3.1 Laws of Chemical Combination when 3.00 g of carbon is burnt in 50.

00 g of
oxygen ? Which law of chemical combinations
Two laws were laid by Antoine L. Lavoisier
will govern your answer ?
3.1.1 Law of conservation of mass Sol. 3g of carbon reacts with 8 g of oxygen to
• Matter can neither be created nor destroyed in a produce 11g of carbon dioxide. If 3g of carbon is
chemical reaction. burnt in 50g of oxygen, then 3g of carbon will react
with 8 g of oxygen. The remaining 42 g of oxygen will
• “In a chemical reaction total mass remains
be left un-reactive. In this case also, only 11g of
conserved i.e. mass of the products remain
carbon dioxide will be formed. The above answer is
equal to the mass of reactants. e.g.
governed by the law of constant proportions.
ice → Water (No change in weight)
DALTON’S ATOMIC THEORY
C + O2 → CO2
• Matter consists of extremely small indivisible
12g + 32g = 44g
particle called atoms.
Q. In a reaction 5.3 g of sodium carbonate • Atom can neither be created nor be destroyed
reacted with 6 g of ethanoic acid. The products during a chemical reaction.
were 2.2 g of carbon dioxide,0.9g water and 8.2 g
of sodium ethanoate. Show that these • All the atoms of a given element have identical
observations are in agreement with the law of mass and chemical properties.
conservation of mass. Sodium carbonate + • Atoms of different elements have different mass &
ethanoic acid → sodium ethanoate + carbon chemical properties.
dioxide + water • Atoms combine in the ratio of small whole numbers
Sol. Sodium Carbonate + Ethanoic acid → Sodium to form compounds.
ethanoate + Carbon dioxide + Water • The relative number and kinds of atoms are
Na2CO3 + CH3COOH → CH3COONa +CO2 +H2O constant in a given compound
5.3g 6g 8.2g 2.2 0.9g • An atom is the smallest particle that takes part in a
Mass of sodium carbonate = 5.3g chemical reaction.
Mass of ethanoic acid = 6g
Mass of sodium ethanoate = 8.2g Q.3 Which postulate of Dalton's atomic theory is
result of the law of conservation of mass?
Mass of carbon dioxide = 2.2
Sol. Atoms are indivisible particles, which can
Mass of water = 0.9g
neither be created nor destroyed in a chemical
Total mass of the reactants = (5.3 + 6) g = 11. 3g reaction.
Total mass of the products = (8.2 + 2.2 + 0.9)= 11.3g
Q.4 Which postulate of Dalton's atomic theory
Total mass of reactants = Total mass of products
can explain the law of definite proportions?
Hence, the given observations are in Sol. The relative number and kind of atoms in a
agreement with the law of conservation of mass. given compound remains constant.
3.1.2 Law of Definite Proportions Q. A 0.24 g sample of compound of oxygen and
• It stated that a chemical compound is always boron was found by analysis to contain 0.096 g if
made up of same elements combined together boron and 0.144 g of oxygen. Calculate
in the same proportion by mass irrespective of percentage composition of compound by weight.
the source. Sol. Mass of boron = 0.096g
• e.g. we may obtain water (H20) from any source Mass of oxygen = 0.144g Mass of sample = 0.24g
like rain, river well, sea, lake etc. but the ratio of Thus, percentage of boron by weight in the
hydrogen & oxygen is of 1:8. Similarly, the ratio of compound = 0.096 × 100% / 0.24 = 40%
carbon & oxygen in CO2 is 12:32 or 3:8 by mass. Thus, percentage of oxygen by weight in the
Q.2 Hydrogen and oxygen combine in the ratio compound = 0.144 × 100%/ 0.24 = 60 %
of 1 : 8 by mass to form water. What mass of 3.2 Atoms
oxygen gas would be required to react
• Atom is the smallest individual particles of the
completely with 3g of hydrogen gas?
matter.
Sol. It is given that the ratio of hydrogen and oxygen • Atoms are very small. Atomic radius is measured in
by mass to form water is 1:8. Then, the mass of nanometers.
oxygen gas required to react completely with 1g of • 1nm = 10—9 m
hydrogen gas is 8g . Therefore, the mass of oxygen
• Radius of hydrogen atom is 10—10 m
gas required to react completely with 3g of hydrogen
gas is 8 × 3g = 24 g. 3.2.1 WHAT ARE THE MODERN DAY SYMBOLS
Q.2 When 3.0 g of carbon is burnt in 8.00 g OF ATOMS OF DIFFERENT ELEMENTS?
oxygen, 11.00 g of carbon dioxide is produced. • Dalton was the first scientist to use the symbols for
What mass of carbon dioxide will be formed elements in a very specific sense.
 • they are also called homo-atomic molecules.
• They are formed by only one type of atoms are
known. E.g. H2, N2, P4 etc.
3.3.2 MOLECULES OF COMPOUNDS
• they are also called hetero-atomic molecules.
• They are formed by different types of atoms. E.g.
CO2, NO2, CH4, HCl etc.
• Berzilius suggested that the symbols of elements
3.3.3 WHAT IS AN ION?
should be made from one or two letters of the
name of the element. • Charged species formed by loss or gain of electron
• IUPAC (International Union of Pure and Applied is called ion. E.g. Na+, Mg2+, Cl–, O2– etc.
Chemistry) approves names of elements, symbols • They are of two types; cation and anion.
and units.
Cation
• Many of the symbols are the first one or two letters
of the element’s name in English. • A positively charged ion is known as cation.
• The first letter of a symbol is always written as a • A cation is formed by the loss of one or more
capital letter and the second letter as a small letter. electrons by an atom.
E.g. (i) hydrogen, H (ii) aluminium, Al and not AL. • The ions of all the metal elements are cations.
• Other symbols of the elements are also taken from • Example: Sodium atom loses 1 electron to form a
their Latin, German or Greek names. E.g. symbol sodium ion, Na+, which is cation :
of iron is Fe from its Latin name ferrum, sodium is
Na from natrium, potassium is K from kalium.

• Cation contains less electrons than parent atom

Anion

3.2.2 ATOMIC MASS • A negatively charged ion is known as anion.

• An anion is formed by the gain of one or more
• It is defined as no. of times mass of an atom of that electrons by an atom.
element is heavier than an1/12th mass of one atom
• The ions of all the non-metal elements are anions.
of carbon-12.
• Example: A chlorine atom gains 1 electron to form
• One atomic mass unit is defined as a mass exactly
equal to 1/12th of the mass of one atom of C-12 a chloride ion, Cl–, which is an anion.
isotope. 1amu=1.66 ×10–24 g or 1/NA g.
3.2.3 HOW DO ATOMS EXIST?
Atoms of most elements do not exist independently. An anion contains more electrons than parent atom.
Atoms form molecules and ions.
Q. Define atomic mass unit.
Sol. 1/12th mass of one atom of carbon - 12 is called
one atomic mass unit. It is written as 'u'
Q. Why is it not possible to see an atom with
naked eyes? • Simple ions. Those ions which are formed from
Sol. The size of an atom is so small that it is not single atoms are called simple ions.
possible to see it with naked eyes. Also, atom of an Example: Sodium ion, Na+, is a simple ion
element does not exist independently. because it is formed from a single sodium atom,
Na.
3.3 What is a Molecule?
• Compound ions or polyatomic ions
• Molecules are formed by the combination of two or A polyatomic ion is a group of atoms carrying a
more atoms. e.g. H2, N2, CO2, NO2, etc charge (positive or negative).
• It is the smallest particle of the matter which always • Example: Ammonium ion NH4+, is a compound ion
have independent existence. which is made up of two types of atoms joined
together, nitrogen and hydrogen.
3.3.1 MOLECULES OF ELEMENTS
Q. What are polyatomic ions? Give examples? Baking Sodium,hydrogen, carbon,
NaHCO3
Sol. A polyatomic ion is a group of atoms carrying a powder oxygen
charge (positive or negative). E.g. ammonium Potassium
2−
ion (NH+ −
4 ), hydroxide ion (OH ), carbonate ion (CO3 ), sulphate
k2SO4 Potassium,sulphur,oxygen
2−
sulphate ion (SO4 ).
Atomicity
The number of atoms constituting a Molecule is known as its atomicity.

3.5 Molecular Mass

3.4 Writing Chemical Formulae
• It is the sum of atomic masses of all elements present
• Chemical formula represents exact no. of atoms of in a molecule.
an element present in a molecule. • It is obtained by multiplying the atomic mass of each
• Chemical formula can be written with the help of the element by the number of its atoms and adding them
valency. together. e.g.
• Valency of the ions is equal to their charge. Molecular mass of methane, (CH4) = (12u) + 4(1)= 16u
Molecular mass of (H2O)= 2 (1) + 16u= 18u
Problem : Calculate molecular mass of
glucose(C6H12O6) molecule.
Solution Molecular mass of glucose (C6H12O6)
= 6(12u) + 12(1u) +6(16u) = 180 u
3.5.2 FORMULA UNIT MASS
• The formula unit mass of a substance is a sum of the
atomic masses of all atoms in a formula unit of a
compound.
• E.g., formula unit mass of NaCl is– 1 × 23 + 1 × 35.5 =
58.5 u
Q. Calculate the molecular masses of H2, O2, Cl2,
Q.4 Write the chemical formula of the CO2, CH4, C2H6, C2H4, NH3, CH3OH.
following: Sol. Molecular mass of H2 = 2 x1 = 2u
(a) Magnesium chloride (b) Calcium oxide Molecular mass of O2 = 2 × 16 = 32u
(c) Copper nitrate (d) Aluminium chloride Molecular mass of Cl2 = 2 × 35.571 u
(e) Calcium carbonate Molecular mass of CO2 = 12 + 2 × 16 = 44 u
Molecular mass of CH4 = 12 + 4 × 1 = 16 u
Sol. (a) Magnesium chloride →MgCl2 Molecular mass of C2H6 = 2×12 + 6×1 = 30u
(b) Calcium oxide →CaO Molecular mass of C2H4 = 2×12 + 4×1= 28u
(c) Copper nitrate →Cu(NO3)2 Molecular mass of NH3 = 14 + 3 × 1 =17 u
(d) Aluminium chloride →AlCl3 Molecular mass of CH3OH = 12+4×1+16 = 32 u
(e) Calcium carbonate →CaCO3 Q. Calculate the formula unit masses of
Q.5 Give the names of the elements present in ZnO, Na2O, K2CO3, given masses of Zn = 65u, Na=
the following compounds: 23u, K =39u, C = 12u, and O = 16u.
(a) Quick lime (b) Hydrogen bromide Sol. Formula unit mass of ZnO = 65 + 16 = 81 u
(c) Baking powder (d) Potassium sulphate Formula unit mass of Na2O = 2 × 23 + 16 = 62u
Chemical Formula unit mass of K2CO3 = 2×39+12+3×16 = 138u
Compound Elements present
formula Q.6 Calculate the molar mass of the following
Quick lime Cao Calcium,oxygen substances:
(a) Ethyne, C2H2 (b) Sulphur molecule, S8
Hydrogen (c) Phosphorus molecule, P4
HBr Hydrogen,bromine
bromide
(d) Hydrochloric acid, HCl (e) Nitric acid, HNO3
Sol.
(a) Molar mass of ethyne, C2H2 = 2×12+2×1 = 28g
(b) Molar mass of sulphur molecule, S8 = 8×32 =
256g
(c) Molar mass of P4 = 4×31 = 124g
(d) Molar mass of HCl = 1 + 35.5 = 36.5g
(e) Molar mass of HNO3 = 1 + 14 + 3 × 16 = 63g
 discharge tube not only cathode ray are produced,
even some rays are also seems to be produced from
the anode side carrying positive charge.
• They are also known as anode rays or canal rays.
Atoms
Discovery of Neutrons
Atoms are the building blocks of matter. It is the
• Subatomic particle having mass almost equal to the
smallest unit of matter that is composed of three sub-
proton (1.675 x 10-24g) & electrically neutral is called
atomic particles: the proton, the neutron, and the
neutron. It was discovered by Chadwick (1932) by
electron.
bombarding a thin sheet of beryllium by -particles.
Cathode ray experiment
4Be + 2He (ά )----→ 6C
9 4 12
+ 0n1
• J. J. Thomson discovered the existence of electrons. • Heaviest & least stable particle of atom.
• He performed a cathode ray tube experiment. • Isotopes are formed as a result of difference in the
• Cathode ray tube is a vacuum-sealed tube with number of only neutrons in the nuclei of atoms.
a cathode and anode on one end. Property Electron Proton Neutron
• A beam created of electrons from cathode travelling
towards the other anode when high pressure Relative
(0.0001mm) and high voltage (10000V) is applied. mass with 1/1837 1 1
H-atom
• The characteristics of cathode rays (electrons) do not
depend upon the material of electrodes and the 9.11 x 10– 1.672 x 10– 1.675 x 10–
Mass in g 28 24 24
nature of the gas present in the cathode ray tube. g g g
Mass in
0.00549 1.0072 1.0087
amu
Charge In 1.672 x 10– 1.672 x 10–
19 19 0
Coulomb C C
Actual
charge 4.8 10–10 4.8 10–10 0
(e.s.u.)
Rutherford’s Nuclear Model of Atom
(–particle scattering experiment)
Electrons Rutherford bombarded very thin gold foil (thickness
100nm) with –particles (He+). This is called –
• Electrons are the negatively charged sub-atomic
particle scattering experiment
particles of an atom.
• The mass of an electron is considered to be
negligible, and its charge is -1.
• The symbol for an electron is e–
• Electrons are extremely small.
• They are found outside the nucleus.
Discovery of Protons

• As atom is electrically neutral, there should be

another subatomic particle present having equal &
opposite charge to electron i.e. +ve charge.
• It is discovered by Goldstein.
• He made hole in the cathode (perforated cathode) &
found that when high voltage is applied in a
 Observations & Conclusion
• Most of the ά particles passed through the gold foil
without any deflection. Hence most of the space in
an atom is empty.
• Few ά –particles were deflected by small angles.
Hence the positive charge in an atom is not
uniformly distributed.
• A very few ά particles (1 in 20,000) bounced back &
• Atom is a very small, spherical & electrically neutral
deflected by nearly 180°. So the volume occupied particle.
by the positively charged particles in an atom is • Positive charge is uniformly distributed throughout the
very small as compared to the total volume of an atom & electrons are embedded into it like seeds in
atom. watermelon or plum pudding in cakes. That’s why it is
Main Points of Model also called plum pudding, raisin pudding or
watermelon model.
(i) The empty space around nucleus is called extra-
nuclear part.. Drawbacks of this model
(ii) The mass of the atom is concentrated in a small • It could not explain the stability of atom as –ve & +ve
region having +ve charge, called nucleus. charge are in direct contact.
(iii) Protons & neutrons are present inside the nucleus • It could not explain no. of experiments like –particle
& called nucleons. scattering experiment of Rutherford.
(iv) +ve charge on nucleus is due to protons. • According to this model electron are stationary but
actually they are moving.
(v) The electrons are moving around nucleus in
circular paths called orbits like solar system in Bohr’s Model of Atom
which planets revolves around the sun.
(vi) Radius of nucleus is 10—15 & is of atom is 10—10.
Dissimilarities between nuclear & solar system
Nuclear model Solar system
They are very small Very large bodies
They are charged Neutral
Revolution of electron around Revolution of planets
nucleus is governed by around sun is governed
centrifugal force. by gravitational force.
Drawbacks of Rutherford Model
• It failed to explain the stability of atom.
• It does not explain the spectrum of the atoms.
Atomic Number and Mass Number (Z)
• An atom consists of heavy positively charged nucleus &
Atomic number (Z) electrons are revolving around it in circular orbits.
= number of protons in the nucleus of an atom
• Electrons revolve only in those orbits which have fixed
= number of electrons in a neutral atom amount of energy & known as energy levels or
e.g. number of protons in the hydrogen nucleus is 1, in stationary states.
sodium atom it is 11, therefore their atomic numbers are • These levels are represented either by letters
1 and 11 respectively. K,L,M,N…. or numbers 1,,2,3,4….
Mass Number (A) • As long as an electron remains in a particular orbit, it
= number of protons (Z) + number of neutrons (n) does not lose or gain energy and its energy remains
= number of electrons + number of neutrons (n) constant. This is called ground state of atom.
= Atomic number + number of neutrons (n) • The change in energy takes place when electron
jumps from one energy level to another energy level.
Thomson Model of Atom This is called excited state of atom.
 No. of Neutrons = 4-2 =2
Electron distribution in different orbits
• No. of electrons in a given shell can be determined
by formula 2n2 where ‘n’ is the number or energy
level energy shell. E.g.
No. of Shell Formula Total No. of
electron in shell
1 2 x 12 2
2 2 x 22 4
2
3 2x3 18
2
4 2x4 32
• The shells are always filled in a step-wise manner
Q.1 On the basis of Thomson’s model of an
from the lower to higher energy levels. Electrons
atom, explain how the atom is neutral as a whole.
are not filled in the next shell unless previous shells
Sol. According to Thomson’s model of the atom,
are filled.
negatively charged particles (electrons) are
embedded in the uniformly distributed positive Valency
charge. These • The outermost shell of an atom is known as the
negative and positive charges are equal in valence shell.
magnitude. Thus, by counterbalancing each other’s • The electrons present in the outermost shell of an
effect, they make an atom neutral. atom are known as the valence electrons.
Q.2 On the basis of Rutherford’s model of an • The combining capacity of the atoms or their
atom, which subatomic particle is present in the tendency to react and form bond with atoms of
nucleus of an atom? the same or different elements is known
Sol. On the basis of Rutherford’s model of an as valency of the atom.
atom, protons (positively-charged particles) are
• It is equal to no. of valance electron if up to 4
present in the nucleus of an atom.
valance electron. If an atom have more than 4
Q.3 Draw a sketch of Bohr’s model of an atom valance electron, subtract them from 8.
with three shells.
Sol. Element Li Be B C N O F Ne
E.C. 2,1 2,2 2,3 2,4 2,5 2,6 2,7 2,8
Valency 1 2 3 4 3 2 1 0
• It generally increases upto middle and then
Q.4 What do you think would be the observation decreases when we go left to right in the periodic
if the α-particle scattering experiment is carried table. E.g.
out using a foil of a metal other than gold? • It remains same along the group. E.g.
Sol. If the α-scattering experiment is carried out
Group-1 Group-2
using a foil of a metal rather than gold, there would
be no change in the observation. In the α-scattering Li 2,1 1 Be 2,2 2
experiment, a gold foil was taken because gold is Na 2,8,1 1 Mg 2,8,2 8
malleable and a thin foil of gold can be easily made. It
K 2,8,8,1 1 Ca 2,8,8,2 2
is difficult to make such foils from other metals.
Q.1 Name the three sub-atomic particles of an Q.1 Write the distribution of electrons in carbon
atom. and sodium atoms?
Sol. The three sub-atomic particles of an atom are: Sol. The total number of electrons in a carbon
(i) Protons atom is 6. The distribution of electrons in carbon atom
(ii) Electrons, and is given by: First orbit or K-shell = 2 electrons
(iii) Neutrons Second orbit or L-shell = 4 electrons
Q.2 Helium atom has an atomic mass of 4 u and Or, we can write the distribution of electrons in
two protons in its nucleus. How many neutrons a carbon atom as 2, 4.
does it have? The total number of electrons in a sodium
Sol. Atomic Mass = No. of proton + No. of atom is 11. The distribution of electrons in sodium
Neutrons atom is given by:
First orbit or K-shell = 2 electrons
Atomic mass of helium = 4
Second orbit or L-shell = 8 electrons
No. of proton = 2 Third orbit or M-shell = 1 electron
4 = 2 + No. of Neutrons
 Or, we can write distribution of electrons in a
sodium atom as 2, 8, 1.
Q.2 If K and L shells of an atom are full, then
what would be the total number of electrons in
the atom?
Sol. The maximum number of electrons that can
occupy K and L-shells of an atom are 2 and 8
respectively. Therefore, if K and L-shells of an atom
are full, then the total number of electrons in the atom
would be (2 + 8) = 10 electrons.

•
Page 35

Q.1 Define atomic mass unit.

Sol. Mass unit equal to exactly one- twelfth the
mass of one atom of carbon - 12 is called one
atomic mass unit. It is written as 'u'

Q.2 Why is it not possible to see an atom

with naked eyes?
Sol. The size of an atom is so small that it is
not possible to see it with naked eyes. Also, atom
of an element does not exist independently.

Page 39

Page 40

Page 42
Q.1 If one mole of carbon atoms weighs 12
gram, what is the mass (in gram) of 1 atom of
carbon?
Sol. One mole of carbon atoms weighs 12g
(Given)
i.e., mass of 1 mole of carbon atoms =
12g
Then, mass of 6.022× 1023 number of
carbon atoms = 12g
Therefore, mass of 1 atom of carbon
= 126.022×1023g
= 1.9926 × 10−23g

Q.2 Which has more number of atoms, 100

grams of sodium or 100 grams of iron (given,
atomic mass of Na = 23u, Fe =56 u)?
Sol. Atomic mass of Na = 23u (Given)
Then, gram atomic mass of Na = 23g
Now, 23g of Na contains =
6.022×1023 number of atoms
Thus, 100g of Na contains
= 6.022×102323×100 number of atoms
= 2.6182 × 1024 number of atoms
Again, atomic mass of Fe = 56u (Given)
Then, gram atomic mass of Fe = 56g
Now, 56 g of Fe contains
= 6.022×1023 number of atoms
Thus, 100 g of
Fe 6.022×102356×100 number of atoms
= 1.0753 × 1024 number of atoms
Therefore, 100 grams of sodium contain
more number of atoms than 100 grams of iron.

ATOMS AND MOLECULES : NCERT Exercise

Questions
Q.1 A 0.24 g sample of compound of Chemical
oxygen and boron was found by analysis to Compound Elements present
formula
contain 0.096 g if boron and 0.144 g
of oxygen. Calculate the percentage Quick lime Cao Calcium,oxygen
composition of the compound by weight.
Sol. Mass of boron = 0.096g(Given) Hydrogen
HBr Hydrogen,bromine
Mass of oxygen = 0.144g (Given) bromide
Mass of sample = 0.24g (Given)
Baking Sodium,hydrogen, carbon,
Thus, percentage of boron by weight in the NaHCO3
powder oxygen
compound = 0.0960.24×100%
= 40% Potassium
Thus, percentage of oxygen by weight in k2SO4 Potassium,sulphur,oxygen
sulphate
the compound = 0.1440.24×100%
= 60 %

Q.2 When 3.0 g of carbon is burnt in 8.00 g Q.6 Calculate the molar mass of the
oxygen, 11.00 g of carbon dioxide is following substances:
produced. What mass of carbon dioxide (a) Ethyne, C2H2
will be formed when 3.00 g of carbon is burnt (b) Sulphur molecule, S8
in 50.00 g of oxygen ? Which law of chemical (c) Phosphorus molecule, P4 (atomic
combinations will govern your answer ? mass of phosphorus = 31)
Sol. Carbon + Oxygen → Carbon dioxide 3g of (d) Hydrochloric acid, HCl
carbon reacts with 8 g of oxygen to produce 11g (e) Nitric acid, HNO3
of carbon dioxide. If 3g of carbon is burnt in 50g Sol. (a) Molar mass of ethyne, C2H2 = 2 × 12
of oxygen, then 3g of carbon will react with 8 g of + 2 × 1 = 28g
oxygen. The remaining 42 g of oxygen will be left (b) Molar mass of sulphur molecule, S8 = 8
un-reactive. In this case also, only 11g of carbon × 32 = 256g
dioxide will be formed. The above answer is (c) Molar mass of phosphorus
governed by the law of constant proportions. molecule,P4 = 4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl =
1 + 35.5 = 36.5g
Q.3 What are polyatomic ions? Give (e) Molar mass of nitric acid, HNO3 = 1 +
examples? 14 + 3 × 16 = 63g
Sol. A polyatomic ion is a group of atoms
carrying a charge (positive or negative).For
example, ammonium ion (NH+4), hydroxide Q.7 What is the mass of --
ion (OH−), carbonate (a) 1 mole of nitrogen atoms?
ion (CO2−3),sulphateion (SO2−4). (b) 4 mole of aluminium atoms (Atomic
mass of aluminium = 27)?
(c) 10 moles of sodium
Q.4 Write the chemical formula of the sulphite (Na2SO3) ?
following: Sol. (a) The mass of 1 mole of nitrogen atoms
(a) Magnesium chloride is 14g.
(b) Calcium oxide (b) The mass of 4 moles of aluminium
(c) Copper nitrate atoms is (4 × 27)g = 108g
(d) Aluminium chloride (c) The mass of 10 moles of sodium
(e) Calcium carbonate sulphite (Na2SO3) is
Sol. (a) Magnesium chloride →MgCl2 10 × [2 × 23 + 32 + 3 × 16]g = 10 × 126g =
(b) Calcium oxide →CaO 1260g
(c) Copper nitrate →Cu(NO3)2
(d) Aluminium chloride →AlCl3
(e) Calcium carbonate →CaCO3 Q.8 Convert into mole.
(a) 12g of oxygen gas
(b) 12g of water
Q.5 Give the names of the elements present (c) 22g of carbon dioxide
in the following compounds: Sol. (a) 32 g of oxygen gas = 1 mole
(a) Quick lime Then, 12g of oxygen gas = 1232 mole =
(b) Hydrogen bromide 0.375 mole
(c) Baking powder (b) 18g of water = 1 mole
(d) Potassium sulphate Then, 20 g of water = 2018 mole = 1.11
Sol. . moles (approx)
(c) 44g of carbon dioxide = 1 mole
 Then, 22g of carbon dioxide = 2244 mole = (b) (i) and (iv)
0.5 mole (c) (ii) and (iii)
(d) (ii) and (iv)
Q.9 What is the mass of : Sol. (d)
(a) 0.2 mole of oxygen atoms? (ii) 20 moles of water = 20 ×18 g = 360 g of water,
(b) 0.5 mole of water molecules? because mass of 1 mole of water is the same as
Sol. (a) Mass of one mole of oxygen atoms = its molar mass, i.e., 18 g. Molarmassis the mass of
16g one mole of a substance.
Then, mass of 0.2 mole of oxygen atoms =
The molecular formula of water is H 2O, which
0.2 × 16g = 3.2g
means that water is made up of two hydrogen and
(b) Mass of one mole of water molecule =
one oxygen atoms.
18g
Then, mass of 0.5 mole of water molecules The standard atomic weight of hydrogen is
= 0.5 × 18g = 9g 1.00794 and that of oxygen is 15.9994. So the
molecular weight will be:
Molecular Weight of H2O = (2 × 1.00794) +
Q.10 Calculate the number of molecules of
(1×15.994) = 18.00988
sulphur (S8) present in 16g of solid sulphur.
Sol. 1 mole of solid sulphur (S8) = 8 × 32g = So the molar mass of water is 18.00988 gm or
256g 0.01800988 Kg.
i.e., 256g of solid sulphur contains = (iv) 1.2044 × 1025 molecules of water contains
6.022 × 1023 molecules 1.2044 × 1025/NA number of moles,
Then, 16g of solid sulpur Where NA = 6.023 × 1023
contains 6.022×1023256×16 molecules Therefore,
= 3.76 × 1022 molecules (approx) 1.2044 × 1025 /6.023×1023
= 0.199 x 102 moles = 20 moles
Q.11 Calculate the number of aluminium
ions present in 0.051g of aluminium oxide.
(Hint: The mass of an ion is the same as that Q.2 Which of the following statements is not true
of an atom of the same element. Atomic mass about an atom?
of Al = 27u) (a) Atoms are not able to exist independently
Sol. 1 mole of aluminium oxide (Al2O3) = 2 × (b) Atoms are the basic units from which molecules
27 + 3 × 16 = 102g and ions are formed
i.e., 102g of Al2O3 = 6.022 (c) Atoms are always neutral in nature
× 1023 molecules of Al2O3 (d) Atoms aggregate in large numbers to form the
Then, 0.051 g of Al2O3 contains matter that we can see, feel or touch
= 6.022×1023102×0.051 molecules Sol. (a)
= 3.011 × 1020 molecules of Al2O3
The number of aluminium
ions (Al3+) present in one molecules of Q.3 The chemical symbol for nitrogen gas is
aluminium oxide is 2. (a) Ni
Therefore, The number of aluminium (b) N2
ions (Al3+) present in (c) N+
3.11 × 1020 molecules (0.051g) (d) N
of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
Sol. (b)
= 6.022 × 1020

Q.4 The chemical symbol for sodium is

Atoms and Molecules : NCERT Exemplar
(a) So
(b) Sd
(c) NA
(d) Na
Multiple Choice Questions
Sol. (d)
Q.1 Which of the following correctly represents 360
g of water?
(i) 2 moles of H20 Q.5 Which of the following would weigh the
highest?
(ii) 20 moles of water
(a) 0.2 mole of sucrose (C12 H22 O11)
(iii) 6.022 × 1023 molecules of water
(b) 2 moles of CO2
(iv) 1.2044 ×1025 molecules of water
(c) 2 moles of CaCO3
(a) (i)
(d) 10 moles of H2O Mass of one atom of oxygen = Atomic mass/N A =
Sol. (c) 16/6.023 × 1023 g
Weight of a sample in gram = number of moles ×
molar mass Q.9 3.42 g of sucrose are dissolved in 18g of water
(a) 0.2 moles of C12H22O11 = 0.2 × 342 = 68.4 g in a beaker. The numbers of oxygen atoms in the
(b) 2 moles of CO2 = 2 × 44 = 88 g solution are
12.011 + 2(16.00) = 12.011 + 32.00 = 44.011g/mol. (a) 6.68 × 1023
(c) 2 moles of CaCO3 = 2 × 100 = 200 g (b) 6.09 × 1022
2(40 + 12 + 3(16) = 2(40 + 12 + 48) = 2 × 100 = (c) 6.022 × 1023
200g (d) 6.022 × 1021
(d) 10 moles of H2O = 10 × 18 = 180 g Sol. (a)
10(1 + 1 + 16) = 10(18) = 10(18) = 180 g

Q.6 Which of the following has maximum number

of atoms? = 3.42 g/342 g mol-1
(a) 18g of H2O = 0.01mol
(b) 18g of O2 1 mol of sucrose( C12H22O11) contains = 11×
(c) 18g of CO2 NA atoms of oxygen, where NA = 6.023×1023
(d) 18g of CH4 0.01 mol of sucrose (C12 H22 O11) contains = 0.01
× 11 × NA atoms of oxygen
Sol. (d)
= 0.11× NA atoms of oxygen
Number of atoms = Mass of substance × Number
of atoms in the molecule/ Molar mass × NA = 18 g/(1x2+ 16)gmol-1
(a) 18 g of water =18 x3/18 ×NA = 3 NA =18 g /18 gmol-1
(b) 18 g of oxygen = 18 x2 /32 × NA = 1.12 NA = 1mol
(c) 18 g of CO2 = 18 x3/44 × NA = 1.23 NA 1mol of water (H2O) contains 1×NA atom of
oxygen
(d) 18 g of CH4 =18 x5 /16 × NA = 5.63 NA
Total number of oxygen atoms =
Number of oxygen atoms from sucrose + Number
Q.7 Which of the following contains maximum
of oxygen atoms from water
number of molecules?
(a) 1g CO2 = 0.11 NA + 1.0 NA = 1.11NA
Number of oxygen atoms in solution = 1.11 ×
(b) 1g N2
Avogadro’s number
(c) 1g H2
= 1.11 × 6.022 ×1023 = 6.68 × 10 23
(d) 1g CH4
Sol. (c)
Q.10 A change in the physical state can be brought
1 g of H2 = ½ x NA = 0.5 NA = 0.5 × 6.022 × 1023 = about
3.011 × 1023
(a) Only when energy is given to the system
(b) Only when energy is taken out from the system
Q.8 Mass of one atom of oxygen is
(c) When energy is either given to, or taken out
from the system
(d) Without any energy change
Sol. (c)

Q.11 Which of the following represents a correct

chemical formula? Name it.
(a) CaCl
(b) BiPO4
(c) NaSO4
(d) NaS
Sol. (b) BiPO4— Both ions are trivalent Bismuth
phosphate(Bi3+- Trivalent anion. anion is an ion
that is negatively charged)
Sol. (a) (PO43- -Trivalent cation. A cation has a net
positive charge)
Q.12 Write the molecular formulae for the following Q.16 Which of the following symbols of elements
compounds are incorrect? Give their correct symbols
(a) Copper (II) bromide (a) Cobalt CO
(b) Aluminium (III) nitrate (b) Carbon c
(c) Calcium (II) phosphate (c) Aluminium AL
(d) Iron (III) sulphide (d) Helium He
(e) Mercury (II) chloride (e) Sodium So
(f) Magnesium (II) acetate (a) Cobalt CO -Incorrect, the correct symbol of
(a) Copper (II) bromide- CuBr2 Cobalt is Co
(b) Aluminium (III) nitrate = Al(NO3) 3 (b) Carbon c - Incorrect, the correct symbol of
(c) Calcium (II) phosphate - Ca3(PO4) 2 Carbon is C
(d) Iron (III) sulphide - Fe2S3 (c) Aluminium AL-Incorrect, the correct symbol of
Aluminium is Al
(e) Mercury (II) chloride - HgCl2
(d) Helium He - Correct
(f) Magnesium (II) acetate- Mg(CH3COO)2
(e) Sodium So- Incorrect, the correct symbol of
Sodium is Na
Q.13 Write the molecular formulae of all the
compounds that can be formed by the combination
of following ions Cu2+, Na+, Fe3+, C1– , SO4 2-, PO43- Q.17 Give the chemical formulae for the following
compounds and compute the ratio by mass of the
CuCl2/ CuSO4/ Cu3 (PO4) 2
combining elements in each one of them. (You
NaCl/ Na2SO4/ Na3 PO4 may use appendix-III).
FeCl3/ Fe2(SO4) 3 / FePO4 (a) Ammonia
(b) Carbon monoxide
Q.14 Write the cations and anions present (if any) (c) Hydrogen chloride
in the following compounds
(d) Aluminium fluoride
(a) CH3COONa
(e) Magnesium sulphide
(b) NaCl
(a) Ammonia
(c) H2
NH3
(d) NH4NO3
N : H×3
Anions
14 : 1 × 3
Cations
14 : 3
(a)
CH3 COO- Na (b) Carbon monoxide
(b) CO
– + C:O
Cl Na
(c) It is a covalent compound 12: 16
(d) 3:4
NO 3 − NH4 + (c) Hydrogen chloride
HCl
Q.15 Give the formulae of the compounds formed H : Cl
from the following sets of elements 1: 35.5
(a) Calcium and fluorine (e)Aluminium fluoride
(b) Hydrogen and sulphur AlF3
(c) Nitrogen and hydrogen Al : F × 3
(d) Carbon and chlorine 27 : 19 × 3
(e) Sodium and oxygen 27 : 57
(f) Carbon and oxygen 9 : 19
(a) Calcium and fluorine - CaF2 (f) Magnesium sulphide
(b) Hydrogen and sulphur- H2S MgS
(c) Nitrogen and hydrogen- NH3 Mg : S
(d) Carbon and chlorine – CCl4 24 : 32
(e) Sodium and oxygen – Na2O 3: 4
(f) Carbon and oxygen- CO2 ; CO
Q.18 State the number of atoms present in each of (ii)Sugar is added, 1 mg at a time, to the beaker
the following chemical species and it is mixed using the spatula. The sugar is
(a) CO3 2– continuously added until it is not able to dissolve in
(b) PO4 3– the water anymore. The amount of sugar that has
been added is calculated and recorded.
(c) P2O5
(iii)The beaker is now placed on the hot plate and
(d) CO
the temperature of the solution is brought to 40°C.
(a) CO3 2– = 1+3 =4 The salt is again added, 10mg a time, until it does
(b) PO4 3– = 1 + 4= 5 not dissolve in the water anymore. The total salt
(c) P2O5 = 2 + 5 = 7 added (including the amount of salt added
(d) CO = 1 + 1 =2 previously) is calculated and recorded
(iv)Step iii is again repeated at 60°C, 80°C and
Q.19 What is the fraction of the mass of water due 100°C and the results are recorded.
to neutrons? The results show that as the temperature of the
The mass of 1 mole of a substance is equal to its water increases, the amount of salt dissolving in
relative atomic or molecular mass in grams. the solution is also higher.
Mass of one mole (Avogadro Number) of neutrons
=1g Q.21 Classify each of the following on the basis of
Mass of one neutron = 1/ Avogadro number(NA) g their atomicity.
Mass of one molecule of water = Molar mass / N A = (a) F2
18/ NA g (b) NO2
The molar mass of water is 18.015 g/mol. This was (c) N2O
calculated by multiplying the atomic weight of (d) C2H6
hydrogen (1.008) by two and adding the result to (e) P4
the weight for one oxygen (15.999) (f) H2O2
Mass of one molecule of water = Molar mass / N A = (g) P4O10
18/ NA g
(H) O3
Avogadro number(NA) =6.022 x 1023 mol¯1
(i) HCl
There are 8 neutrons in one atom of oxygen
(j) CH4 (k) He (l) Ag
Number of neutrons in oxygen= number of oxygen
Atomicity is the number of atoms present in a
- Atomic number of oxygen
molecule
Oxygen's atomic weight= 15.9994
(a) 2- diatomic
increases with an increase in temperature.
Therefore the mass is 16 (b) NO2 = 1+ 2 = 3. Triatomic
Therefore number of neutrons= 16 - 8 = 8 (c)N2O = 2 + 1 = 3. Triatomic
Mass of one neutron = 1/ Avogadro number(NA) g (d) C2H6 = 2 + 6 = 8 Octa atomic
Mass of 8 neutrons = 8/ Avogadro number(NA) g (e) P4 = 4 Tetra
Fraction of mass of water due to neutrons = 8/18 g (f) H2O2 = 2 + 2 = 4. Tetra
(g) P4O10 = 4 + 10 = 14
Q.20 Does the solubility of a substance change (h) O3 = 3
with temperature? Explain with the help of an (i) HCl = 1+ 1 = 2
example (j) CH4 = 1+ 4= 5
Solubility refers to the ability of a given volume of (k) 1 (monatomic (inert gases)do not combine and
solvent to dissolve a substance. exist as monoatomic gases)
The solubility of a given solute in a given solvent (l) Polyatomic.
typically depends on temperature. Solubility of
solids and liquids usually increase as the Q.22 You are provided with a fine white coloured
temperature increases, but the solubility of gases powder which is either sugar or salt. How would
decrease with increasing temperature. you identify it without tasting?
For example, (i)On heating the powder, if the white powder is
(i)Take some salt in a bowl. Bring the temperature sugar, it will melt down to a liquid form (sucrose
of the water to 20°C by adding some ice cubes to has a decomposition point and melting point at
the water. Once the temperature of the water has temperatures between 190 to 192 degrees
stabilized, 100ml of the water is measured using Celsius) and turned a light brown colour. Once the
the measurement cylinder and poured a beaker. sugar had been heating for awhile, it should have
turned black and started producing wisps of (d) C2H5OH
smoke. (e) NH3
Salt (sodium chloride) has a melting point of 841 (f) Ca(OH)2
degrees Celsius and 1545.8 degrees Fahrenheit. (a) CaCO3
If we don’t heat it to that point nothing much
Ca: C : O × 3
happens.
40 : 12 : 16 × 3
(ii)Alternatively, the powder may be dissolved in
water and checked for its conduction of electricity. 40: 12 : 48
If it conducts electricity, it is a salt. When table salt 10 : 3 : 12
is dissolved in water, the solution conducts very (b) MgCl2
well, because the solution contains ions. The ions Mg : Cl × 2
come from the table salt, whose chemical name is 24: 35.5 × 2
sodium chloride. Sodium chloride contains sodium 24: 71
ions, which have a positive charge, and chloride
ions, which have a negative charge. Because (c) H2SO4
sodium chloride is made up of ions, it is called an Hx2:S:O×4
ionic substance. 2: 32 : 16 × 4
Sugar is made of uncharged particles called 2 : 32 : 64
molecules. When sugar is dissolved in water, the 1: 16: 32
solution does not conduct electricity, because (d) C2H5OH
there are no ions in the solution.
C×2:H×6:O
12 × 2 : 1 × 6 : 16
Q.23 Calculate the number of moles of magnesium
24 : 6 : 16
present in a magnesium ribbon weighing 12 g.
Molar atomic mass of magnesium is 24g mol–1. 12 : 3 : 8
Number of moles = w/ atomic weight = 12/24 (e) NH3
= 0.5 mol N:H×3
14 : 1 × 3
Q.24 Verify by calculating that 14: 3
(a) 5 moles of CO2 and 5 moles of H2O do not have (f) Ca(OH)2
the same mass. Ca : O × 2 : H × 2
(b) 240 g of calcium and 240 g magnesium 40 : 16 × 2 : 1 × 2
elements have a mole ratio of 3:5. 40 : 32 : 2
(a) CO2 has molar mass = 12.011 + 2(16.00) = 20 : 16 : 1
12.011 + 32.00 = 44.011g/mol.
5 moles of CO2 have molar mass = 44 × 5 = 220 g Q.26 Calcium chloride when dissolved in water
Molar mass of H2O = H × 2 = 1.01588 x 2 = 2. dissociates into its ions according to the following
01588 g equation. CaCl2 (aq) → Ca2 + (aq) + 2Cl– (aq)
O × 1 = 1 x 15.9994 g = 15.9994 g Calculate the number of ions obtained from
Molar mass of H2O = 2. 01588 + 15.9994 = CaCl2 when 222 g of it is dissolved in water
18.0153 g/mol. Molar mass of CaCl2 = 40.078 + 35.453 × 2 =
5 moles of H2O have mass = 18 × 5 g = 90 g 110.984 g/mol
(b) Number of moles = w/ atomic weight 1 mole of calcium chloride = 111g
Atomic weight of Ca= 40 amu ∴ 222g of CaCl2 is equivalent to 2 moles of CaCl2
Number of moles in 240g Ca metal 240/ 40 = 6 Since 1 formula unit CaCl 2 gives 3 ions(one Ca2 +
Number of moles in 240g of Mg metal 240/ 24 = 10 cation and two Cl- anions),
Atomic weight of Mg = 24amu Therefore, 1 mol of CaCl2 will give 3 moles of ions
Ratio 6:10 2 moles of CaCl2 would give 3 × 2 = 6 moles of
3: 5 ions.
No. of ions = No. of moles of ions × Avogadro
number
Q.25 Find the ratio by mass of the combining
elements in the following compounds. (You may = 6 × 6.022 ×1023 = 36.132 ×1023 = 3.6132
use Appendix-III) ×1024 ions
(a) CaCO3
(b) MgCl2 Q.27 The difference in the mass of 100 moles each
of sodium atoms and sodium ions is 5.48002 g.
(c) H2SO4
Compute the mass of an electron.
A sodium atom and ion, differ by one electron. Thus, Krish’s container is heavy
Sodium loses one electron to form sodium ion. (b) Both the bags have same number of atoms as
For 100 moles each of sodium atoms and ions they have same number of moles of atoms
there would be a difference of 100 moles of
electrons. Mass of 100 moles of electrons= Q.32 Fill in the missing data in the Table .
5.48002 g
As 1 mole of electron have 6.022 ×
1023 electrons(1 mole = 6.022 ×1023 electrons)
Therefore mass of 6.022 × 1023 electrons =
5.48002/ 100 g
Mass of one electron = 5.48002/ 100 × 6.022 ×10
= 9.1 ×10-28 g = 9.1×10-31 kg

Q.28 Cinnabar (HgS) is a prominent ore of

mercury. How many grams of mercury are present
in 225 g of pure HgS? Molar mass of Hg and S are
200.6 g mol–1 and 32 g mol–1 respectively.
Molar mass of HgS = 200.6 + 32 = 232.6 g mol–1
1molecule of HgS contains 1 atom of Hg
232.6 g of HgS contains 200.6 g of Hg
Therefore, Mass of Hg in 225 g of HgS =( 200.6 /
232.6) × 225 = 194.04g

Q.29 The mass of one steel screw is 4.11g. Find Q.33 The visible universe is estimated to contain
the mass of one mole of these steel screws. 1022 stars. How many moles of stars are present in
Compare this value with the mass of the Earth the visible universe?
(5.98 × 1024kg). Which one of the two is heavier Number of moles of stars = 10 22 / 6.023 ×1023 =
and by how many times? 0.0166 moles
One mole of screws weigh = 2.475 ×10 24g =
2.475×1021 k
Q.34 What is the SI prefix for each of the following
Mass of the Earth/ Mass of 1 mole of screws = multiples and submultiples of a unit?
5.98 ×1024 kg / 2.475 ×1021 kg = 2.4 x 103
(a) 103
Therefore, we can say that the mass of earth is
(b) 10–1
2.4×103 times the mass of screws.
(c) 10–2
The earth is 2400 times heavier than one mole of
screws. (d) 10–6
(e) 10–9
Q.30 A sample of vitamin C is known to contain (f) 10–12
2.58 ×1024 oxygen atoms. How many moles of (a) 103 = 1000= kilo
oxygen atoms are present in the sample? (b) 10–1 =1/10= 0.1= deci
1 mole of oxygen atoms = 6.023×10 23 atoms (c) 10–2 =1/100 = 0.01= centi
Therefore, (d) 10–6 = 0.000 001= micro
Number of moles of oxygen atoms = 2.58 × (e) 10–9 =0.000 000 001 = nano
1024/6.023 ×1023 = 4.28 mol (f) 10–12=0.000 000 000 001 = pico
4.28 moles of oxygen atoms
Q.35 Express each of the following in kilograms
Q.31 Raunak took 5 moles of carbon atoms in a (a) 5.84 × 10–3 mg
container and Krish also took 5 moles of sodium (b) 58.34 g
atoms in another container of same weight. (c) 0.584g
(a) Whose container is heavier? (d) 5.873 × 10-21g
(b) Whose container has more number of atoms? (a) 5.84 × 10–3 mg = 5.84 ×10–9 kg
(a) Mass of sodium atoms carried by Krish = (5 (b) 58.34 g =5.834 ×10–2 kg
×23) g = 115 g
(c) 0.584g =5.84 ×10–4 kg
Atomic weight of Na = 23
(d) 5.873×10-21g=5.873 ×10–24 kg
While mass of carbon atom carried by Raunak = (5
×12) g = 60g
Q.36 Compute the difference in masses of 103 Number of moles of gold = Mass of gold/ Atomic
moles each of magnesium atoms and magnesium weight of gold
ions. (Mass of an electron = 9.1 × 10 –31 kg) Atomic weight of gold = 196.967 grams/mol= 197
A Mg2+ ion and Mg atom differ by two electrons. grams/mol
103 moles of Mg2+ and Mg atoms would differ by Therefore,
103 × 2 moles of electrons Number of moles of gold = 0.9/197 = 0.0046
Mass of 2 ×103 moles of electrons = 2×103 × 6.023 One mole of gold contains = 6.022 ×10 23
×1023 × 9.1 ×10–31 kg ∴ 0.0046 mole of gold will contain =0.0046 × 6.022
= 2 × 6.022 × 9.1 × 10–5kg = 109.6004 ×10–5 kg = ×1023 = 2.77 ×1021
1.096 × 10–3kg
Q.40 What are ionic and molecular compounds?
Q.37 Which has more number of atoms? 100g of Give examples.
N2 or 100 g of NH3 Chemical elements can join with each other to form
No of moles of atoms = w/ atomic weight chemical compounds.
100 g of N2 = 100/ 2 x 14 moles Ionic compounds: Ionic compounds are formed
100 g of N2 =100/28 moles by the attraction between positive and negative
Number of molecules = 100/ 28 × 6.022 ×1023 ions. Positively charged ions are known as cations
Molar mass of N2= 2 × molar mass of monatomic and negatively charged ions are known as anions.
N = 2 × 14.0067 Ionic compounds are formed. The compounds
formed by the transfer of electrons are called as
Molar mass of N2 = 28.01340
ionic compound. The bond found in these
Molar mass of N2 = 28 compounds are ionic in nature.
To find number of molecules, number of moles X These compounds are pure substances which are
multiply by Avogadro's number formed be a metal and a non-metal. Cations are
Number of molecules = 100/ 28 × 6.022 ×1023 usually formed by metal atoms and the anions are
Number of atoms in N2 = 1 + 1 = 2 = 2 ×100/ 28 × formed by nonmetal atoms. They are good
6.022 ×1023 = 43.01×1023 conductors of electricity when in a molten or
(ii) 100 g of NH3 = 100/ 17 moles aqueous state, have high melting point, soluble in
To find number of molecules, number of moles × water. Ionic compounds exist as crystals.
Avogadro's number Examples of ionic compounds: sodium
Number of molecules = 100/ 17 × 6.022 × chloride, sodium bromide, potassium chloride etc.
1023 molecules Ex : - 2Na + Cl2 → 2Na+ Cl- → 2NaCl (sodium
Number of atoms in NH3 = 1 + 3 = 4 = 100/ 17× chloride- common salt.) Sodium is a group 1 metal,
6.022 ×1023 × 4 atoms = 141.69 ×1023 thus forms a +1 charged cation. Chlorine is a
nonmetal, and has the ability to form a -1 charged
Therefore,NH3 would have more atoms anion.
Molecular Compounds: Molecular compounds
Q.38 Compute the number of ions present in 5.85 are formed by uncharged atoms. They are also
g of sodium chloride. called as covalent compounds, as they are
Atomic weight; Na = 22.99 g, Cl = 35.45 g formed by sharing of electrons between the two
Atomic weight of NaCl= 23 + 35.5 = 58.5g. atoms and the elements are held together by
5.85 g of NaCl = 5.85/ 58.5 = 0.1moles covalent bonds.
Each NaCl particle is equivalent to one Na + and These, are the pure substances, formed by non-
one Cl– = 2 ions metals. It has a low melting point and cannot
Total moles of ions = 0.1 × 2= 0.2 moles conduct electricity regardless of state. Molecular
compounds can exist as either solid, liquid or
Number of ions= total number of moles of ions X
gaseous state. Eg: carbon dioxide, carbon
Avogadro's number
monoxide, hydrogen chloride, water, methane etc.
Therefore,
Ex: 2C + O2 → 2CO ( Carbon monoxide)
No. of ions= 0.2 × 6.022 ×1023
1.2042 ×1023 ions
Q.41 Compute the difference in masses of one
mole each of aluminium atoms and one mole of its
Q.39 A gold sample contains 90% of gold and the ions. (Mass of an electron is 9.1×10 –28 g). Which
rest copper. How many atoms of gold are present one is heavier?
in one gram of this sample of gold? Mass of 1 mole of aluminium atom = the molar
Given, the sample contains 90% of Gold (Au). mass of aluminium = 27 g mol–1
One gram of gold sample will contain 90% of Gold An aluminium atom loses three electrons to
i.e. 90/100=0.9g of gold become an ion, Al3+
For one mole of A13+ ion, three moles of electrons
are to be lost Q.44 Fill in the blanks
Therefore, total number of electrons lost =3 (a) In a chemical reaction, the sum of the masses
The mass of three moles of electrons = 3 × of the reactants and products remains unchanged.
(9.1×10–28) × 6.022×1023 g This is called __________.
= 27.3 × 6.022 ×10–5 g (b) A group of atoms carrying a fixed charge on
= 164.400 ×10–5 g them is called ________.
= 0.00164 g (c) The formula unit mass of Ca 3 (PO4)2 is
Molar mass of Al3+ = molar mass of aluminium – _________.
mass of electrons (d) Formula of sodium carbonate is _______ and
(As it loses electrons to become Al 3+) =(27– that of ammonium sulphate is ________.
0.00164) g mol–1 = 26.9984 g mol–1 (a) Law of conservation of mass.
Therefore, difference in masses of one mole each (b) ions.
of aluminium atoms and one mole of its ions (c) 310
= 27 – 26.9984 = 0.0016 g 3 × atomic mass of Ca+ 2 × atomic mass of
phosphorus + 8 × atomic mass of oxygen) = 310
Q.42 A silver ornament of mass ‘m’ gram is 3 × 40 + 2 × 31 + 8 × 16 = 120 + 62 + 128 = 310
polished with gold equivalent to 1% of the mass of (d) Na2 CO3 (NH4) 2 SO4
silver. Compute the ratio of the number of atoms of
gold and silver in the ornament. Q.45 Complete the following crossword puzzle
Mass of silver = ‘m’ g … (i) Figure by using the name of the chemical
Mass of gold= m/100 g … (ii) elements. Use the data given in Table.
Number of atoms of silver(Ag) = Mass / Atomic
mass × Avogadro's number(NA)
= m/108 × NA (from (i)
Atomic weight of silver = 108 g/mol
Number of atoms of gold(Au) = Mass / Atomic
mass × Avogadro's number(NA)
= m/100 ×197 × NA (from (ii)
Ratio of number of atoms of gold to silver = Au : Ag
= m/100 × 197 × NA : m/108 × NA
= 108 : 100 × 197
= 108 : 19700
= 1 : 182.41

Q.43 A sample of ethane (C2H6) gas has the same

mass as 1.5 ×1020 molecules of methane (CH4).
How many C2H6 molecules does the sample of gas
contain?
Mass of 1 molecule of CH4 = 16 g/ NA
The atomic weight of carbon is 12, the atomic
weight of hydrogen is 1
weight of CH4 = 12 + 4 × 1 = 16
Mass of 1.5 ×1020 molecules of methane = 1.5
×1020 × 16/ NA
Mass of 1 molecule of C2H6 =30 g/ NAC
C- 12 ; H-1 ; Therefore C2H6= 2 × 12 + 6 × 1 = 30
Mass of molecules of C2H6 is = Mass of
CH4(given)
Therefore,
Mass of molecules of C2H6 =1.5 ×1020 x 16/ NA
Number of molecules of C2H6 = {1.5 ×1020 x 16/
NA}/ 30 g/ NA
= {1.5 ×1020 ×16/ NA} × NA/30
= 0.8 × 1020