Mole Concept

Mole Concept
1. CLASSIFICATION OF MATTER :
Matter

Pure
Mixture
Substance

Homogeneous Heterogeneous Elements Compounds

Mixture Mixture

Fig: Classification of matter

A mixture may be homogeneous or heterogeneous. In a homogeneous mixture, the components
completely mix with each other and its composition is uniform throughout. Sugar solution, and air are thus, the
examples of homogeneous mixtures. In contrast to this, in heterogeneous mixtures, the composition is not
uniform throughout and sometimes the different components can be observed. For example, the mixtures of
salt and sugar, grains and pulses along with some dirt (often stone) pieces, are heterogeneous mixtures.
Pure substances have characteristics different from the mixtures. They have fixed composition, whereas
mixtures may contain the components in any ratio and their composition is variable. Copper, silver, gold, water,
glucose are some examples of pure substances.
Pure substances can be further classified into elements and compounds. An element consists of only one
type of particles. These particles may be atoms or molecules.
Some elements such as sodium or copper, contain single atoms held together as their constituent particles
whereas in some others, two or more atoms combine to give molecules of the element. Thus, hydrogen,
nitrogen and oxygen gases consist of molecules in which two atoms combine to give their respective molecules.
This is illustrated in Fig. When two or more atoms of different elements combine, the molecule of a compound
is obtained. The examples of some compounds are water, ammonia, carbon dioxide,

Atoms of different elements

an atom of another atom a molecule of

hydrogen (H) of hydrogen (H) hydrogen (H2)
A representation of atoms and molecules
sugar etc. The molecules of water and carbon dioxide are represented in Fig.

A depiction of molecules of water and carbon dioxide

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Chemistry

2. ATOMIC / MOLECULAR MASSES:

2.1 Atomic mass unit (or amu) :
th
1
The atomic mass unit (amu) is equal to   mass of one atom of carbon -12 isotope.
 12 

1
 1amu = × mass of one C-12 atom  mass of one nucleon is C-12 atom.
12
1
= 1.66 × 10-24 gm or 1.66 × 10-27 kg = NA = Avogadro’s Number = 6.023 × 1023
NA
• one amu is also called on Dalton (Da).
• Today, amu has been replaced by ‘u’ which is known as unified mass
2.2 Relative atomic mass :
One of the most important concept come out from Dalton’s atomic theory was that of relative atomic mass
or relative atomic weight. This is done by expressing mass of one atom with respect to a fixed standard.
Dalton used hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the
reference. Therefore relative atomic mass is given as
(a) On hydrogen scale :

Mass of one atom of an element

Relative atomic mass (R.A.M.) =
mass of one hydrogen atom

(b) On oxygen scale :

Mass of one atom of an element

Relative atomic mass (R.A.M.) = 1  mass of one oxygen atom
16
(c) On carbon scale : (Present Standard)
The present standard unit which was adopted internationally in 1961, is based on the mass of one
carbon-
12 atom.
Mass of one atom of an element
Relative atomic mass (R.A.M.) = 1  mass of one C - 12 atom
12
Note : Relative atomic mass is nothing but nearly equal to the number of nucleons present in the atom.
2.3 Atomic mass :-
It is the mass of 1 atom of a substance. It is expressed in amu.
Atomic mass = R.A.M. × 1 amu
2.4 Average Atomic Mass :
The atomic masses of the elements have been determined accurately during the recent years using an instrument
called “mass spectrometer”. It is found that in a number of cases, atoms of the same element possess
different masses (called isotopes). Thus, in such cases, the atomic mass of the element is taken as the average
value. For example, ordinary chlorine is a mixture of two isotopes with atomic masses 35 u and 37 u and they
35  3  37 1
are present in the ratio of 3 : 1. Hence, the average atomic mass of chlorine would be = 35.5 u
3 +1
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 Mole Concept
Thus the atomic mass of an element is the average relative mass of its atoms as compared with an atom of
carbon taken is 12.
Alternatively the average relative atomic mass of an element can be calculated from the ‘fractional abundances’
of the isotopes of that element.
Fractional abundance of an isotope is the fraction of the total number of atoms that is comprised of that
particular isotope.
The mass of elements having different isotopes is taken as an average of the relative atomic masses of all the
isotopes.
eg. Isotopes Fractional abundance Percentage abundance
20
Ne 0.9051 90.5%
21
Ne 0.0027 0.27%
22
Ne 0.0022 0.22%
At.Mass of I ×% of I+At. mass of II isotope ×% of II
Average atomic mass =
100
Average atomic mass of Ne = (20 × 0.9051) + (21 × 0.0027) + (22 × 0.0022) = 20.179 u
Average Atomic mass can be in fractions as it is the average of the mass of various isotopes in the
particular proportions.
2.5 Relative molecular mass

mass of one molecule of the substance

Relative molecular mass =
1
 mass of one C - 12 atom
12
2.6 Molecular mass = Relative molecular mass × 1 amu
2.7 Formula Mass :
Ionic compound does not contain discrete molecules as their constituent unit in such compounds positive &
negative entities are arranged in a three dimensional structure.
In case of ionic complexes such as NaCl, formula mass is calculated instead of molecular mass. (As in solid
state, NaCl does not exist as single entity)
Formula mass of NaCl = At. mass of Na + At. Mass of Cl
= 23.0 a.m.u. + 35.5 a.m.u. = 58.5 a.m.u.

3. MOLE CONCEPT :
Mole Concept is essential tool for the fundamental study of chemical calculations. This concept is simple but its
application requires a through practice. The 14th Geneva Conference on weight & measures adopted mole as
a seventh basic SI unit of the amount of a substance. There are many ways of measuring the amount of
substance, weight and volume being the most common, but basic unit of chemistry is the atom or a molecule and
to measure the number of molecule is more important.
(a) Mole :
The mole is Latin means heap or mass or pile. A mole of atom is collection of atoms whose total mass is
the number of grams equal to the atomic mass. As equal number of moles of different element contain
equal number of atoms, it is convenient to express amount of the elements in terms of moles. In SI unit,
mole (symbol, mol) was used for the amount of a substance.
A mole is the amount of a substance that contains as many entities (Atoms, Molecules, ions or
any other particles) as there are atoms is exactly 12g of C-12 isotopes.
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Chemistry
As mass of a carbon – 12 was determined as 1.992648 × 10 –23 gm. So the no. of atoms in 12 gm i.e. 1 mole
12
of carbon will be = = 6.0221367 × 1023 atoms/mol
1.992648 1023
(b) Avogadro’s Number
Just as a pair means two, a dozen means twelve objects, a score means twenty objects. We can define a
mole as a definite number of particles viz. atoms, molecules, ions, electrons, or protons etc. This definite
number is called Avogadro’s Number equal to 6.023 × 1023. The value of Avogadro’s Number depends
on the atomic weight scale.
The mole of a substance always contain the same number of entities (no matter what the substance may
be)
A mole of hydrogen atoms mean 6.022 × 1023 atoms of hydrogen whereas a molecule of hydrogen mean
6.022 × 1023 molecules of hydrogen or 2 × 6.022 × 1023 atoms of hydrogen (because each of hydrogen
molecule contains 2 atoms of hydrogen)
In modern practice, gram - molecule & gram - atom are termed as a mole of molecule & a mole of
atom respectively e.g.
1 gm molecule of O2 = 1 mole O2
1 gm atom of oxygen = 1 mole of oxygen atom. Number of moles of a substance can be calculated by
various means. Thus 1 mole may be defined in any one of the three ways.
(c) Different definitions of mole:-
I st Definition (in terms of mass) :
A mole is defined as that amount of the substance which has mass equal to gram atomic mass if
the substance is atom or gram molecular mass if the substance is molecule.
1 mole of oxygen atoms = 16 g ( At. Mass of O = 16 u)
1 mole of oxygen molecules = 32 g ( molecular mass of O2 = 32 u)
1 mole of CO2 molecules = 44 g ( molecular mass of CO2 = 44 u)
Molar mass is the mass of 1 mole of a substance in gms.
2 nd Definition (in terms of number) :
A mole is defined as the amount of the substance which contains Avogadro’s number
(6.022 × 1023) of atoms if the substance is atom or Avogadro’s number (6.022 × 1023) of molecules
if the substance is molecule :
1 mole of oxygen atoms = 6.022 × 1023 atoms of oxygen
1 mole of oxygen molecules = 6.022 × 1023 molecules of oxygen
3 rd Definition (in terms of volume) :
In case of gases, a mole is defined as that amount of the gas which has a volume of 22.4 litres at
STP.
e.g. 1 Mole of oxygen gas = 22.4 litres of oxygen at STP
1 Mole of CO2 gas = 22.4 litres of CO2 at STP
(d) In terms of S.I. unit we may define mole as :
Amount of the substance which contains as many elementary entities as there are atoms in exactly 0.012
kg (i.e. 12g) of carbon - 12 isotope. The elementary entities must be specified i.e. whether they are atoms,
molecule, ions, electrons or any other entity.

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 Mole Concept
Thus we can express 1 mol as:-

Mass of the Substance in Grams

The number of moles of any substance =
Molecular mass

Volume of gas at NTP in litres

The number of moles of a gas at NTP =
22.4
So a mole is nothing but a collection of 6.023 × 1023 particles (atoms, molecules or ions) as well as it
is equal to atomic weight (in gm), molecular weight (in gm) and ionic weight (in gm) (for atoms,
molecules & ions respectively.)
Examples
(a) 1 gm atom Na = 23 gm = mass of 6.023 × 1023 atoms of sodium
(b) 1 gm molecule CH4 (1mole) = mass of 6.023 × 1023 CH4 molecule = 6.023 × 1023 × 5 atoms
( 1 molecule of CH4 contains 5 atoms)
(c) 1 gm ion = ionic weight (in gm) = mass of 6.023 × 1023 ions
1 gm ion N–3 = 14 gm N–3 ion = mass of 6.023 × 1023 N–3 ions.
(d) 1 gm molecule glucose (C6H12O6) = 180 gm glucose = mass of 6.023 × 1023 molecules of glucose

4. MOLAR MASSES :
(a) Gram Atomic Mass :
The atomic mass of an element expressed in gram is called gram atomic mass of the element.
OR
It is also defined as mass of 6.022 × 1023 atoms.
OR
It is also defined as the mass of one mole atoms.
For example for oxygen atom :
Atomic mass of ‘O’ atom = mass of one ‘O’ atom = 16 amu
gram atomic mass = mass of 6.02 × 1023 ‘O’ atoms
= 16 amu × 6.02 × 1023 = 16 x 1.66 × 10-24 g × 6.02 × 1023 = 16 g
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Chemistry

1
( 1.66 × 10-24 × 6.02 × 1023  1) ( amu = N )
A

(b) Gram Molecular Mass :

The molecular mass of a substance expressed in gram is called the gram-molecular mass of the
substance. OR
23
It is also defined as mass of 6.02 × 10 molecules
OR
It is also defined as the mass of one mole molecules.
For example for ‘O2’ molecule.
Molecular mass of ‘O2’ molecule = mass of one ‘O2’ molecule = 2 x mass of one ‘O’ atom
= 2 × 16 amu = 32 amu
gram molecular mass = mass of 6.02 × 1023 ‘O2’ molecules = 32 amu × 6.02 × 1023
= 32 × 1.66 × 10-24 gm × 6.02 × 1023 = 32 gm
(c) Gram Formula Mass:-
The formula mass of a substance expressed in gram is called gram formula mass of the substance
OR
23
It is also defined as mass of 6.022 × 10 formula units
OR
It is also defined as the mass of one mole formula units
For example ‘NaCl’.
Formula mass of NaCl = 58.54
Gram formula mass of NaCl = Mass of 6.022 × 1023 ‘NaCl’ Formula units
= 58.5 amu × 6.022 × 1023
= 58.5 × 1.66 × 10–24 gm × 6.022 × 1023 = 58.5 gm

Example 1. In three moles of ethane (C2H6), calculate the following: (NCERT)

(i) Number of moles of carbon atoms. (ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution (i) 1 mole of C2H6 contains 2 moles of carbon atoms.
 Number of moles of carbon atoms in 3 moles of C 2H6 = 2 × 3 = 6
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.
 Number of moles of carbon atoms in 3 moles of C2H6 = 3 × 6 = 18
(iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane.
 Number of molecules in 3 moles of C2H6 = 3 × 6.023 × 10 23 = 18.069 × 1023

5. APPLICATIONS OF AVOGADRO’S NUMBER AND MOLE CONCEPT :

(i) To calculate the mass of an atom or a molecule of a substance :
Atomic mass in grams
Mass of an Atom =
Avogadro's number

Molecular mass in grams

Mass of a molecule =
Avogadro's number
(ii) To calculate the number of atoms or molecules in a given weight of an element or a
compound :
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 Mole Concept

Number of atoms or Molecules =

Mass of the substance in grams × Avogadro’s Number
Atomic or molecular mass
(iii) To calculate the number of molecules present in a given volume of the gas at NTP:

Number of molecule present in ‘V’, litres of gas = Volume of gas in litres at NTP × Avogadro’s Number
22.4
(iv) Weight volume relation of gases :
Mass of gas in grams
The volume occupied by a given weight of a gas at NTP =  22.4 L
Molecular Mass

Example 1. What is the volume occupied by one carbon tetrachloride molecule at 200C ? The density of carbon
tetrachloride is 1.6 g cm-3 at 200C.
Solution Molecular mass of CCl4 =1× 12 + 4 × 35.5 = 154 amu
1 mole of CCl4(6.023 × 1023) molecules weight = 154g
154 g ×1molecule
Mass of 1 CCl4 molecule = 6.023×10 23 molecules

Mass 154 g
Volume of 1 CCl4 molecule = = 6.023×1023 ×1.6 g cm –3 =1.598 × 10-22 cm3.
Density

6. APPLICATION OF AVOGADRO’S LAW :

6.1 To find the relationship between molecular mass and vapour density of a gas.

Density of the gas

Vapour density of a gas =
Density of hydrogen

Mass of certain volume of the gas at S.T .P. Mass of n molecules of the gas
= =
Mass of same volume of H 2 at S.T.P. Mass of n molecules of H 2

Mass of 1 molecule of the gas Molecular mass

= =
Mass of 1 molecule of H 2 (i.e.2 atoms of H) 2
or Molecular mass = 2 × Vapour density
Vapour density is also called ‘relative density’ of the gas as it tells us how heavy the gas is with respect to
hydrogen).
6.2 To find the relationship between mass and volume of gas :
Molecular mass = 2 × Vapour density

Mass of certain volume of the gas at S.T .P. Mass of 1L the gas at S.T.P.
=2× =2×
Mass of same volume of H 2 at S.T.P. Mass of of 1L of H 2 at S.T.P.

2
= × Mass of 1L of the gas at S.T.P.. = 22.4 × Mass of 1 L of the gas at S.T.P.
0.089 g
= Mass of 22.4 L of the gas at S.T.P.
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Chemistry
Thus, 22.4 litres of any gas at S.T.P. weight equal to the molecular mass of the gas expressed
ingrams. This is called Gram - Molecular Volume (G.M.V.) Law
Earlier the STP conditions (i.e. Standard temperature and pressure) were taken as 1 atm and 0 0C. However,
now these are taken as 1 bar and 00C. Under these conditions, instead of 22.4 L, we have 22.7 L. (1 atm = 1.01
bar)

Absolute density
Specific gravity =
Denisty of pure water at 4 0 C
Density of pure water = 1 gm/ml . Specific gravity = Absolute density

Example 1. Use the data given in the following table to calculate the molar mass of naturally occurring argon
isotopes: (NCERT)
Isotope Isotopic molar mass Abundance
–1
36
Ar 35.96755 gmol 0.337%
–1
38
Ar 37.96272 gmol 0.063%
–1
40
Ar 39.9624 gmol 99.600%
Solution Molar mass of argon

 0.337   0.063   99.60  

  35.96755     37.96272     39.9624 
–1
  gmol = 39.947 gmol–1
  100   100   100  
Example 2. Which one of the following will have largest number of atoms? (NCERT)
(i) 1 g Au (s) (ii) 1 g Na (s) (iii) 1 g Li (s) (iv) 1 g of Cl2(g)
1
Solution 1 g of Au (s)  mol of Au (s)
197

6.022  10 23 1
 atoms of Au (s) = 3.06 × 1021 atoms of Au (s) 1 g of Na (s)  mol of Na (s)
197 23

6.022  10 23
 atoms of Na (s) = 0.262 × 1023 atoms of Na (s)
23

1
= 26.2 × 1021 atoms of Na (s) 1 g of Li (s)  mol of Li (s)
7

6.022  10 23
 atoms of Li (s) = 0.86 × 1023 atoms of Li (s)
7

1
= 86.0 × 1021 atoms of Li (s) 1 g of Cl2 (g)  mol of Cl2 (g)
71
(Molar mass of Cl2 molecule = 35.5 × 2 = 71 g mol –1)

6.022  10 23
 atoms of Cl2 (g) = 0.0848 × 1023 atoms of Cl2 (g)
71
= 8.48 × 1021 atoms of Cl2 (g)
Hence, 1 g of Li (s) will have the largest number of atoms.

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 Mole Concept
Example 3. What will be the mass of one 12C atom in g? (NCERT)
Solution 1 mole of carbon atoms = 6.023 × 1023 atoms of carbon
= 12 g of carbon
12g
 Mass of one 12C atom  = 1.993 × 10–23 g
6.022  1023
Example 4. In three moles of ethane (C2H6), calculate the following: (NCERT)
(i) Number of moles of carbon atoms.
(ii) Number of moles of hydrogen atoms.
(iii) Number of molecules of ethane.
Solution (i) 1 mole of C2H6 contains 2 moles of carbon atoms.
 Number of moles of carbon atoms in 3 moles of C 2H6 = 2 × 3 = 6
(ii) 1 mole of C2H6 contains 6 moles of hydrogen atoms.
 Number of moles of carbon atoms in 3 moles of C 2H6 = 3 × 6 = 18
(iii) 1 mole of C2H6 contains 6.023 × 1023 molecules of ethane.
 Number of molecules in 3 moles of C2H6
= 3 × 6.023 × 10 23 = 18.069 × 1023
Example 5. 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. (NCERT)
23
Solution (i) 1 mole of Ar = 6.022 × 10 atoms of Ar
 52 mol of Ar = 52 × 6.022 × 1023 atoms of Ar = 3.131 × 1025 atoms of Ar
(ii) 1 atom of He = 4 u of He OR
1
4 u of He = 1 atom of He 1 u of He  atom of He
4

52
52u of He  atom of He = 13 atoms of He
4
(iii) 4 g of He = 6.022 × 1023 atoms of He

6.022  1023  52
 52 g of He  atoms of He = 7.8286 × 1024 atoms of He
4
Example 6. A metal M of atomic weight 54.94 has a density of 7.42 g/cc. Calculate the apparent volume occupied
by one atom of the metal.
54.94
Solution 7.42 g of metal M occupies volume of 1cc.  54.94 g occupies a volume of = = 7.404 cc.
7.42
Since the weight of 1 mole of atoms is the atomic weight and 1 mole of atom contains Avogadro No.
of atoms.
Vol. of 1mole
Volume occupied by 1 atom =
Avogadro's. Number.

7.404
-23
=
6.022×10 23 = 1.23 × 10 cc

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Chemistry
Example 7. Calculate the volume in litres of 20 g hydrogen gas at STP

Mass 20 gm
No. of moles of hydrogen gas = =
Solution
Molecular mass 2 gm = 10 mol
volume of hydrogen gas at STP = 10 x 22.4 Lt. = 224 lt
Y-map : Interconversion of mole - volume, mass and number of particles.

Number lt Volume at STP

N .4
A 22 lt
N .4
A 22
Mole

mole. wt. mole. wt.

At. wt. At. wt.

Mass

Example 8. Vapour density of a gas with respect of oxygen gas is 6. What will be its Molecular weight.
Solution M.Wt. = 6 × 32 = 192 (it tells us how heavy the gas is with respect to oxygen)

7. LAW OF CHEMICAL COMBINATION :

(i) The law of conservation of mass :
In a chemical change total mass remains conserved.
i.e. mass before reaction is always equal to mass after reaction.
(ii) Law of definite proportion or constant composition :
This law was first stated by Proust and established by Richard.
The law states that
“A chemical compound always contains the same elements combined together in a definite
proportion by mass”.
eg: Pure water always contains hydrogen and oxygen combined together in the proportion 1 : 8 by mass
Limitations :
(i) The law is not applicable if an element exists in different isotopes which may be involved in the
formation of the compound. For example, in the formation of the compound CO2, if C-12 isotope
combines, the ratio of C : O is 12 : 32. But if C - 14 isotope combines, the ratio of C : O is 14 : 32
(ii) The elements may combine in the same ratio but the compounds formed may be different. For
example, in the compounds C2H5OH and CH3OCH3 (both having same molecular formula viz. C2H6O
the ratio of C : H : O = 24 : 6 : 16 = 12 : 3 : 8 by mass.
(iii) Law of multiple proportions :
This law was stated by John Dalton as ‘When two elements combine together to form two or more
compounds, the different masses of one of these which combine with fixed mass of the other element,
bear a simple integral ratio to one another.
eg : In H2O and H2O2 , the masses of oxygen which combine with 2 grams of hydrogen are 16 and 32 i.e.
a simple ratio of 1 : 2

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 Mole Concept
Compound of Nitrogen and Oxygen :
The elements nitrogen and oxygen combine to produce as many as 5 oxide of nitrogen, viz., Nitrous oxide,
Nitric oxide, Nitrogen trioxide, Nitrogen tetroxide & Nitrogen pentoxide.

Compound Nitrous Nitric Nitrogen Nitrogen Nitrogen

oxide (N2O) oxide (NO) trioxide (N2O3) tetroxide (N2O4) Pentoxide (N2O5)

Nitrogen 28 14 28 28 28 (parts by mass)

Oxygen 16 16 48 64 80 (part by mass)

Fixing the mass of nitrogen as 14 parts, the masse of oxygen in these five oxides are 8, 16, 24, 32 and 40
parts respectively. These masses bear a simple ratio of 1 : 2 : 3 : 4 : 5 to one another.
(iv) Law of reciprocal Proportions :
This law was proposed by Ritcher as ‘The ratio of two different elements A and B combining separately
with a fixed mass of a third element C, is either the same or a simple multiple of the proportion in which A
and B combine directly with each other.’
The law is now called law of combining weights or equivalent proportions.eg :
2 : 16 (or)
12 : 4 H
4 : 32
CH4 H 2O
C O
12 : 32
CO2
(v) Gay Lussac’s law of combining volumes :
This law states that ‘Whenever gases combine, they do so in volumes which bear a simple ratio to each
other and also to the products if they are also gaseous, provided all the volumes are measured under the
same conditions of temperature and pressure.
Or Gases always react together in simple whole number ratio of their volumes measured under similar
conditions of temperature and pressure. This is also known as law of definite proportions by volume.
H 2 + Cl 2  2HCl 2
e.g. 1vol. 1vol. 2vol.

(vi) Avogadro’s Law :

An atom is the smallest principle of an element which can take part in a chemical reaction. It may or may
not be capable of independent existence. A molecule is the smallest particle of an element or a compound
which is capable of independent existence.
Since the smallest particle of a gas which can exist independently is the molecule and not the atom. So the
volume of a gas must be related to the number of molecule (rather than atoms) present in it. He thus put
forward his hypothesis known as Avogadro’s hypothesis. This states that
“Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of
molecules.”

Hydrogen + Chlorine  Hydrogen chloride gas

1 vol. 1vol. 2 vol.
(By experiment)
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Chemistry

n molecules + n molecules  2 n molecules

(By Avogadro's Law)

1 1
molecule + molecule  1molecule
2 2
(by diving throughout by 2 n)

Example 1. If ten volumes of dihydrogen gas react with five volumes of dioxygen gas, how many volumes of water
vapour would be produced ?
Solution 2H 2 (g) + O 2 (g)  
 2H 2 O(g)
2vol 1vol 2vol
10vol 5vol 10vol

 10 volumes of water will be produced during the reaction.

Example 2. Calculate the mass percent of different elements present in sodium sulphate (Na2SO4). (NCERT)
Solution The molecular formula of sodium sulphate is Na2SO4.
Molar mass of Na2SO4= [(2 × 23.0) + (32.066) + 4 (16.00)]
= 142.066 g
Mass of that element in the compound
Mass percent of an element  ×100
Molar mass of the compound

46.0 g
 mass percent of sodium   100  32.379 =32.4 %
142.066 g

32.066 g
Mass percent of sulphur   100 = 22.57 = 22.6%
142.066 g

64.0 g
Mass percent of oxygen = 142.066 g  100 = 45.049 = 45.05 %

Example 3. 1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in
steam gave 2.50 g of its oxide. Show that these results illustrate the law of constant proportion.
Solution In the first sample of the oxide,
Wt. of metal = 1.80 g, Wt. of oxygen = (3.0 – 1.80) g = 1.2 g

wt.of metal 1.80g

 wt.of oxygen  1.2g  1.5

In the second sample of the oxide, Wt. of metal = 1.50 g, Wt. of oxygen = (2.50 – 1.50) g = 1 g.

wt.of metal 1.50 g

 wt.of oxygen  1g  1.5

Thus, in both samples of the oxide the proportions of the weights of the metal and oxygen a fixed.
Hence, the results follow the law of constant proportion.

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 Mole Concept

8. PERCENTAGE COMPOSITION :
Percentage composition of a compound is the number of grams of each element in one hundred gram of a
compound.
Mass of element in a compound ×100
% composition of a compound =
Total mass of sample

2 × At.wt. of Chromium
eg. %age of Cr in K2Cr2O7 is = Molecular wt. of K Cr O ×100
2 2 7

9. EMPIRICAL AND MOLECULAR FORMULA :

We have just seen that knowing the molecular formula of the compound we can calculate percentage composition
of the elements. Conversely if we know the percentage composition of the elements initially, we can calculate
the relative number of atoms of each element in the molecules of the compound. This give us the empirical
formula of the compound. Further if the molecular mass is known then the molecular formula can easily be
determined.
The empirical formula of a compound is a chemical formula showing the relative number of atoms in the
simplest ratio. An empirical formula represents the simplest whole number ratio of various atoms present in a
compound.
The molecular formula gives the actual number of atoms of each element in a molecule. The molecular
formula shows the exact number of different types of atoms present in a molecule of a compound.
The molecular formula is an integral multiple of the empirical formula.
i.e. molecular formula = empirical formula × n
molecular formula mass
where n =
empirical formula mass
Let look at the example to calculate empirical & molecular formula:-
An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition.
C = 40.684% ; H = 5.085% and O = 54.228%
The molecular weight of the compound is 118 gm. Calculate the molecular formula of the compound.
Step-1
To calculate the empirical formula of the compound.

Element Symbol Percentage At. mass Relative no. of Simplest Simplest

Percentage
of element of element atoms atomic ratio no. atomic ratio
At.Mass

40.687 3.390
Carbon C 40.687 12  3.390 1 2
12 3.389

5.085 5.085
Hydrogen H 5.085 1  5.085  1.5 3
1 3.389

54.228 3.389
Oxygen O 54.228 16  3.389 1 2
16 3.389

 Empirical Formula is C2H3O2

[D -13]
Chemistry
Step-2
To calculate the empirical formula mass.
The empirical formula of the compound is C2H3O2.
 Empirical formula mass = (2 × 12) + (3 × 1) + (2 × 16) = 59
Step-3
To calculate the value of ‘n’
Molecular mass 118
n= = =2
Empirical formula mass 59
Step-4
To calculate the molecular formula of the salt.
Molecular formula = n × (Empirical formula) = 2 × C2H3O2 = C4H6O4
Thus the molecular formula is C4H6O4

Example 1. How much copper can be obtained from 100 g of copper sulphate (CuSO4)? (NCERT)
Solution 1 mole of CuSO4 contains 1 mole of copper.
Molar mass of CuSO4 = (63.5) + (32.00) + 4(16.00) = 63.5 + 32.00 + 64.00 = 159.5 g
159.5 g of CuSO4 contains 63.5 g of copper.
63.5  100 g
 100 g of CuSO4 will contain of copper..
159.5

 Amount of copper that can be obtained from 100 g CuSO  63.5  100 = 39.81 g
4
159.5
Example 2. Determine the molecular formula of an oxide of iron in which the mass percent of iron and oxygen
are 69.9 and 30.1 respectively. Given that the molar mass of the oxide is 159.69 g mol –1.
(NCERT)
Solution Mass percent of iron (Fe) = 69.9% (Given)
Mass percent of oxygen (O) = 30.1% (Given)
69.90
Number of moles of iron present in the oxide  = 1.25
55.85
30.1
Number of moles of oxygen present in the oxide  = 1.88
16.0
1.25 1.88
Ratio of iron to oxygen in the oxide, = 1.25 : 1.88  : = 1 : 1.5 = 2 : 3
1.25 1.25
 The empirical formula of the oxide is Fe2O3.
Empirical formula mass of Fe2O3 = [2(55.85) + 3(16.00)] g
Molar mass of Fe2O3 = 159.69 g
Molar mass 159.69 g
n  
Emprical formula mass 159.7 g = 0.999 = 1 (approx)
Molecular formula of a compound is obtained by multiplying the empirical formula with n. Thus, the
empirical formula of the given oxide is Fe2O3 and n is 1.
Hence, the molecular formula of the oxide is Fe2O3.
[D-14]
 Mole Concept
Example 3. A pure sample of ethylamine (C2H5NH2) was quantitatively analysed for nitrogen. 0.0275 g of the pure
sample was found to contain 0.007 g nitrogen. What is the mass % nitrogen in ethylamine ?

Mass of N 0.007g
Solution Mass % N = ×100 = ×100 = 25.5% N
Mass of sample 0.0275g
Example 4. 0.45 g of an organic-compound containing only carbon, hydrogen and nitrogen on combustion gave 1.1g
CO2 and 0.3g water. What is % C, % H and % N in the organic-compound ?
Solution Mass of organic-compound, w = 0.45 g Mass of CO2 , w1 = 1.1g Mass of water, H2O, w2 = 0.3 g
1 mol C  1 mol CO2 12 g  44 g CO2

12g Mass of CO 2 12g 1.1g

(a) Percentage of carbon = × ×100 = × ×100 = 66.7%C
44g Mass of sample 44g 0.45g
(b) 2 mol H  1 mol H2O 2g H  18g H2O

2g Mass of H 2 O 2g 0.3g
Percentage of hydrogen = × ×100 = × ×100 = 7.4% H
18g Mass of Sample 18g 0.45g
(c) % N = 100 – [% C + % H] = 100 - [66.7 + 7.4] = 25.9 % N
Example 5. A chemist added an excess of sodium sulphate to a solution of a soluble barium compound to precipitate
all of the barium ion as barium sulphate, BaSO4. How many grams of barium ion are in a 458 mg
sample of the barium compound if solution of the sample gave 513 mg BaSO4 precipitate. What is the
mass percentage of barium in the compound ?
Solution 1 mol Ba+2 = 1 mol BaSO4

0.513
No. of moles of Ba+2 = = 2.2 × 10-3 mol
233
1 mol of Ba weight = 137 g
2.2 × 10-3 mol of Ba weight = 137 × 2.2 × 10-3 = 0.301 g
0.301
Mass %age of Ba in the compound = ×100= 65.9 %
0.458
Example 6. Formula of a metal sulphate is MSO4. 5H2O. Find out At. Wt. of metal if % of S in hydrated salt is 14.2

100
Solution 14.2 gm ‘S’ = 100 gm salt. ; 32 gm S = ×32 = 226 = molecular weight
14.2
Atomic weight of M = Molecular wt of salt – M wt. of SO 2-
4 – 5 × M. Wt of H2O

= 225 – 96 – 90 = 40 amu
Example 7. 0.66 g of an organic compound containing C, H and O gave on combustion 0.968 g of CO 2 and 0.792 g
of H2O . Calculate percentage of O in the compound. (C = 12, H = 1, O = 16)
0.968
Solution Mole of C in CO2 = 1 × Mole of CO2 =1× = 0.022
44
Wt. of C = 0.022 × 12 = 0.264 g
0.792
Mole of H in H2O = 2 × Mole of H2O = 2 × = 0.088
18
[D -15]
Chemistry
Weight of H = 0.088 × 1 = 0.088 g.
Total weight of C and H in the compound = (0.264 + 0.088)g = 0.352 g
 Weight of O in the compound = (0.66 – 0.352)g = 0.308 g
0.308
 % of O in the compound = × 100 = 46.67%
0.66

10. STOICHIOMETRYAND STOICHIOMETRIC CALCULATIONS:

The word ‘stoichiometry’ is derived from two Greek words - stoicheion (meaning element) and metron (meaning
measure). Stoichiometry, thus deals with the calculation of masses (sometimes volumes also) of the reactants
and the products involved in a chemical reaction. Before understanding how to calculate the amounts of reactants
required or those produced in a chemical reaction, let us study what information is available from the balanced
chemical equation of a given reaction. Let us consider the combustion of methane. A balanced equation for this
reaction is as given below:
CH 4 (g) + 2O 2 (g) 
 CO 2 (g) + 2 H 2 O(g)
Here, methane and dioxygen are called reactants and carbon dioxide and water are called products. Note that
all the reactants and the products are gases in the above reaction and this has been indicated by letter (g) in the
brackets next to its formula. Similarly, in the case of solids and liquids, (s) and (1) are written respectively.
The coefficients 2 for O2 and H2O are called stoichiometric coefficients. Similarly the coefficient for CH 4 and
CO2 is one in each case. They represent the number of molecules(and moles as well) taking part in the reaction
or formed in the reaction.
Thus, according to the above chemical reaction.
 One mole of CH4(g) reacts with two moles of O2(g) to give one mole of CO2(g) and two mole of
H2O(g)
 One molecule of CH4(g) reacts with 2 molecules of O2(g) to give one molecule of CO2(g) and 2
molecules of H2O(g)
 22.4 L of CH4(g) reacts with 44.8 L of O2(g) to give 22.4 L of CO2(g) and 44.8 L of H2O(g).
 16 g of CH4(g) reacts with 2 × 32 g of O2(g) to give 44 g of CO2(g) and 2 × 18 g of H2O(g).
From these relationships, the given data can be interconverted as follows:
mass  moles  no. of molecules
10.1 Stoichiometry of Chemical Equations :
These are classified under three categories :
(i) Problems involving mass-mass relationship
(ii) Problems involving mass-volume relationship
(iii) Problems involving volume-volume relationship
In general, following steps are taken into consideration :
(i) Write the balanced chemical equation.
(ii) Write down the relative number of moles or the relative masses of the reactants and products below the
respective formula.
(iii) In the case of gaseous reactants or products, write down 22.4 litre at S.T.P. below the formula instead of
1 mole.
(iv) Write down the actual quantities of the reactants or products given in question and find out the weight/
volume of the unknown substance by unitary method.

[D-16]
 Mole Concept
10.1.1 Calculations involving mass - mass relationship :
In these problems, the mass of one of the reactants / products is to be calculated if that of other reactants /
products are given.
e.g. ZnCO3 (s)   ZnO (s) + CO2 (g)
Calculate the mass of zinc oxide formed by heating 0.5 mole of zinc carbonate.
1 mole of zinc carbonate on heating gives = 81g of ZnO
0.5 mole of zinc carbonate will give = 81 × 0.5 = 40.5 g of ZnO
10.1.2 Mass - Volume relationship :
In these problems, mass or volume of one of the reactants or products is calculated from the mass or volume of
other substances.
e.g :
Calculate the volume of carbon dioxide at NTP evolved by strong heating of 20 g calcium carbonate.
CaCO3 
 CaO + CO2
1 mole = 22.4 litre at NTP
100g of CaCO3 evolve carbon dioxide = 22.4 litre.
22.4
20g of CaCO3 will evolve carbon dioxide = × 20 = 4.48 litre
100
10.1.3 Volume-volume relationship :
In these problems, the volume of one of the reactants / products is given and that of the other is to be calculated.
eg. 1 litre of oxygen at NTP is allowed to react with three times of carbon monoxide at NTP. Calculate the
volume of each gas found after the reaction.

2CO + O 2 
 2CO 2
2 Vol 1 Vol 2 Vol

1 volume of O2 reacts with 2 volume of CO

or 1 litre of O2 reacts with 2 litre of CO
Thus, 1 litre of CO remains unreacted.
1 volume of O2. produces CO2 = 2 volume
or, 1 litre of O2. produce CO2 = 2 litre
Thus, gaseous mixture after the reaction consists. Volume of CO = 1 litre Volume of CO 2 = 2 litre
10.1.4 Theoretical yield :
The amount of a product that is formed when the limiting reactant is completely used up is called the theoretical
yield of that product.
Percentage Yield :
In general, when a reaction is carried out in the laboratory we do not obtain actually the theoretical amount of
the product. The amount of the product that is actually obtained is called the actual yield. Knowing the actual
yield and theoretical yield the percentage yield can be calculated as :

Actual yield
% yield =
Theoretical yield × 100

[D -17]
Chemistry

Example 1. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction,
CaCO3(s) + 2 HCl(aq)  CaCl 2(aq) + CO2(g) + H2O(l)
What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl? (NCERT)
Solution 0.75 M of HCl  0.75 mol of HCl are present in 1 L of water
 [(0.75 mol) × (36.5 g mol –1)] HCl is present in 1 L of water
 27.375 g of HCl is present in 1 L of water
Thus, 1000 mL of solution contains 27.375 g of HCl.
27.375g
 Amount of HCl present in 25 mL of solution   25mL = 0.6844 g
1000 mL
From the given chemical equation,
CaCO3(s) + 2HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l)
2 mol of HCl (2 × 36.5 = 71 g) react with 1 mol of CaCO 3 (100 g).
100
 Amount of CaCO3 that will react with 0.6844 g   0.6844g = 0.9639 g
71
Example 2. Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous
hydrochloric acid according to the reaction (NCERT)
4HCl (aq) + MnO2(s)  2H2O(l) + MnCl 2(aq) + Cl2(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Solution 1 mol [55 + 2 × 16 = 87 g] MnO2 reacts completely with 4 mol [4 × 36.5 = 146 g] of HCl.
 5.0 g of MnO2 will react with

146g
  5.0g of HCl = 8.4 g of HCl
87 g
Hence, 8.4 g of HCl will react completely with 5.0 g of manganese dioxide.
Example 3. In the thermite process Cr2O3 (s) + 2A(s)  Al 2 O3 (s)  2Cr(s) , when 18.7 g of Cr2O3 were heated
with excess aluminium, 10.8 g of Cr were isolated from the products. What was the percentage yield ?
Solution Cr2O3(s) + 2Al(s)   Al2O3(s) + 2Cr(s)
Atomic mass, Cr = 52, Al = 27
18.7 g
No. of moles of Cr2O3 = 152 g mol-1 = 0.123 mol

No. of moles of Cr = 2 × Cr2O3 = 2 × 0.123 = 0.246 mol

Theoretical mass obtained = 0.246 × 52 g = 12.79 g, Actual mass = 10.8g
10.8
 Percentage yield = 12.79  100 = 84.4%

11. LIMITING REAGENT :

The reactant which is consumed first and limits the amount of product formed in the reaction, and is therefore,
called limiting reagent.

[D-18]
 Mole Concept
Limiting reagent is present in least stoichiometric amount and therefore controls amount of product.
The remaining or left out reactant is called the excess reagent.
When you are dealing with balanced chemical equation then if number of moles of reactants are not in the ratio
of stoichiometric coefficient of balanced chemical equation, then there should be one reactant which is limiting
reactant.
11.1 How to find limiting reagent :
Step : I Divide the given moles of reactant by the respective stoichiometric coefficient of that reactant.
Step : II See for which reactant this division come out to be minimum. The reactant having minimum value
is limiting reagent for you

Example 1. Calculate the amount of carbon dioxide that could be produced when (NCERT)
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Solution The balanced reaction of combustion of carbon can be written as:
C + O2   CO2
(i) As per the balanced equation, 1 mole of carbon burns in 1 mole of dioxygen (air) to produce 1
mole of carbon dioxide.
(ii) According to the question, only 16 g of dioxygen is available. Hence, it will react with 0.5 mole of
carbon to give 22 g of carbon dioxide. hence, it is a limiting reactant.
(iii) According to the question, only 16 g of dioxygen is available. It is a limiting reactant. Thus, 16 g of
dioxygen can combine with only 0.5 mole of carbon to give 22 g of carbon dioxide.
Example 2. In a reaction (NCERT)
A + B2  AB2
Identify the limiting reagent, if any in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B (ii) 2 mol A + 3 mol B2
(iii) 100 atoms of A + 100 molecules of B (iv) 5 mol A + 2.5 mol B2
(v) 2.5 mol A + 5 mol B
Solution A limiting reagent determines the extent of a reaction. It is the reactant which is the first to get
consumed during a reaction, thereby causing the reaction to stop and limiting the amount of products
formed.
(i) According to the given reaction, 1 atom of A reacts with 1 molecule of B. Thus, 200 molecules
of B will react with 200 atoms of A, thereby leaving 100 atoms of A unused. Hence, B is the
limiting reagent.
(ii) According to the reaction, 1 mol of A reacts with 1 mol of B2. Thus, 2 mol of A will react with
only 2 mol of B2. As a result, 1 mol of B2 will not be consumed. Hence, A is the limiting reagent.
(iii) According to the given reaction, 1 atom of A combines with 1 molecule of B2. Thus, all 100
atoms of A will combine with all 100 molecules of B. Hence, the mixture is stoichiometric where
no limiting reagent is present.
(iv) 1 mol of atom A combines with 1 mol of molecule B2. Thus, 2.5 mol of B2 will combine with only
2.5 mol of A. As a result, 2.5 mol of A will be left as such. Hence, B is the limiting reagent.
(v) According to the reaction, 1 mol of atom A combines with 1 mol of molecule B. Thus, 2.5 mol of
A will combine with only 2.5 mol of B2 and the remaining 2.5 mol of B will be left as such.
Hence, A is the limiting reagent.
[D -19]
Chemistry
Example 3. Dinitrogen and dihydrogen react with each other to produce ammonia according to the following
chemical equation: (NCERT)
N2(g) + H2(g)  2NH3(g)
(i) Calculate the mass of ammonia produced if 2.00 × 10 3 g dinitrogen reacts with 1.00 × 10 3 g of
dihydrogen.
(ii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?
Solution (i) Balancing the given chemical equation,
N2(g) + 3H2(g)  2NH3(g)
From the equation, 1 mole (28 g) of dinitrogen reacts with 3 mole (6 g) of dihydrogen to give 2
mole (34 g) of ammonia.

6g
2.00 × 103 g of dinitrogen will react with  2.00  103 g dihydrogen i.e.,
28g
2.00 × 103 g of dinitrogen will react with 428.6 g of dihydrogen.
Given,
Amount of dihydrogen = 1.00 × 103 g
Hence, N2 is the limiting reagent.  28 g of N2 produces 34 g of NH3.

34g
Hence, mass of ammonia produced by 2000 g of N2  28g  2000g = 2428.57 g

(ii) N2 is the limiting reagent and H2 is the excess reagent. Hence, H2 will remain unreacted.
(iii) Mass of dihydrogen left unreacted = 1.00 × 103 g – 428.6 g = 571.4 g
Example 4. Zinc and hydrochloric acid react according to the reaction.
Zn (s) + 2HCl (aq) 
 ZnCl2 (aq) + H2(g)
If 0.30 mol Zn are added to hydrochloric acid containing 0.52 mol of HCl, how many moles of H 2 are
produced ?
Solution The reaction is Zn (s) + 2 HCl (aq) 
 ZnCl2 (aq) + H2(g)
Thus, 1 mole of zinc reacts with 2 moles of HCl.
 0.30 mole of zinc will react with HCl = 2 × 0.30 = 0.60 mol
But we have only 0.52 mole of HCl. Hence, zinc cannot react completely and hence is not a limiting
reactant. HCl is limiting reagent.
Again, 2 moles of HCl react with zinc of 1 mol and produce 1 mole of H2
 0.52 mol of HCl will produce H2
1
= × 0.52 mol = 0.26 mol
2
Example 5. 3.0 g of H2 react with 29.0 g of O2 to form H2O.
(i) Which is the limiting reactant ?
(ii) Calculate the maximum amount of H2O that can be formed.
(iii) Calculate the amount of the reactant left unreacted.
Molecular mass of H2 = 2.016.
[D-20]
 Mole Concept

Solution 2 H2 + O2 
 2H 2O
2 × 2.016 32g 2 × (2.106 + 16)
= 4.032 g = 36.032
32
3 g of H2 require O2 = × 3 = 23.8 g
4.032
Thus, O2 (29 g) is present in excess. Hence, H2, is the limiting reactant.
36.032
H2O formed = × 3 g = 26.8 g. O2 left unreacted = 29 – 23.8 = 5.2 g
4.032
Example 6. If 20.0 g of CaCO3 is treated with 20.0 g of HCl, how many grams of CO2 be produced according to
the reaction.
CaCO3 (s) +2HC(aq)  CaC 2 (aq)+H 2O()+CO 2 (g)
Solution In this problem, CO2 can be calculated by using either of the reactants CaCO3 or HCl. So, first of all we
are to calculate the limiting reactant.
CaCO3 (s) + 2HC(aq)  CaC 2 (aq)+H 2O()+CO 2 (g)
1 mol 2 mol
40 + 12 + 3 × 16 2 × (1 + 35.5)
100 g = 73 g
According to above equation. 100 g of CaCO3 require 73 g of HCl
73
20 g of CaCO3 require = × 20 = 14.6 g. But amount of HCl actually present = 20.0 g
100
Therefore, CaCO3 is limiting reactant and HCl is excess reactant.
Now let us calculate the amount of CO2 produced when entire quantity of limiting reactant reacts.
CaCO3 (s) + 2HC(aq)  CaC 2 (aq)+H 2O()+CO 2 (g)
1 mol 1 mol
100 g 44 g
20 g ?
100 g of CaCO3 produces CO2 = 44 g
44
20 g of CaCO3 will produces CO2 = × 20 = 8.80 g
100

12. PRINCIPLE OF ATOM CONSERVATION (POAC) :

POAC is conservation of mass. Atoms are conserved, moles of atoms shall also be conserved in a chemical
reaction (but not in nuclear reactions.)
This principle is fruitful for the students when they don’t get the idea of balanced chemical equation in the
problem.
The strategy here will be around a particular atom. We focus on an atom and conserve it in that reaction. This
principle can be understand by the following example.

[D -21]
Chemistry
Consider the decomposition of KClO3 (s)  KCl (s) + O2 (g) (unbalanced chemical reaction)
Apply the principle of atom conservation (POAC) for K atoms.
Moles of K atoms in reactant = moles of K atoms in products.
or moles of K atoms in KClO3 = moles of K atoms in KCl.
Now, since 1 molecule of KClO3 contains 1 atom of K
or 1 mole of KClO3 contains 1 mole of K, similarly, 1 mole of KCl contains 1 mole of K.
Thus, moles of K atoms in KClO3 = 1 × moles of KClO3
and moles of K atoms in KCl = 1 × moles of KCl.
wt.of KClO3 in g wt.of KCl in g
 moles of KClO3 = moles of KCl or =
mol.wt.of KClO 3 mol.wt.of KCl
The above equation gives the mass-mass relationship between KClO3 and KCl which is important in stoichiometric
calculations.
Again, applying the principle of atom conservation for O atoms.
moles of O in KClO3 = 3 × moles of KClO3
moles of O in O2 = 2 × moles of O2

wt.of KClO 3 vol.of O 2 at NTP

 3 × moles of KClO3 = 2 × moles of O2 or 3 × = 2×
mol.wt.of KClO 3 standard molar vol.(22.4 lt .)
The above equation thus gives the mass-volume relationship of reactants and products.

Example 1. 25.4 g of iodine and 14.2 g of chlorine are made to react complete to yield a mixture of ICl and ICl 3.
Calculate the number of moles of ICl and ICl3 formed.

Solution I + 2 Cl2 
2  ICl + ICl3
1mole 2moles 1mole 1mole
2127 2(2×35.5)
 254 g =142g

1mol
(a) 254 g I2 form 1 mol ICl ; 25.4 g I2 form =
254 g × 25.4 g = 0.1 mol ICl

1mol
(b) 254 g I2 form 1 mole ICl2 ; 25.4 g l2 form = × 25.4 g = 0.1 mol ICl3
254 g
Example 2. A sample of HCl has 20% hydrogen chloride. How many grams of this sample is needed to completely
react with 50 g CaCO3 ?

Solution CaCO3 + 2HCl 

 CaCl 2 +H 2 O+CO 2
100g 73g

100 g CaCO3  73 g HCl

73
 50 g of CaCO3  100 × 50 = 36.5 g HCl.

20 g HCl  100 g sample

 36.5 g HCl  182.5 g HCl sample

[D-22]
 Mole Concept
Example 3. A solution containing 1.5 g MI formed 2.35 g of silver iodide with excess silver nitrate solution. What is
the atomic mass of metal M ?
Cl = 35.5u ; Ag = 108u
Solution MI + AgNO3   AgI + MNO3
Molecular mass Ml = M + 127 amu ; Molecular mass of AgI = 108 + 127 = 235 amu
235
(M + 127) g MI produces = 235 g AgI ; 1.5 g MI produces = × 1.5 g AgI
M+127
235×1.5
=
M +127 = 2.35 or M + 127 = 150 M = 150 – 127 = 23 amu

Example 4. How much volume sulphur dioxide at NTP will be obtained by completely burning 10 g of
pure sulphur ?
Solution The reaction involved is : S + O2  SO2
i.e. one mole of S combines with one mole of O2 to produce one mole of SO2.
According to reaction stoichiometry
1 mol S = 1 mol SO2
32 g S = 22.4 L SO2
22.4
10 g S =  10 = 7L . Thus volume of SO2 obtained is 7 L
32

13. SOLUTIONS :
A mixture of two or more substances can be a solution. We can also say that “a solution is a homogeneous
mixture of two or more substances” Homogeneous means ‘uniform throughout’. Thus a homogeneous mixture
i.e. a solution, will have uniform composition throughout.
Properties of a solution :
• The solute in a solution does not settle down even after the solution is kept undisturbed for some time.
• In a solution, the solute particle cannot be distinguished from the solvent particles or molecules even
under a microscope. In a true solution, the particles of the solute disappear into the space between the
solvent molecules.
• The components of a solution cannot be separated by filtration.
13.1 Concentration terms :
The following concentration terms are used to expressed the concentration of a solution. These are
• Molarity (M)
• Molality (m)
• Mole fraction (x)
• % calculation
• ppm
• Normality (N)
Remember that all of these concentration terms are related to one another. By knowing one concentration
term you can also find the other concentration terms. Let us discuss all of them one by one.

[D -23]
Chemistry
13.1.1 Molarity (M) :
The number of moles of a solute dissolved in 1 L (1000 ml) of the solution is known as the molarity
of the solution.
number of moles of solute
i.e., Molarity of solution =
volume of solution in litre
Let a solution is prepared by dissolving w gm of solute of mol. wt. M in V ml water.
w
 Number of moles of solute dissolved =
M
w
 V ml water have mole of solute
M
w×1000 w×1000
 1000 ml water have M × V  Molarity (M) = (Mol. wt of solute)×V
ml ml

Number of millimoleof solute

• Molarity of solution may also be given as (M) = Total volume of solution in ml (V )
ml

Number of milimoles of solute = (Molarity of solution × V ml)

• Molarity is a unit that depends upon temperature. It varies inversely with temperature.
Mathematically : Molarity decreases as temperature increases.

1 1
Molarity  
temperature volume
Effect of dilution on Molarity:-
If a particular solution having volume V1 and molarity M1 is diluted upto volume V2 mL than
M 1V1 = M 2V 2
M2 : Resultant molarity
Mixing of solution & resultant molarity:-
If a solution having volume V1 and molarity M1 is mixed with another solution of same solute having volume V2
mL & molarity M2
then M1V1 + M2V2 = MR (V1 + V2)

M1V1 +M 2 V2
MR = Resultant molarity =
V1 +V 2
13.1.2 Molality (m) :
The number of moles of solute dissolved in 1000 gm (1kg) of a solvent is known as the molality of
the solution.
number of moles of solute
i.e., molality = ×1000
mass of solvent in gram
Let Y gm of a solute is dissolved in X gm of a solvent. The molecular mass of the solute in M0. Then
Y/M0 mole of the solute are dissolved in X gm of the solvent. Hence
Y
Molality = M ×X ×1000
0

• Molality is independent of temperature changes.

[D-24]
 Mole Concept
13.1.3 Mole fraction (x) :
The ratio of number of moles of the solute or solvent present in the solution and the total number of moles
present in the solution is known as the mole fraction of substances concerned.
Let number of moles of solute in solution = n
Number of moles of solvent in solution = N
n N
 Mole fraction of solute (x1) = n + N  Mole fraction of solvent (x2) = n + N

also x1 + x2 = 1
Mole fraction is a pure number. It will remain independent of temperature changes.
13.1.4 Formality (F) :
It is defined as the number of gram formula masses of solute dissolved per litre of the solution.
Number of gram formula mass of solute
Formality = Volume of solution in litres
This term is used for ionic solute like NaCl.

13.1.5 Percentage Terms:-

(a) Weight - weight percent (w/W) :
Weight of solute present in 100g of the solution.

weight of solute (g)

Weight percent   100 .
weight of solution (g)

(b) Volume - volume percent (v/V) :

(In liquid - liquid solution)
Volume of solute in ml present in 100ml of the solution is called volume - volume percentage.

volume of solute (mL)

Volume - volume percentage   100
volume of solution (mL)

(c) Weight - volume percent (w/V) :

Weight of solute in g present in 100mL of the solution is called weight - volume percentage.

weight of solute (g)

weight - volume percentage   100
volume of solution (ml)
13.1.6 ppm (Part per million)
The parts of the component per million parts (106) of the solution.

w
ppm   106 where w = weight of solute, W = weight of solvent
w  W

[D -25]
Chemistry

Example 1. A sample of drinking water was found to be severely contaminated with chloroform, CHCl 3, supposed
to be carcinogenic in nature. The level of contamination was 15 ppm (by mass).
(i) Express this in percent by mass. (NCERT)
(ii) Determine the molality of chloroform in the water sample.
Solution (i) 1 ppm is equivalent to 1 part out of 1 million (106) parts.
15
 Mass percent of 15 ppm chloroform in water   100  1.5  10 –3%
106
(ii) 100 g of the sample contains 1.5 × 10–3 g of CHCl3.
 1000 g of the sample contains 1.5 × 10–2 g of CHCl3
1.5  10 –2 g
 Molality of chloroform in water 
Molar mass of CHCl3
Molar mass of CHCl 3 = 12.00 + 1.00 + 3(35.5) = 119.5 g mol –1
 Molality of chloroform in water = 0.0125 × 10 –2 m = 1.25 × 10–4 m
Example 2. If the density of methanol is 0.793 kg L–1, what is its volume needed for making 2.5 L of its 0.25 M
solution? (NCERT)
Solution Molar mass of methanol (CH3OH) = (1 × 12) + (4 × 1) + (1 × 16)
= 32 g mol –1 = 0.032 kg mol–1

0.793kg L–1
Molarity of methanol solution  = 24.78 mol L–1
0.032 kg mol –1
(Since density is mass per unit volume) Applying, M1V1 = M2V2
(24.78 mol L–1) V1 = (2.5 L) (0.25 mol L–1)
V1 = 0.0252 L V1 = 25.22 mL
Example 3. What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in enough water
to make a final volume up to 2 L? (NCERT)
Solution Molarity (M) of a solution is given by,
Number of moles of solute Mass of sugar/molar mass of sugar
= =
Volume of solution in Litres 2L

20 g / 12  12   1 22   11 16   g 20 g / 342 g 0.0585mol

   = 0.02925 mol L–1
2L 2L 2L
 Molar concentration of sugar = 0.02925 mol L–1
Example 4. Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375 molar aqueous
solution. Molar mass of sodium acetate is 82.0245 g mol–1 (NCERT)
Solution 0.375 M aqueous solution of sodium acetate
= 1000 mL of solution containing 0.375 moles of sodium acetate
0.375
 Number of moles of sodium acetate in 500 mL   500 = 0.1875 mole
1000
Molar mass of sodium acetate = 82.0245 g mole–1 (Given)
 Required mass of sodium acetate = (82.0245 g mol–1) (0.1875 mole) = 15.38 g
[D-26]
 Mole Concept
Example 5. Calculate the concentration of nitric acid in moles per litre in a sample which has a density,
1.41 g mL–1 and the mass per cent of nitric acid in it being 69%. (NCERT)
Solution Mass percent of nitric acid in the sample = 69% [Given]
Thus, 100 g of nitric acid contains 69 g of nitric acid by mass.
Molar mass of nitric acid (HNO3)
= {1 + 14 + 3 (16)} g mol–1 = 1 + 14 + 48 = 63 g mol–1
69g
 Number of moles in 69 g of HNO3 = = 1.095 mol
63g mol –1

Volume of 100 g of nitric acid solution = Mass of solution

density of solution

100 g
 = 70.92 mL = 70.92 × 10–3 L
1.41g mL–1

mole 1.095 mole = 15.44 mol/L

Concentration of nitric acid = =
volume in L 70.92 ×10 –3 L
 Concentration of nitric acid = 15.44 mol/L

Mole Concept :
(a) Definition of one mole : One mole is a collection of that many entities as there are number of atoms
exactly in 12 gm of C-12 isotope.

1 1g
(b) 1u = 1 amu =   of mass of 1 atom of = = 1.66 × 10–24 g
 12  NA
(c) For Elements :
• 1 g atom = 1 mole of atoms = NA atoms.
• 1 g atomic mass (GAM) = mass of NA atoms in g.
Mass ( g )
• Mole of atoms =
GAM or molar mass
(d) For molecules :
• 1 g molecule = 1 mole of molecule = NA molecule
• 1 g molecular mass (GMM) = mass of NA molecule in g.
Mass( g )
• Mole of molecule =
GMM or molar mass
(e) For ionic compounds :
• 1 g formula unit = 1 mole of formula unit = NA formula unit.
• 1 g formula mass (GFM) = mass of NA formula unit in g.
Mass ( g )
• Mole of formula unit =
GFM or molar mass

[D -27]
Chemistry

Contains 6.022 × 1023 particles

(f) 1 mole of a substance Weighs as much as molecular weight/atomic/ionic weight in grams.
If it is a gas, one mole occupies a volume of 22.4 L at 1 atm & 273 K

(g) Average or mean atomic mass : Average atomic mass of element.

A1 x1  A2 x2  .......
AAvg. 
x1  x2  ......
Here A1, A2 are isotopic mass of element and x1, x2 are natural abundance of isotopes.
(h) Average or mean molar mass : The average molar mass of the different substance present in the
M1n1  M2n2  .......
container. MAvg. 
n1  n2  ......
Here M1, M2 are molar mass of substances and n1, n2 are mole of substances present in the container.
Empirical Formula, Molecular Formula :
(a) Empirical formula : Formula depicting constituent atom in their simplest ratio.
Molecular formula : Formula depicting actual number of atoms in one molecule of the compound.
(b) Relation between molecular formula and empirical formula :
Molecular mass
n
Empirical Formula mass
(c) Densities :
Mass
• Density = , Unit: g/cc
Volume

Density of any substance

• Relative density 
Density of reference substance

Density of any substance

• Specific gravity 
Density of water at 4°C
• Vapour density : Ratio of density of vapour to the density of hydrogen at similar pressure and temperature.
Density of vapour at some temperature and pressure
Vapour density 
Density of H 2 gas at same temperature and pressure

Molecular mass
Vapour density 
2
Laws of Chemical Combination :
Chemical reactions take place according to certain laws. These laws are called the Laws of Chemical Combination.
“These are no longer useful in chemical calculations now but gives an idea of earlier methods of analysing and
relating compounds by mass.”
• Law of Conservation of Mass [Lavoisier (1774)]
During any physical or chemical change, the sum of masses of all substances present in reactions
vessel remain conserved.
• Law of Constant Composition or Definite Proportions [Proust (1799)]
In a given chemical compound, the elements are always combined in the same proportions by mass.
[D-28]
 Mole Concept
• Law of Multiple Proportions [Dalton (1803)]
Whenever two elements form more than one compound, the different masses of one element that
combine with the same mass of the other element are in the ratio of small whole numbers.
• Law of Reciprocal Proportions [Richter (1792)]
When two elements combine separately with a fixed mass of a third elements then the ratio of their
masses in which they do so is either same or some whole number multiple of the ratio in which they
combine with each other.
• Gay-Lussac’s Law of combining volumes
According to Gay-Lussac’s law of combining volume, when gases react together, they always do so in
volumes which bear a simple ratio to one another and to the volumes of the products, if these are also
gases, provided all measurements of volumes are done under similar conditions of temperature and
pressure.
• Avogadro’s Law
The volume of a gas (at fixed pressure and temperature) is proportional to the number of moles (or
molecules of gas present). Mathematically we can write
V n
Or Equal volumes of all gases under similar conditions of temperature and pressure contain equal
number of molecules.
Stoichiometry :
Stoichiometry pronounced by (“stoy - key - om - e - tree”) is the calculations of the quantities of reactants and
products involved in a chemical reaction. Following methods can be used for solving problems.
(a) Mole Method (Balancing is required)

For Ex. : 2KClO3   2KCl + 3O2
Mole of KClO3 Mole of KCl Mole of O 2
 
2 2 3
(b) Principle of Atom Conservation (P.O.A.C.) method (Balancing is not required)

For Ex. : KClO3   KCl + O2
POAC for K : 1 × mole KClO3 = 1 × mole of KCl
POAC for Cl : 1 × mole KClO3 = 1 × mole of KCl
POAC for O : 3 × mole KClO3 = 2 × mole of O2
(c) Limiting Reagent : It is very important concept in chemical calculation. It refers to reactant which is
present in minimum stoichiometry quantity for a chemical reaction. It is reactant consumed fully in a
chemical reaction. So all calculations related to various products or in sequence of reactions are made
on the basis of limiting reagent.
(d) Calculation of Limiting Reagent : Divide given moles of each reactant by their stoichiometric
coefficient, the one with least ratio is limiting reagent.
Percentage Yield :
Actual yield
The percentage yield of product   100
Theoretical maximum yield
Concentration Terms
(a) For solutions (homogeneous mixture) :
• If the mixture is not homogeneous, then none of them is applicable.
[D -29]
Chemistry

 w  Wt. of solute
(i) % by mass    ×100
 W  Wt. of solution
[X% by mass means 100 gm solution contains X gm solute ;  (100 – X) gm solvent]

(ii) %   
w Wt. of solute
×100
 V  Volume of solution

w
[ X %   ] means 100 mL solution contains X gm solute]
V 

 V  Volume of solute
(iii) %    ×100
 V  Volume of solution
• For gases % by volume is same as mole%
Moles of solute
(iv) Mole %  ×100
Total moles

Moles of solute
(v) Mole fraction of solute (X) 
Total moles

Moles of solute
(vi) Molarity (M) 
Volume of solution (in litre)

Moles of solute
(vii) Molality (m) 
Mass of solvent (in kg)

Moles of solute Moles of solute

(viii) Parts per million (ppm)   106   106
Mass of solvent Mass of solution

No. of formula unit

(ix) Formality (F) 
Volume of solution (in litre)
(b) (i) On adding solvent in a solution (dilution) : Number of mole of solute remains constant
MfVf = MiVi
(ii) Mixing of two solutions of same solute
MfVf = M1V1 + M2V2 + ....

[D-30]
 Mole Concept

(A) BASICS OF MOLE CONCEPT: 14. Calculate the molecular mass of the following:
1. Find the relative atomic mass of ‘O’ atom and its (NCERT)
atomic mass. (i) H2O (ii) CO2 (iii) CH4
2. How many atoms of oxygen are there in 16g oxygen. 15. Calculate the atomic mass (average) of chlorine
3. The molecular mass of H2SO4 is 98 amu. using the following data: (NCERT)
Calculate the number of moles of each element in % Natural Abundance Molar Mass
294 g of H2SO4. 35 75.77 34.9689
Cl
4. Calculate the number of molecules of dinitrogen oxide 37 24.23 36.9659
Cl
(N2O) in 0.044 kg of the gas 16. Calculate the number of atoms present in 5.6 litres
5. What is the mass, in grams of 0.25 mol copper ? if a
Cu = 63.5 amu (i) monoatomic and
6. A sample of phosphorus has 0.25 moles of P4 (ii) diatomic gas at NTP ?
molecules.
(i) How many P4 molecules are there ?
(B) PERCENTAGE WEIGHT OF ELEMENTS
(ii) How many P atoms are there ? IN COMPOUNDS, MASS % OF
(iii) How many moles of P atoms are there in the COMPOUNDS IN SAMPLE, MOLECULAR
sample ? FORMULA & EMPIRICAL FORMULA :
(iv) What is the mass of the sample ? 1. Calculate the mass percent of different elements
7. What is the mass of carbon present in 0.5 mole of present in sodium sulphate (Na2SO4). (NCERT)
K4 [Fe(CN)6]? 2. Determine the empirical formula of an oxide of
8. How many moles of gold are there in 49.25 g gold iron which has 69.9% iron and 30.1% dioxygen
rod ? atomic mass of gold = 197 amu.. by mass. (NCERT)
9. Arrange the following in order of their increasing 3. Acetylene and benzene both have the empirical
masses in grams ? formula CH. The molecular masses of acetylene
(i) One atom of silver and benzene are 26 and 78 respectively. Deduce
(ii) One gram – atom of nitrogen, their molecular formula.
(iii) One mole of calcium 4. A crystalline hydrated salt on being rendered
anhydrous, loses = 45.6% of its weight. The
(iv) One moles of oxygen molecules
percentage composition of anhydrous salt is : Al =
(v) 1023 atoms of carbon (vi) One gram of iron. 10.5%, K = 15.1% , S = 24.8% and oxygen = 49.6%.
10. The volume of a gas is 1.12 x 10–7 cm3 at NTP. Find the empirical formula of the anhydrous and
Calculate the number of molecules of the gas. crystalline salt.
11. How many H atoms are present in 25.6 gm of urea 5. What is the percentage composition of each element
[(NH 2 ) 2 (CO)] having molar mass of in zinc phosphate Zn3(PO4)2 ?
60 g/mol ? [Zn = 65.5, P = 31, O = 16]
12. The volume of a drop of water is 0.04 ml. How
many H 2 O molecules are there in a drop of
water ? Density of water = 1.0 g/ml. (C) LAWS OF CHEMICAL COMBINATION :
13. Chlorophyll, the green colouring matter of plants 1
contains 2.68% of magnesium by weight. Calculate 1. H 2 ( g )  O2 ( g )  H 2O ()
2
the number of magnesium atoms in 2.00g of
chlorophyll (at. Mass of Mg = 24) Prove the relation follow law of conservation of
mass?

[D -31]
Chemistry

2. The following data are obtained when dinitrogen N2H4 (l) + N2F4 (g)   2N2(g) + 4 HF(g)
and dioxygen react together to form different 2.25 g HF is obtained. What is the percent yield of
compounds: (NCERT) the reaction ?

Mass of dinitrogen Mass of dioxygen 7. KClO3 on heating decomposes to KCl and O2.

(i) 14 g 16 g 3

KClO3   KCl + 2 O 2
(ii) 14 g 32 g
What is the volume of oxygen at STP liberated by
(iii) 28 g 32 g thermal decomposition of 0.1 mol of KClO3 ?
(iv) 28 g 80 g 8. 0.5 L of CO2 is passed over red hot coke. The volume
becomes 700 ml. Find the composition of the product
Which law of chemical combination is obeyed by if all volumes are measured at NTP.
the above experimental data?Give its statement.
CO2(g) + C(s) 
 2CO(g)
(E) STRENGTH OF SOLUTIONS :
(D) STOICHIOMETRY :
1. 149 gm of potassium chloride (KCl) is dissolved in
1. Three mole of Na2CO3 is reacted with 6 moles of 10 Lt of an aqueous solution. Determine the molarity
HCl solution. Find the volume of CO2 gas produced of the solution (K = 39, Cl = 35.5)
at STP. The reaction is
2. 255 gm of an aqueous solution contains 5 gm of
Na2CO3 + 2HCl 
 2 NaCl + CO2 + H2O urea. What is the concentration of the solution in
terms of molality. (Mol. wt. of urea = 60u)
2. 8 gm of methane is burnt with 4.48L of O2 at STP.
Find out the volume of CO2 gas produced at STP 3. What is the weight percentage of NaCl solution in
and also the weight of CO2 gas. which 20g NaCl is dissolved in 60g of water.
(A) 22.4 L, 44 g (B) 2.24 L, 4.4 g (A) 10% (B) 5%
(C) 1.12 L, 22 g (D) 44.8 L, 88 g (C) 25% (D) 15%
3. 1 g of silver ore is dissolved in HNO3 and Ag 4. A solution is prepared by mixing of 10ml ethanol
precipitated by HCl. The mass of obtained AgCl is
with 120ml of methanol. What is volume percentage
0.7880 g. Calculate the % amount of Ag in the ore.
of ethanol:-
4. 27.6g K2CO3 was treated by a series of reagents
so as to convert all of its carbon to (A) 10% (B) 7.7%
K 2Zn 3[Fe(CN) 6] 2. Calculate the weight of the (C) 20% (D) 15%
product.
[Mol. wt. of K 2CO3 = 138u and mol. wt. of K2Zn3 5. What is weight volume percentage of a solution in
[Fe(CN)6]2 = 698u] which 7.5g of KCl is dissolved in 100mL
5. Following two reactions are primarily responsible of the solution–
for the production of iron : (A) 7.5% (B) 92.5%
2C (s) + O2 (g) 
 2CO (g) (C) 50% (D) none
6. Calculate the Molality of a solution containing 20.7
Fe2O3 (s) +3CO (g)   2 Fe (s) + 3CO2 (g) g of potassium carbonate (K2CO3) dissolved in 500
How many kilograms of oxygen are required to ml of solution. (Assume density of the solution = 1g/
produce 200 kg of iron ? ml)
Fe = 56 amu 7. What is the concentration of sugar (C12H22O11) in
6. 3.2 g Hydrazine, N 2 H 4 and 5.6g dinitrogen mol L–1 if its 20 g are dissolved in enough water to
tetrafluoride, N 2 F 4 react according to the makes a final volume up to 2L ?
equation :

[D-32]
 Mole Concept
8. Calculate the Molality of a solution of ethanol in 15. Calculate (i) molarity and (ii) molality of sulphuric
water in which the mole fraction of ethanol is 0.040. acid solution of specific gravity 1.198 containing 27%
9. Calculate the Molarity of water if its density is 1000 H2SO4 by weight.
kg/m3. 16. 4.450 g 100 per cent sulphuric acid was added to
10. What is the molarity of a barium chloride solution 82.20 g water and the density of the solution was
prepared by dissolving 3.50 g of BaCl2.2H2O in found to be 1.029 g/cc at 250C and 1 atm pressure.
enough water to make 500 mL of solution ? (At Calculate (a) the weight per cent, (b) the mole
mass Ba = 137, Cl = 35.5, H = 1, O = 16) fraction (c) the mole per cent, (d) the molality, (e)
the molarity (f) the normality of sulphuric acid in the
11. Concentration of glucose (C6H12O6) in normal blood
solution under these conditions.
is approximately 90 mg per 100 mL. What is the
molarity of the glucose solution in blood ? 17. The density of a 3 M sodium thio sulphate solution
is 1.23 g per ml. Calculate
12. The sterile saline solution used to rinse contact lenses
can be made by dissolving 400mg of NaCl in sterile (a) the percentage by weight of sodium thio sulphate
water and diluting to 100 mL. What is the molarity (b) the mole fraction of sodium thio sulphate
of the solution ? (c) the molalities of Na+ and S2O32– ions
13. A solution contains 25% water, 25% ethanol and 18. Commercially available concentrated hydrochloric
50% acetic acid by mass. Calculate the mole fraction acid contains 38% HCl by masses .
of each component.
(a) What is the Molarity of this solution ? The density
14. A sample of NaOH weighing 0.40 g is dissolved in is 1.19 g cm–3
water and the solution is made to 50.0 cm3 in
(b) What volume of concentrated hydrochloric acid
volumetric flask. What is the Molarity of the resulting
is required to make 1.00 L of 0.10 M
solution.
HCl ?

[D -33]
Chemistry

(A) BASICS OF MOLE CONCEPT: 8. The molecular weight of hydrogen peroxide is 34.
1. Boron has two isotopes 10B and 11B whose relative What is the unit of gm molecular weight
abundances are 20% and 80% respectively. Atomic
weight of Boron is (A) g (B) mol
(A) 10 (B) 11 (C) g mol– 1 (D) mol g– 1
(C) 10.5 (D) 10.8 9. The total number of g-molecules of SO2Cl2 in
2. Neon has two isotpoes Ne20 and Ne22. If atomic 13.5 g of sulphuryl chloride is
weight of Neon is 20.2, the ratio of the relative (A) 0.1 (B) 0.2
abundances of the isotopes is
(C) 0.3 (D) 0.4
(A) 1 : 9 (B) 9 : 1
(C) 70 % (D) 80 % 10. The number of moles of sodium oxide in 620 g of
it is
3. Choose the correct statement
12
(A) 1 mol (B) 10 moles
The use C scale has superseded the older scale
(C) 18 moles (D) 100 moles
of atomic mass based on 16 O isotope, one important
advantage of the former being 11. 2g of oxygen contains number of atoms equal to
that in
(A) The atomic masses on 12
C scale became
(A) 0.5g of hydrogen (B) 4g of sulphur
whole numbers.
12
(C) 7g of nitrogen (D) 2.3g of sodium
(B) C is more abundant in the earth’s crust than
16
12. The number of molecules of SO2 present in 64g
O
of SO2 is
(C) The difference between the physical and
chemical atomic masses got narrowed down (A) 6.0 × 1023 (B) 3 × 1023
significantly (C) 12 × 1023 (D) 3 × 1010
(D) 12
C is situated midway between metals and 13. The number of sodium atoms in 2 moles of sodium
non-metals in the periodic table ferrocyanide Na4 [Fe(CN)6] is :-
4. Weight of oxygen in 32.2 g Na2SO4. 10H2O is
(A) 12 × 1023 (B) 26 × 1023
(A) 20.8 gm (B) 22.4 gm
(C) 34 × 1023 (D) 48 × 1023
(C) 2.24 gm (D) 2.08 gm
14. Number of neutrons present in 1.7 grams of
5. The volume occupied by 4.4 g of CO2 at STP is
ammonia is -
(A) 22.4 L (B) 2.24 L
(C) 0.224 L (D) 0.1 L (A) NA (B) NA/10 × 4
6. The total number of protons in 10 g of calcium (C) (NA/10) × 7 (D) NA × 10 × 7
carbonate is (N0 = 6.023 × 1023) 15. 5.6 litre of oxygen at NTP is equivalent to –
(A) 1.5057 × 1024 (B) 2.0478 × 1023 (A) 1 mole (B) ½ mole
(C) 3.0115 × 1024 (D) 4.0956 × 1024 (C) ¼ mole (D) 1/8 mole
7. The number of molecules in 16 g of methane is 16. The number of atoms present in 16 g of oxygen is
23 23
(A) 3.0 × 10 (B) 6.02 × 10 (A) 6.02 × 1011.5 (B) 3.01 × 1023
16 16 (C) 3.01 × 1011.5 (D) 6.02 × 1023
(C) × 1023 (D) × 1023
6.02 3.0
[D-34]
 Mole Concept

17. Given that one mole of N2 at NTP occupies 22.4 26. 8 gms of O2 has the same number of molecules as
litre the density of N2 is - (A) 7 gm CO (B) 14 gm of CO
(C) 22 gm of CO2 (D) 44 gms of CO2
(A) 1.25 g/L (B) 0.80 g/L
(C) 2.5 g/L (D) 1.60 g/L
(B) PERCENTAGE WEIGHT OF ELEMENTS
18. If 32 g of O2 contains 6.022 × 1023 molecules at IN COMPOUNDS, MASS % OF
NTP then 32 g of S, under the same conditions, COMPOUNDS IN SAMPLE,
will contain:- MOLECULAR FORMULA & EMPIRICAL
(A) 6.022 × 1023 S (B) 3.011 × 1021 S FORMULA :
(C) 1 × 1023 S (D) 12.044 × 1023S 1. A hydrocarbon contains 75% of carbon. Then its
19. 4.4 g of an unknown gas occupies 2.24 litres of empirical formula is -
volume at STP. The gas may be :- (A) CH4 (B) C2H4
(A) N2O (B) CO (C) C2H6 (D) C2H2
(C) CO2 (D) A & C both 2. A compound of X and Y has equal mass of them.
20. Maximum number of atoms are present in If their atomic weights are 30 and 20 respectively.
(A) 11.2 lit. of SO2 at STP Molecular formula of that compound (its mol. wt.
(B) 22.4 lit. of Helium at STP is 120) could be -
(C) 2.0 gms. of hydrogen (A) X2Y2 (B) X3Y3
(D) 11.2 litres of methane at STP (C) X2Y3 (D) X3Y2
21. Which of the following has highest mass 3. An oxide of sulphur contains 50% of sulphur in it.
(A) 50 gms of iron Its empirical formula is -
(B) 5 moles of nitrogen (A) SO2 (B) SO3
(C) 1 gm atom of silver (C) SO (D) S2O
(D) 5 × 1023 atoms of carbon 4. 1 lt. of a hydrocarbon weighs as much as one
22. 1.5 moles of oxygen atoms are present in litre of CO2. Then the molecular formula of the
(A) 0.5 moles of BaCO3 hydrocarbon is -
(B) 1 mole of BaCO3
(A) C3H8 (B) C2H6
(C) 2 moles of BaCO3
(C) C2H4 (D) C3H6
(D) 0.25 moles of BaCO3
5. A hydrocarbon contains 80% of carbon, and its
23. Which of the follwing contains less number of
V.D. = 15 then the hydrocarbon is -
molecules
(A) 11 gms of CO2 (B) 32 gm of SO2 (A) CH4 (B) C2H4
(C) 2 gms of hydrogen (D) 4 gms of helium (C) C2H6 (D) C2H2
24. The gas having same number of molecules as 16g. 6. Empirical formula of glucose (Molecular Formula
of oxygen is C6H12O6) is -
(A) 16g. of O3 (B) 16g. of SO3
(A) C 6H12O6 (B) C3H6O3
(C) 48g. of SO3 (D) 1gm of hydrogen (C) C2H4O2 (D) CH2O
25. The no. of electrons present in one mole of Azide
7. Percentage of Se in peroxidase anhydrous enzyme
 
ion are N 3 is 0.5% by weight (at.wt. = 78.4) then min.mol. wt.
(A) 21N (B) 20N of peroxidase anhydrous enzymes is :-
(C) 22N (D) 43N (A) 1.568 × 104 (B) 1.568 × 103
(C) 15.68 (D) 2.136 × 104
[D -35]
Chemistry
8. An oxide of metal M has 40% by mass of oxygen. 4. One part of an element A combines with two parts
Metal M has atomic mass of 24. The empirical of another element B. Six parts of the element C
formula of the oxide combines with four parts of the element B. If A
(A) M2O (B) M2O3 and C combine together the ratio of their weights
(C) MO (D) M3O4 will be governed by -
(A) Law of definite proportion
9. A gas is found to contain 2.34 grams of Nitrogen
(B) Law of multiple proportion
and 5.34 grams of oxygen. Simplest formula of the
compound is - (C) Law of reciprocal proportion
(D) Law of conservation of mass
(A) N2O (B) NO
(C) N2O3 (D) NO2 5. Among the following pairs of compounds, the one
that illustrates the law of multiple proportions is
10. 2.2 g of a compound of phosphorous and sulphur
(A) NH3 and NCl3 (B) H2O and D2O
has 1.24 g of 'P' in it. Its empirical formula is -
(C) CuO and Cu2O (D) CS2 and FeSO4
(A) P2S3 (B) P3S2
(C) P3S4 (D) P4S3
11. Which of the following compounds has same (D) STOICHIOMETRY :
empirical formula as that of glucose:- 1. Volume of CO2 obtained at STP by the complete
(A) CH3CHO (B) CH3COOH decomposition of 9.85 g. BaCO3 is –

(C) CH3OH (D) C2H6 BaCO3 

 BaO + CO2
(At. wt of Ba = 137)
(C) LAW OF CHEMICAL COMBINATION : (A) 2.24 lit (B) 1.12 lit
1. Which one of the following pairs of compounds (C) 0.84 lit (D) 0.56 lit
illustrate the law of multiple proportions ? 2. If 0.5 mol of BaCl2 is mixed with 0.1 mole of
(A) H2O, Na2O (B) MgO, Na2O Na 3 PO 4 , the maximum number of mole of
(C) Na2O, BaO (D) SnCl2, SnCl4 Ba3(PO4)2 that can be formed is:-

2. In the reaction N2 + 3H2 

 2NH3, ratio by 3BaCl2 + 2Na3PO4 
 Ba3(PO4)2 + 6NaCl
volume of N 2, H2 and NH3 is 1 : 3 : 2. This (A) 0.7 (B) 0.05
illustrates law of - (C) 0.30 (D) 0.10
(A) Definite proportion 3. 12 lit. of H2 and 11.2 lit. of Cl2 are mixed and
(B) Multiple proportion exploded. The composition by volume of mixture
(C) Reciprocal proportion is–
(D) Gaseous volumes
H2 + Cl2 
 2HCl
3. Different proportions of oxygen in the various
oxides of nitrogen prove the law of - (A) 24 lit. of HCl (g)
(A) Equivalent proportion (B) 0.8 lit. Cl2 and 20.8 lit. HCl (g)

(B) Multiple proportion (C) 0.8 lit. H2 and 22.4 lit. HCl (g)
(D) 22.4 lit. HCl (g)
(C) Constant proportion
4. If 1/2 moles of oxygen combine with aluminium to
(D) Conservation of matter
form Al2O3 then weight of Aluminium metal used
in the reaction is (Al = 27)–
[D-36]
 Mole Concept

3 1 2
2Al + O 2 
 Al2O3 (C) mol (D) mol
4 3
2
11. The impure 6 g of NaCl is dissolved in water and
(A) 27 g (B) 18 g then treated with excess of silver nitrate solution.
(C) 54 g (D) 40.5 g The weight of precipitate of silver chloride is found
5. 1.2 g of Mg (At mass 24) will produce MgO equal to be 14 g. The % purity of NaCl solution would be
:
to -
(A) 95% (B) 85%
(A) 0.05 mol (B) 40 g
(C) 75% (D) 65%
(C) 40 mg (D) 4 g
12. The mass of N2F4 produced by the reaction of 2.0
6. The moles of O2 required for reacting with 6.8 g g of NH3 and 8.0 g of F2 is 3.56 g. What is the per
of ammonia cent yield?
(......NH3 + ......O2  ......NO + ......H2O) is- 2NH3 + 5F2 
 N2F4 + 6HF
(A) 5 (B) 2.5 (A) 79.0 (B) 71.2
(C) 1 (D) 0.5 (C) 84.6 (D) None of these
7. For the reaction : A + 2B  C 13. Calculate the weight of lime (CaO) obtained by
5 mole of A and 8 mole of B will produce - heating 200 kg of 95% pure lime stone (CaCO3) :
(A) 104.4 kg (B) 105.4 kg
(A) 5 mole of C (B) 4 mole of C
(C) 212.8 kg (D) 106.4 KG
(C) 8 mole of C (D) 12 mole of C
8. How much amount of zinc is required to react with (E) STRENGTH OF SOLUTIONS :
dilute H2SO4 for obtaining 224 ml hydrogen at STP- 1. The amount of anhydrous Na 2 CO 3 present in 250
ml of 0.25 M solution is
Zn + dil. H2SO4 
 ZnSO4 + H2
(A) 6.225 g (B) 66.25 g
(A) 0.65 g (B) 6.5 g
(C) 65 g (D) 0.065 g (C) 6.0 g (D) 6.625 g

9. 0.5 mole of H2SO4 is mixed with 0.2 mole of 2. The molarity of 0.006 mole of NaCl in 100 ml
Ca(OH)2. The maximum number of moles of CaSO4 solution is
formed is - (A) 0.6 (B) 0.06
(C) 0.006 (D) 0.066
Ca(OH)2 + H2SO4 
 CaSO4 + 2H2O
3. Molarity is expressed as
(A) 0.2 (B) 0.5
(A) Gram/litre (B) Moles/litre
(C) 0.4 (D) 1.5
(C) Litre/mole (D) Moles/1000 gms
10. How many moles of potassium chlorate to be heated
4. How much of NaOH is required to neutralise 1500
to produce 11.2 litre oxygen at NTP ?
cm 3 of 0.1 M HCl (At. wt. of Na =23)

3 (A) 4 g (B) 6 g

KClO3   KCl + O 2 (C) 40 g (D) 60 g
2
5. When the concentration is expressed as the number
1 1 of moles of a solute per litre of solution it known as
(A) mol (B) mol
2 3
(A) Normality (B) Molarity
(C) Mole fraction (D) Mass percentage

[D -37]
Chemistry
6. The molarity of a solution made by mixing 50ml of 12. Molar solution means 1 mole of solute present in
conc. H2SO4 (72M) with 50 ml of water is (A) 1000g of solvent (B) 1 litre of solvent
(A) 36 M (B) 18 M (C) 1 litre of solution (D) 1000g of solution
(C) 9 M (D) 6 M 13. What volume of 0 .8 M solution contains 0.1 mole
7. With increase of temperature, which of these of the solute
changes [AIEEE 2002] (A) 100 ml (B) 125 ml
(A) Molality (C) 500 ml (D) 62 .5 ml
(B) Weight fraction of solute 14. Mole fraction (X ) of any solution is equal to
(C) Fraction of solute present in water
No. of moles of solute
(D) Mole fraction (A) Volume of solution in litre
8. 2.0 molar solution is obtained , when 0.5 mole solute No. of gram equivalent of solute
is dissolved in (B) Volume of solution in litre
(A) 250 ml solvent (B) 250 g solvent No. of moles of solute
(C) 250 ml solution (D) 1000 ml solvent (C) Mass of solvent in kg

9. The number of moles present in 2 litre of 0.5 M No. of moles of any constituent
(D)
NaOH is Total no. of moles of all constituents
(A) 0.5 (B) 0.1 15. When WB gm solute (molecular mass MB) dissolves
(C) 1 (D) 2 in WA gm solvent. The molality M of the solution is

10. The molality of a solution is W M W 1000

(A) W  1000 (B) M  W
B B B

A B A
(A) Number of moles of solute per 1000 ml of
the solvent W 1000 WA  M B
(C) W  M
A
(D) W  1000
B B B
(B) Number of moles of solute per 1000 gm of
the solvent 16. 1 M HCl and 2 M HCl are mixed in volume ratio of
4 : 1. What is the final molarity of HCl solution?
(C) Number of moles of solute per of the
(A) 1.5 (B) 1
solution
(C) 1.2 (D) 1.8
(D) Number of gram equivalents of solute per
17. Decreasing order (first having highest and then
of the solution
others following it) of mass of pure NaOH in each
11. The number of moles of a solute in its solution is 20 of the aqueous solution :
and total number of moles are 80. The mole fraction (i) 50 g of 40% (w/W) NaOH
of solute is (ii) 50 mL of 50% (w/V) NaOH [dsoln. = 1.2 g/mL]
(A) 2.5 (B) 0.25 (iii) 50 g of 15 M NaOH [dsoln. = 1 g/mL]
(C) 1 (D) 0.75 (A) i, ii, iii (B) iii, ii, i
(C) ii, iii, i (D) ii, i, iii

[D-38]
 Mole Concept

1. One mole of CO2 contains 11. The total volume of mixture of 2 gms of helium and
(A) 3 gram atoms of CO2 7 gms of nitrogen under S.T.P. conditions is
(B) 18 × 1023 CO2 molecules (A) 22.4 lit (B) 11.2 lit
(C) 6 × 1023 O atoms (D) 6 × 1023 C atoms (C) 16.8 lit (D) 5.6 lit
2. The number of atoms present in 4.25 grams of NH3 12. 4.6  10 atoms of an element weight 13.8 gms.
22

is approximately The atomic mass of the elements is

(A) 1 × 1023 (B) 1.5 × 1023 (A) 290 (B) 180
(C) 2 × 1023 (D) 6 × 1023 (C) 34.4 (D) 104
3. The atomicity of a species is x and its atomic weight
13. How many moles of magnesium phosphate,
is y. The molecular weight of the species is
Mg3(PO4)2 will contain 0.25 mole of oxygen atoms?
(A) x + y (B) y + x
(C) xy (D) x - y (A) 0.02 (B) 3.125 × 10–2
4. The number of molecules present in 4.4 g of CO2 (C) 1.25 × 10–2 (D) 2.5 × 10–2
gas is 14. Given that the abundance of isotopes 54Fe, 56Fe and
57
(A) 6.023 × 1023 (B) 5.023 × 1023 Fe are 5%, 90% and 5%,m respectively, the atomic
(C) 6.023 × 1024 (D) 6.023 × 1022 mass of Fe is :
5. The number of moles present in 24.5gms of H2SO4 (A) 55.85 (B) 55.95
is (C) 55.75 (D) 56.05
(A) 2.5 (B) 0.5
15. An element, X has the following isotopic composition
(C) 4 (D) 0.25
6. The number of carbon atoms present in 2.8 gms of 200 X: 90%, 199X : 8.0% 202X : 2.0%
carbon monoxide are The weighted average atomic mass of the naturally
(A) 3.01 × 1023 (B) 3.01 × 022 occuring element X is closest to
23
(C) 6.02 × 10 (D) 6.02 x 1022 (A) 201 u (B) 202 u
7. The density of water is 1 gm/ml. The number of (C) 199 u (D) 200 u
molecules present in 1 litre of water are
16. Volume occupied by one molecule of water
(A) 18 (B) 18 × 1000 (density = 1 g cm–3) is
(C) 6.023 × 1023
(A) 9.0 × 10–23 cm3 (B) 6.023 × 10–23 cm3
(D) 55.55 × 6.023 × 1023
(C) 3.0 × 10–23 cm3 (D) 5.5 × 10–23 cm3
8. 0.5 moles of a gas (Mol.wt.20) occupies 11.2 litres
at STP. The volume occupied by 0.25 mole of a 17. The number of atoms in 0.1 mole of a triatomic gas
lighter gas (Mol. Wt.=(B)) at STP will be is (NA = 6.02 × 1023 mol–1)
(A) 11.2 lit (B) 5.6 lit (A) 6.026 × 1022 (B) 1.806 × 1023
(C) 2.8 lit (D) 22.4 ltrs (C) 3.600 × 1023 (D) 1.800 × 1022
9. Weight of 1 atom of Hydrogen is 18. The percentage of Carbon in CO2 is
(A) 1.66 × 10-24 gm (B) 1023 gm (A) 27.27% (B) 29.27%
(C) 1022 gm (D) 1024 gm (C) 30.27% (D) 26.97%
10. Which one of the following pairs of gases contain 19. Haemoglobin contain 0.33% of Fe by weight. If 1
the same number of molecules at S.T.P. molecule of Haemoglobin contains two Fe atoms,
(A) 11 grams of CO2 and 14 grams of CO the molecular weight of Haemoglobin will be (at.
(B) 14 grams of C2H4 and 16 grams of methane wt. of Fe = 56) nearly
(C) 16 grams of oxygen and 17 grams of Hydrogen (A) 67000 (B) 34000
sulphide
(C) 17000 (D) 20000
(D) 4 grams of helium and 4 grams of hydrogen

[D -39]
Chemistry
20. The empirical formula of a gaseous compound is 28. Among the following pairs of compounds, The one
CH2. The density of the compound is 1.25 gm/lit. that illustrates the law of multiple proportions is :
at S.T.P. The molecular formula of the compound (A) NH3 and NCl3 (B) H2S and SO2
is X
(C) CuO and Cu2O (D) CS2 and FeSO4
(A) C2H4 (B) C3H6
29. One part of an element (A) combines with two parts
(C) C6H12 (D) C4H8 of another element (B). 6 parts of element (C)
21. An oxide of nitrogen contains 36.8% by weight combines with 4 parts of (B). If A and (C) combine
of nitrogen. The formula of the compound is together the ratio of their weights, will be governed
by :
(A) N2O (B) N2O3
(A) law of definite proportion
(C) NO (D) NO2
22. A compound contains 28% of Nitrogen and 72% of (B) law of multiple proportion
metal by weight. Three atoms of metal combines (C) law of reciprocal proportion
with Two atoms of Nitrogen. The atomic weight of (D) law of conservation of mass
the metal is
30. Which of the following is the best example of law
(A) 12 (B) 24 of conservation of mass.
(C) 36 (D) 48 (A) 12 gm of carbon combines with 32 gm of
23. 3.0 gms of an organic compound on combustion gives oxygen to form 44 gm of CO2.
8.8 gm of CO2 and 5.4gm of water. The empirical (B) When 72 gm of carbon is heated in vaccum
formula of the compound is and no change in mass takes place.
(A) CH3 (B) C2H4 (C) The weight of a piece of platinum is the same
(C) C2H2 (D) C2H6 before and after heating in air.
24. One gram of a certain compound has 0.2729 gram (D) 2 gm of Hydrogen combines with 32 gm of Oxygen
of carbon and 0.7271 grams of oxygen. The formula
31. Irrespective of the source, pure sample of water
of the compound is
always yields 88.89% mass of oxygen and 11.11%
(A) CO (B) CO2 mass of hydrogen. This is explained by the law of
(C) C3O2 (D) C4O2 (A) Conservation of mass
25. On analysis, a certain compound was found to
(B) Constant proportions
contain 254 grams of Iodine and 80 grams of oxygen.
The formula of the compound is (C) Multiple proportions (D) Constant volume.
(A) IO (B) I2O 32. Avogadro’s number is the number of molecules
(C) I5O3 (D) I2O5 present in
26. In a hydrocarbon, mass ratio of hydrogen and carbon (A) 1 g of molecule (B) 1 atom of molecule
is 1 : 3, the empirical formula of hydrocarbon is (C) gram molecular mass (D) 1 litre of molecule
(A) CH4 (B) CH2 33. 20 gms of sulphur on burning in air produced 11.2 its
(C) C2H (D) CH3 of SO2 at STP. The percentage of unreacted sulphur
27. The % composition of four hydro carbons is as (A) 80% (B) 20%
follows: (C) 60% (D) 40%
(i) (ii) (iii) (iv) 34. 23g of sodium will react with ethyl alcohol to give
%C 75 80 85.7 91.3 (A) 1 mole of H2 (B) 1/2 mole of H2
%H 25 20 14.3 8.7 (C) 1 mole of O2 (D) 1 mole of NaOH
The data illustrates the law of 35. How much volume of CO2 at S.T.P is liberated by
(A) Constant proportion the combustion of 100 cm 3 of propane
(B) Conservation of mass (C3H8) ?
(C) Multiple Proportions (A) 100 cm3 (B) 200 cm3
(D) Reciprocal Proportions. (C) 300 cm3 (D) 400 cm3
[D-40]
 Mole Concept
36. The composition of residual mixture will be if 30 g (A) 1 (B) 1.5
of Mg combines with 30 g of oxygen : (C) 2 (D) 2.5
(A) 40g MgO + 20g O2 44. What is the volume (in litres) of oxygen required at
(B) 45g MgO + 15g O2 S.T.P. to completely convert 1.5 moles of sulphur
(C) 50g MgO + 10g O2 into sulphur dioxide?
(D) 60g MgO only S + O2   SO2
37. In a reaction if 7 grams of CO is completely oxidised (A) 11.2 (B) 22.4
to CO2, the weight of CO2 formed in grams is (C) 33.6 (D) 44.8
45. What is the volume (in litres) of oxygen at STP
1
CO + O2 
 CO2 required for complete combustion of 32 gms of CH4?
2 (Molecular wt. Of CH4 is 16)
(A) 14 (B) 11 CH4 + O2   CO2 + H2O (unbalanced)
(C) 44 (D) 30 (A) 44.8 (B) 89.6
38. To get 5.6 lit of CO2 at STP, weight of CaCO3 to be (C) 22.4 (D) 179.2
decomposed is 46. In the reaction

CaCO3   CaO + CO2 2AI (s)  6HCl(aq)  2Al3+ (aq )  6Cl – (aq)  3H 2 (g)
(A) 100 gm (B) 50 gm
(A) 6L HCl (aq) is consumed for every 3 L H2
(C) 25 gm (D) 75 gm produced
39. The volume of H2 at STP required to completely
reduce 160 gms of Fe2O3 is (B) 33.6 L H 2(g) is produced regardless

Fe2O3 + 3H2 
 2Fe + 3H2O temperature and pressure for every moles that
reacts
(A) 3 × 22.4 litres (B) 2 × 22.4 litres
(C) 22.4 litres (D) 11.2 litres (C) 67.2 L H 2(g) at STP is produced for every mole
40. The weight of SO2 formed when 20 gms of sulphur of Al that reacts
is burnt in excess of O2 is
(D) 11.2 L H 2(g) at STP is produced for every mole
S + O2 
 SO2
(A) 32 gm (B) 64 gm of HCl(aq) consumed
(C) 40 gm (D) 60 gm 47. What volume of oxygen gas (O2) measured at 0oC
41. If 12.0 lit of H2 and 8.0 lit of O2 are allowed to and 1 atm, is needed to burn completely 1 L of propane
react, the O2 left unreacted will be gas (C3H8) measured under the same conditions?
C3H8 + O2   CO2 + H2O (unbalanced)
1
H2 + O 
 H2O (A) 7 L (B) 6 L
2 2
(C) 5 L (D) 10 L
(A) 4.0 lit (B) 6.0 lit 48. How many moles of lead (II) chloride will be formed
(C) 1.0 lit (D) 2.0 lit from a reaction between 6.5 g of PbO and 3.2 g of
42. The volume in lit of CO2 liberated at S.T.P when HCl ?
10g of 90% pure lime is heated completely is PbO + 2HCl   PbCl2 + H2O
CaCO3 
 CaO + CO2 (A) 0.044 (B) 0.333
(A) 2.016 (B) 20.16 (C) 0.011 (D) 0.029
(C) 2.24 (D) 22.4 49. 10 g of hydrogen and 64 g of oxygen were filled in a
steel vessel and exploded. Amount of water
43. The no. of moles of CO2 produced when 3 moles of
produced in this reaction will be
HCl reacts with excess of CaCO3 is
(A) 2 moles (B) 3 moles
CaCO3 + HCl 
 CaCl2 + H2O + CO2
(C) 4 moles (D) 1 mole
[D -41]
Chemistry
50. The reaction of calcium with water is represented by (A) 0.202 (B) 0.158
the equation (C) 0.176 (D) 0.221
Ca + 2H2O   Ca(OH)2 + H2 56. The density of a 56.0% by weight aqueous solution
What volume of H2 at STP would be liberated when of 1-propanol (CH3CH2CH2OH) is 0.8975 g/cm3.
8 gm of calcium completely reacts with water? What is the mole fraction of the compound?
(A) 0.2 cm3 (B) 0.4 cm3 (A) 0.292 (B) 0.227
(C) 2240 cm3 (D) 4480 cm3 (C) 0.241 (D) 0.276
51. An aqueous solution of glucose is 10% (w/v). The 57. Calculate the mass of anhydrous HCl in 10 mL of
volume in which 1 mole of glucose is dissolved, will concentrated HCl (density = 1.2 g/ mL) solution
be having 37% HCl by weight.
(A) 18L (B) 9L (A) 4.44 g (B) 4.44 mg
–3
(C) 0.9L (D) 1.8L (C) 4.44 × 10 mg (D) 0.444 µg
52. In 1200 g solution, 12 g urea is present. If density of 58. The lead nitrate, Pb(NO3)2, in 25 mL of 0.15 M
the solution is 1.2 g/ml, then the molarity of the solution reacts with all of the aluminium sulphate,
solution is Al2(SO4)3, in 20 mL of solution. What is the molar
concentration of the Al2(SO4)3?
(A) 0.2 M (B) 10 M
(C) 0.167 M (D) 12 M 3Pb(NO3)2(aq) + Al2(SO4)3(aq)   3PbSO4(s)
53. –1
The density (in g mL ) of a 3.60 M sulphuric acid + 2Al(NO3)3(aq)
solution that is 29% (H2SO4 molar mass = 98 g (A) 6.25 × 10–2 M (B) 2.421 × 10–2 M
mol–1) by mass will be: (C) 0.1875 M (D) None of these
(A) 1.22 (B) 1.45 59. Concentrated HNO3 is 63% HNO3 by mass and has
(C) 1.64 (D) 1.88 a density of 1.4 g/mL. How many millimetres of this
solution are required to prepare 250 mL of a 1.20 M
54. A 5.2 molal aqueous solution of methyl alcohol,
HNO3 solution?
CH3OH is supplied. What is the mole fraction of
methyl alcohol in the solution? (A) 18.0 (B) 21.42
(C) 20.0 (D) 14.21
(A) 0.100 (B) 0.190
60. How many millilitres of 0.1 M H2SO4 must be added
(C) 0.086 (D) 0.050
to 50 mL of 0.1 M NaOH to give a solution that has
55. Wood’s metal contains 50.0% bismuth, 25.0% lead, a concentration of 0.05 M in H2SO4 ?
12.5% tin and 12.5% cadmium by weight. What is the
(A) 400 mL (B) 200 mL
mole fraction of tin?
(C) 100 mL (D) None of these
(Atomic weights : Bi = 209, Pb = 207, Sn = 119, Cd
= 112)

[D-42]
 Mole Concept

EXERCISE-0 mass after reaction= mass of 1 mole water = 18gm

(A) BASICS OF MOLE CONCEPT:
2. (a) the law of multiple proportions.
1. 16, 16 amu
2. NA
3. H S O (D) STOICHIOMETRY :
6 mole 3 mole 12 mole 1. 67.2 L
4. 6.023 x 1023 molecules 2. (B)
5. 15.875 g 3. 59.3 %
6. (i) 1.506 ×1023 P4 molecules 4. 11.6g
(ii) 6.023 × 1023 atoms 5. 85.71 × 103 g = 85.71 kg O2
(iii) 1 mol (iv) 31.0 g 6. 52.3%
7. 36 g 7. 3.36 litre
8. 0.25 mol 8. Volume of unreduced CO2 = 0.3 L
9. (i) 1.793 × 10–22 g (ii) 14.0 g (iii) 40.0 g
(iv) 32.0 g (v) 1.993 g (E) STRENGTH OF SOLUTIONS :
(vi) One atom of silver < one gram of iron is < 1. 0.2 M
10 23 atoms of carbon < one gram - atom of 2. 0.332
nitrogen < one mole of oxygen < one mole of 3. (C)
calcium 4. (B)
10. 3.01 × 1012 molecules 5. (A)
11. 1.0278 × 1024 H atoms 6. 0.313 m
12. 1.3384 × 1021 molecules 7. 0.029 mol L–1
13. 1.3 × 1021 atoms 8. 2.315 m
14. (i) 18.02 u (ii) 44.01 u (iii) 16.043 u 9. Molarity = 55.56 M
15. 35.4527 u 10. 0.028 M
16. 2 × 1.5 × 1023 atoms. 11. 0.005 M
12. 0.0684 M
(B) PERCENTAGE WEIGHT OF ELEMENTS 13. Mole fraction of ethanol = 0.196
IN COMPOUNDS, MASS % OF Mole fraction of acetic acid = 0.301
COMPOUNDS IN SAMPLE, 14. 0.2 mol L–1
MOLECULAR FORMULA & 15. (i) 3.30 M
EMPIRICAL FORMULA : (ii) 3.77 m
1. Mass percent of sodium : 32.4% 16. (a) Weight per cent = 5.14
Mass percent of sulphur : 22.6% (b) Mole fraction = 0.0454
Mass percent of oxygen: 45.05% Mole of solvent = 0.0098
2.  2:3 (c) Mole per cent = 0.98
3. C 6H 6 (d) Molality = 0.552
4. KAlS2O8.12H2O. (e) Molarity = 0.539
5. (a) 50.84% Zn (b) 16.04% P (c) 33.12% O (f) Normality = 1.078 N
17. (a) (37.92 %(w/w)
(C) LAWS OF CHEMICAL COMBINATION : (b) 0.065
(c) 3.865 m
1 18. (a) 12.38 M
1. mass before reaction = mass of 1 mole H2 (g) +
2 (b) 8.1 cm3
mole O2 (g) = 2 + 16 = 18 gm

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Chemistry

EXERCISE- 1 7. (B) 8. (A) 9. (A)

(A) BASICS OF MOLE CONCEPT: 10. (B) 11. (A) 12. (D)
1. (D) 2. (B) 3. (A)
13. (D)
4. (B) 5. (B) 6. (C)
7. (B) 8. (C) 9. (A)
10. (B) 11. (B) 12. (A) (E) STRENGTH OF SOLUTIONS :
1. (D) 2. (B) 3. (B)
13. (D) 14. (C) 15. (C)
4. (B) 5. (B) 6. (C)
16. (D) 17. (A) 18. (A) 7. (C) 8. (C) 9. (C)
19. (D) 20. (D) 21. (B) 10. (B) 11. (B) 12. (C)
22. (A) 23. (A) 24. (D) 13. (B) 14. (D) 15. (B)
25. (C) 26. (A) 16. (C) 17. (B)

(B) PERCENTAGE WEIGHT OF ELEMENTS EXERCISE- 2

IN COMPOUNDS, MASS % OF 1. (D) 2. (D) 3. (C)
COMPOUNDS IN SAMPLE, 4. (D) 5. (D) 6. (D)
MOLECULAR FORMULAC & 7. (D) 8. (B) 9. (A)
EMPIRICAL FORMULA : 10. (C) 11. (C) 12. (B)
13. (B) 14. (B) 15. (D)
1. (A) 2. (C) 3. (A)
16. (C) 17. (B) 18. (A)
4. (A) 5. (C) 6. (D) 19. (B) 20. (A) 21. (B)
7. (A) 8. (C) 9. (D) 22. (B) 23. (A) 24. (B)
25. (D) 26. (A) 27. (C)
10. (D) 11. (B) 28. (C) 29. (C) 30. (A)
31. (B) 32. (C) 33. (B)
34. (B) 35. (C) 36. (C)
(C) LAW OF CHEMICAL COMBINATION : 37. (B) 38. (C) 39. (A)
1. (D) 2. (D) 3. (B) 40. (C) 41. (D) 42. (A)
4. (C) 5. (C) 43. (B) 44. (C) 45. (B)
46. (D) 47. (C) 48. (D)
49. (C) 50. (D) 51. (D)
(D) STOICHIOMETRY : 52. (A) 53. (A) 54. (C)
55. (C) 56. (D) 57. (A)
1. (B) 2. (B) 3. (C) 58. (A) 59. (B) 60. (C)
4. (B) 5. (A) 6. (D)

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