Deptt. Of Applied Sciences
Govt. Polytechnic College For Girls Patiala
Presented ByDr. Raman Rani Mittal
M.Sc., M.Phil, Ph.D.
(Chemistry)
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Contents
 Mole Concept
• Atomic & Gram atomic mass
• Molecular mass & Gram molecular
mass
• Mole concept & its importance
 Solutions
• Methods of expressing
concentrations of solutions
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Atomic & Molecular Mass
• Matter is made up of atoms & molecules
• Mass of matter is due to atoms & molecules
• Mass of an atom is called as atomic mass &
mass of molecule is termed as molecular mass
• The actual mass of an individual atom or
molecule is extremely small. This mass can not
be expressed in grams.
• To express the masses of atoms & molecules,
the unit called Atomic Mass Unit (a.m.u.) is
introduced.
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• The atomic mass unit (amu) may be defined as
one-twelfth of the actual mass of an atom of
carbon (Carbon-12 isotope).
• The atomic mass of an element tell us as to how
many times an atom of the element is heavier
than 1/12th of an atom of Carbon (C-12)
Atomic mass =
Mass of an atom of an element
1/12 x mass of an atom of carbon(C-12)
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• For example, atomic mass of oxygen is 16
a.m.u. It means that an atom of oxygen is
16 times heavier than 1/12th of mass of
carbon atom (C-12)
The atomic mass of an element is the
average relative mass of its atoms as
compared with an atom of carbon taken
as 12 a.m.u.
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Gram Atomic Mass
Gram atomic mass
It is the quantity of an element whose
mass in grams is numerically equal to its
atomic mass.
In simple words, atomic mass of an element
expressed in grams is the gram atomic
mass or it is also called gram atom
For example, the atomic mass of oxygen is 16
a.m.u. and thus gram atomic mass of
oxygen is 16 g.
Number of gram-atom
Mass in grams
Gram atomic mass
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Molecular mass
Molecular mass of a substance ( element or
compound) may be defined as the average
relative mass of a molecule of the substance
as compared with mass of an atom of carbon
(C12)taken as 12 a.m.u.
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In other words,
Molecular mass expresses as to how many
times a molecule of the substance is
heavier than 1/12th of the mass of an atom
of carbon.
For example,
A molecule of CO2 is 44 times heavier than
1/12th of the mass of carbon atom.
Hence, molecular mass of CO2 is 44 a.m.u.
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The molecular mass is obtained by adding
together the atomic masses of various atoms
present in a molecule.
For example,
The molecular formula of carbon dioxide is
CO2.
Hence , its molecular mass is
= Atomic mass of carbon + 2 x (Atomic mass
of oxygen)
= 12 + 2 x 16
= 44 a.m.u.
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Gram Molecular Mass
Gram molecular mass is the quantity of a
substance whose mass in grams is numerically
equal to its molecular mass.
In other words, molecular mass of a substance
expressed in grams is called gram molecular
mass. It is also known as gram molecule.
For example, the molecular mass of oxygen is 32
and therefore, its gram molecular mass is 32 g.
Mass in grams
No. of gram molecule
Gram molecular mass
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Example : Calculate the molecular mass of calcium
carbonate.
Solution :
Molecular mass of calcium carbonate
(CaCO3)
= 1 x atomic mass of Ca + atomic mass of C
+ 3 x atomic mass of O
= 1 x 40 + 12 + 3 x 16
= 40 + 12 + 48
= 100 amu
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Mole Concept
• In chemistry we use a unit mole to count
particles (atoms, ions or molecules).
• A mole is a collection of 6.023 x 1023 particles
irrespective of their nature.
• The no. 6.023 x 1023 is called Avogadro’s
number and is denoted by No.
1 Mole = 6.023 x 1023 particles
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For example,
1 mole of atoms = 6.023 x 1023 atoms
1 mole of molecules = 6.023 x 1023 molecules
1 mole of ions = 6.023 x 1023 ions
1 mole of electrons = 6.023 x 1023 electrons
1 mole of protons = 6.023 x 1023 protons
While using the term mole, it is important to
indicate the kind of particles involved.
1 mole of H atoms = 6.023 x 1023 atoms of H
1 mole of H molecules = 6.023 x 1023 molecules of
Hydrogen
The mole is also related to the mass of the substance
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1. Mole & Gram Atomic Mass
• Mass of one mole of atoms of any element in
grams is equal to its gram atomic mass or one
gram atom.
One mole of atoms = 6.023 x 1023atoms
= Gram atomic mass of
the element
For example,
The mass of 6.023 x 1023 atoms of oxygen
is 16 grams.
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2. Mole & Gram Molecular Mass
Mass of one mole of molecules of any substance
in grams is equal to its gram molecular mass.
One mole of molecules = 6.023 x 1023 molecules
= Gram molecular mass
For example,
• The mass of 6.023 x 1023 molecules of sulphur
dioxide (1 mole) is equal to 64 grams.
• Similarly, the mass of 6.023 x 1023 molecules of
carbon dioxide (CO2) is equal to 44 grams.
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3. Mole concept for Ionic compounds
Mass of one mole of formula units of any ionic
compound in grams is equal to its gram formula
mass.
One mole of formula units = 6.023 x 1023 formula units
= Gram formula mass
For example,
The mass of 6.023 x 1023 formula units of
NaCl or 6.023 x 1023 Na+ ions and 6.023 x 1023
Cl- ions (one mole of NaCl) is equal to 58.5 g or
1 formula mass of NaCl.
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4.Mole in terms of Volume
One mole of any gas at S.T.P. (0oC and 760 mm
pressure) occupies 22.4 litres. This volume is
known as molar volume.
For example,
1 mole (2 grams) of H2 gas = 22.4 litres at S.T.P.
1 mole (28 grams) of N2 gas = 22.4 litres at S.T.P.
1 mole (32 grams) of O2 gas = 22.4 litres at S.T.P.
1 mole (44 grams) of CO2 gas = 22.4 litres at S.T.P.
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One mole of a gas = 22.4 litres at S.T.P.
= 6.023 x 1023 molecules
= Gram molecular mass
For example,
1 mole H2 gas = 22.4 litres at S.T.P.
= 6.023 x 1023 molecules of H2
=2g
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To calculate number of moles
(a)For Elements
Number of moles
No. of atoms of the element
6.023 x 1023
Mass of the element in grams
Gram atomic mass
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Example: Calculate the number of moles in 22 g of
CO2.
Solution: Molecular mass of CO2 = 12+2x16
= 44 amu
Gram molecular mass of CO2 = 44 g
Since, 44 g of CO2 make = 1 mole
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.∙. 22 g of CO2 will make
6.023 x 1023g
= 0.5 mole
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(b) For Compounds
No. of moles
No. of molecules of the compound
6.023 x 1023
Mass of the compound in grams
Gram molecular mass
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(c) For an assembly of things
Number of moles
(d)
Number of things
6.023 x 1023
1 mole = gram molecular mass
= 6.023 x 10 23
= 22.4 L at S.T.P.
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Importance of Mole concept
1. To calculate mass of one atom of element
Gram atomic mass
Mass of an atom of element =
6.023 x 10 23
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Example : What is the mass of single atom of
Hydrogen ?
Solution :
Atomic mass of Hydrogen = 1.008 amu
Gram atomic mass of hydrogen = 1.008 g
Thus 6.023 x 1023 atoms of
hydrogen have mass = 1.008 g
1.008
Hence, a single atom of H will have mass =
6.023 x 1023
= 1.66 x 10- 24g
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Example : What is the mass of single atom of
Carbon?
Solution :
Atomic mass of Carbon =12 amu
Gram atomic mass of carbon = 12 g
Thus 6.023 x 1023 atoms of Carbon
have mass = 12 g
Hence, a single carbon atom will
have mass =
12
6.023 x 1023
= 1.99 x 10- 23g
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2. To calculate mass of one molecule of the substance.
Mass of one molecule
of substance
Gram molecular mass
6.023 x 10 23
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Example: What is the mass of a single molecule of
hydrogen?
Solution:
Molecular mass of H2 = 2 amu
Gram molecular mass of H2 = 2 g
Thus, 6.023 x10 23 molecules of H2
have mass = 2g
.∙. A single molecule of H2 will
have mass =
2
6.023 x10 23
= 3.3 x 10 - 24 g
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Example: Calculate the mass of a single molecule of
water?
Solution:
Molecular mass of H2O = 2x1 + 16
=18 amu
Gram molecular mass of H2O = 18 g
Thus, 6.023 x10 23 molecules of H2O
have mass = 18g
.∙. A single molecule of H2O will
have mass =
18
6.023x1023
= 2.99 x 10-23g
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3. To calculate the number of atoms in a given
mass of the element.
No. of atoms in a given mass of element
Mass of element in grams
x 6.023x1023
Gram atomic mass
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Example: Calculate the number of atoms in 20g
of calcium (At. Mass of Ca = 40).
Solution: Atomic mass of Ca = 40 amu
Gram atomic mass of Ca = 40 g
Thus 40 g of Ca contains = 6.023 x1023 atoms
6.023 x1023
.∙. 20g of Ca will contain
x20
40
= 3.01 x 1023atoms
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4. To calculate the number of molecules in a given
mass of substance.
No. of molecules in a given mass of substance
Mass of substance in grams
X 6.023x1023
Gram molecular mass
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Example: Calculate the number of molecules of
methane in 0.80 g of methane.
Solution: Molecular mass of methane (CH4)
= 1x12 + 4x1
= 16 amu
Gram molecular mass of CH4= 16 g
Thus 16g of CH4 contain = 6.023 x1023 molecules
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6.023
x10
.∙. 0.80g of CH4 will contain
x 0.80
16
= 3.01x1022 molecules
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5. To calculate the volume occupied by a given
mass of the gas at S.T.P.
Volume occupied by a given
mass of a gas at S.T.P.
Mass of gas in grams
x22.4
Gram molecular mass of gas
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Example: Calculate the volume occupied by 16 g
of oxygen at S.T.P.
Solution:
Molecular mass of O2 = 32 amu
Gram molecular mass of O2 = 32 g
Now 32 g of O2 at S.T.P. occupy volume
= 22.4 L
.∙. 16g of O2 at S.T.P. would occupy volume
22.4
=
x16
32
= 11.2 litres
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Solutions
• Solution is a homogeneous mixture of two or
more pure substances
• Composition of solution can be varied within
certain limits.
• Solution results by dissolving a solute in a
solvent.
Solute + Solvent
Solution
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• The substance which is dissolved & is present
in lesser quantity is called solute.
• The substance in which solute is dissolved & is
present in greater quantity is called solvent.
For example,
2g of sugar is dissolved into 50 ml of water
to form a solution.
In this case sugar is solute & water is
solvent.
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Methods of expressing Concentration of a solution
The concentration of solution is defined as the
amount of solute present in the given quantity
of the solution.
It can be expressed in the following ways:
1. Strength
2. Molarity
3. Molality
4. Mole fraction
5. Normality
6. Mass percentage
7. Volume percentage
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1. Strength: It is the amount of solute in grams
dissolved per litre of the solution.
Thus,
Mass of solute in grams
Strength of solution =
Volume of solution in litres
If ‘a’ grams of solute is dissolved in V ml of a
given solution, then
a x 1000
Strength =
V
It is expressed in g/L.
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2. Molarity (M):
The number of moles of solute dissolved
per litre of the solution is called molarity.
Number of moles of solute
Molarity =
Volume of solution in litres
It is convenient to express volume in ml.
So that
Number of moles of solute
x1000
M =
Volume of solution in ml
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And number of moles of solute can be
calculated from the given mass of solute
which is dissolved
Given mass
Moles of solute =
Gram molecular mass
Thus,
Given mass x 1000
M=
Gram mol. mass x V
Molarity changes with temperature.
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Example: Calculate the molarity of the solution
containing 0.5g of NaOH dissolved in 500 ml
Solution: Given mass of NaOH = 0.5g
Mol.mass of NaOH = 23 + 16 + 1= 40
0.5
No. of moles of NaOH =
= 0.0125
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Volume of solution, V = 500ml
No. of moles of NaOH x 1000
Now,
Molarity =
V
0.0125 x 1000
=
= 0.025 M
500
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3. Molality: The number of moles of solute
dissolved per kg of the solvent is called molality.
No. of moles of solute
Molality (m) =
Mass of solvent in kg
No. of moles of solute
=
x1000
Mass of solvent in grams
If ‘a’ grams of the soute is dissolved in W gram
of the solvent, then
a
1000
m=
x
mol. Mass of solute W
Molality does not changes with temperature.
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Example: Calculate the molality of an aqueous
solution containing 4g of urea (Mol.mass = 60)
in 500 g of water.
Solution: Given mass of urea = 4g
Mol. mass of urea = 60
Mass of water , W = 500g
Given mass 1000
Since,
Molality (m) =
x
Mol. mass W
4 1000
=
x
60 500
= 0.133 m
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4. Mole fraction (x):
It is the ratio of the number of moles of a
component to the total number of moles
of all the components (solute & solvent)
in the solution.
Suppose a solution contains
nA moles of solute (A) and
nB moles of solvent (B).
Then,
nA
Mole fraction of solute (xA) =
nA + nB
nB
Mole fraction of solvent (xB) = nA + nB
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The sum of mole fractions of all the
components is always equal to one.
nA
nB
xA + xB =
+
=1
nA + nB
nA + nB
Mole fraction being a ratio, is dimensionless
property.
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Example: A solution containing 23g of ethanol &
90g of water. What is the mole fraction of
ethanol & water in solution?
Solution:
Mass of ethanol (C2H5OH) = 23g
Molecular mass of ethanol = 2x12 + 1x6 + 16
= 46
23
.∙. Number of moles of ethanol =
= 0.5
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Mass of water = 90g
Molecular mass of water = 18
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.∙.
.∙.
.∙.
90
Number of moles of water =
18
=5
Total number of moles = 0.5 + 5 = 5.5
0.5
Mole fraction of ethanol =
5.5
= 0.09
Mole fraction of water = 1 – 0.09
= 0.91
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5. Normality (N):
It is the number of moles of gram equivalents
of the solute dissolved per litre of the given
solution.
No. of gram equivalents of solute
Normality (N) =
Volume of solution in litres
Or
= No. of g.equivalents of solute x1000
Volume of solution in ml
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Gram equivalents of solute can be calculated
as
Mass of solute
a
Gram equivalents =
=
Equivalent mass Eq. mass
where ‘a’ is the mass of the solute in grams
present in V ml of a given solution
Thus,
a
1000
N=
x
Eq. mass V
Like molarity, normality of a solution also
changes with temperature.
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6. Mass percent(w/w):
It is equal to the weight of solute present per
100 g of the solution.
Weight of solute
%Mass =
x100
Weight of solution
Weight of solute in grams
=
x100
Wt. of solute + Wt. of solvent
Mass (weight) percent can be expressed as w/w.
For example, a 5%(w/w) solution of NaCl means
a solution containing 5g of NaCl in 100g of the
solution or 95 g of the solvent.
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7. Volume %age:
It is equal to the volume of the component
present per 100 parts of the volume of the
solution.
For example,
VA and VB are volumes of the components A &
B respectively in a solution, then
Volume of A
Volume %age of A =
X 100
Vol. of A + Vol. of B
It is expressed as v/v.
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