g
 Weight of metal = 2 – 0.4023 = 1.5977 g
Weight of oxygen which combines with 1.5977 g of metal = 0.4023 g
 Weight of oxygen which combines with 1 g of metal =
5977
.
1
4023
.
0
g = 0.2518 g
Step 3. Compare the weights of oxygen which combine with the same weight of metal in the
two oxides.
The weights of oxygen which combine with 1 g of metal in the two oxides are respectively
0.1259 g and 0.2518 g. These weights are in the ratio 0.1259 : 0.2518 or 1 : 2.
Since this is a simple ratio, so the above results establish the law of multiple proportions.
4.4 GAY LUSSAC’S LAW OF COMBINING VOLUMES
Gay Lussac investigated a large number of chemical reactions occurring in gases and as a
result of his experiments ,Gay Lussac found that there exists a definite relationship among the
volumes of gaseous reactants and products. Hence, he put forward a generalization known as the
Gay Lussac’s Law of combining volumes. This may be stated as follows when gases react they do so
in volumes which bear a simple ratio to one another and to the volumes of products, if these are
also gases, provided all measurements are done under similar conditions of temperature and
pressure.
Example: One volume of hydrogen and one volume of chlorine always combine to form two
volumes of hydrogenchloride.
H2(g) + Cl2 (g) 2HCl(g)
1vol 1vol 2vol
The ratio between the volumes of the reactants and the products in this reaction is simple i.e. 1:1:2.
Hence, it illustrates the law of combining volumes.
4.5 AVOGADRO’S LAW
It states that equal volumes of gases at the same temperature and pressure contain
equal number of molecules. It means that 1 ml of hydrogen, oxygen, ammonia, or a mixture of gases
taken at the same temperature and pressure contains the same number of molecules. Avogadro’s law
can prove that simple elementary gas molecules like hydrogen and oxygen are diatomic.
Consider the experimental result,

1 volume of hydrogen + 1 volume of chlorine  2 volumes of hydrogen chloride at the
same temperature and pressure.
Let 1 volume contains ‘n’ molecules. Then ‘n’ molecules of hydrogen and ‘n’ molecules of
chlorine gives ‘2n’ molecules of hydrogen chloride.
Canceling the common ‘n’, we have 1 molecule of hydrogen and 1 molecule of chlorine gives 2
molecules of hydrogen chloride.
A molecule of hydrogen chloride should contain at least 1 atom of hydrogen and
1 atom of chlorine. Two molecules of hydrogen chloride should contain at least
2 atoms of hydrogen and 2 atoms of chlorine and these should have come from
1 molecule of hydrogen and 1 molecule of chlorine respectively. Thus Avogadro’s hypothesis
enables us to establish that hydrogen and chlorine molecules must contain
at least 2 atoms.
Key points
1. Law of conservation of mass: Mass is neither created nor destroyed in a chemical reaction.
2. Law of definite proportion: A sample of pure compound always consists of the same elements combined
together in the same proportions by mass.
3. Law of multiple proportion: When two elements combine to form two or more chemical compounds, then
the ratio of mass of one of these elements which combine with a fixed mass of the other bears a simple
ratio to one another.
4. Gay Lussac’s law of gaseous volumes: When gases react they do so in volumes which bear a simple
ratio to one another and to the volumes of products, if these are also gases, provided all measurements
are done under similar conditions of temperature and pressure.
5. Avogadro’s law: Equal volumes of gases at the same temperature and pressure contain equal number
of molecules.
A formula is a symbolic representation of a molecule of a substance, which tells us the
number and kinds of atoms of various elements present in its molecule. e.g. formula of sulphuric acid
is H2SO4 i.e. each molecule of sulphuric acid consists of two atoms of hydrogen, one atom of sulphur
and four atoms of oxygen.
5.1 CALCULATION OF PERCENTAGE COMPOSITION FROM FORMULA
The percentage of any element or constituent in a compound is the number of parts by mass
of that element or constituent present in 100 parts by mass of the compound. It can be determined
by the following two steps
1. Calculation of the molecular mass
Calculate the molecular weight of the compound from its formula by adding the atomic
masses of the elements present.
2. Calculation of the percentage composition of the constituents
The percentage can be calculated by using the following relation:
% of the element =
compound
the
of
weight
Molecular
100
element
the
of
mass
by
parts
of
.
No 
Illustration 4
Question: Fe2(SO4)3 is used in water and sewage treatment to aid the removal of suspended
impurities. Calculate the mass percentage of iron, sulphur and oxygen in this
compound.
Solution: Step 1:
Molecular weight of Fe2(SO4)3 = (56 × 2) + (32 × 3) + (16 × 12) = 400
PERCENTAGE COMPOSITION AND MOLECULAR FORMULA
5

Step 2:
% of Fe = 100
)
SO
(
Fe
of
weight
Molecular
Fe
of
weight
by
parts
of
Numer
3
4
2

= 100
400
2
56


= 28%.
% of S = 100
)
SO
(
Fe
of
weight
Molecular
S
of
weight
by
parts
of
Numer
3
4
2

= 100
400
3
32


= 24%.
% of O = 100
)
SO
(
Fe
of
weight
Molecular
O
of
weight
by
parts
of
Numer
3
4
2

= 100
400
12
16


= 48%.
5.2 EMPIRICAL AND MOLECULAR FORMULA
Whereas, molecular formula of the compound represents the true formula of a compound;
the empirical formula of a compound is the simplest formula that shows the ratio of the
number of atoms of each kind in the compound. For example, the molecular formula of
hydrogen peroxide is H2O2, showing that a molecule of hydrogen peroxide consists of two
hydrogen atoms and two oxygen atoms or the ratio of number of hydrogen atoms to that of
oxygen atoms is 2 : 2 or 1 : 1. Therefore, the simplest or empirical formula for hydrogen
peroxide is HO.
5.2.1 Calculation of the empirical formula
The empirical formula of a chemical compound can be deduced from a knowledge of the
(a) percentage composition of different elements.
(b) atomic masses of the elements.
1. Calculation of the relative number of atoms or atomic ratio
Divide the percentage of each element by its atomic mass. This gives the relative number
of atoms or the atomic ratio of the various elements present in one molecule of the
compound.
Atomic ratio =
element
same
the
of
mass
Atomic
element
an
of
Percentage
.
2. Evaluating the simplest atomic ratio
Divide the atomic ratio obtained in step 1 by the smallest quotient or the least value from
amongst the values obtained for each element. This gives the simplest atomic ratio.
3. Calculate the simplest whole number ratio
The simplest atomic ratios as calculated in step 2 are generally whole numbers.
If they are not, then
(a) round off the values to the nearest whole number.
(b) multiply all the simplest atomic ratios by a suitable integer.
4. Deducing the empirical formula
Write the symbols of the various elements side by side. Now insert the numerical value of
the simplest whole number ratio of each element as obtained in step 3 at the lower right
hand corner of each symbol. This gives the empirical formula of the compound.
Illustration 5
Question: An inorganic salt gave the following percentage composition: Na = 29.11%,
S = 40.51% and O = 30.38%. Calculate the empirical formula of the salt.

Solution: Calculation of empirical formula.
Element Symbol % of element
At. mass
of
element
Relative no. of
atoms =
mass
Atomic
Percentage
Simplest
atomic ratio
Simplest
whole
number
atomic
ratio
Sodium Na 29.11 23
23
11
.
29
= 1.266
266
.
1
266
.
1
= 1
2
Sulphur S 40.51 32
32
51
.
40
= 1.266
266
.
1
266
.
1
= 1
2
Oxygen O 30.38 16
16
38
.
30
= 1.898
266
.
1
898
.
1
= 1.49
~

 1.5
3
Thus, the empirical formula is Na2S2O3.
5.2.2 Calculation of the molecular formula
The molecular formula of a compound can be deduced from its
(a) empirical formula
(b) molecular mass
The determination of molecular formula involves the following steps :
(i) Calculation of the empirical formula from its percentage composition.
(ii) Calculation of empirical formula mass by adding the atomic masses of all the atoms
present in the empirical formula.
(iii)Determination of the molecular mass of the compound from the given data.
(iv)Determination of the value of ‘n’ by using the relation,
n =
mass
formula
Empirical
mass
Molecular
, where n is any integer such as 1,2,3…… etc.
(v) Determination of the molecular formula by using the relation
Molecular formula = n × empirical formula.
Chemical stoichiometry deals with the determination of quantities of reactants or products of
a chemical reaction. The word “stoichiometry” is derived from greek word “stoicheion” meaning
element and “metron” meaning measure.
6.1 THE MOLE
The concept of amount of a substance is confined to the chemical measurements. The amount
of substance of a system is proportional to the number of elementary entities (which may be atoms
or molecules or ions or specified group of such particles) of that substance present in the system.
Let us take elements Ag, Mg and Hg with masses equal to their atomic masses in grams, and
then to our surprise, each element contains equal number of atoms. This is not only limited to atoms
but also applicable to the molecules. For example, if we have molecules like CO2, NO2 and SO2 with
masses equal to their molecular masses in grams, then they would also contain equal number of
molecules. This specified number of atoms or molecules is referred to as a “mole”.
Thus, a system containing a specified number (6.023  1023) of elementary entities is said to
contain 1 mole of the entities. Thus 1 mole of an iron sample means that the sample contains 6.023 
1023 atoms of iron. Similarly, 1 mole of NaCl crystal contains 6.023  1023 ion pairs (Na+Cl
).
STOICHIOMETRY
6

This specific number 6.023  1023 elementary entities is called Avogadro number (NAV). The
SI unit for amount of substance is the mole. One mole is defined as the amount of a substance that
contains as many entities (atoms, molecules or other particles) as there are in exactly 12 g of 12C
isotope.
Number of moles of species =
Number
Avogadro
species
that
of
molecules
/
atoms
of
.
No
The mass of specific number (6.023  1023) of elementary entities is equal to atomic mass for
atoms and molecular mass for molecules.
Let ‘M’ g/mole be the molecular mass of a species. Thus ‘M’ g be the mass of 1 mole
(equal to the mass of 6.023  1023 molecules) of the species. Then, ‘x’ g of the species contain






 x
M
1
mole. Hence
Number of moles of a species =
M
w
(g/mole)
mass
molecular
or
Atomic
(grams)
taken
mass

It is also known that one mole of a gas at STP occupies a volume of 22.4 litres. Thus, if a gas
occupies ‘x’ L at STP, then the number of moles of the gas can be calculated by dividing the actual
volume occupied by the gas at STP with the volume occupied by 1 mole of the gas at STP.
Thus, number of moles of a gas =
STP
at
gas
the
of
mole
1
by
occupied
Volume
STP
at
gas
by
occupied
Volume
.
The volume of gas and the number of moles of gas at temperature and pressure other than
the STP can be related by ideal gas equation, PV = nRT.
Illustration 6
Question: Calculate the number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He
Solution: (i) 1 mole of Ar contains 6.023 × 1023 atoms
 52 mole of Ar contains atoms
= 52 × 6.023 × 1023
= 3.132 × 1025 atoms.
(ii) 1 atom of He contains 4 u
 52 u of He contains = 1
4
52
 = 13 atoms.
(iii) 1 mol of He  4 g = 6.023 × 1023 atoms
 52 g of He contains atoms = 23
10
023
.
6
g
4
g
52

 = 7.8299 × 1024 atoms.
6.2 BALANCING A CHEMICAL EQUATION
According to the law of conservation of mass, a balanced chemical equation has the same
number of atoms of each element on both sides of the equation. Many chemical equations can be
balanced by trial and error. Let us take the reactions of a few metals and nonmetals with oxygen to
give oxides
4Fe(s) + 3O2(g)  2Fe2O3(s) (a) balanced equation
2Mg(s) + O2(g)  2MgO(s) (b) balanced equation
P4(s) + O2(g)  P4O10(s) (c) unbalanced equation

Equations (a) and (b) are balanced since there are same number of metal and oxygen atoms
on each side of equations. However, equation (c) is not balanced. In this equation, phosphorus atoms
are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on the left of
oxygen on the left side of the equation to balance the oxygen atoms appearing on the right side of the
equation.
P4(s) + 5O2(g)  P4O10(s) balanced equation
Now let us take combustion of propane, C3H8. This equation can be balanced in steps.
Step 1. Write down the correct formulae of reactants and products. Here propane and
oxygen are reactants and carbon dioxide and water are products.
C3H8(g) + O2(g)  CO2(g) + H2O(l) unbalanced equation
Step 2. Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore,
three CO2 molecules are required on the right side.
C3H8(g) + O2(g)  3CO2(g) + H2O(l) unbalanced equation
Step 3. Balance the number of H atoms: Since on the left there are 8 hydrogen atoms in the
reactants however, each molecule of water has two hydrogen atoms, so four molecules of water will
be required for eight hydrogen atoms on the right side.
C3H8(g) + O2(g)  3CO2(g) + 4H2O(l) unbalanced equation
Step 4. Balance the number of O atoms: There are ten oxygen atoms on the right side
(3 × 2 = 6 in CO2 and 4 × 1 = 4 in H2O). Therefore, five O2 molecules are needed to supply the required
ten oxygen atoms.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(l)
Step 5. Verify that the number of atoms of each element is balanced in the final equation. The
equation shows three carbon atoms, eight hydrogen atoms and ten oxygen atoms on each side.
All equations that have correct formulae for all reactants and products can be balanced.
Always remember that subscripts in formulae of reactants and products cannot be changed to
balance an equation.
6.3 QUANTITATIVE INFORMATION FROM A BALANCED CHEMICAL EQUATION
Let us now consider a balanced chemical equation as
MnO2 + 4HCl  MnCl2 + 2H2O + Cl2
The quantitative information drawn from this balanced chemical equation is
(a) The molar ratio in which the two reactants (MnO2 and HCl) are reacting is 1 : 4.
(b) The molar ratio between any two products can also be known i.e. moles of H2O
produced would be double the moles of MnCl2 produced.
(c) The initial moles of MnO2 and HCl to be taken in vessel for the reaction to occur not
necessarily be 1 and 4 respectively (or in the molar ratio of 1 : 4).
(d) We can start reaction with MnO2 and HCl taken in any molar ratio, but moles of the two
reacting will always be in the molar ratio of 1 : 4.
(e) The balanced chemical equation should follow the law of conservation of mass.
6.4. STOICHIOMETRIC CALCULATIONS
Let us consider the same chemical system as considered above with initial composition
(in terms of mole) as .
n
and
,
n
,
n
,
n
,
n 2
2
2
2 Cl
O
H
MnCl
HCl
MnO





Let the

HCl
n is four times that of .
n 2
MnO

When the reaction occurs, these mole numbers change as the reaction progresses. The mole
numbers of the various species do not change independently but the changes are related by the
stoichiometric coefficients in the chemical equation. Let after time ‘t’ from the commencement of the
reaction, the moles of MnO2 reacting be ‘x’, then the moles of HCl reacting in the same time interval
be ‘4x’ since MnO2 and HCl react in the molar ratio of 1 : 4.

Thus, after time t, the composition of the system would be
2
MnO
n = 
2
MnO
n  x
nHCl =

HCl
n  4x
2
MnCl
n = 
2
l
MnC
n + x
O
H2
n = 
O
H2
n + 2x
2
Cl
n = 
2
l
C
n + x
The algebric signs,  and + indicates that the reactants are consumed and the products are
produced.
In general, mole numbers of various species at any time would be given as
t
n = 
i
n + 
i
 x
where 
i
n is the initial amount, x is the degree of advancement and i
 is the stoichiometric
coefficient which will be given a negative sign for the reactants and a positive sign for the products.
After long time interval from the commencement of reaction i.e. after infinite time, i.e., when
the reaction is 100% complete, the composition of the system would be
2
MnO
n = o, HCl
n = 0
2
Cl
n
M
n = 
2
MnCl
n + 
2
MnO
n = 
2
MnCl
n +
4
nHCl

O
H2
n = 
O
H2
n + 
2
nO
M
n
2 = 
O
H2
n +
2
nHCl

2
Cl
n = 
2
Cl
n + 
2
MnO
n = 
2
Cl
n +
4
nHCl

6.5 THE LIMITING REAGENT
Let the initial moles of MnO2 and HCl be 
2
MnO
n and

HCl
n respectively and

HCl
n  .
n
4 2
MnO

Thus,
in the given chemical reaction, after infinite time, one of the reactant will be completely consumed
while the other would be left in excess. Thus, the reactant which is completely consumed when a
reaction goes to completion and which decides the yield of the product is called the limiting reagent.
For example, if in the given case

HCl
n > ,
n
4 2
MnO

and there is no MnCl2 and H2O in the
beginning, then
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O
Initially, t = 0 
2
MnO
n

HCl
n 0 0 0
At t =  0

HCl
n  
2
MnO
n
4

2
MnO
n

2
MnO
n

2
MnO
n
2
Thus, MnO2 is the limiting reagent and the yield of all the products is governed by the amount
of MnO2 taken initially.
Similarly, if in the given case

HCl
n < 
2
MnO
n
4 and no MnCl2, Cl2 and H2O are present initially, then
MnO2 + 4HCl  MnCl2 + Cl2 + 2H2O

Initially, t = 0 
2
MnO
n

HCl
n 0 0 0
t =  
2
MnO
n 
4
nHCl

0
4
nHCl

4
nHCl

2
nHCl

Here, HCl would become the limiting reagent and the products yield are decided by the
amount of HCl taken initially.
6.6 ANALYSIS BASED ON MASS AND VOLUME RELATIONSHIPS
This analysis section is broadly classified into three heads.
(a) Massmass relationship
(b) Massvolume relationship and
(c) Volumevolume relationship
6.6.1 Massmass Relationship
This relates the mass of a species (reactant or product) with the mass of another species
(reactant or product) involved in a chemical reaction.
Let us consider a chemical reaction,
CaCO3(s) 


CaO(s) + CO2(g).
Let the mass of CaCO3 taken be ‘x’ g and we want to calculate the mass of CaO obtained by
heating ‘x’ g CaCO3. Then the moles of CaCO3 taken would be
1
M
x
(where M1 represents the molar
mass of CaCO3). According to the balanced reaction, the molar ratio of CaCO3 and CaO is 1 : 1, so same
number of moles 







1
M
x
of CaO would be formed. Now for converting the moles of CaO into mass of
CaO obtained, we need to multiply the moles of CaO with the molar mass of CaO.
Let the molar mass of CaO be M2, so the mass of CaO obtained by heating ‘x’ g of CaCO3 would
be .
g
M
M
2
1









x
6.6.2 Massvolume Relationship
This establishes the relationship between the mass of a species (reactant or product) and the
volume of a gaseous species (reactant or product) involved in a chemical reaction.
Let us take ‘x’ g of CaCO3 in a vessel of capacity ‘V’ L and the vessel is heated so that CaCO3
decomposes as
CaCO3(s) 


CaO(s) + CO2(g)
We want to find out the volume of CO2 evolved at STP by heating ‘x’ g of CaCO3. Then
Moles of CaCO3 =
1
M
x
Moles of CO2 evolved =
1
M
x
(since molar ratio of CaCO3 and CO2 is 1 : 1)
 Volume of CO2 evolved at STP = L
4
.
22
M1









x
But, if the volume of CO2 evolved is to be calculated at pressure ‘P’ atm and temperature ‘T’ K.
Then, Volume of CO2 evolved at pressure ‘P’ and temperature ‘T’ =
P
RT
M1

x
(Using PV = nRT)

Illustration 7
Question: Oxygen is prepared by catalytic decomposition of potassium chlorate (KClO3).
Decomposition of potassium chlorate gives potassium chloride (KCl) and oxygen
(O2). How many moles and how many grams of KClO3 are required to produce
2.4 mole O2? g/mole)
122.5
(M 3
KClO 
Solution: Decomposition of KClO3 takes place as
2KClO3(s)  2KCl(s) + 3O2(g)
2 mole KClO3 gives 3 mole of O2
 3 mole O2 is formed by 2 mole of KClO3.
 2.4 mole O2 will be formed by 





 4
.
2
3
2
mole KClO3 = 1.6 mole KClO3.
Mass of KClO3 = Number of moles × Molar mass = 1.6 × 122.5 = 196 g.
Illustration 8
Question: If 20 g of CaCO3 is treated with 20 g of HCl, how many grams of CO2 can be
generated according to the following equation?
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
Solution: CaCO3(s) + 2HCl(aq)  CaCl2(aq) + H2O(l) + CO2(g)
1 mol 2 mol 1 mol
100 g 73 g 44 g
Let CaCO3(s) be completely consumed in the reaction.
 100 g CaCO3 gives 44 g CO2.
 20 g CaCO3 will give g
20
100
44
 CO2 = 8.8 g CO2.
Let HCl be completely consumed
 73 g HCl give 44 g CO2
 20 g HCl will give 2
CO
g
20
73
44
 = 12.054 g CO2.
Since, CaCO3 gives least amount of product CO2, hence CaCO3 is the limiting reactant.
Illustration 9
Question: For the reaction,
CaO + 2HCl  CaCl2 + H2O
1.23 g of CaO is reacted with excess of hydrochloric acid and 1.85 g of CaCl2 is
formed. What is the percent yield?









 100
lly
theoretica
produced
product
the
of
mass
actually
produced
product
the
of
mass
yield
%
Solution: The balanced equation is
CaO + 2HCl  CaCl2 + H2O
1 mol 1 mol
56 g 111g
56 g of CaO produce CaCl2 = 111 g

1.23 g of CaO produce CaCl2 = 23
.
1
56
111
 = 2.438
Thus, theoretical yield = 2.438 g
Actual yield = 1.85 g
Percent yield = 100
438
.
2
85
.
1
 = 75.88%.
6.6.3 Volumevolume Relationship
This relationship deals with the volume of a gaseous species (reactant or product) with the
volume of another gaseous species (reactant or product) involved in a chemical reaction.
Let us consider the reaction, N2(g) + 3H2(g)  2NH3(g). We are given ‘x’ L of N2 at pressure
‘P’ atm and temperature ‘T’ K and we want to know the volume of H2 required to react with it at
another pressure P atm and temperature T K, then
Moles of N2 =
RT
Px
Moles of H2 required =
RT
P
3 x


 Volume of H2 required at pressure P atm and temperature T K
= L
T
P
PT
3
P
RT
RT
P
3















 x
x
Illustration 10
Question: 1 litre mixture of CO and CO2 is taken. This is passed through a tube containing red
hot charcoal. The volume now becomes 1.6 litre. The volumes are measured under
the same conditions. Find the composition of mixture by volume.
Solution: Let there be ‘x’ ml of CO in the mixture. Hence there will be (1000  x)ml CO2. The reaction of
CO2 with red hot charcoal may be given as,
CO2(g) + C(s)  2CO(g)
1000  x
– 2000  2x
Total volume of the gas becomes = x + 2(1000  x)
x + 2000  2x = 1600
x = 400 ml
 Volume of CO = 400 ml
and volume of CO2 = 600 ml.
A solution is defined as a homogenous mixture of two or more chemically non-reacting
substances, the relative amount of which can be varied upto a certain limit. If a solution consists of
only two components it is called a binary solution. The component present in smaller amount is
called the solute while, the other present in larger amount is called the solvent.
The concentration of a solution can be expressed in a number of ways as follows:
1. Mass percent or weight percent (w / w%) 2. Mole fraction
3. Molarity 4. Molality
7.1 MASS PERCENT
It is obtained by using the following relation: Mass percent = 100
solution
of
Mass
solute
of
Mass

7.2 MOLE FRACTION
CONCENTRATIONJ OF SOLUTION
7

It is the ratio of number of moles of a particular component to the total number of moles of
the solution. If a substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA and nB
respectively; then the mole fractions of A and B are given as
Mole fraction of A =
solution
of
moles
of
Number
A
of
moles
of
Number
=
B
A
A
n
n
n

Mole fraction of B =
solution
of
moles
of
Number
B
of
moles
of
Number
=
B
A
B
n
n
n

7.3 MOLARITY (M)
It is defined as the number of moles of solute present in one litre of the solution.
Molarity (M) =
)
litres
in
(
solution
of
Volume
solute
of
moles
of
number
Let the weight of solute be ‘w’ g, molar mass of solute be ‘M1’g/mol and the volume of
solution be ‘V’ litre.
Hence, number of moles of solute =
solute
of
mass
molar
or
Atomic
solute
of
weight
=
1
M
w
 M =
)
litres
in
(
V
1
M
w
1

 Number of moles of solute =
1
M
w
= M  V (in litres)
Note: When a solution is diluted, the moles of solute donot change but molarity changes. On
taking out a small volume of solution from a larger volume, the molarity of solution donot change but
moles change proportionately.
7.4 MOLALITY (m)
It is defined as the number of moles of solute present in 1 kg of solvent. It is denoted
by ‘m’.
Molality (m) =
)
kg
in
(
solvent
the
of
mass
solute
of
moles
of
number
Illustration 11
Question: A solution of oxalic acid C2H2O4.2H2O is prepared by dissolving 0.63 g of the acid in
250 cm3 of the solution. Calculate molarity of the solution.
Solution: Molar mass of oxalic acid = 126 g/mol.
 250 cm3 or
1000
250
L = 0.25 L of the solution contains 0.63 g oxalic acid.
 Molarity of the solution =
25
.
0
1
126
63
.
0
 = 0.02 M
Illustration 12
Question: Calculate the molarity of H2O in pure water. (density H2O = 1 g/cm3)
Solution: 1 L of pure water = 1000 g (assuming density =1.0 g/cm3)

 Number of moles in 1 L of pure water =
18
1000
= 55.55.
Key points
Adding a solvent to a solution, a process known as dilution, decreases the concentration (molarity) of
the solution without changing the total moles of solute present in the solution.
1. Weight of oxygen in one mole each of Fe O and FeO is in the simple ratio of:
a) 3 : 2 b)1 : 2 c) 2 : 1 d)3 : 1
2. Equivalent weight of a bivalent metal is 37.2. The molecular weight of its chloride is
a) 412.2 b)216 c) 145.4 d)108.2
3. 0.0833 mole of carbohydrate of empirical formula CH O contain 1 g of hydrogen. The molecular
formula of the carbohydrate is
a) C H O b)C H O c) C H O d)C H O
4. The equivalent weight of Zn(OH) in the following reaction is equal to its,
Zn(OH) + HNO ⟶ Zn(OH)(NO ) + H O:
a)
Formula wt.
1
b)
Formula wt.
2
c) 2 × formula wt. d)3 × formula wt.
5. 5.85 g of NaCl are dissolved in 90 g of water. The mole fraction of NaCl is:
a) 0.1 b)0.01 c) 0.2 d)0.0196
6. 2.76 g of silver carbonate on being strongly heated yield a residue weighing
a) 2.16 g b)2.48 g c) 2.64 g d)2.32 g
7. A solution contains Na CO and NaHCO .10 mL of the solution required 2.5 mL of 0.1 𝑀H SO
for neutralization using phenolphthalein as indicator. Methyl orange is then added when a
further 2.5 mL of 0.2 𝑀H SO was required. The amount of Na CO in 1 litre of the solution is:
a) 5.3 g and 4.2 g b)3.3 g and 6.2 g c) 4.2 g and 5.3 g d)6.2 g and 3.3 g
8. The volume occupied by one molecule of water (density 1 g cm ) is:
a) 18 cm b)22400 cm c) 6.023 × 10 cm d)3.0 × 10 cm
9. 510 mg of a liquid on vaporization in Victor meyer’s apparatus displaces 67.2 cm of air at
(STP). The molecular weight of the liquid is:
a) 130 b)17 c) 170 d)1700
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1

10. What volume of 6 M HCL should be added to 2 M HCL to get 1 L of 3 M HCL?
a) 0.25 L b)1.00 L c) 0.75 L d)2.50 L
11. The normality of one molar sodium carbonate solution is:
a) 2 b)1 c) 0.5 d)1.5
12. If H SO ionises as H SO + 2H O → 2H O + SO , then total number of ions produced by 0.1 M
H SO will be
a) 9.03 × 10 b)3.01 × 10 c) 6.02 × 10 d)1.8× 10
13. 𝑊 of an element combines with oxygen forming 𝑊 g of its oxide. The equivalent weight of the
element is:
a) [𝑊 / 𝑊 ] × 8 b)
𝑊
𝑊 − 𝑊
× 8 c)
𝑊 − 𝑊
𝑊
× 8 d)
𝑊
𝑊 − 𝑊
× 8
14. A sample of ammonium phosphate (𝑁𝐻 ) 𝑃𝑂 contains 6.36 moles of hydrogen atoms. The
number of moles of oxygen atom in the sample is
(atomic mass of N = 14.04, H = 1, P = 31, O = 16)
a) 0.265 b)0.795 c) 2.12 d)4.14
15. To neutralise 20 mL of 𝑀/ 10 NaOH, the volume of 𝑀/20 HCl needed is:
a) 10 mL b)30 mL c) 40 mL d)20 mL
16. 𝐴, 𝐸, 𝑀and 𝑛are the atomic weight, equivalent weight, molecular weight and valence of an
element. The correct relation is:
a) 𝐴 = 𝐸 × 𝑛 b)𝐴 = 𝑀/𝐸 c) 𝐴 = 𝑀/𝑛 d)𝑀 = 𝐴 × 𝑛
17. Which one of the following set of units represents the smallest and largest amount of energy
respectively?
a) J and erg b)erg and cal c) Cal and eV d)eV and L-atm
18. The number of atoms present in a 0.635 g of Cu piece will be
a) 6.023 × 10 b)6.023 × 10 c) 6.023 × 10 d)6.023 × 10
19. What volume of hydrogen gas, at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g
of elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?
a) 89.6 L b)67.2 L c) 44.8 L d)22.4 L
20. The numerical value of 𝑁/𝑛(where 𝑁 is number of molecules is 𝑛 moles of gas) is:
a) 8.314 b)6.02 × 10 c) 1.602 × 10 d)1.66 × 10
21. In the relationship molecular formula = empirical formula × 𝑛. The 𝑛 may have:
a) Any value
b)Zero value
c) Only positive integer value
d)None of the above
22. 10 g CaCO on heating gives 5.6 g CaO and …. g CO .
a) 4.4 b)5.6 c) 6.5 d)4.2
23. Which of the following changes with increase in temperature?
a) Molality
b)Weight fraction of solute
c) Fraction of solute present in water
d)Mole fraction
24. On combustion of 4 g of the methane, 10.46 kJ of heat is liberated. Heat of combustion of
methane is
a) 83.68 kJ b)10.46 kJ c) 41.84 kJ d)20.93 kJ

25. A gas is found to have the formula (CO) . Its VD is 70. The value of 𝑥 must be:
a) 7 b)4 c) 5 d)6
26. Choose the wrong statement.
a) 1 mole means 6.023× 10 particles
b)Molar mass is mass of one molecule
c) Molar mass is mass of one mole of a substance
d)Molar mass is molecular mass expressed in grams
27. The term standard solution is used for the solutions whose:
a) Normality is known b)Molarity is known c) Strength is known d)All of these
28. The ratio of mole fraction of a solute and a solvent in a binary solution is:
a) Ratio of their wt. b)One c) Ratio of their mole d)Zero
29. If in a reaction HNO is reduced to NO, the mass of HNO absorbing one mole of electrons would
be
a) 21.0 g b)36.5 g c) 18.0 g d)31.5 g
30. At STP 5.6 litre of a gas weighs 60 g. The vapour density of gas is:
a) 60 b)120 c) 30 d)240
31. The number of atoms present in 16 g of oxygen gas is:
a) 6.02 × 10 . b)3.01 × 10 c) 3.01 × 10 . d)6.02 × 10
32. On analysis a certain compound was found to contain iodine and oxygen in the ratio of 254 g of
iodine (at. mass 127) and 80 g oxygen (at. mass 16). What is the formula of the compound?
a) IO b)I O c) I O d)I O
33. The vapour density of a volatile chloride of a metal is 95 and the specific heat of the metal is 0.13
cal/g. The equivalent weight of the metal will be:
a) 6.0 b)12.3 c) 18.6 d)24.5
34. The equivalent weight of a certain trivalent element is 20. Molecular weight of its oxide is
a) 152 b)56 c) 168 d)68
35. Gram molecular volume of oxygen at STP is
a) 3200 cm b)5600 cm c) 22400 cm d)11200 cm
36. Two elements 𝑋 (at. Wt. 75) and 𝑌 (at. wt. 16) combine to give a compound having 75.8% of𝑋.
The formula of compound will be
a) 𝑋𝑌 b)𝑋 𝑌 c) 𝑋𝑌 d)𝑋 𝑌
37. The amount of oxalic acid (hydrated) required to prepare 500 mL of its 0.1 𝑁 solution is:
a) 0.315 g b)6.3 g c) 3.15 g d)63.0 g
38. The equivalent weight of KMnO for acid solution is
a) 79 b)52.16 c) 158 d)31.6
39. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using
diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of
dichromate is
a) 3 b)4 c) 5 d)6
40. A mixture of 𝐶𝐻 , 𝑁 and 𝑂 is enclosed in a vessel of one litre capacity at 0°𝐶. The ratio of
particle pressures of gases is 1 : 4 : 2. Total pressure of the gaseous mixture is 2660 mm. the
number of molecules of oxygen present in the vessel is
a)
6.02 × 10
22.4
b)6.02 × 10 c) 22.4 × 10
d)1000
41. 𝑥 g of 𝐴𝑔 was dissolved in HNO and the solution was treated with excess of NaCl when 2.87 g of

𝐴𝑔𝐶𝑙 was precipitated. The value of 𝑥 is
a) 1.08 g b)2.16 g c) 2.70 g d)1.62 g
42. One mole electron means:
a) N electrons
b)6.023 × 10 electrons
c) 0.55 mg electrons
d)All of these
43. A signature, written in carbon pencil weights 1 mg. What is the number of carbon atoms present
in the signature?
a) 5.02 × 10 b)5.02 × 10 c) 6.02 × 10 d)0.502 × 10
44. The minimum quantity of H S needed to precipitate 63.5 g of Cu will be nearly:
a) 63.5 g b)31.75 g c) 34 g d)20 g
45. An unknown element forms an oxide. What will be the equivalent weight of the element if the
oxygen content is 20% by weight?
a) 16 b)32 c) 8 d)64
46. Cortisone is a molecular substance containing 21 atoms of carbon per molecule. The molecular
weight of cortisone is 360.4. what is the percentage of carbon in cortisone?
a) 59.9% b)75% c) 69.98% d)None of these
47. Which mode of expressing concentration is independent of temperature?
a) Molality b)Per cent by weight c) Mole fraction d)All of these
48. An ion is reduced to the element when it absorbs 6 × 10 electrons. The number of equivalent
of ion is:
a) 0.1 b)0.01 c) 0.001 d)0.0001
49. The volume of 0.1 𝑀H SO required to neutralise 30 mL of 2.0 𝑀 NaOH is:
a) 100 mL b)300 mL c) 400 mL d)200 mL
50. The law of definite proportions is not applicable to nitrogen oxide because
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
51. 1.520 g of hydroxide of a metal on ignition gave 0.995 g of oxide. The equivalent weight of metal
is
a) 1.52 b)0.995 c) 190 d)9
52. A hydrocarbon contains 10.5 g carbon and 1 g hydrogen. Its 2.81 g has 1L volume at 1 atm and
127 C, hydrocarbon is
a) C H b)C H c) C H d)None of the above
53. 1 mole of methyl amine on reaction with nitrous acid gives at NTP
a) 1.0 L of nitrogen b)22.4 L of nitrogen c) 11.2 L of nitrogen d)5.6 L of nitrogen
54. The weight of sulphuric acid needed for dissolving 3 g magnesium carbonate is:
a) 3.5 g b)7.0 g c) 1.7 g d)17.0 g
55. When a metal is burnt, its weight is increased by 24 per cent. The equivalent weight of the metal
will be:
a) 25 b)24 c) 33.3 d)76
56. A metal oxide is reduced by heating it in a stream of hydrogen. It is found that after complete
reduction, 3.15 g of oxide yielded 1.05 g of metal. From the above data we can say that
a) The atomic weight of metal is 8 b)The atomic weight of metal is 4
c) The equivalent weight of metal is 4 d)The equivalent weight of metal is 8
57. The ratio of amounts of H S needed to precipitate all the metal ions from 100 mL of 1 M AgNO

and 100mL of CuSO , will be
a) 1 : 1 b)1 : 2 c) 2 : 1 d)None of these
58. The mole fraction of NaCl in a solution containing 1 mole of NaCl in 1000 g of water is:
a) 0.0177 b)0.001 c) 0.5 d)0.244
59. Which is correct for Na HPO ?
a) It is not an acid salt b)Eq. wt. = c) Ox. no. of P is + 3 d)All of these
60. How many g of NaOH will be needed to prepare 250 mL of 0.1𝑀 solution?
a) 1 g b)10 g c) 4 g d)6 g
61. If the specific heat of a metallic element is 0.214 cal/g, the atomic weight will be closest to:
a) 66 b)12 c) 30 d)65
62. An ore contains 1.34% of the mineral argentite, 𝐴𝑔 𝑆, by mass. How many gram of this ore
would have to be processed in order to obtain 1.00 g of pure solid silver, 𝐴𝑔?
a) 74.6 g b)85.7 g c) 107.9 g d)134.0 g
63. In which of the following numbers all zeros are significant?
a) 0.500 b)30.000 c) 0.00030 d)0.0050
64. Weight of an atom of an element is 6.644 × 10 g. What will be the number of g atom of that
element in 40 kg?
a) 10 b)10 c) 1.5 × 10 d)None of these
65. In a compound 𝐴 𝐵 ∶
a) Mole of 𝐴 = mole of 𝐵 = mole of 𝐴 𝐵
b)Eq. of 𝐴 = Eq. of𝐵 = Eq. of𝐴 𝐵
c) 𝑌 × 𝑋mole of 𝐴 = 𝑌 × 𝑋mole of 𝐵 = (𝑋 + 𝑌)𝑋 mole of 𝐴 𝐵
d)𝑌 × 𝑋mole of 𝐴 = 𝑌 × 𝑋mole of 𝐵
66. One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (Valency
=2) combines with 4 g of bromine. The equivalent weight of calcium is
a) 10 b)20 c) 40 d)80
67. A bivalent metal has an equivalent mass of 32. The molecular mass of the metal nitrate is
a) 182 b)168 c) 192 d)188
68. 12 g of Mg (at. wt. = 24 ) will react completely with an acid to give:
a) One mole of H b)Half mole of H c) One mole of O d)None of these
69. The atomic weight of a metal (𝑀) is 27 and its equivalent weight is 9, the formula of its chloride
will be:
a) 𝑀Cl b)𝑀Cl c) 𝑀 Cl d)𝑀Cl
70. 1.60 g of a metal were dissolved in HNO to prepare its nitrate. The nitrate on strong heating
gives 2 g oxide. The equivalent weight of metal is:
a) 16 b)32 c) 48 d)12
71. 5.85 g of NaCl dissolved in H O and solution is made upto 500 mL. The molarity is:
a) 0.1 b)0.2 c) 1.0 d)0.117
72. Which property of an element is not variable?
a) Valence b)At. wt. c) Eq. wt. d)None of these
73. The oxide of an element possesses the formula 𝑀 O . If the equivalent weight of the metal is 9,
then the atomic weight of the metal will be:
a) 9 b)18 c) 27 d)54
74. 0.7 g of Na CO ∙ 𝑥H O were dissolved in water and the volume was made to 100 mL, 20 mL of
this solution required 19.8 mL of 𝑁/10 HCl for complete neutralisation. The value of 𝑥 is:

a) 7 b)3 c) 2 d)5
75. The specific heat of an element of atomic weight 32 is likely to be:
a) 0.25 cal/g b)0.24 cal/g c) 0.20 cal/g d)0.15 cal/g
76. Number of atoms in 560 g of Fe (atomic mass 56 g mol ) is
a) Twice that of 70 g N b)Half that of 20 g H c) Both are correct d)None of these
77. A 400 mg iron capsule contains 100 mg of ferrous fumarate, (CHCOO) Fe. the percentage of iron
present in it is approximately
a) 33% b)25% c) 14% d)8%
78. Equal weights of Zn metal and iodine are mixed together and 𝑙 is completely converted toZnl .
What fraction by weight of original Zn remains unreacted? (Zn = 65, I = 127)
a) 0.34 b)0.74 c) 0.84 d)Unable to predict
79. An aqueous solution containing 6.5 g of NaCl of 90% purity was subjected to electrolysis. After
the complete electrolysis, the solution was evaporated to get solid NaOH. The volume of 1 M
acetic acid required to neutralise NaOH obtained above is
a) 1000 cm b)2000 cm c) 100 cm d)200 cm
80. Which of the following is correct?
a) Mole fraction of I + mole fraction of II = 1
(if only two components are present)
b)
Mole fraction of I
Mole fraction of II
=
mole of I
mole of II
(if only two components are present)
c) Mole fraction of solute =
d)All of the above
81. The number of significant figures in Avogadro’s number is
a) Four b)Two c) Three d)Can be any of these
82. A gas has a vapour density 11.2. The volume occupied by 1g of the gas at NTP is
a) 1 L b)11.2 L c) 22.4 L d)4 L
83. A metal nitride, 𝑀 𝑁 contains 28% of nitrogen. The atomic mass of metal, 𝑀 is
a) 24 b)54 c) 9 d)87.62
84. An oxide of iodine (I = 127) contains 25.4 g of iodine for 8 g of oxygen. Its formula could be:
a) I O b)I O c) I O d)I O
85. 20 g of an acid furnishes 0.5 moles of H O ions in its aqueous solution. The value of 1 g eq. of
the acid will be:
a) 40 g b)20 g c) 10 g d)100 g
86. 10 mL of gaseous hydrocarbon on combustion gives 40 mL of CO (g) and 50 mL of H O (vap).
The hydrocarbon is:
a) C H b)C H c) C H d)C H
87. 10 mL of concentrated H SO (18 𝑀)is diluted to one litre. The approximate molecular of the
dilute acid is:
a) 18 𝑀 b)180 𝑀 c) 0.18 𝑀 d)1.8 𝑀
88. Which represents per cent by strength?
a)
wt. of solute
volume of solution
× 100

b)
wt. of solute
volume of solution
× 100
c)
volume of solute
volume of solution
× 100
d)All of the above
89. An alkaloid contains 17.28% of nitrogen and it’s molecular mass is 162. The number of nitrogen
atoms present in one molecule of alkaloid is
a) 5 b)4 c) 3 d)2
90. 6.02 × 10 molecules of urea are present in 100 mL of its solution. The molarity of urea
solution is:
a) 0.1 b)0.01 c) 0.02 d)0.001
91. What volume of H at 273 K and 1 atm will be consumed in obtaining 21.6 g of elemental boron
(at. mass 10.8) from the reduction of boron trichloride with H ?
a) 44.8 L b)22.4 L c) 89.6 L d)67.2 L
92. In a metal chloride, the weight of metal and chlorine are in the ratio of 1:2. The equivalent
weight of the metal will be:
a) 71 b)35.5 c) 106.5 d)17.75
93. KMnO (mol.wt.= 158) oxidizes oxalic acid in acid medium to CO and water as follows
5C O + 2MnO + 16H → 10CO + 2Mn + 8H O
What is the equivalent weight of KMnO ?
a) 158 b)31.6 c) 39.5 d)79
94. How many H-atoms are present in 0.046 g of ethanol?
a) 6 × 10 b)1.2 × 10 c) 3 × 10 d)3.6 × 10
95. The pair of species having same percentage of carbon is:
a) CH COOH and C H O
b)CH COOH and C H OH
c) HCOOCH and C H O
d)C H O and C H O
96. The maximum number of molecules is present in:
a) 15 L of H gas at STP b)5 L of N gas at STP c) 0.5 g of H gas d)10 g of O gas
97. If one mole of ethanol (C H OH) completely burns to carbon dioxide and water, the weight of
carbon dioxide formed is about:
a) 22 g b)45 g c) 66 g d)88 g
98. How many moles of MgIn S can be made from 1 𝑔 each of 𝑀𝑔, in and S? (Atomic mass : Mg =
24, , In = 114.8, S = 32)
a) 6.47 × 10 b)3.0× 10 c) 9.17× 10 d)8.7× 10
99. One g of a mixture of Na CO and NaHCO consumes 𝑦 equivalent of HCl for complete
neutralisation. One g of the mixture is strongly heated, then cooled and the residue treated with
HCl How many equivalent of HCl would be required for complete neutralization?
a) 2𝑦 equivalent b)𝑦 equivaletnt c) 3𝑦/4 equivalent d)3𝑦/2 equivalent
100.An organic compound containing C and H has 92.3% of carbon, its empirical formula is
a) CH b)CH c) CH d)CH

1. In which of the following pairs do the two species resemble each other most closely in chemical
properties?
a) H and H b) O and O c) Mg and Mg d) N and N
2. Number of moles of K Cr O reduced by 1 mol of Sn is
a)
1
3
b)
1
6
c)
2
3
d)1
3. A mixed solution of potassium hydroxide and sodium carbonate required 15 mL of an N/20 HCl
solution when titrated with phenolphthalein as an indicator. But the same amount of the
solution when titrated with methyl orange as indicator required 25 mL of the same acid. the
amount of KOH present in the solution is
a) 0.014 g b)0.14 g c) 0.028 g d)1.4 g
4. The volume strength of 1.5 N H O solution is
a) 4.8 b)8.4 c) 3.0 d)8.0
5. A solution of 200 ppt is
a) 2 × 10 ppm b)2 × 10 ppb c) Both (a) & (b) d)None of these
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 2

6. Which gives ppt with K CrO ?
a) Hg , Pb , Ag , Ba b)Pb , Ag , Ba c) Ag , Ba d)Pb , Ba
7. Mass of one atom of the element 𝐴 is 3.9854 × 10 g. How many atoms are contained in 1 g of
the element 𝐴?
a) 2.5092 × 10 b)6.022 × 10 c) 3.9854 × 10 d)1.66 × 10
8. 1.056 g of the Snwas treated with 1.947 g of I . After reaction was over, 0.601 g of Sn was
recovered. Thus, empirical formula of the compound formed is (Sn = 119, I = 127)
a) SnI b)SnI c) SnI d)SnI
9. 0.4 mole of HCl and 0.2 mole of CaCl were dissolved in water to have 500 mL of solution, the
molarity of Cl ion is
a) 0.8 M b)1.6 M c) 1.2 M d)10.0 M
10. Match the substances given in 𝐵 based on reactions given in 𝐴 select correct answer from the
alternate
𝑨(Reaction) 𝑩(Substances)
I A white, waxy
solid, normality
stored under
water because it
spontaneously
inflames in air
𝑃 HNO
II A viscous liquid
that reacts with
table sugar,
giving a charred
mass
𝑄 Cl
III An acid that
reacts with
copper metal,
releasing brown
fumes
𝑅 P
IV A pale greenish
yellow gas that
dissolves in
aqueous NaOH
to give a solution
used as a bleach
𝑆 H SO
Codes
I II III IV
a) 𝑅𝑄𝑅𝑆
b)𝑆𝑅𝑄𝑃
c) 𝑅𝑆𝑃𝑄
d)𝑆𝑅𝑃𝑄
11. A mixture contains Cu , Al and Ni . Following steps have been adopted but written in
disorder
I: Filter, boil of H S gas and add NH Cl,heat and add NH OH
II: Filter, add NH OH and pass H S gas
III: Pass H S gas into acidified solution of mixture
𝑆𝑡𝑒𝑝𝑠𝑤𝑖𝑙𝑙𝑏𝑒𝑢𝑠𝑒𝑑𝑖𝑛𝑡ℎ𝑒𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛g𝑜𝑟𝑑𝑒𝑟

a) I, II, III b)III, I, II c) III, II, I d)I, III, II
12. The reaction between yttrium metal and dilute HCl produces H (g) and 𝑌 ions. The molar ratio
of yttrium to that hydrogen produced is
a) 2:3 b)3:2 c) 1:2 d)2:1
13. A 500 g toothpaste contains 0.2 g fluoride. The concentration of fluoride in terms of ppm is
a) 100 b)250 c) 400 d)450
14. 44.8 L of CO at NTP is obtained by heating 𝑥 g of pure CaCO . 𝑥 is
a) 100 g b)200 g c) 50 g d)44.8 g
15. A hydrate of Na SO has 50% water by mass. It is
a) Na SO  5H O b)Na SO  6H O c) Na SO  7H O d)Na SO  2H O
16. Potassium selenate is isomorphous with potassium sulphate and contains 50.0% of Se. Find the
atomic weight of Se
a) 142 b)71 c) 47.33 d)284
17. Oxalic acid (H C O )forms two series of salt HC O and C O .If 0.9 g of oxalic acid is in 100 mL
solution, HC O and C O have normality respectively
a) 0.1 N, 0.1 N b)0.1 N, 0.2 N c) 0.2 N, 0.2 N d)0.2 N, 0.1 N
18. In basic medium, CrO reacts with S O resulting in the formation of Cr(OH) ⊖
and SO .
How many mL of 0.1 M Na CrO is required to react with 40 mL of 0.25 MNa S O ?
a) 240.2 mL b)24.02 mL c) 266.67 mL d)26.67 mL
19. Volume of H C O ∙ 2H O solution to prepare 0.10 M from 1.575 g of it is
a) 125 mL b)250 mL c) 500 mL d)1000 mL
20. A sodium salt of an unknown anion when treated with MgCl gives white precipitate only
boiling. The anion is
a) SO b)HCO c) CO d)NO
21. 𝐴(colourless salt)
∆
→ 𝐵 + 𝐶 + 𝐷; 𝐷 ⎯ 𝐸
Gas 𝐶 turns solution 𝐸 milky. 𝐵 burns with blue flame. 𝐴 also decolourlses MnO /H . Thus,
𝐴, 𝐵, 𝐶, 𝐷 and 𝐸 are
𝐴𝐵𝐶𝐷𝐸
a) CaC O COCO CaOCa(OH) b) CaC O CO COCaOCa(OH)
c) CaCO CaOCOCO Ca(OH) d) CaOCl Cl O CaOCa(OH)
22. Fe(OH) and Cr(OH) can be separated using
a) NaOH b)NaOH + H O c) Both (a) and (b) d)None of these
23. Maximum number of moles of PbSO that can be precipitated by mixing 20.00 mL of
0.1 M Pb(NO ) and 30.00 mL of 0.1 M Na SO will be
a) 0.002 b)0.003 c) 0.005 d)0.001
24. When 10 mL of ethyl alcohol (density = 0.7893 g mL ) is mixed with 20 mL of water (density
0.9971 g mL ) at 25C, the final solution has a density of 0.9571 g mL . The percentage change
in totalvolume on mixing is
a) 3.1% b)2.4% c) 1% d)None of these
25. Which has maximum number of millimoles of Cl ion?
a) 0.208 g BaCl b)100 mL of 0.1 M BaCl
c) 0.745 g KCl d)Equal
26. For H PO , the correct choice is
a) H PO is dibasic and reducing b)H PO is dibasic and non-reducing
c) H PO is tribasic and reducing d)H PO is tribasic and non-reducing

27. A mixture of Na C O and KHC O ∙ H C O required equal volumes of 0.2 M KMnO and 0.2 M
NaOH separately for complete titration. The mole ratio of Na C O and KHC O . H C O in the
mixture is
a)
2
11
b)
11
2
c)
5
2
d)
7
2
28. NH + OCl⊖
→ N H + Cl⊖
On balancing the above equation in basic solution, using integral coefficient, which of the
following whole numbers will be the coefficient of N H ?
a) 1 b)2 c) 3 d)4
29. 0.05 g of a piece of metal in dilute acid gave 24.62 mL of H at 27℃ and 760 mm pressure. The
𝐸𝑤of metal is
a) 25 b)12.5 c) 50 d)37.5
30. How many moles of O will be liberated by one mole of CrO is the following reaction:
CrO + H SO → Cr (SO ) + H O + O
a) 4.5 b)2.5 c) 1.25 d)None
31. A on reaction with dil. H SO gives a colourless pungent gas that can turn Cr O /H green.
Green colour is due to formation of
a) CrO b)Cr c) CrO d)CrO Cl
32. When 80 mL of 0.20 M HCl is mixed with 120 mL of 0.15 M KOH,the resultant solution is the
same as a solution of
a) 0.16 M KCl and 0.02 M HCl b)0.08 M KCl
c) 0.08 M KCl and 0.01 M KOH d)0.08 M KCl and 0.01 M HCl
33. 5.6 g of a metal forms 12.7 g of metal chloride. Hence equivalent weight of the metal is
a) 127 b)254 c) 56 d)28
34. Which of the following salts does not give positive test for nitrate ion?
a) KNO b)NaNO c) Pb(NO ) d)Mg(NO )
35. 27 g of Al will react completely with …….. g of O
a) 8 g b)10 g c) 24 g d)49 g
36. Test tube A contains ZnCl aqueous solution while test tube B contains aq CdCl solution. On
passing H S gas
a) ZnS is precipitated b)CdS is precipitated
c) Both (a) and (b) are precipitated d)None of the above is precipitated
37. A solution of a metal ion when treated with KI gives a red precipitate which dissolves in excess
KI to give a colourless solution. Moreover, the solution of metal ion on treatment with a solution
of cobalt (III) thiocyanate gives rise to a deep blue crystalline precipitate. The metal ion is
a) Pb b)Hg c) Cu d)Co
38. Mixture is initially orange in colour. When solution is prepared in dil. Acid, it changes to dark
brown colour. Mixture contains
a) HgI , Cr O b)I , Cr O c) I , Cu d)I , SO
39. When 100 mL of 0.1 M Ba(OH) is neutralized with a mixture of 𝑥 mL of 0.1 M HCl and y mL of
0.2 M H SO using methyl orange indicator, what is value of 𝑥 and 𝑦?
a) 200, 100 b)100, 200 c) 300, 200 d)200, 300
40. 𝐴 → 𝐵 (gas) used by dentist. Hence, 𝐴 is
a) NH Cl b)NH NO c) NH NO d)NH OH
41. Borax on heating strongly above its melting point melts to a liquid, which then solidifies to a

transparent mass commonly known as borax-bead. The transparent glassy mass consist of
a) Sodium pyroborate b)Boric anhydride
c) Sodium metaborate d)Boric anhydride and sodium metaborate
42. Consider the ionisation of H SO as follow:
H SO + 2H O → 2H O
+ SO
The total number of ions furnished by 100 mL of 0.1 M H SO will be
a) 1.2 × 10 b)0.12 × 10 c) 0.18 × 10 d)1.8 × 10
43. White ppt of 𝐴 on reaction with 𝑎𝑞NH are blackened. Select correct statement about 𝐴
a) 𝐴 is also called calomel b)𝐴 reacts with aq NH forming HgNH Cl
c) 𝐴 changes to grey on reaction with SnCl d)All of the above are correct statements
44. What volume of 0.05 M K Cr O in acidic medium is needed for complete oxidation of 200 mL of
0.06 M FeC O solution?
a) 1.2 mL b)1.2 L c) 120 mL d)800 mL
45. PbO oxidises MnO (black) to
a) MnO b)MnO c) Mn d)Mn O
46. H S would separate the following at pH < 7
a) Zn , Co b)Cu , Cd c) Cu , Cr d)Cu , As
47. 10 mLof H O solution (volume strength = x) required 10 mLof .
MnO solution in acidic
medium. Hence, x is
a) 0.56 b)5.6 c) 0.1 d)10.0
48. Two substances I and II of carbon and oxygen have respectively 72.73% and 47.06% oxygen.
Hence, they follow
a) Law of multiple proportion b)Law of reciprocal proportion
c) Law of definite proportion d)Law of conservation of mass
49. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using
diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of
dichromate is
a) 3 b)4 c) 5 d)6
50. A sample of copper sulphate pentahydrate contains 3.782 g of Cu. How many grams of oxygen
are in the sample?
a) 0.952 g b)3.809 g c) 4.761 g d)8.576 g
51. 4.4 g of CO contains how many litres of CO at STP?
a) 2.4 L b) 2.24 L c) 44 L d) 22.4 L
52. The equivalent weight of a certain trivalent element is 20. Molecular weight of its oxide is
a) 152 b) 56 c) 168 d) 68
53. 4.2 g of a metallic carbonate MCO was heated in a hard glass tube and CO evolved was found to have
1120 mL of volume at STP. The 𝐸𝑤 of the metal is
a) 12 b) 24 c) 18 d) 15
54. Mass of one N − atom is
a) 14 u b) 7 u c) 14 g d) 7 g
55. 5.3 g of 𝑀 CO is dissolved in 150 mL of 1 N HCl.Unused acid required 100 mL of 0.5 N NaOH.Hence,
equivalent weight of 𝑀 is
a) 23 b) 12 c) 24 d) 13
56. A mixture on heating gave a gas used as an anaesthetic, 1.1 g of gas occupies 0.56 L at NTP. Mixture

contains
a) NaNO + NH Cl b) NaNO + NH Cl
c) CaCO + MgCO d) NH Cl + Na SO
57. Which is temperature independent?
a) Mass per cent b) Volume per cent
c) Mass/volume per cent d) Molarity
58. In the flowing equation
What volume of 0.2 M Na CrO solution is required just to react with 30 mL of 0.2 M Na S O solution
a) 40 mL b) 80 mL c) 20 mL d) 60 mL
59. 0.3 g platinichloride of an organic diacidic base left 0.09 g of platinum on ignition. The molecular weight of
the organic base is
a) 120 b) 240 c) 180 d) 60
60. Passing H S gas into a mixture of Mn , Ni , Cu and Hg ions in an acidified aqueous solution
precipitates
a) CuS and HgS b) MnS and CuS c) MnS and NiS d) NiS and HgS
61. Equivalent weight of H PO (molecular weight= 𝑀) when it disproportionates into PH and H PO is
a) 𝑀 b)
𝑀
2
c)
𝑀
4
d)
3𝑀
4
62. NaOH is formed according to reaction
2Na +
1
2
O → Na O
Na O + H O → 2NaOH
To make 4g NaOH, Na required is
a) 4.6 g b) 4.0 g c) 2.3 g d) 0.23 g
63. 40 mL of 0.05 M solution of Cr is oxidized to CrO by 20 mL of H O . H O is
a) 0.15 M
b) 0.30 M
c) 0.10 M
d) 0.20 M
64. Mass of one atom of 𝑋 is 6.66 × 10 g. Hence, number of moles of atom 𝑋 in 40 kg is
a) 10 mol b) 10 mol c)
40 × 10
6.66 × 10
mol d)
40 × 10
6.022 × 10
mol
65. 0.7 g sample of iron ore was dissolved in acid. Iron was reduced to +2 state and it required 50
ml of M/50 KMnO solution for titration. The percentage of Fe and Fe O in the ore is
a) 40 % Fe, 55.24%, Fe O b) 55.24 % Fe, 40 %, Fe O
c) 8 % Fe, 11 %, Fe O d) 11 % Fe, 8 %, Fe O
66. To prepare a solution that is 0.50 M KCl starting with 100 mL of 0.40 M KCl
a) Add 0.75 g KCl b) ADD 20 mL of water
c) Add 0.10 mol KCl d) Evaporate 10 mL water
67. How many moles of MnO ⊖
ion will react with 1 mol of ferrous oxalate in acidic medium?
a)
1
5
b)
2
5
c)
3
5
d)
5
3
68. K [Fe(CN) ] can be used to detect one or more out of Fe , Fe , Zn , Cu , Cd

a) Fe , Fe b) Fe , Zn , Cu c) All but Fe d) All but Fe
69. A mixture upon adding conc. H SO gives orange red fumes. It may contain the anion pair
a) CrO + Cl b) Br + Cl c) NO + Cl d) CrO + NO
70. 34 g of H O is present in 1120 mL of solution. This solution is called
a) 10 vol solution b) 20 vol solution c) 34 vol solution d) 32 vol solution
71. N + 3H → 2NH
Molecular weight of NH and N are 𝑥 are 𝑥 , respectively. Their equivalent weights are 𝑦 and 𝑦 ,
respectively. Then (𝑦 – 𝑦 )is
a)
2𝑥 – 𝑥
6
b) (𝑥 – 𝑥 ) c) (3𝑥 – 𝑥 ) d) (𝑥 – 3𝑥 )
72. In the mixture of (NaHCO + Na CO ), volume of HCl required is 𝑥 mL with phenolphthalein
indicator and then 𝑦 mL with methyl orange indicator in same titration. Hence, volume of HCl
for complete reaction of Na CO is:
a) 2𝑥 b) 𝑦 c)
𝑥
2
d) (𝑦 − 𝑥)
73. HF attacks glass (Na SiO ) forming
a) Na SiF b) Na SiF c) Na SiF d) H SiO
74. When 2 g of gas A is introduced into an evacuated flask kept of 25℃, the pressure was found to be 1
atmosphere. If 3 g of another gas B is then added to the same flask, the pressure becomes 1.5 atm.
Assuming ideal behavior, the ratio of molecular weights (𝑀 : 𝑀 ) is
a) 1:3 b) 3:1 c) 2:3 d) 3:2
75. If equal volumes of 1 M KMnO and 1 M K Cr O solutions are allowed to oxidise Fe (II) to Fe (III) in
acidic medium, then Fe (II) oxidized will be
a) More by KMnO b) More by K Cr O c) Equal in both cases d) Can’t be determined
76. The simplest formula of a compound containing 50% of an element X (atomic weight 10) and 50% of
element Y (atomic weight 20) is:
a) XY b) X Y c) XY d) X Y
77. Number of millilitres of a 1.6% BaCl (𝑤/𝑉) solution which required to precipitate the sulphur as
BaSO in a 0.60 g sample that contains 12% S is
a) 58.50 mL b) 14.63 mL c) 29.25 mL d) 21.00 mL
78. All the oxygen in a 0.5434 g sample of a pure oxide of iron is removed by reduction in a stream
of H . The loss in weight is 0.1210 g. hence, formula of the iron oxide is (Fe = 56)
a) FeO b) Fe O c) Fe O d) FeO
79. A certain metal sulphide 𝑀S is used extensively as a high temperature lubricant. If 𝑀S is 40.00% by
mass sulphur, atomic mass of 𝑀 is
a) 60 b) 96 c) 100 d) 80
80. The weight of 1 L of ozonised oxygen at STP was found to be 1.5 g. When 100 mL of this mixture at STP
was treated with turpentine oil, the volume was reduced to 90 mL. The molecular weight of ozone is
a) 49 b) 47 c) 46 d) 47.9
81. The empirical formula of a compound of carbon with hydrogen is CH .1 L of this gaseous
compound has mass equal to that of 1 L N under standard state. Thus, molecular formula of
the compound is
a) C H b) C H c) C H d) C H
82. The concentration of 10% CH COOH in mol L is
a) 10 b) 0.83 c) 1 d) 1.67

83. KCl + conc. H SO + K Cr O
∆
→ (𝑋) ⎯⎯⎯ (𝑌). (𝑋) is reddish brown coloured gas soluble in
NaOH forming (𝑌), (𝑋) and (𝑌) are
a) Cr OCl , Na CrO b) Cr O Cl , Na CrO c) CrO Cl, Na CrO d) CrO Cl , Na CrO
84. In the titration of 100 mL of 0.01 M CH COOH with 0.01 M NaOH, [H O ] = 𝐾 when
a) 100 mL of NaOH has been added b) 75 mL of NaOH has been added
c) 50 mL of NaOH has been added d) 25 mL of NaOH has been added
85. By H O /OH , Cr(OH) changes to
a) CrO b) Cr O c) CrO d) [Cr(OH) ]
86. 800 g of a 40% solution by weight was cooled. 100 g of solute precipitated. The percentage
composition of remaining solution is
a) 31.4% b) 20.0% c) 23.0% d) 24%
87. The purity of H O in a given sample is 85%. Calculate the weight of impure sample of H O which
requires 10 mL of M/5 KMnO solution in a titration in acidic medium
a) 2 g b) 0.2 g c) 0.17 d) 0.15 g
88. 𝐴 is a colourless substance. Aqueous solution of 𝐴 gives reddish-orange ppt with KI; ppt dissolves
in excess of KI forming a colourless solution. If NH Cl and NaOH solution is added to this
colourless solution reddish brown ppt is formed. Substance 𝐴 is
a) Epsom salt b) Mohr’s saslt c) Calomel d) Corrosive sublimate
89. K [Fe(CN) ] can be used to detect some ions out of Cd , Cu , Zn , Fe , Pb and Bi .
Exclude ions are
a) Fe , Bi , Pb b) Cu , Zn , Bi c) Bi , Pb d) Fe , Pb
90. In the estimation of nitrogen by Kjeldahl’s method, 2.8 g of an organic compound required 20
millimoles of H SO for the complete neutralisation of NH gas evolved. The percentage of nitrogen
in the sample is
a) 20% b) 10% c) 40% d) 30%
91. Mixture of 1 mole of Na CO and 2 moles of NaHCO forms 1 mole of CO . Thus, per cent yield of CO is
a) 25% b) 50% c) 75% d) 100%
92. NH is formed in the following steps
I. Ca + 2C → CaC 50% yield
II. CaC + N → CaCN + C100% yield
III. CaCN + 3H O → 2NH + CaCO 50% yield
To obtain 2 moles NH , calcium required is
a) 1 mol b) 2 mol c) 3 mol d) 4 mol
93. When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium
hydroxide, the ratio of volume of hydrogen evolved is
a) 1:1 b) 1:2 c) 2:1 d) 9:4
94. Acidified MnO can be decolourised by
a) CaC O b) H O c) FeSO d) All of these
95. An element A (atomic weight = 12) and B (atomic weight = 35.5) combines to form a compound X.
If 4 mol of B combines with 1 mol of A to give 1 mol of X. The weight of 1 mol of X would be
a) 47.5 g b) 74.0 g c) 154.0 g d) 148.8 g
96. The density of 1 M solution of NaCl is 1.0585 g/mL.The molality of the solution is
a) 1.0585 b) 1.00 c) 0.10 d) 0.0585
97. KMnO reacts with oxalic acid according to the equation

2MnO + 5C O + 16H → 2Mn + 10CO + 8H O
Here 20 mL of 0.1 M KMnO is equivalent to
a) 20 mL of 0.5 M C H O b) 50 mL of 0.1 M C H O
c) 50 mL of 0.5 M C H O d) 20 mL of 0.1 M C H O
98. Which of the following substances contains greatest mass of chlorine?
a) 5.0 g Cl b) 0.5 mol Cl c) 0.10 mol KCl d) 30.0 g MgCl
99. What mass of CO could be formed by the reaction of 16 g CH with 48 g ofO ?
CH + 2O → CO + 2H O
a) 44 g b) 33 g c) 16 g d) 24 g
100. At 100℃ and 1 atm, if the density of the liquid water is 1.0 g cm and that of water vapour is
0.00006 g cm , then the volume occupied by water molecule in 1 L steam at this temperature is
a) 6 cm b) 60 cm c) 0.6 cm d) 0.06cm
1) d 2) c 3) d 4) a
5) d 6) a 7) a 8) d
9) c 10) a 11) a 12) d
13) b 14) c 15) c 16) a
17) d 18) d 19) b 20) b
21) c 22) a 23) c 24) c
25) c 26) b 27) d 28) c
29) a 30) b 31) d 32) d
33) b 34) c 35) b 36) d
37) c 38) d 39) d 40) a
41) b 42) d 43) d 44) c
45) b 46) c 47) d 48) c
49) b 50) c 51) d 52) b
53) b 54) a 55) c 56) c
57) b 58) a 59) d 60) a
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 1 (ANSWERS)

61) c 62) b 63) b 64) a
65) b 66) b 67) d 68) b
69) d 70) b 71) b 72) b
73) c 74) c 75) c 76) c
77) d 78) b 79) c 80) d
81) a 82) a 83) a 84) c
85) a 86) d 87) c 88) c
89) d 90) b 91) d 92) d
93) b 94) d 95) a 96) a
97) d 98) c 99) b 100) a
1 (d)
Wt. of O inFe O and FeO is 48 ∶ 16
2 (c)
Equivalent weight of bivalent metal=37.2
∴ Atomic weight of metal=37.2 × 2 = 74.4
∴ Formula of chloride=𝑀Cl
Hence, molecular weight of chloride
𝑀Cl = 74.4 + 2 × 35.5
= 145.4
3 (d)
∵ 0.0833 mole of carbohydrate has hydrogen=1 g
∴ 1 mole of carbohydrate has hydrogen
= .
= 12 g
Given, empirical formula of carbohydrate (CH O)has 2 g of hydrogen.
∴ 𝑛 =
12
2
= 6
∴ Molecular formula of carbohydrate is
(CH O) = (CH O) = C H O
4 (a)
Eq. wt. Zn(OH) =
. .
= ;
Acidity of Zn(OH) = 1; only one OH is replaced.
5 (d)
M. f. =
5.85/58.5
5.85
58.5
+
90
18
= 0.0196
6 (a)
2Ag CO
∆
⎯
⎯ 4 Ag + 2CO + O
2 × 276 g 4 × 108 g (𝑠)
∵ 2 × 276 g of Ag CO gives=4 × 108 g Ag
∴ 1 g of Ag CO gives=
×
×
∴ 276 g of Ag CO gives=
× × .
×
=2.16 g
7 (a)

For phenolphthalein:
1
2
Meq. of Na CO = 2.5 × 0.1 × 2 = 0.5
For methyl orange:
Meq. of Na CO + Meq. of NaHCO
= 2.5 × 0.2 × 2 = 1.0
∴ Meq. of NaHCO = 0.5 and Meq. of Na CO = 1.0
∴ × 1000 = 0.5 /
× 1000 = 1
∴w = 0.042 g in 10 mL ∴𝑤 = 0.053 g in 10 mL
∴𝑤 = 4.2 g in 1 litre = 5.3 g in 1 litre
8 (d)
∵ 18 g water has 𝑁 molecules
∴1 g water has molecules
or molecules occupy volume = 1cm 𝑑 =
∴ 1molecule occupies volume
= = . ×
≈ 3 × 10 cm
9 (c)
𝑚 =
𝑤𝑅𝑇
𝑃𝑉
=
510 × 10 × 0.0821 × 273
1 × 67.2/1000
= 170
10 (a)
Suppose the volume of 6 M HCL required to obtain 1 L of 3 M
HCl = 𝑥 L
∴ volume of 2 N HCl required = (1 − 𝑥) L
Applying the molarity equation
𝑀 𝑉 + 𝑀 𝑉 = 𝑀 𝑉
6M HCl + 2 MHCl 3M HCl
6𝑥 + 2(1 − 𝑥) = 3 × 1
4𝑥 = 1
𝑥 = 0.25 L
Hence, volume of 6M HCl required = 0.25 L
and volume of 2M HCl required = 0.75 L
11 (a)
𝑁 = 𝑀 × acidity = 1 × 2 = 2 (Na CO is diacidic base)
12 (d)
1 mole of H SO gives = 3 moles of ions or 3 × 6.023 × 10 ions
∴ 0.1 mole of H SO will give = 0.1 × 3 × 6.023 × 10 ions
= 1.8 × 10 ions
13 (b)
Eq. of element = Eq. of oxygen or
𝑊
𝐸
=
𝑊 − 𝑊
8
14 (c)
1 mole of (𝑁𝐻 ) 𝑃𝑂 contains 12 moles of hydrogen atoms.
∴ 12 moles of hydrogen atoms ≡ 1 mole of (NH ) PO
∴ 1 moles of hydrogen atom = mole of (NH ) PO

∴ 6.36 moles of hydrogen atom = × 6.36
=
.
mole of (NH ) PO
1 mole of (NH ) PO = 4 moles of oxygen
So,
.
mole of (NH ) PO =
× .
= 2.12mol
15 (c)
Meq. of HCl= Meq. of NaOH;
Thus, × 𝑉 = 20 ×
𝑉 = 40 mL
16 (a)
Molecular weight = Eq. wt. × valence factor
17 (d)
Smallest and largest amount of energy respectively eV and L-atm.
1 eV = 1.6 × 10 J
1 L − atm = 101.325 J
18 (d)
∵ 63.8 g of Cu has atoms = 6.023× 10
∴ 1𝑔of Cu has =
. ×
.
∴ 0.635 𝑔 of Cu has =
. ×
.
× 0.635
= 6.023 × 10 atoms
19 (b)
2BCl + 3H → 2B + 6HCl
2 mol 3 mol 2 mol
21.6 g=2 mol
21.6 g B= 2 mol B≡ 3 mol H
𝑝𝑉 = 𝑛𝑅𝑇
∴ 𝑉 =
𝑛𝑅𝑇
𝑃
=
3 × 0.0821 × 273
1
= 67.2 L
20 (b)
𝑁
𝑛
=
𝑁 .× 𝑛
𝑛
= 𝑁 .
21 (c)
𝑛𝑖𝑠an integer.
22 (a)
Conservation of mass should be noticed.
23 (c)
The volume of water changes with temperature.
24 (c)
∵ Amount of heat evolved on combustion of 4 g of methane=10.46 kJ
∴ The amount of heat evolved on combustion of one mole of methane (𝑖𝑒, 16 g of CH )
=
.
× 16 = 41.84kJ
25 (c)
Mol. wt. = 70 × 2 = 140;
(CO)𝑥,∴(12 +16). 𝑥 = 140 ∴ 𝑥 = 5
28 (c)

Mole fraction of solute = ;
Mole fraction of solvent = ;
29 (a)
We have HNO →
+2
𝑁𝑂
Change in oxidation number = 3
Equivalent mass of HNO = = 21 g eq
30 (b)
5.6 litre = 60 g
∴ 22.4 litre = 240 g = mol. wt.
∴ Vapour density = 𝑀/2 = 120
31 (d)
32 g O contains 2𝑁 atoms.
33 (b)
Mol. wt. of metal chloride = 95 × 2 = 190
At. wt. of metal =
.
.
= 49.23
Let the metal chloride be 𝑀Cl
Then 49.23 + 𝑛 × 35.5 = 190
∴ n = 3.9 ≈ 4;
∴Eq. wt. of metal =
.
= 12.3
34 (c)
Atomic weight of element,
𝑀 =equivalent weight×valency
= 20 × 3
= 60
Molecular formula of its oxide=𝑀 𝑂
Hence, molecular weight of oxide
= 2 × 60 + 3 × 16
= 120 + 48 = 168
35 (b)
Gram molecular volume of oxygen at STP is 5.6L or 5600 cm .
36 (d)
Element Percentage At.
Wt.
Moles Simple
st
Ratio
𝑋
𝑌
75.8
24.2
75
16
75.8
15
= 1
24.2
16
= 1.5
2
3
∴ The formula of the compound is𝑋 𝑌 .
37 (c)
Meq. of oxalic acid = 500 × 0.1 = 50

 × 1000 = 50
∵ 𝑤 = × ∵ 𝐸 =
= 3.15 g
38 (d)
In acidic medium following reaction takes place.
8H + 5𝑒 + MnO → Mn + 4H O
∴ Equivalent weight of KMnO in acidic medium
=
molecular weight of KMnO
5
=
158
5
= 31.6
39 (d)
6Fe + Cr O + 14H → 6Fe + 2Cr + 7H O
+6
Cr O
→ Cr
𝑥-factor=6
Mohr’s salt, FeSO . (NH ) SO . 6H O
oxidation; Fe → Fe
𝑥-factor=1
Mole ratio is reverse of 𝑥-factor ratio. Therefore, one mole of dichromate required=6 moles
of Mohr’s salt.
40 (a)
Particle pressure of oxygen = × 2660
= 760 mm
Thus, 1 L oxygen gas is present at 0°𝐶 and 760 mm pressure.
∴ Number of oxygen molecules =
. ×
.
41 (b)
2Ag + 2HNO ⟶ 2AgNO + H
2AgNO + 2NaCl ⟶ 2AgCl + NaNO
AgCl ≡ AgNO ≡ Ag
143.5g 170 g 108g
∵ 143.5 g AgCl is obtained from Ag = 108g
∴ 2.87 g AgCl is obtained from Ag =
× .
.
= 2.16g
42 (d)
1 mole is defined as the amount of matter that contains as many as objects (atoms,
molecule, electron, proton or whatever, objects we are considering) as the number of atoms
in exactly 12g of C , 𝑖. 𝑒., Avogadro’s number.
43 (d)
∵ Number of atoms present in 12 g carbon
= 6.023 × 10
∴ No. of atoms present in 1 mg carbon
=
6.023 × 10 × 1
12 × 1000
= 0.502 × 10
44 (c)

Meq. of H S = Meq. of Cu
𝑤
34/2
× 1000 =
63.5
63.5/2
× 1000
45 (b)
Given that, oxygen contents in element oxide is 20% by weight.
Hence, element contents in element oxide is 80% by weight.
Then, equivalent weight of unknown element= × 8
∴ Equivalent weight of unknown element=32
46 (c)
Molecular weight of cortisone = 360.4
Molecular weight of 21 carbon atom = 21 × 12 = 252
% of carbon in cortisone =
×
.
= 69.9%
47 (d)
The terms which involves only weights in their formula
𝑒. g. molality =
wt. of solute × 1000
mol. wt. of solute × wt. of solvent
are independent of temperature. On the other hand, since, volume change with
temperature, the terms having volume in their formula
𝑒. g. molality =
wt. of solute × 1000
mol. wt. of solute × volume of solvent
are dependent on temperature.
48 (c)
6 × 10 electron ≡1 equivalent.
49 (b)
Meq. of H SO = Meq. of NaOH
𝑉 × 0.1 × 2 = 30 × 2.0 × 1
∴ 𝑉 = 300 mL
50 (c)
Nitrogen shows variable valency and thus, have variable equivalent weight.
51 (d)
𝐸
𝐸 + 𝐸
>
𝐸
𝐸 + 𝐸
1.520
𝐸 + 17
=
0.995
𝐸 + 8
or 𝐸 = 9
52 (b)
Given, mass of C=10.5 g
H=1.0 g
𝑝=1atm
𝑉= 1 L
𝑇 = 127℃=127+273=400 K
Mass of gas=2.81 g
Weight of C + weight of hydrogen=10.5+1.0=11.5 g
∴ % of carbon=
.
.
× 100 = 91.3%

∴ % of hydrogen=
.
.
× 100 = 8.7%
Ele
men
t
% At.
weigh
t
Ratio of
atoms
Simplest
ratio
C
H
91.
3
8.7
12
1
91.3/12=
7.61
8.7/1=8.
7
7.61/7.61=
1
8.7/7.61
=1.14×7=
8
From gas equation, 𝑝𝑉 = 𝑛𝑅𝑇
or 𝑛 =
mass
mole mass
=
𝑝𝑉
𝑅𝑇
or 2.81/mole mass=
×
. ×
=92
Empirical formula wt.=C H
∴ Empirical formula=7 × 12 + 8 × 1
= 92
𝑛 =
molecular wt.
empirical formula wt.
=
92
92
= 1
Molecular formula=𝑛 (empirical formula)
= 1 (C H )
= C H
53 (b)
CH − NH + HNO → CH OH + N + H O
1 mole of methyl amine gives 1 mole N
𝑖. 𝑒., 22.4 L of nitrogen at NTP.
54 (a)
Meq. of MgCO = Meq. of H SO
 /
× 1000 = × 1000 ;
𝑤 = 3.5 g
55 (c)
Eq. of metal = Eq. of oxide
100
𝐸
=
24
8
∴ 𝐸 = 33.3
57 (b)
100 mL of 1 M AgNO ≡ 0.1 mol AgNO
100 mL of 1 M CuSO = 0.1 mol CuSO
2AgNO + H S → Ag S + 2HNO
2 mol 1 mol
0.1 mol 0.05 mol
CuSO + H S → CuS + H SO
1 mol 1 mol

0.1 mol 0.1 mol
∴ Ratio of the amounts of H S needed=0.05:0.1=1:2
58 (a)
Mole fraction = = 0.0177
59 (d)
H PO is dibasic acid; thus, Na HPO is normal salt of Eq. wt. = 𝑀/2
60 (a)
Meq. of NaOH = 250 × 0.1 = 25
∴ × 1000 = 25
∴ w = 1 g
61 (c)
At. wt. × specific heat ≈ 6.4
62 (b)
Ag S ≡ 2Ag
248g 2 × 108g
2 × 108 g Ag is obtained from Ag S = 248 g
1 g Ag will be obtained from Ag S =
×
×
=
248
216
g
But, the ore contains only 1.34% Ag S.
Thus, 1 g Ag is obtained from ore = × .
g
= 85.68 g
64 (a)
Number of atoms in 40 kg=
×
. ×
(∵ Weight of an atom=6.644 × 10 g)
= 6.02 × 10
∴ Number of gram atoms of element in 40 kg
=
6.02 × 10
6.02 × 10
= 10
66 (b)
Since, 1 g hydrogen combines with 80 g bromine, the eq. wt. of bromine = 80
∵ 4 g bromine combines with Ca = 1g
∴ 80 g bromine will combine with Ca =
×
= 20g
∴ Eq. wt. of Ca is 20 g.
67 (d)
Atomic mass of the metal=32 × 2 = 64
Formula of metal nitrate=𝑀(NO )
∴ Molecular mass=64+28+96=188
68 (b)
Mg + 2HCl ⟶MgCl + H
24 g Mg gives one mole H
69 (d)
Valence of 𝑀 = = 3,

Thus, formula of chloride is 𝑀Cl .
70 (b)
Eq. of metal = Eq. of oxide
1.6
𝐸
=
2
𝐸 + 8
; 𝐸 = 32
71 (b)
𝑀 =
5.85 × 1000
58.5 × 500
= 0.2
72 (b)
Valence of an element is variable say it is 2 and 3 in FeCl and FeCl respectively. Also
equivalent weight =
.
and thus, it is also variable.
73 (c)
At. wt. = Eq. wt.× 3 (valence = 3)
74 (c)
Meq. of Na CO ∙ 𝑥H Oin 20 mL=19.8×
∴ Meq. of Na CO ∙ 𝑥H Oin 100 mL = 19.8× × 5
∴ × 1000 =19.8× × 5
or
.
/
× 1000 =
.
∴ 𝑀 = 141.41
∴ 23 × 2 + 12 + 3 × 16 + 18𝑥 = 141.41
∴ 𝑥 = 2
75 (c)
At. wt. × specific heat = 6.4
76 (c)
Moles of Fe= = 10
Moles of N= = 5
Moles of H= = 20
Equal number of moles have equal number of atoms.
Hence, number of atoms in 560 g of Fe is twice that of 70 g N and is half that of 20 g of H.
77 (d)
Molecular mass of (𝐶𝐻𝐶𝑂𝑂) 𝐹𝑒=170
∴ In 100 g (𝐶𝐻𝐶𝑂𝑂) 𝐹𝑒, iron present = × 100 mg
= 32.9 mg
Since, this quantity of Fe is present in 400 mg of capsule,
∴ % of Fe in capsule =
.
× 100 = 8.2%
78 (b)
By the equation
Zn + I ⟶ Znl
Initial moles (if x be the wt. 0
Of Zn and I each initially)
No. of moles at the end − 0
Of reaction

So, fraction of Zn unreacted = = 0.74
79 (c)
Weight of pure NaCl=6.5 × 0.9 = 5.85 g
No. of equivalent of NaCl=
.
.
= 0.1
No. of equivalent of NaOH obtained=0.1
Volume of 1 M acetic acid required for the neutralisation of
NaOH =
0.1 × 1000
1
= 100 cm
82 (a)
Given vapour density=11.2
Molecular weight=2 × 11.2 = 22.4
∴22.4 g of gas occupies=22.4 L at STP
∴ 1 g of gas occupies=
.
.
× 1 = 1 L at STP
83 (a)
In the given metal nitride, nitrogen present is 28% that means, the nitride contains 28 g
nitrogen and 72 g metal.
Moles of metal =
Moles of nitrogen = = 2
⟹ Molar ratio, 𝑀: 𝑁 = : 2 = 3: 2
72
𝑥
= 3
∴ 𝑥 = 24
84 (c)
g atom of I=
.
= 0.2
g atom of oxygen = = 0.5
∴Ratio of g atoms I : O : : 2 : 5
85 (a)
0.5 mole of H O = 20 g; Also H O is monovalent, thus
Mol. wt. = Eq. wt.
∴1 mole of H O = 40 g
86 (d)
C H + 𝑎 +
𝑏
4
O ⟶ 𝑎CO + (𝑏/2)H O
10 Excess - -
0 10𝑎5𝑏
10𝑎 = 40𝑎 = 4
5𝑏 = 50𝑏 = 10
87 (c)
Milli mole of H SO
( .)
= Milli mole of H SO
( .)
10 × 18 = 𝑀 × 1000

∴ 𝑀 = 0.18
89 (d)
100 g alkaloid contains nitrogen=17.28 g
∴ 162 g alkaloid will contain nitrogen
=
17.28 × 162
100
g
= 27.9 g ≈ 28 g
Atomic weight of nitrogen=14
So, number of atoms of nitrogen present in one molecular of alkaloid= = 2
90 (b)
𝑀 =
moles of urea
volume in litre
=
6.02 × 10
6.02 × 10 ×
100
1000
= 0.01 𝑀
91 (d)
2BCl + 3H ⟶ 2B + 3HCl
2 × 10.8g𝐵 ≡ 3 × 22.4 LH
21.6g B ≡
× . × .
× .
= 67.2L H
92 (d)
Eq. of metal = Eq. of chlorine
= .
∴ 𝐸 =
.
= 17.75
93 (b)
5C O +
+7
2MnO
+ 16H → 10CO + 2Mn + 8H O
Equivalent weight=
=
158
5
= 31.6
94 (d)
Mol. wt. of C H OH
= 12 × 2 + 1 × 5 + 16 + 1 = 46 g
∵ 46 g of C H OH has hydrogen atoms
= 6 × Avogadro number
∴ 0.046 g of C H OH has hydrogen atoms
=
6 × 6.023 × 10 × 0.046
46
= 3.6 × 10 atoms of hydrogen.
95 (a)
Both have same empirical formula CH O.
96 (a)
Moles of H = .
= 0.67
Moles of N = .
= 0.22
Moles of H =
.
= 0.25
Moles of O = = 0.31
Larger is number of mole, more is number of molecule.
97 (d)

C H OH + 3O ⟶ 2CO + 3H O
2 mole CO is formed.
99 (b)
2NaHCO
∆
→ Na CO + H O + CO
Na CO
∆
→ Na CO
The no. of equivalent of NaHCO = No. of equivalent of Na CO formed. Thus , same
equivalent of HCl will be used.
100 (a)
Element %ntage
atomic wt.
Simplest
ratio
C
H
92.3
12
= 7.69
7.7
1
= 7.70
7.69
7.69
= 1
7.70
7.69
= 1
∴ Empirical formula=CH
1) a 2) a 3) a 4) b
5) c 6) a 7) a 8) d
9) b 10) c 11) b 12) a
13) c 14) b 15) c 16) a
17) b 18) c 19) a 20) b
21) a 22) b 23) a 24) a
25) b 26) a 27) b 28) a
IMPORTANT PRACTICE QUESTION SERIES FOR IIT-JEE EXAM – 2 (ANSWERS)

29) a 30) d 31) b 32) c
33) d 34) c 35) c 36) b
37) b 38) b 39) a 40) c
41) d 42) c 43) d 44) b
45) b 46) c 47) d 48) a
49) d 50) d 51) b 52) c
53) a 54) a 55) a 56) b
57) a 58) b 59) b 60) a
61) d 62) c 63) a 64) a
65) a 66) a 67) c 68) d
69) a 70) a 71) a 72) a
73) c 74) a 75) b 76) b
77) c 78) a 79) b 80) c
81) b 82) d 83) d 84) c
85) c 86) a 87) b 88) d
89) c 90) a 91) d 92) d
93) a 94) d 95) c 96) b
97) b 98) b 99) b 100) d
1 (a)
H and H are isotopes. Thus, they resemble very closely in their chemical properties
2 (a)
Cr O = Sn (Sn → Sn + 2e )
(𝑛 = 6) (𝑛 = 2)
1 Eq = 1 Eq
1
6
mol =
1
2
mol
∴ 1 mol of Sn = mol of Cr O
=
1
3
mol ofCr O
3 (a)
KOH + Na CO
(𝑥 mmol) (𝑦 mmol)
i. (𝑥 × 1) + (𝑦 × 2) × = × 15
ii. (𝑥 × 1) + (𝑦 × 2) = × 25 ⇒ 𝑦 = 0.5 and 𝑥 = 0.25
⇒ KOH = 𝑥 mmoles = 0.25 × 10 × 56 g = 0.014 g
4 (b)
2H O → 2H O + O
68 g 22.4 L (at STP)
Mass of H O in 1.5 N solution = 𝐸𝑤 of H O × 1.5 N
= 17 × 1.5 = 25.5 g
So, volume strength of 1.5 N H O solution
=
22.4 L × 25.5 g
68.0
= 8.4 L
7 (a)

3.9854 × 10 g = 1 atom
Thus, 1 g =
1
3.9854 × 10
atoms
= 2.5092 × 10 atoms
8 (d)
Sntaken = 1.056 g
Sn unreacted = 0.601 g
Sn reacted = 1.056 − 0.601 = 0.455
=
0.455
119
mol = 0.004 mol
I reacted = 1.947 g =
1.947
254
mol = 0.008 mol
Thus, Sn ∶ l = 0.004 ∶ 0.008 = 1 ∶ 2
Thus, empirical formula is Sn(l ) orSnl
9 (b)
0.4 mol HCl ≡ 0.4 mol Cl
0.2 mol CaCl = 0.4 mol Cl
Total moles = 0.8 mol Cl in 0.5 L solution
Thus, molarity = 1.6 M
10 (c)
I. P + O → P O
II. C H O + H SO → 12C + H SO ∙ 11H O
conc. charred
mass
III. Cu + HNO → NO
brown fumes
IV. Cl + 2NaOH → NaClO + NaCl + H O
Bleaching agent
12 (a)
2Y → 2Y + 6e (Y → Y + 3e , (𝑛 factor = 3)
6H
+ 6e → 3H 2H
+ 2e → H , 𝑛 factor = 2
1 eq if Y = 1 eq of H
(𝑛 = 3)(𝑛 = 2)
1
3
mol Y =
1
2
mol H
∴ Y: H = 2: 3

13 (c)
0.2 × 10
500
= 400
14 (b)
CaCO → CaO + CO
100 g = 1 mol 22 4 L at NTP
200 g 44.8 L at NTP
15 (c)
Na SO : H O ≡ 50: 50 (Mw of Na SO = 126)
Mole∶ ∶
Ratio: 1:7
16 (a)
K SO andK SeO are isomorphous
K SeO
⇒ 39 × 2 + 𝑥 + 64 = 142 + 𝑥
(142 + 𝑥)g of K SeO ⇒ 𝑥 g of Se
100 g of K SeO ⇒
𝑥
142 + 𝑥
× 100
∴
𝑥
142 + 𝑥
× 100 = 50
∴ 𝑥 = 142
17 (b)
H C O ⇌ H + HC O ⇌ H + C O
I II
Upto stage I-monobasic acid
II-dibasic acid
18 (c)
S O → 2SO + 8e
CrO ≡ S O
mEq = mEq
𝑉 × 0.1 × 3 ≡ 40 mL × 0.25 × 8
𝑉 = 266.67 mL
19 (a)
H C O ∙ 2H O = 126 g mol
1.575 g H C O .2H O =
1.575
126
= 0.0125 mol
Let the volume be = 𝑉L

∴
0.0125 mol
𝑉L
= 0.10 M
∴ 𝑉 =
0.0125
0.1
= 0.125 L
= 125 mL
20 (b)
MgCl + 2HCO ⟶ Mg(HCO )
Mg(HCO )
∆
⎯ MgCO + H O + CO
21 (a)
𝐴 decolourises MnO /H
Thus, 𝐴 is CaC O
CaC O ⟶ CaO + CO + CO
𝐴𝐷𝐶𝐵
𝐵 burns with blue flame
CaO + H O ⟶ Ca(OH)
𝐸
Ca(OH) ⎯ CaCO
𝐸 milk
23 (a)
20.00 mL of 0.1 M Pb = 2 × 10 mol Pb
30.00 mL of 0. M SO = 3 × 0 mol SO
Pb + SO → PbSO
Pb is in the limiting a quantity and every 1 mole of Pb gives equal moles of PbSO
hence, PbSO formed = 0.002 mol
24 (a)
Total weight of alcohol and water
= 10 × 0.7893 + 20 × 0.9971
Volume of mixture =
× . × .
.
= 29.08 mL
Change in volume = (20 + 10) − 29.08
= 0.92 mL
% change in volume =
. ×
= 3.06%
≈ 3.1%
25 (b)
(a) 0.208 g BaCl =
.
mole
= 1 × 10 mole
= 1 millimoleBaCl

= 2 millimolesCl
(b) 100 mL of 0.1 M BaCl = 100 × 0.1 millimolesBaCl
= 100 × 0.2 millimoles Cl
= 20 millimoles
(c) 0.745 g KCl =
.
.
mole KCl
= 0.01 mole
= 0.01 × 1000 millimoles
= 10 millimoles
26 (a)
H PO has structure
27 (b)
Na C O ; KHC O ∙ H C O
𝑥 mmol 𝑦 mmol
i. 𝑥 × 0 + 𝑦 × 3 = 0.2 × 𝑉
ii. 𝑥 × 2 + 𝑦 × 4 = (0.2 × 5) × 𝑉 ⇒ =
28 (a)
2NH + OCl⊝
→ N H + Cl⊝
+ H O
29 (a)
Volume of H at STP = 24.62 × = 22.40 mL
22400 mL of H at STP = 1 mole = 2 Eq of H
22.4 mL of H = × 22.4
= 0.002 Eq of H
= 0.002 Eq of metal
∴
Weight
𝐸𝑤
= 0.002
0.05
0.002
= 𝐸𝑤 = 25
30 (d)
Balance the equation :
2CrO + 3H SO → Cr (SO ) + 3H O +
7
2
O
2 mol CrO ⇒
7
2
mol of O

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