Basic Concepts of Chemistry.pdf

BASIC CONCEPTS OF CHEMISTRY
No. of Sessions –10
1
SESSION – 0
AIM
 To introduce units and dimensions
THEORY
PHYSICAL QUANTITIES AND THEIR S.I. UNITS:
Chemists describe the behaviour of chemical substances on the basis of physical and chemical
properties. The measurements of chemical properties involve chemical reactions, whereas the
measurement of physical properties does not involve any chemical reactions. The common physical
properties are mass, length, time, volumes, temperature, density, etc., among these mass, lengths and
time are fundamental physical quantities.
(1) Mass tells us about the quantity of matter. Mass is measured with the help of analytical balance.
(2) The size of the object is measured in terms of length, area and volume. Length refers to one dimension,
area to two dimensions and volume to three dimensions of space.
(3) Time helps us to know how long it takes for a process to occur.
The units of physical quantities depend on three basic units i.e., units of mass, length and time. Since
threes are independent units and cannot be derived from any other units, they are called fundamental
units. But these three fundamental units of cannot describe all the physical quantities. Thus seven basic
units of measurement namely mass length, time, temperature, electric current, luminous intensity and
amount of substance are taken as basic units. All the other units can be derived from them are called
derived units. The units of area, volume, force, work, density, velocity, energy, etc., are all derived
units.
S.I. UNITS:
The S.I system has seven basic units from which all other units are derived.
Physical Quantity Name of unit Symbol
Length Meter m
Mass Kilogram kg
Time Second s
Temperature Kelvin K
Electric current ampere A
Luminous intensity candela cd
Amount of substance mole mol

2
PREFIXES:
The SI units of some of the physical quantities are either too small or too large. To change the order of
magnitude, these are expressed by using prefixes before the names of the basic units.
Multiple Prefix Symbol Multiple Prefix Symbol
10-1
Deci d 101
deca da
10-2
Centi c 102
hecta h
10-3
Milli m 103
kilo k
10-6
Micro μ 106
mega M
10-9
Nano n 109
giga G
10-12
Pico p 1012
tera T
10-15
Femto f 1015
peta P
10-18
Atto a 1018
exa E
10-21
Zepto z 1021
Zetla Z
10-24
Yocto y 1024
yotla Y
DERIVED UNITS:
The units of different physical quantities can be derived from the seven basic units. These are called
derived units because these are derived from the basic units. For deriving these units, we can multiply
or divide the symbols for units as if they are algebraic quantities
Example:
1) Volume = Lengthx Breadth xHeight
If units of length are m, then volume = m xm x m = m3
2) Area = Length x Breadth = mx m = m2
3) Density = =
4) Acceleration = =
.
= m.s-2
5) Force = Mass x acceleration = Kg. m.s-2
6) Pressure = =
. .
= Kg.m-1. s-2
7) Energy, work = Force x distance = Kg. m.s-2x m = Kg.m2.s-2 = Joule
8) Electric charge = current x time = A. S = Coulomb
9) Electric potential = =
. .
.
= Joule. A-1.s-1 = Volt
UNITS AND DIMENSIONAL ANALYSIS
Convertion of Units: It is frequently necessary to convert one set of units to another in calculations. This
can be done by a method called conversion factor method or also called dimensional analysis. In order
to use this method, we write the units with every number and carry the units through the calculations,
treating them as algebraic quantities. For interconversion of the units of time, we know that
1= or 1=
These equalities are called unit conversion factor or convertion factor or simply unit factor.
Basic Concepts of Chemistry : Chemistry (Module 1)

Arihant Academy: Chemistry (Module 1)
3
Examples
1) To find the number the seconds in 5 min.
5 min = 5 min = 300 sec
2) To convert 0.74 A° into picometre.
1A° = 10-10 m or 1=
10−10
1
0.74 A° = 0.74A° x = 0.74 x 10-10 m
1 pm = 10-12 m or 1 =
∴ 0.74 × 10 = 0.74 × 10 = 0.74 × 10 = 74 ∴
Convertion of litre - atmosphere to joule
1L = 10-3 m3 or 1 =
1L atm = 1 L atm x = 10-3m3 atm
1 atm = 101, 325 Pa or 1 =
,
10-3 m3 atm = 10-3 m3 atm x
,
= 101, 325 Pa 10-3 m3 = 101.325 Pa.m3
But, Pa = 2
101.325 Pa.m3 = 101.325 .m3
2= 101.325 N.m = 101.325 J ( 1 J = N.m)

4
SESSION – 1
AIM
 To introduce Measurement and Significant figures
 To understand chemical classification of matter
THEORY
Precision and Accuracy: To express the results of different scientific measurements two terms accuracy
and precision are commonly used.
“Accuracy is a measure of the difference between the experimental value and the true value.”
Smaller the difference between the experimental value and the true value, larger is the accuracy.
Accuracy expresses the correctness of measurement.
“Precision is expressed as the difference between a measured value and the arithmatic mean value for a
series of measurements.”
Precision refers for the closeness of the set of values obtained from identical measurement of a quantity.
Example: Three students were asked to determine the mass of piece of metal where mass is known to
be 0.520 g. Data obtained by each student are recorded as follows.
Measurements (g) Average (g)
(1) (2) (3)
Student A 0.521 0.515 0.509 0.515
Student B 0.516 0.515 0.514 0.515
Student C 0.521 0.520 0.520 0.520
The data for student A are neither very precise nor accurate. The individual values differ widely from one
another and the average value is not accurate. Student B was able to determine the mass more precisely.
The values deviate, but little from one another. However, the average mass is still not accurate. The data
for student C is both precise and accurate.
SCIENTIFIC NOTATION (OR) EXPONENTIAL NOTATION
In scientific notation, the large or small numbers are expressed as a number between 1.000 and 9.999
multiplied or divided by 10, an appropriate no.of times.
Example 1) 138.42 = 1.384 × 102
2) 0.00013842 = 1.3842 × 10-4
In scientific notation, a no. is generally expressed in the form of N × 10n. Where, N is a no between1.000
and 9.999 and n is exponent.
(a) To transform a number larger than 9.999... to scientific rotation, the decimal point there is only one
nonzero digit before the decimal point. If the decimal point is moved x places to the left, then exponent
n = x

5
Ex: 138.42 = 1.3842 × 102
21.654 = 2.1654 × 101 1395.2 = 1.3952 × 103
(b) To transform a number smaller than 1 to scientific notation, the decimal point is moved to the right until
there is one nonzero digit before the decimal point. If the decimal point is moved ‘y’ places to the right,
the exponent, n = – y
Ex: 0.00013482 = 1.3482 × 10-4
0.00549 = 5.49 × 10-3
0.1641 = 1.641 × 10-1
SIGNIFICANT FIGURES
Every measurement gives an inexact number because, every such measurement is uncertain to some
extent. This is mainly due to the skill of the observer and limitation of the instrument. To express the
results in an accurate way, we express generally those digits which are known with certainty. This is
done is terms of significant figures.
The significant figures in a number are all the certain digits plus one doubtful digit. The digits in a
properly recorded measurement are knows as significant figures. The greater the number of significant
figures in a reported result, smaller is the uncertainty and greater the precision.
Rules for determining number of significant figures:
1) To determine the number of significant figures in a measurement, read the number from left to right and
count all of the digits, starting with the first digit that is not zero.
Ex: The number of significant figures in 1.887 = 4
The number of signficant figures in 12.612 = 5
2) When a number is greater than 1, all the zeros to the right of the decimal point are significant
The number of significant figures in 42.000 = 5
3) For a number less than l, only zeros to the right of the first significant digit at significant. But the zeros
to the left of the first significant digit are not significant
4) A zero becomes significant if it comes in between two non - zero digits

6
5) When adding or subtracting, the number of decimal places in the answer should be equal to the number
of decimal places in the number with the least number of decimal places.
Ex: 3.21 (3 singificant figures 2 decimal places)
1.5 (2 singificant figures 1 decimal places)
21.402 (5 significant figures 3 decimal places)
Since the term 1.5 involved in addition, has only one decimal place, the overall answer of 26.112 should
be reported as 26.1.
6) In multiplication and division, the number of significant figures in the answer should be same as that in
the number with least number of significant figures.
Ex: 1.25 × 3.3761 = 4.220125 Since the term 1.25 has only 3 significant figures, the answer should be
4.22
7) When a number is rounded off the number of significant figures is reduced. The last digit retained is
increased by 1, only if the following digit is 5, and is left as such if the following digit is 4.
Ex: 12.696, 18.35 and 13.93 are reported as 12.7, 18.4, 13.9 respectively when rounded off to three
significant figures
CHEMICAL CLASSIFICATION OF MATTER:
Anything that has mass and occupies space is called matter. Matter can be classified in two ways:
(A) Physical classification of matter
(B) Chemical classification of matter
(A) Physical classification of matter: Depending upon the physical state of matter, it can be classified into
solid, liquid and gaseous state.
(B) Chemical classification of matter: On the basis of chemical composition, matter is being classified into
three types: elements, compounds and mixtures
Element: A pure substance which can neither be decomposed into simpler substance nor built from
simpler substances by ordinary physical or chemical methods is called element.
Ex: hydrogen, oxygen, nitrogen, sulphur etc.
Compound: A substance which is obtained by the combination of two or more elements in a definite
proportion by weight and it can be decomposed into respective elements by suitable chemical method is
called a compound. The properties of the compound are totally different from the elements from which
these are formed: A compound can be organic or inorganic.
Mixture: A combination of two or more elements or compounds in any proportion so that the
components do not lose their identity is called mixture. Mixtures are of two types.

7
(a) Homogeneous mixtures: These have same composition throughout the sample.
Ex: air, alloys, gasoline, etc
(b) Heterogeneous mixtures: These consist of two or more phases which have different compositions.
Ex: a mixture of sand and salt.
CLASS EXERCISE
1] Write the significant figures for the following:
a) 1.02 b) 30.9 c) 231.9 d) 0.540
e) 0.096 f) 0.070 g) 1.0 h) 6.023
i) 6.023x1023
j) 1.0 x 103
2] The answer to the following problem in standard exponential form is:
(2.0 x 1013
) + (1.5 x 1014
)
a) 3.5 x 1013
b) 3.5 x 1014
c) 2.15 x 1013
d)1.7x 1014
3] 24.8 ÷ 12.4 =? The correct answer to this problem in proper number of significant digit is
a) 2 b) 2.0 c) 2.00 d) 2.000
4] 152. 06 x 0.24 = 36.499. The correct answer to this problem in proper number of significant digits is
a) 36.4944 b) 36.494 c) 37 d) 36
5] 1280  2.0 =? The correct answer to this problem in proper number of significant digits is
a) 64 b) 6.40 x 102
c) 640.0 d) 640
6] The correctly reported answer of the addition of 29.4406, 3.2 and 2.25 will have significant figures
a) Three b) Four c) Two d) Five
Matter
Physical Chemical
Solids Liquids Gases Mixtures
Elements Compounds Hetergeneous
Homogeneous
inorganic Organic
Pure substance

8
HOME EXCERSICE
1] Which one of the following is not an element?
a) Diamond b) Graphite c) Silica d) Ozone
2] Which one of the following statements is incorrect?
a) All elements are homogeneous
b) Compounds made up of a number of elements are heterogeneous
c) A mixture is not always heterogeneous
d) Air is heterogeneous mixture
3] The correctly reported answers of the addition of 294.406, 280.208 and 24, will be
a) 598.61 b) 599 c) 598.6 d) 598.614
4] A mixture that can be separated by sublimation is
a) AgCl + NaCl b) BaCl2 + NaCl c) HgCl2 + NaCl d) MgCl2 + NaCl
5] Two students X and Y report the mass of the same substance as 7.0 g and 7.00 g respectively. Which of
the following statement is correct?
a) Both are equally accurate b) X is more accurate than Y
c) Y is more accurate than X d) both are inaccurate scientifically
NCERT Text Book Questions: 1.16, 1.18, 1.19, 1.20, 1.31.

9
SESSION – 2
AIM
 To introduce laws of Chemical Combination
 To understand Dalton’s atomic theory
THEORY
LAWS OF CHEMICAL COMBINATION
1) Law of conservation of mass: It states that “matter is neither created nor destroyed during any
physical or chemical change”
This law is also called the Law of indestructibility of matter. The following experiments illustrate the
truth of this law.
(a) When matter undergoes a physical change: A piece of ice (solid water) is taken in a small conical
flask. It is well corked and weighed. The flask is now heated gently to melt the ice (solid) into water
(liquid).
⎯
The flask is again weighed. It is found that there is no change in the Weight though a physical change
has taken place.
(b) When matter undergoes a chemical change: The following chemical changes illustrate the law.
Ex: Decomposition of Mercuric oxide: 100 g of mercuric oxide when heated in a closed tube,
decomposed to produce 92.6 g of mercury and 7.4 g of oxygen gas,
i.e. total mass of products = 100 g:
( ) → ( ) + ( )
100g 92.6g 7.4 g
Thus, during the above decomposition reaction, matter is neither gained nor lost.
2) Law of definite proportion (Law of constant proportion): If states that “Any pure compound however
made contains the same elements in the fixed ratio of their weights”.
Ex: Pure water contains 2 gm of hydrogen and 16 gm of oxygen i.e., the ratio of hydrogen and oxygen
in pure water is 1: 8.
Ex: A sample of CO2 may be prepared in the laboratory by (a) heating lime stone (CaCO3), (b) by
burning coal in air, (c) by the action of dilute hydrochloric acid on marble, (d) by heating sodium
bicarbonate. In each case, it is found that CO2 is made up of the same elements. i.e., carbon and
oxygen, combined together in the same fixed ratio of 12: 32 or 3: 8 by mass.
a) CaCO ⎯ CaO + CO 
b) C + O  CO 
c) CaCO + 2HCl  CaCl + CO  + H O
d) 2NaHCO  Na CO + CO  + H O

10
Limitations of Law of Constant Composition:
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 isotopes combines, the ratio of
C:O is 14 : 32.
Sample Problem 1. 6.488g of lead combine directly with 1.002 g of oxygen to form lead peroxide (PbO2).
Lead peroxide is also produced by heating lead nitrate and it was found that the percentage of oxygen
present in lead peroxide is 13.38 percent. Use these data to illustrate the law of constant composition.
Solution: Step1. To calculate the percentage of oxygen in first experiment.
Mass of peroxide formed = 6.488 + 1.002 = 7.490g.
7.490 g of lead peroxide contain 1.002 g of oxygen
∴ 100g of lead peroxide will contain oxygen, =
1.002
7.490
× 100 = 13.38 ,
i.e. oxygen present = 13.38%
Step2. To compare the percentage of oxygen in both the experiments.
Percentage of oxygen in PbO2 in the first experiment = 13.38
Percentage of oxygen in PbO2 in the second experiment = 13.38
Since the percentage composition of oxygen in both the samples of PbO2 is identical, the above data
illustrate the law of constant composition.
3) Law of multiple proportions: When two elements combine to form two or more than two compounds,
the weight of one of the element which combines with a fixed weights of the other, will be in a simple
numerical ratio.
Ex: The weight of Oxygen that will combine with 12gm of carbon in CO and CO2 is in the ratio of 1: 2
Ex: The weight of Oxygen that will combine with 32 gm of sulphur in SO2 and SO3 is the ratio of 2: 3
Sample Problem 2. Carbon is found to form two oxides, which contain 42.9% and 27.3% carbon
respectively. Show that these figures illustrate the law of multiple proportions.
Solution: Step1. To calculate the percentage composition of carbon and oxygen in each of the two
oxides
First oxide Second oxide
Carbon 42.9% 27.3%
Oxygen 57.1 % 72.7%
Step2. To calculate the masses of carbon which combine with a fixed mass, i.e., one part by mass of
oxygen in each of the two oxides:
In the first oxide, 57.1 parts by mass of oxygen combine with carbon = 42.9 parts.
 1 part by mass of oxygen will combine with carbon=
.
.
= 0.751
In the second oxide. 72.7 parts by mass of oxygen combine with carbon = 27.3 parts.
 1 part by mass of oxygen will combine with carbon=
.
.
= 0.376

11
Step3. To compare the masses of carbon which combine with the same mass of oxygen in both the
oxides:
The ratio of the masses of carbon that combine with the same mass of oxygen (1 part) is
0.751: 0.376 or 2: 1
Since this is a simple whole number ratio, of the above data illustrate the law of multiple proportions.
4) Law of Reciprocal Proportions:
The ratio of the masses of two elements A and B which combine separately with a fixed mass of the
third element C is either the same or some simple multiple of the ratio of the masses in which A and B
combine directly with each other.
This law may be illustrated with the help of the following examples:
(1) The elements C and O combine separately with the third element H to form CH4 and H2O and they
combine directly with each other to form CO2, as shown in Fig 1.7
In CH4, 12 parts by mass of carbon combine with 4 parts by mass of hydrogen. In H2O, 2 parts by mass
of hydrogen combine with 16 parts by masses of oxygen. Thus, the masses of C and O which combine
with fixed mass of hydrogen (say 4 parts by mass) are 12 and 32, i.e., they are in the ratio 12 : 32 or 3 :
8.
Now, in CO2, 12 parts by mass of carbon combine directly with 32 parts by mass of oxygen, i.e. they
combine directly in the ratio 12: 32 or 3: 8 which is the same as the first ratio.
(2) The elements H and O combine separately with the third element S to form H2S and SO2 and they
combine directly with each other to from H2O as shown in Fig. 1.8. As shown in the Fig, the masses of
H and O which combine with the fixed mass of S, viz, 32 parts are 2 and 32, i.e. they are in the ratio 2:
32 or 1:16. When H and O combine directly to form H2O, the ratio of their combining masses is 2:16 or
1:
The two ratios are related to each other as : i.e., they are simple multiple of each other.

12
Sample Problem 3. Ammonia contains 82.35% of nitrogen and 17.65% of hydrogen. Water contains
88.90% of oxygen and 11.10% of hydrogen. Nitrogen trioxide contains 63.15% of oxygen and 36.85%
of nitrogen. Show that these data illustrate the law of reciprocal proportions.
Solution: In NH3. 17.65 g of H combine with N = 82.35 g
∴1 g of H combines with N=
.
.
= 4.67
In H2O, 11.10 g of H combine with O = 88.90 g 1 g of H combine with O =
.
.
= 8.01
∴ Ratio of the masses of N and O which combine with fixed mass (=1g) of H =4.67:8.01=1: 1.72
In N2O3, ratio of masses of N and O which combine with each other = 36.85: 63.15 = 1: 1.71
Thus, the two ratios are the same. Hence, it illustrates the law of reciprocal proportions.
5) Gay - Lussac’s law of combining volumes: Under similar conditions of temperature and pressure,
whenever gases react together, the volumes of the reacting gases as well as products are in a simple
whole number ratio
Ex: 2H2 + O2 → 2H2O Volume ratio is 2: 1:2
2 Vol 1 Vol 2 Vol
Ex: N2 + 3H2 → 2NH3 Volume ratio 1: 3:2
1 Vol 3 Vol 2 Vol
6) Avogadro’s Law: Equal volumes of all gases under similar conditions of temperature and pressure
contains equal number of molecules
DALTON’ ATOMIC THEORY: The main postulates of this theory are:
1) Matter is made up of extremely small individual particles called atoms.
2) Atoms of the same element are identical in all respects.
3) Atoms of the different elements are different in all respects and have different masses and chemical
properties.
4) Atom is the smallest unit that takes part in chemical combinations.
5) Atoms of two or more element combine in a simple whole number ratio to form compound atoms.
6) Atoms can neither be created nor destroyed during any physical and chemical change.
7) Chemical reactions involve only combination, separation or rearrangement of atoms.

13
CLASS EXERCISE
1] 12 g carbon combines with 64g sulphur to form CS2. 12 g carbon also combine with 32 g
oxygen is form CO2. 10 g sulphur combines with 10g oxygen to form SO2. These data illustrate the
a) Law of multiple proportions b) Law of definite proportions
c) Law of reciprocal proportions d) Law of gaseous volumes.
2] Which of the following data illustrates the law of conservation of mass?
a) 56 g of CO reacts with 32 g of oxygen to produce 44 g of CO2
b) 1.70 g of AgNO3 reacts with 100 mL of 0.1 NaCl to produce 1.435 g of AgCl and 0.85 g
of NaNO3
c) 12 g of C is heated in vacuum and on cooling there is no change in mass.
d) None of the above.
3] If law of conservation of mass was to hold true, then 20.8 g of BaCl2 on reaction with 9.8 g
of H2SO4, will produce 7.3 g of HCl and BaSO4 equal to
a) 11.65 g b) 23.3 g c) 25.5 g d) 30.6 g
4] 1.5 g of hydrocarbon on combustion in excess of oxygen produces 4.4 g of CO2 and 2.7 g of H2O, the
data illustrates
a) Law of conservation of mass b) Law of multiple proportions
c) Law of constant composition d) Law of reciprocal proportions
HOME EXERCISE
1] The law of multiple proportions is illustrated by
a) Carbon monoxide and carbon dioxide b) potassium bromide and potassium chloride
c) Water and heavy water d) calcium hydroxide and barium hydroxide
2] Hydrogen and oxygen combine to form H2O2 and H2O containing 5.93% and 11.2%
hydrogen respectively. The data illustrates
a) Law of conservation of mass b) Law of constant proportions
c) Law of reciprocal proportions d) Law of multiple proportions
3] Oxygen combines with two isotopes of carbon 12
C. And14
C to from two samples of carbon dioxide the
data illustrates
a) Law of conservation of mass b) Law of multiple proportions
c) Law of reciprocal proportions d) none of these

14
4] 4.4 g of an oxide of nitrogen gives 2324 L of nitrogen and 60 g of another oxide of nitrogen
gives 22.4 L of nitrogen at S.T.P. The data illustrates
a) Law of conservation of mass b) Law of constant proportions
c) Law of multiple proportions d) Law of reciprocal proportions
5] Which one of the following pair of substances illustrates law of multiple proportions?
a) CO, CO2 b) NaCl, NaBr c) H2O, D2O d) MgO, Mg (OH)2
NCERT Text Book Questions: 1.21

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SESSION – 3
AIM
 To introduce the terms Atomic Mass, Molecular Mass, Mole, etc
THEORY
Atom: The smallest particle of an element that can take part in chemical bonding but generally
Cannot exist freely as such.
Molecule: The smallest particle of a substance (element or compound) which has independent existence.
Atomic Mass: The number which indicates how many times the mass of one atom of the element
a heavier than
-24
also called gram atom.
Ex: The atomic mass of Oxygen = 16 amu
The gram atomic mass of Oxygen = 16 gm
Molecular Mass: The number which indicates how many times the mass of one molecule of a substance is
heavier than
1
12
ℎ
part of mass of C12.
Gram molecular mass or Gram molecule: The molecular mass of an element or compound expressed in
grams is called its gram molecular mass. It is also called gram molecule.
Ex: The molecular mass of Oxygen = 32 amu
The gram molecular mass of Oxygen = 32 gm
One gram molecule of Oxygen = 32 gm of Oxygen
Element Atomic Mass Element Atomic mass Element Atomic Mass
H 1 Ne 20 K 39
He 4 Na 23 Ca 40
Li 7 Mg 24 Cr 52
Be 9 Al 27 Mn 55
B 11 Si 28 Fe 56
C 12 P 31 Cu 63.5
N 14 S 32 Zn 65.3
O 16 Cl 35.5 Br 80
F 19 Ar 40 Ag 108
Calculation of Molecular Mass:
Ex.1 H2SO4: (2 x 1) + (1 x32) + (4 x 16) = 98
Ex.2. Glucose (C6H12O6): (6 x 12) + (12 x 1) + (6 x 16) = 180
Ex.3 Oxalic acid (H2C2O4.2H2O): (2 x 1) + (2 x 12) + (4 x 16) + 2 (18) = 126
Gram atomic mass: The atomic mass of an element expressed in grams is called its gram atomic mass.It is
of the mass of C–12 is knows as atomic mass unit (amu). It is equal to 1.66 x 10 gm

16
Mole: THE amount of substance that contains the same number of chemical entities (atoms,
molecules, ions), as the number of atoms present in 12 gm of C12.
Avogadro number: The number of atoms - present in one gram atom of an element or the number of
molecules present in one gram molecule (1 mole) of a substance is called Avogadro number. It is equal to
6.023 x 1023
Thus a mole contains 6.023 x 1023
units. These units can be atoms, molecules, ions, electron or anything
else.
1 mole of hydrogen atoms means 6.023 x 1023
hydrogen atoms.
1 mole of hydrogen molecules means 6.023 x 1023
hydrogen molecules
1 mole of potassium ions means 6.023 x 1023
potassium ions
1 mole of electron ions means 6.023 x 1023
electrons.
How much does one mole weigh? That depends on the nature of particles (units). The mass of one mole
atoms of any element is exactly equal to the atomic mass in grams (gram atomic mass or gram atom) of that
element. For example, the atomic mass of aluminum is 27 amu. One amu is equal to 1.66 x 10-27
kg. One
mole of aluminium contains 6.023 x 1023
aluminium atoms
Mass of one atom aluminium = 27 x 1.66 x 10-24
g
Mass of one mole aluminium = 27 x 1.66 x 10-24
x 6.023 x 1023
= 27 g
This is the atomic mass of aluminium in grams or it is one gram atomic mass or one gram atom of
aluminium. Similarly, the mass of 6.023 x 1023
molecules (1 mole) of a substance is equal to its molecular
mass in grams or gram molecular mass or gram molecule
For example, molecular mass of water is 18 amu.
Thus, mass of one mole of water = 18 x 1.66 x 10-24
x 6.023 x 1023
= 18 g
This is the molecular mass of water 1 grams or one gram molecular mass or one gram molecule
Gram molar volume: The volume occupied by one mole of a gas at STP is called gram molar volume. It
is equal to 22.4 lit = 22,400 ml
Standard Temperature and Pressure (STP OR NTP): The temperature of 273 K and a pressure of 1 atm
are taken as STP conditions
Formula weight: The formula weight of a substance is the total mass of all atoms present in the chemical
formula of the substance.
Molar Mass: The mass of one mole of a substance is called its molar mass. The units of molar mass are
gm/mole or Kg/mole. Therefore, the molar mass is equal to atomic mass or molecular mass expressed
in grams depending upon whether the substance contains atoms or molecules.
1 mole atom = 6.02 x 1023 atoms = Gram Atomic weight = 22.4 litres at STP (only for gas) (or 1
gram atom)
Ex: mole He = 6.02 x 1023 atoms = 4 gram = 22.4 litres at STP
1 mole molecule = 6.02 x 1023 molecules = Gram Molecular weight = 22.4 litres at STP
(Only for gas)
Similarly,
Ex: 1 mole CO2 = 6.02 x 1023 molecules = 44 gram = 22.4 litres at STP.

17
CLASS EXCERSICE
1] Mass of the one atom of the element X is 1.66 x 10-26
g. Number of atoms in 1 g of the
element is
a)
1.66×10−26
0
b) 1.66 x 1025
c) 1.66 x 10-26
x No d) 6.024 x 1025
2] The number of molecules in 16 g of methane is
a)0.1NA b) NA c) 2NA d) 0.2 NA
3] The mass of 112 cm3
of O2 gas at STP is
a) 0.16 g b) 0.8 g c) 0.08 g d) 1.6 g
4] The flask A and B of equal size contain 2 g of H2 and 2 g of N2 respectively at the same temperature.
The number of molecules in flask A is:
a) Same as those in flask B b) Less than those in flask B
c) Greater than those in flask B d) exactly double than those in flask B
5] Which of the following has the largest number of atoms?
a) 0.5 g atom of Cu b) 0.635 g of Cu
c) 0.25 moles of Cu atom d) 1 g of Cu
6] One litre of a gas is at a pressure of 10-6
mm of Hg at 250
C. How many molecules are present in the
vessel?
a) 3.2 x 106
b) 3.2 x 1013
c) 3.2 x 1010
d) 3 x 104
HOME EXCERSICE
1] The containers P and Q of equal volume (1litre each) contain 6 g of O2 and SO2 respectively
at 300 K and 1 atmosphere. Then.
a) No. of molecules in P is less than that in Q
b) No. of the molecules in Q is less than that in P
c) No. of molecules in P and Q are same
d) either (a) or (b)
2] Number of moles in 1 m3
gas at NTP is
a) 4.46 b) 44.6 c) 446 d) 4460
3] 80 g of oxygen contains as many atoms as in
a) 10 g of hydrogen b) 5 g of hydrogen
c) 80 g of hydrogen d) 1 g hydrogen
4] The number of molecules in 18 mg of water in terms of Avogadro number N is
a) 10-3
N b) 10-2
N c) 10-1
N d) 10 N
5] How many times an atom of sulphur is heavier than an atom of carbon?
a) 32 times b) 12 times c) 8/3 times d) 12/32 times
NCERT Text Book Questions: 1. 1, 1. 30

18
SESSION – 4
AIM
 To understand Mole Concept upto greater extent by working critical numericals.
THEORY
Sample Problem 1:
Calculate the number of molecules present (i) in 34.20 grams of cane sugar (C12H22O11) (ii) in one litre of
water assuming that the density of water is 1 g/cm3
. (iii) In one drop of water having mass 0.05 g.
Solution:
(i) 1 mole of C12H22O11= 342 g
[Molecular mass of cane sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16 = 342amu]
= 6.022 x1023
, molecules
Now, 342 g of cane sugar contain 6.022 x1023
molecules
∴ 34.2 g of cane sugar will contain molecules
. ×
34.2 = 6.022× 10 molecules
(ii) 1 mole of water = 18 g = 6.022 x 1023
molecules
Mass of 1 litre of water = Volume x density = 1000 mL x 1 g mL-1
= 1000g
Now, 18 g of water contains = 6.022 x 1023
molecules
1000 g of water will contain
. × ×
= 3.346 × 1025
molecules
(iii) 1 mole of H2O = 18 g = 6.022 x 1023
molecules
Mass of 1 drop of water = 0.05 g
Now, 18 g of H2O contain = 6.022 x 1023
molecules
0.05 g of H2O will contain =
. ×
× 0.05 = 1.673 × 1021
molecules.
Sample Problem 2: Calculate the number of atoms of the constituent elements in 53 g of Na2CO3.
Solution: By mole concept.
1 mole of Na2CO3 = Gram formula mass of Na2CO3 = 2 x 23 + 12 + 16 x 3 = 106 g
Now, 106 g of Na2CO3 = 1 mole  53 g of Na2CO3= 0.5 mole of Na2CO3
But 1 mole of Na2CO3 contains 2 moles of Na+
ion or 2 x 6.022 x 1023
Na+
ions.
∴ 0.5 mole of Na2CO3 will contain 2 x 6.022 x 1023
x 0.5 Na+
ions = 6.022 x 1023
Na+
ions
Again, 1 mole of Na2CO3 contains 1 mole of carbon atoms = 6.022 x 1023 Na+
ions
∴ 0.5 mole of Na2CO3 will contain = 6.022 x 1023
x 0.5 carbon atoms = 3.011 x 1023
carbon atoms
Further, 1 mole of Na2CO3 contains 3 moles of oxygen atoms or 3 x 6.022 x 1023
oxygen atoms.
∴ 0.5 mole of Na2CO3 will contain = 3 x 6.022 x 1023
x 0.5 oxygen atoms = 9.033 x 1023
oxygen atoms.
Sample Problem 3: Calculate the number of molecules present in 350 cm3
of NH3 gas at 273 K and 2
atmosphere pressures.
Solution: First of all, we have to determine the volume of the gas at STP.
Given conditions At STP
V1 = 350 cm3
V2=?

19
T1 = 273 K T2 = 273 K
P1 = 2 atmospheres P2 = 1 atm
Applying gas equation: = we get
×
=
×
or =
350×2
273
×
273
1
= 700
By mole concept, 1 mole of NH3 = 6.022 x 1023
molecules = 22400 cm3
at STP
Thus, 22400 cm3
of NH3 at STP contain 6.022 x 1023
molecules
∴700 cm3
of NH3
at ATP will contain
. ×
× 700 = 1.882 × 1022
molecules
Sample Problem 4: (i) Assuming the density of water to be 1g/cm3
, calculate the volume occupied by one
molecule of water.
(ii) Assuming the water molecule to be spherical, calculate the diameter of the water molecule.
Solution: (i) 1 mole of H2O = 18 g = 18 cm3
= 6.022 x 1023
molecules of H2O
Thus, 6.022 x 1023
molecules of H2O have volume = 18 cm3
(density of H2O = 1 g/cm3
)
 1 molecule of H2O will have volume =
. ×
= 2.989 x 10-23
cm3
(ii) As water molecules is assumed to be spherical. If R is its radius, then its volume will be =
2.989 × 10 or = 7.1336 x 10-24
or R = (7.133) 1/3
x 10-8
= 1.925 x 10-8
cm
Take n = (7.133)1/3
 log n= log 7.122 = × 0.8533 = 0.2844
N = Antilog 0.2844 = 1.925
∴ Diameter of water molecule = 2 x 1.925 x 10-8
cm = 3.85 x 10-8
cm
CLASS EXERCISE
1] A sample of CaCO3 contains 3.01 x 1023
ions of Ca2+
and CO3
2-
. The mass of the sample is:
a) 100 g b) 50 g c) 200g d) 5 g
2] How many moles of carbon atoms and hydrogen atoms respectively are present in 3 moles of ethane?
a) 6, 18 b) 6, 12 c) 2, 18 d) 2, 6
3] Population of a country is approx. 100 million. If one mole of rupees is distributed equally among all
the peoples, one person will get rupees approximately
a) 1015
b) 6.22 x 1014
c) 6.022 x 1015
d) 10 million
4] What mass of carbon monoxide has the same number of oxygen atom as are present in 22 g CO2?
a) 14 g b) 280 g c) 28 g d) 56 g
5] Number of atoms of oxygen present in 10.6 g of Na2CO3 will be
a) 60.02 x 1022
b) 12.04 x 1022
c) 1.806 x 1023
d) 31.80 x 1028
6] The number of gram molecules of oxygen in 6.02 x 1024
CO molecules is
a) 10 g molecules b) 5 g molecules c) 1 g molecules d) 0.5 molecules
7] The numerical value of (where N is the number of molecules in a given sample of gas
and n is thenumber of moles of the gas) is
a) 8.314 b) 6.02 x 1023
c) 0.0821 x10-24
d) 1.66 x 10-19

20
HOME EXCERSICE
1] If 3.01 x 1020
molecules are removed from 98 mg of H2SO4, then the number of moles of
H2SO4 left are
a) 0.1 x 10-3
b) 0.5 x 10-3
c) 1.66 x 10-3
d) 9.95 x 10-2
2] The maximum number of molecules are present in
a) 5 L of N2 gas at STP b) 0.5 g of H2 gas c) 10 g of O2 gas d) 15 L of H2 gas at STP
3] Which has the maximum number of atoms?
a) 6.022 x 1021
molecules of CO2 b) 22.4 L of CO2 at N.T.P.
c) 0.44 g of CO2 d) 1 molecules of ozone
4] How many H-atoms are present in 0.04 g of ethanol?
a) 6 x 1020
b) 1.2 x 1021
c) 3 x 1021
d) 3.6 x 1021
5] A heavy element has atomic number X and mass number Y. Correct relation between X and
Y is
a) X > Y b) X < Y c) X = Y d) X = Y (1 – Y)
6] One mole of CO2 contains:
a) 3 g atoms of CO2 b) 18.1 x 1023
molecules of CO2
c) 6.02 x 1023
atoms of O d) 6.02 x 1023
atoms of C.
7] How many moles of Al2 (SO4)3 would be present in 50 g of the substance?
a) 0.083 mole b) 0.952 mole c) 0.481 mole d) 0.140 mole
8] An alloy of iron (54.7%) nickel (45%) and manganese (0.3%) has a density of 8.17 g/cm3
.
How many iron atoms are there in a block of alloy measuring 210 cm x 20 cm x 15 cm?
a) 6.02 × 10 b) 1.44 × 10 c) 3.03 × 10 d) 6.02 × 10
9] An analysis of Pyrex glass showed 12.9% B2O3, 2.2 % Al2O3, and 3.8%. Na2O, 0.4% K2O
and remaining is SiO2. What is the ratio of silicon to boron atoms in the glass?
a) 7.3: 1 b) 5:1 c) 2:7 d) 1:5
NCERT Text Book Questions: 1. 10, 1. 28, 1.33

21
SESSION – 5
AIM
 To calculate percentage composition.
 To introduce Empirical and Molecular Formulae.
THEORY
CALCULATION OF PERCENTAGE COMPOSITIONS FROM FORMULA
The percentage of any element or constituent in a compound is the number of parts by mass of that
element or constituent present in 100 parts by mass of the compound. It can be calculated by the
following two steps:
Step1. Calculate the molecular mass of the compound from its formula its formula by adding the atomic
masses of the elements present.
Step2. Calculate the percentage of the element or the constituent by supplying the following relation:
Percentage of the element or constituent =
.
.
× 100
Sample Problem 1: Calculate the percentage compositions of the various elements in MgSO4.
Solution: Mol. mass of MgSO4 = 24 + 32 + 4 x 16 = 120
% of Mg =
.
.
× 100 = ×100=20%
% of S =
.
.
× 100 = ×100=26.67%
% of O =
.
.
× 100 = ×100=53.33%
Sample Problem 2: Calculate the percentage of water of crystallisation in the sample of blue vitriol
(CuSO45H2O).
Solution: Mol. mass of CuSO4.5 H2O = 6.3 + 32 + 4 x 16 + 5 x 18 = 249.5
No. of parts by mass of H2O = 5 x 18 = 90 ∴ % of H2O =
.
× 100 = 36.07%
EMPIRICAL FORMULA AND MOLECULAR FORMULA
The formula which gives the actual number of atoms of various elements present in one molecule
of the compound is called molecular formula.
The formula which gives the relative number of atoms of various elements in one molecule of the
compound is called empirical formula.
Ex: Molecular formula of benzene is C6H6. It means that one molecule of benzene has six carbon
atoms and six hydrogen atoms.
Empirical formula of benzene is CH. It means that carbon and hydrogen are present in the ratio of 1:1
in benzene.
Compound Molecular Formula Emperical formula
Hydrogen Peroxide H2O2 HO
Benzene C6H6 CH
Glucose C6H12O6 CH2O

22
Relation between emperical formula and molecular formula
Molecular Formula = Emperical Formula x n
n =
Note: Molecular mass = 2 x Vapour Density
DETERMINATION OF THE EMPERICAL FORMULA OF A COMPOUND
1) Divide the percentage of each element by its atomic mass.
2) Divide the result obtained in the above step by the smallest value among them to get the simplest ratio
of various atoms.
3) If any number obtained above is not a whole number, then multiply all the numbers by a suitable
integer to get whole number ratio. This ratio is the simplest whole number ratio.The empirical formula
of the compound written the help of this ratio.
Sample problem1: Chemical analysis of a carbon compound gave the following percentage composition by
weight of the element present. Carbon = 10.06%, hydrogen = 0.84 % and chlorine = 89.10 % calculate
the molecular formula of the compound, if its molecular weight is 119.5.
Sol. (1) Dividing the percentage composition of each element by their respective atomic masses
Carbon Hydrogen chlorine
. . .
.
= 0.84 = 0.84 = 2.51
(2) Dividing the above values by the smallest number among them
.
.
.
.
.
.
1 1 3
(3) The simplest ratio of various atoms C: H: Cl = 1: 1: 3
Empirical formula = CHCl3
Molecular formula = empirical formula x n
n = =
.
.
= 1
Molecular formula = CHCl3 1 = CHCl3
CLASS EXERCISE
1] An alkaloid contains 17.28% of nitrogen and its molecular mass is 162. The number of nitrogen atoms
present in one molecule of alkaloid is
a) Five b) Four c) Three d) Two
2] A compound of Se and Cl has 52.1% Se. The empirical formula is (At. mass of Se = 79.Ou)
a) SeCl b) SeCl2 c) SeCl3 d) SeCl4
3] 3.0 x 1022
atoms of X and 6.0 x 1022
atoms of Y are present in a compound. The empirical formula is
a) XY b) XY2 c) XY3 d) X2Y3

23
4] An aqueous solution containing 100 grams of dissolved MgSO4 is fed to a crystalliser where 80% of
the dissolved salt crystallizes out as MgSO4.6H2O) crystals. How many grams of the hexahydrate salt
crystals are obtained from the crystallizer?
a) 80 b) 152 c) 120 d) 100
5] A purified cytochrome protein was found to contain 0.376% iron. What is the minimum molecular
massof the protein?
a) 14,800 u b) 1480 u c) 148,000 u d) 148 u
6] Element X (Atomic mass = 75 and element Y (Atomic mass = 16) combine to give a compound having
75.8% X. The formula of the compound is:
a) XY b) X2Y c) X2Y2 d) X2Y3
7] A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass’s 99.
What is the empirical formula?
8] An organic substance containing C, H and O, gave the percentage composition as:
C= 40.687%, H = 5.085% and 0 = 54.228%. The vapour density of the compound is 59. Calculate the
molecular formula of the compound.
HOME EXCERSICE
1] The hydrated salt Na2SO4 xH2O on heating loses 55.9% weight and becomes anhydrous. The value of x
is
a) 3 b) 6 c) 7 d) 10
2] Caffeine has a molecular mass of 194. If it contains 28.9% by mass of nitrogen, number of atoms of
nitrogen in one molecular of it is
a) 1 b) 2 c) 3 d) 4
3] Manganese forms non-stoichiometric oxides having general formula MnOx. The value of x
for a compound that analysed 63.70% Mn is (at. mass Mn = 54.94 u)
a) 1.958 b) 1.859 c) 1.898 d) 2.859
4] 1.625 g of anhydrous ZnSO4 was placed in moist air. After few days its weight was found to be 2.857
g. What is the molecular formula of hydrated salt?
a) XnSO4. H2O b) ZnSO4.3H2O c) ZnSO4.7H2O d) ZnSO4. 10H2O
5] A compound containing sodium. Sulphur, hydrogen and oxygen gave the following results on analysis:
Na = 14.28%, S= 9.92% H = 6.20%. Calculate the molecular formula of the anhydrous compound. If
all the atoms of hydrogen in the compound are present in combination with oxygen as water of
crystallization what is the structure of the crystalliser salt? The molecular mass of the crystalline salt is
322.
a) . b) . 4 c) . 7 d) . 10
6] A compound containing C, H and O gave the following analytical data
C = 40.0% and H =- 6.67%Calculate the molecular formula of the compound if its molecular mass is
180.
a) b) c) d)
NCERT Text Book Questions: 1.2, 1. 3, 1. 8, 1.34

24
SESSION – 6
AIM
 To introduce Stoichiometric Calculations
THEORY
STOICHIOMETRY OF CHEMICAL REACTIONS
A General Approach: One of the most important aspects of a chemical equation is that when it is
written in the balanced form, it gives quantitative relationships between the various reactants and
products in terms of moles, masses, molecules and volumes. This is called stoichiometry (Greek word,
meaning ‘to measure an element’). The coefficients of the balanced chemical equation are called
stoichiometric coefficients. For example, a balanced chemical equation along with the quantitative
information conveyed by it is given below:
CaCO3 + 2HCl → CaCl2 + H2O + CO2
1 Mole 2 Moles 1Moles 1Mole 1Mole
40+12+3×16 2(1+35.5) 40+2×35.5 2×1+16 12+2×16
=100g =73g =111g =18g =44g
Or 22.4 litres at STP
Thus,
(i) 1 mole of calcium carbonate react with 2 moles of hydrochloric acid to give 1 mole of calcium chloride,
1 mole of water and 1 mole of carbon dioxide.
(ii) 100 g of calcium carbonate react with 73 g hydrochloric acid to give 111 g of calcium Chloride, 18 g of
water and 44 g (or 22.4 litres at STP) of carbon dioxide.
The quantitative information conveyed by a chemical equation helps in a number of calculations. The
problems involving these calculations may be classified into the following different types:
(1) Mass-Mass Relationships, i.e., mass of one of the reactants or product is given and the mass of some
other reactant or product is to be calculated.
(2) Mass-Volume Relationship, i.e, mass/volume of one of the reactants or products is given and the
volume/mass of the other is to be calculated.
(3) Volume-Volume Relationship, i.e, volume of one of the reactants or the products is given and the
volume of the other is to be calculated.
The general method of calculations for all the problems of the above types consists of the following
steps:
(i) Write down the balanced chemical equation
(ii) Write the relative number of moles or the relative masses (gram atomic or molecular masses) of the
reactants and the products below their formulae.
(iii) In case of a gaseous substance, write down 22.4 litres at STP below the formula in place of 1 mole 7

25
Sample Problem 1: Calculate the mass of iron which will be converted into its oxide (Fe3O4) by the action
of 18 g of steam on it.
Solution: The chemical equation representing the reaction is:
3Fe + 4H2O → Fe3O4+ 4H2
3×56 =168g 4×18=72g
Thus, 72 f of steam reacts with 168 g of iron
∴ 18 g of steam will react with × 18 = 42 g of iron ∴Mass of iron required = 42 g.
Sample Problem 2: What mass of slaked lime would be required to decompose completely 4 grams of
ammonium chloride and what would be the mass of each product?
Solution: The equation representing the decomposition of NH4CI by slaked lime, i.e. Ca (OH)2 is:
Ca(OH)2 + 2NH4Cl → CaCl2 + 2NH3 + 2H2O
40+2(1+16) 2(14+4+35.5) 40+2×35.5 2(14+3×1) 2(2×1+16)
=74g =107g =111g =34g =36g
(i) To calculate mass of Ca (OH)2 required to decompose 4g of NH4Cl.
From the above equation, 107 g of NH4Cl are decomposed by 74 g of Ca (OH)2
∴ 4 g of NH4Cl will be decomposed by Ca (OH)2=
74
107
× 4 = 2.766
Thus, the mass of slaked lime required = 2.766 g.
(ii) To calculate the mass of CaCl2 formed.
107 g of NH4Cl when reacted with Ca (OH)2 produce 111 g of CaCl2.
∴ 4 g of NH4Cl when reacted with Ca (OH)2 will produce CaCl2=
111
107
× 4 = 4.15
Hence the mass of CaCl2 produced = 4.15 g.
(iii) To calculate the mass of NH3 produced.
107 of react to form 34 of
∴ 4 of react to form = 4 × = 1.271
(iv) To calculate the mass of H2O formed
107 g of NH4Cl react with Ca (OH)2 to yield 36 g of H2O
∴ 4 g of NH4Cl when reacted with =
36
107
× 4 = 1.345
So the mass of H2O formed = 1.3458 g.
Sample Problem 3: 1.5 g of an impure sample of sodium sulphate dissolved in water was treated with
excess of barium chloride solution when 1.74 g of BaSO4was obtained as dry precipitate. Calculate the
percentage purity of the sample.
Solution: Given: 1.5 g of impure Na2SO4
Treated with
Treated with 1.74 g of BaSO4
The chemical equation representing the reaction is:
Na2SO4 + BaCl2 → BaSO4 + 2NaCl
2×23+32+4×16=142g 137+32+4×16=233g

26
Step1. To calculate the mass of Na2SO4 which produces 1.74 g of BaSO4? From the chemical equation, 233
g of BaSO4 are produced from Na2SO4 = 142 g
∴ 1.74 g of it would be obtained from NaSO4 =
142
233
× 1.74 = 1.06
This is the mass of pure Na2SO4 present in 1.5 g of impure sample.
Step2. To calculate the percentage purity of impure sample.
1.5 g impure sample contains pure Na2SO4 = 1.06 g
∴100 g of the impure sample will contain pure Na2SO4 = 1.06
1.5
× 100 = 70.67
Thus, percentage purity of impure sample = 70.67
Sample Problem 4: Current market prices of A1, Zn and Fe scraps per kg are Rs. 20, Rs. 16 and Rs.3
respectively. If H2 is to be prepared by the reaction of one of these metals with H2SO4, which would be
the cheapest metal to use? Which would most expensive?
Solution: The various chemical reactions involved are given below:
(i) 2Al + 3H2SO4 → Al2(SO4)3 + 3H2
2×27=57g 3×2=6g
(ii) Zn +H2SO4 → ZnSO4 + H2
65g 2g
(iii) Fe + H2SO4 → FeSO4 + H2
56g 2g
Let us suppose that the amount of hydrogen to be prepared = 100g.
Step1. To calculate the cost of preparation of 100 g of H2 from A1
6 g of H2 is prepared from A1 = 54 g
∴100g of H2 will be obtained from A1= × 100
Cost of 1000 g of A1 = Rs. 20  cost of 900 g of A1 = × 900 = . 18
Step 2. To calculate the cost of preparation of 100 g of H2 from Zn
2 g of H2 is produced from Zn = 65 g ∴ 100 g of H2 will be obtained from Zn
= × 100 = 3250
Cost of 1000 g of Zn = Rs.16/- ∴cost of 3250 g of Zn= × 3250 = . 52
Step3. To calculate the cost of preparation of 100 g of H2 from Fe
2 g of H2 produced from Fe = 56 g ∴ 100 g of H2 will be obtained from
Fe = × 100 = 2800
Cost of 1000 g of Fe = Rs.3/- ∴ Cost of 2800 g of Fe = × 2800 = 8.40
Thus, Fe is the cheapest and Zn is the most expensive metal to use of the preparation of H2.

27
Sample Problem5: Calculate the amount of lime, Ca (OH)2, required to remove hardness of 50,000 litres of
well water which has been found to contain 1.62 g of calcium bicarbonate per 10 litre.
Solution:
(i) Calculation of total Ca (HCO3)2 present.
10 L of water contain Ca (HCO3)2 = 1.62 g
 50,000 L of water will contain Ca (HCO3)2 =
.
× 50000 = 8100
(ii) Calculation of lime required. The balanced equation for the reaction involved is:
Ca (HCO3)2 + Ca (OH)2→ 2CaCO3 + 2 H2O
1 mole 1 mole
162 g Ca (HCO3)2 require lime = 74 g
∴8100 g Ca (HCO3)2 will require lime = × 8100 = 3700 g = 3.7 kg
Sample Problem 6:1.0 g of a mixture of carbonates of calcium and magnesium gave 240 cm3
of CO2 at
STP. Calculate the percentage composition of the mixture.
Solution: Mass of mixture of carbonates of Ca and Mg taken = 1.0 g
Suppose the mass of CaCO3 = x g Mass of MgCO3 = (1-x) g
The chemical equations involved are:
CaCO3→CaO + CO2
40+12+3×16=100g 22400cm3
at STP
MgCO3 → MgO + CO2
24+12+3×16=84 g 22400cm3
at STP
Step1. To calculate the volume of CO2 evolved at STP from x g of CaCO3.
100 g of CaCO3 evolve CO2 at STP = 22400 cm3
∴x g of CaCO3 will evolve CO2 at STP = ×xcm3
= 224 xcm3
Step2. To calculate the volume of CO2 evolved at STP from, (1 – x) g of MgCO3.
84 g of MgCO3 will evolve CO2 at STP = × (1 − )cm3
= (1 − )cm
Step3. To calculate the value of x
Total volume of CO2 evolved at STP = 224 x+ × (1 − )cm
But total volume of CO2 evolved at STP = 240 cm3
(Given)
∴ 224 x+ × (1 − ) = 240 or 672 x + 800 – 800x = 720 or 128x = 80
∴ = 5
8
Step4. To calculate the percentage composition of the mixture
∴ Percentage of CaCO3 =
5
8×1
× 100 = 62.5∴ Percentage of MgCO3 = 100 – 62.5 = 37.5

28
Sample Problem 7: 3.0 g of H2 reacts 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.
(ii) Calculate the amount of the reactant left unreacted. Molecular mass of H2 = 2.016
Solution: 2H2 + O2→ 2H2O
2×2 = 4.0g 32 g 2× (2+16) =36
3 g of H2 require O2 = × 3 = 24
Thus, O2(29g) is present in excess. Hence, H2 is the limiting reactant
H2O formed = × 3 = 27
O2 left unreacted = 29 − 24 = 5.0
CLASS EXERCISE
1] 2.79 gm of iron is completely converted into rust (Fe2O3). Weight of the oxygen in the rust is
a) 2 g b) 1.1g c) 3g d) 1.8 g
2] What is the volume (in lit) of CO2 liberated at STP, when 2.12 grams of sodium carbonate (mol. wt.
106) is treated with excess of dilute HCl?
a) 2.28 b) 0.448 c) 44.8 d) 22.4
3] 2 Moles of BaCl2 react with excess of dil.H2SO4. The no. of moles of BaSO4 formed is
a) 3 moles b) 4 moles c) 2 moles d) ½ mole
4] Number of moles of KClO3 required to produced 5.6 litres of O2 at STP is
a) 1/6 b) 1/8 c) ½ d) 1/3
5] 10 gram of a mixture of CaCO3 and Na2CO3 on ignition suffered a loss in weight of 2.2 gram. The mass
ratio of CaCO3 and Na2CO3 is
a) 1 : 1 b) 1 : 1.4 c) 1.4 : 1 d) 1.75 : 1
6] How many grams of CaCO3 are required to produce carbon dioxide that is sufficient for the conversion
of 0.1 mole sodium carbonate to sodium bicarbonate?
a) 1 gram b) 10 gram c) 100 gram d) 1000 gram
7] The mass of CaO that shall be obtained by heating 20 kg of 90 % pure lime stone is
a) 11.2 kg b) 8.4 kg c) 10.08 kg d) 16.8 kg
8] The mass of 70% H2SO4 by mass required for neutralisation of 1 mol of NaOH is
a) 49 g b) 98 g c) 70 g d) 34.3 g
9] If 0.50 mole of BaCl2is mixed with 0.20 mole of Na3PO4. The maximum number of moles of Ba3(PO4)2
that can be formed is
a) 0.70 b) 0.53 c) 0.20 d) 0.10
10] One litre of a mixture of CO and CO2 is passed over red hot coke when the volume increased to 1.6
litre under the same conditions of temperature and pressure. The volume of CO in the original mixture
is
a) 400 ml b) 600 ml c) 500 ml d) 800 ml

29
HOME EXERCISE
1] The loss in weight when 10.5 gram of MgCO3 is heated in an open crucible is
a) 4.4 gram b) 5.5gram c) 2.2 gram d) 8.9 gram
2] When 10 ml of H2 and 12.5 ml of Cl2are allowed to react, the final mixture contains under
the same conditions
a) 22.5 ml of HCl b) 12.5 ml of HCl
c) 20 ml of HCl d) 2.5 ml of Cl2& 20 ml HCl
3] A mixture of 20 ml of CH4 and 20 ml of O2 at STP is exploded and cooled to room temperature. If the
reaction between the two substances is written as CH4+2O2 CO2+ 2H2O. The final volume of the gas
mixture at STP is
a) 16 ml b) 14 ml c) 26 ml d) 20 ml
4] 2 mol of H2S and 11.21 L of SO2 at N.T.P. react to form x moles of sulphur according to the following
equation, SO2 + 2H2S  3S + 2H2O, x is
a) 1.5 b) 3 c) 11.2 d) 6
5] ‘x’ grams of calcium carbonate was completely burnt in air. The weight of the solid residue formed is
28 g. What is the value of ‘x’ grams?
a) 144 b) 200 c) 150 d) 50
6] The weight of magnesium that will be required to produce just sufficient hydrogen to combine with all
the oxygen that can be obtained by the complete decomposition of 24.5 gram of Potassium chlorate is
a) 10.4 gram b) 14.4 gram c) 9.32 gram d) 12.4 gram
NCERT Text Book Questions: 1.4, 1.7, 1.23, 1.24, 1.26, 1.36

30
SESSION – 7
AIM
 To apply equivalent concept in solving stoichiometric problems.
THEORY
EQUIVALENT WEIGHT OF ELEMENT
It is the weight of element that combines with or displaces 8 parts by weight of oxygen, 35.5 parts by
weight of chlorine or 1 part by weight of hydrogen.
Eg. 1: Mg + Cl2 → Mg Cl2
1 mole 1 mole
24 gram 71 grams
71 gram Cl2 → 24 gram Mg
35.5 gram Cl2 → 12 gram Mg
E.W. of Mg = 12 = =
.
Eg. 2: 4A l + 3O2 → 2Al 2 O3
4 mole 3 mole
4 x 27 gram 3 x 32 gram
3 x 32 gram O2 → 4 x 27 gram Al
8 gram O2 → ×
×
x 8 = 9 gram Al
E.W. of Al = 9 = =
.
. =
Equivalent weight of acid:
It is the weight of acid that gives one mole H+
ions
Eg.1: H2 SO4 → 2H+ +
1 mole 2 mole
98 gram
2 mole H+ → 98 gram H2 SO4
1 mole H+ → = 49 gram H2SO4
E.W. of H2SO4 = 49 = =
.

31
Eg.2: H3PO4 → 3H+ +
3 mole H+ → 98 gram H3PO4
1 mole H+ → = 32.66
E.W. of H3PO4= 32.66 =
98
3
=
.
Basicity: It is number of H+ ions given by 1 molecule of acid.
Eq.wt of acid =
.
+
2 → Ca2+
1 mole 2 mole
74 gram
2 mole OH → 74gram Ca(OH)2
1 mole OH → = 37gram Ca(OH)2
E.W. of base =
.
Acidity: It is the number of OH 
given by 1 molecule of base.
Equivalent weight of ion:
It is the weight of ion that contains 1 mole charge
Eq.1: SO4
2- 1 mole SO4
2- →2 mole charge →96 gram
1 mole charge →
96
2
= 48 gram
E.W of SO4
2- = 48 =
96
2
=
Eq.2: PO 1 mole PO → 3 mole charge → 95 grams
1 mole charge → = 31.66 gram
E.W of PO = 31.66 = =
ℎ
ℎ
E.W. of ions =
Eg. Ca (OH)
1 mole 3 mole 98 gram
Equivalent weight of base:
It is the weight of base that gives one mole OH
. or reacts with 1 mole H
+ 2OH 

32
Equivalent weight of ionic compounds:
It is the weight of compound that contains 1 mole positive and 1 mole negative charge
Ex. Al Cl3 → Al3+ + 3Cl-
1 mole
133.5 gram
133.5 gram AlCl3→3 mole positive & 3mole negative charge
1 mole charge → = 44.5 g AlCl3
E.W. of AlCl3= 44.5 =
.
E.W. of ionic compound =
Gram equivalent weight: When equivalent weight is expressed in grams, it is called gram equivalent
weight.
Number of equivalents/ Number of gram equivalents: Analogous to number of moles, number of
gram equivalents has been defined to make stoichiometric calculations easy
Number of gram equivalents =
Law of chemical equivalence: In any chemical reaction, the number of equivalents of all reactants and
products are equal.
Eg. 2 Mg + O2 → 2 Mg O
Moles 2 mole 1 mole 2 mole
Wt in gram 48 gram 32 gram 80 gram
E.W =
. .
= 12 = 8 = 20
No. of Equivalent =
.
= 4 = 4 = 4
Relation between number of moles and equivalent:
No. of Moles =
ℎ
. .
No. of Equivalents =
.
. .
=
.
. . /
x → Valency, Acidity, Basicity
=
. . /
×
. .
=
No. of Equivalents = No. of Moles × x

33
Determination of equivalent mass:
(i) Hydrogen displacement method: Calculate the mass of the metal which displaces 1.008 parts by
mass of hydrogen.
Eq. mass = × 1.008 =
( )
× 11200
(ii) Oxide formation or reduction of the oxide method: Calculate the mass of the metal which combines
with or displaces 8 parts by mass of oxygen.
Eq.mass = × 8 = ( ) 2
× 5600
(iii) Chloride formation method: Calculate the mass of the element which combines with or displaces 35.5
parts by mass of chlorine
Eq.mass = × 35.5
Methods of determining atomic mass:
(i) By application of the relation: At.mass = Eq.mass × Valency
Knowing approx.atomic mass and exact equivalent mass, first valency is calculated and then the exact
atomic mass.
(ii) Dulong and Petit’s method: For solid elements (except Be, B, C and Si), according to Dulong and
Petit’s law
At.mass ×Specific Heat =6.4 approx.
∴ Approx. atomic mass =
.
.
Exact atomic mass = Eq.wt ×Valency
Where, Valency =
.
.
(Take nearest whole no.)
CLASS EXERCISE
1] What is the equivalent weight of H3PO4 in the reaction?
Ca (OH)2 + H3PO4→CaHPO4 + 2H2O
a) 98 b) 49 c) 3.66 d) 24.5
2] A chloride of an element contains 49.5% chloride. The specific heat of the element is 0.056. Calculate
the equivalent mass, valency and atomic mass of the element
3] On dissolving 2.0 g of metal in H2SO4, 4.51g of the metal sulphate was formed. The specific heat of the
metal is 0.057. What is the valency of the metal and exact atomic mass?
4] The molecular mass of chloride, MCl is 74.5 the equivalent mass of the metal M will be:
a) 39.0 b) 74.5 c) 110.0 d) 35.5

34
HOME EXERCISE
1] One gram of the chloride was found to contain 0.835 g of chlorine. Its vapour density is 85. Calculate
itsmolecular formula.
a) b) c) d)
2] The oxide of an element contains 32.33% of the element and the vapour density of its chloride is79.
Calculate the atomic mass of the element.
a) 8 b) 32 c) 10 d) 15.28
3] A1, g of an element gives A2 g of its oxide. The equivalent mass of the element is
a) × 8 b) × 8 c) × 8 d) (A2 –A1) 8
4] The oxide of an element possesses the molecular formula M2O3. If the equivalent mass of the metal is
9, its atomic mass is
a) 27 b) 18 c) 9 d) 4.5

35
SESSION – 8
AIM
 To introduce various methods of expressing Concentration of solutions.
THEORY
INTRODUCTION
A solution is a homogeneous mixture of two or more chemically non reacting substance whose
composition can be varied within certain limits.
Solution made up of; two components are called binary solutions, three components are ternary
solutions, four components are quaternary solutions.
Components of binary solutions are solute and solvent.
The componet which is major ingredient and whose physical state is the same as that of solution is
called solvent.
METHODS OF EXPRESSING THE CONCENTRATION OF A SOLUTION
The concentration of a solution is defined as the amount of solute present in the given quantity of the
solution (or solvent). Solutions containing relatively high concentration of solute are called
concentrated solutions, while those of relatively low concentrations of solute are called dilute solutions
The concentration of a solution can be expressed in a number of ways. The important methods are
1. Mass Percentage: It is the amount of solute in grams present in 100 grams of the solution
Mass percentage × 100
2. Volume Percentage: It is the volume of the solute present per 100 parts by volume of solution.
Volume percentage × 100
3. Parts per million (ppm): It is the amount of the solute in grams present in 106 grams of the solution.
ppm of ‘x’= × 106
4. Molarity: Molarity is the number of moles of the solute present in one litre of the solution. It is denoted
by ‘M’. The units of molarity are moles/litre.
When one mole of the solute is dissolved in 1 litre solution, the solution is called one molar (or 1 M)
solution.
Mathematical Relations:
1. Molarity =
.
( )
=
.
. . .
×
1
( )
=
.
. . .
×
1000
( )

36
2. No. of moles = M.V.(litres)
No. of millimoles = M.V.(ml)
3. Calculation of molarity, when % mass of solution is given
=
10 × %
4. Calculation of molarity, when % mass of solution and density are given (only for competitive exams)
=
10 × % ×
5. When a solution of molarity M1 and volume V1 is diluted with a solvent to get a solution of molarity
M2 and volume V2, then
M1V1 = M2V2
6. Calculation of resultant molarity, when two or more solutions of the same substance are mixed.
=
+ + + … … … .
M1 and V1, M2 and V2, M3 and V3 are the molarities and volumes of the individual solutions.
7. When two solutions of different substance are mixed together (titrated against each other), then
=
M1, V1, and M2, V2, are molarities and volumes of the two solutions n1 and n2 are the no. of moles of
the two substances involved.
Note:
Semimolar: molarity is 0.5; Decimolar: molarity is 0.1
Centimolar: molarity is 0.01; Millimolar: molarity is 0.001
Molarity changes with temp because volume of solution changes with temp.
CLASS EXERCISE
1] A solution is prepared by adding 2 g of a substance. A to 18g of water. The mass percent of the solute is
2] 0.5 moles of a solute is present in 500 ml of the solution. Then its molarity is
a) 1 M b) 10-3
M c) 10-2
M d) 10-1
M
3] 100 milli moles of a solute is present in 200 ml of the solution. Then its molarity is
a) 2 M b) 1 M c) 0.5 M d) 1.5 M
4] The number of milli moles of solute present in 10ml of decimolar solution is
a) 1 b)10-3
c) 10-2
d) 10-1
5] 2 gms of NaOH is present in 1 litre of the solution. The molarity of the solution is
a) 0.5 M b) 0.05 M c) 0.1 M d) 0.005 M
6] NaOH solution is labelled as 10% by volume. then the molarity of NaOH solution is
a) 2 M b) 2.5 M c) 4 M d)1M
7] The volume of water to be added to 100ml of 0.5M urea solution in order to make it decimolar is
a) 500 ml b) 400 ml c) 600 ml d) 50 ml

37
8] 100 ml of 1 M HCl, 200 ml of 2 M HCl and 300 ml of 3 M HCl are mixed with enough water to get 1
M solution.The volume of water to be added is
a) 600 ml b) 700 ml c) 800 ml d) 125 ml
9] 10.6 g of a substance of molecular weight 106 was dissolved in 100 ml. 10 ml of this solution was
pipetted out into a 1000 ml flask and made up to the mark with distilled water. The molarity of the
resulting solution is
a) 1 M b) 10-2
M c) 10-3
M d) 10-4
M
10] Concentrated aqueous sulphuric acid has 98%. H2SO4 by mass and has a density of 1.84 g cm-3
What
volume of concentrated acid is required to make 5.0 litre of 0.5 M H2SO4 solution
HOME EXERCISE
1] The number of moles of solute present in 10 ml of decimolar solution is
a) 1 b) 10-2
c) 10-3
d)10-1
2] 4.9 g of H2SO4 is present in 500 ml of the solution. The molarity of the solution is
a) 0.1M b) 0.2 M c) 0.02 x 10-2
M d) 0.05 M
3] The volume of water to be added to convert 10 ml of deca molar HCl solution to decimolar solution is
a) 99 ml b) 100 ml c) 1000 ml d) 990 ml
4] 100 ml of 1 M HCl, 200 ml 2 M HCl and 300 ml 3 M HCl are mixed.The Molarity of the resulting
solution is
a) 1 M b) 2.66 M c) 2.33 M d) 4.25 M
5] The volumes of 1 M HCl and 5 M HCl to be mixed to get 2 lit of 2M HCl are
a) 1 lit and 1 lit b) 1.5 lit and 0.5 lit
c) 1.25 lit and 0.75 lit d) 1.33 lit and 0.66 lit
6] The concentration of 100 ml solution containing ‘x’ grams of Na2CO3 is yM. The values of x and y are
a) 2.12, 0.05 b) 1.06, 0.2 c) 1.06, 0.1 d) 2.12, 0.1
7] In a reaction vessel, 0.184 g of NaOH is required to be added for completing the reaction. How many
millilitres of 0.15 M NaOH solutions should be added for this requirement?
a) 100 b) 68 c) 154 d) 30.6
NCERT Text Book Questions: 1.5, 1.6, 1.11, 1.12, 1.25, 1.35

38
SESSION –9
AIM
 To discuss different methods of expressing concentration of solution [Contd.]
THEORY
5. Normality: It is the number of gram equivalents of the solute present in one litre of the solution. It is
denoted by ‘N’. The units of normality are gram equivalents/litre. When one gram equivalent weight of
a solute is dissolved in 1 litre solution, the solution is called one normal (1 N) solution.
Mathematical Relations:
1. Normality =
.
( )
=
.
. . .
×
1
( )
=
.
. . .
×
1000
( )
2. No. of equivalents = N.V.(litres)
No. of milli equivalents = N.V.(ml)
3. Calculation of normality, when % mass of solution is given =
×%
4. Calculation of normality, when % mass of solution and density are given (only for competitive exams)
=
10 × % ×
5. When a solution of Normality N1 and volume V1 is diluted with a solvent to get a solution of Normality
N2 and volume V2, then
N1V1 = N2V2
6. Calculation of resultant Normality, when two or more solutions of the same substance are mixed.
=
+ + + … … … .
7. When two solutions of different substance are mixed together (titrated against each other), then
N1V1 = N2V2 (or)
. .
= . . ( )
i.e. no. of equivalents of both the substance should be equal.
8. Calculation of resultant normality when two solutions of different substances are mixed together. Let
Na and Va be the normality and volume of acid, Nb and Vb be the normality and volume of base.
(i) If NaVa > NbVb  acidic (pH < 7)
=
−
(ii) If NbVb > NaVa  basic (pH > 7)
=
−
(iii) If NaVa = NbVb  neutral

39
9. Relation between Normality and Molarity
=
wt
G. E. W.
×
1
( )
=
wt
G. M. W.
×
1
( )
Normality
=
G. M. W.
G. E. W.
=
G. M. W.
. . ./
 Normality = Molarity × n factor
n factor is basicity for acids, acidity for bases, total +ve or –ve charges for salts, change in oxidation no.
for oxidant & reductant.
CLASS EXERCISE
1] 10 milli equivalents of solute are present in 5ml of an aqueous solution. Then its Normality is
a) 1 N b) 0.5 N c) 2 N d) 0.25 N
2] 9.8 g of Orthophosphoric acid is dissolved in water and the solution is made upto 2 litres with distilled
water. The Normality of the solution is
a) 0.75 N b) 0.05 N c) 0.3 N d) 0.15 N
3] 20 ml of 0.1 N FeSO4 solution will be completely oxidised by ------- ml of 0.05N KMnO4 solution in
acidic medium.
a) 20 ml b) 10 ml c) 40 ml d) 80 ml
4] The weight of KMnO4 that can oxidise 100 ml of. 0.2 M oxalic acid in acidic medium is:
a) 1.58 g. b) 1.264 g. c) 12.64 g. d) 15.8 g.
5] 0.25g.of an acid was exactly neutralised by 40 ml of 0.125 N base. The equivalent
weight of the acid is
a) 100 b) 50 c) 75 d) 25
6] Calculate the molarities and normalities of the solutions obtained by mixing
i) 100 ml of 0.2 M H2SO4 with 50 ml of 0.1M HCl
ii) 100 ml of 0.2 M H2SO4 with 50 ml of 0.1 M NaOH
HOME EXERCISE
1] The number of milli equivalents of solute present 10 ml of 0.1 N solutions are
a) 1 b) 10 c) 100 d) 0.1
2] 250 ml of a solution of Na2CO3 contains 4.24g of solute. Its Normality is
a) 0.16 N b) 0.32 N c) 0.64 N d) 0.8 N
3] The weight of crystalline Oxalic acid (H2C2O4. 2H2O) required to prepare 100 ml of 0.05N
solution is
a) 0.1575 g b) 1.575 g c) 0.315 g d) 0.63 g
4] The Molarity of 200 ml of HCI solution which can neutralise 10.6g of anhydrous Na2CO3 is
a) 0.1 M b) 1M c) 0.6 M d) 0.75 M

40
5] The volume of 0.05M K2Cr2O7 solution which can oxidise 200 ml of 0.1 M FeSO4 solution in acidic
medium is
a) 200 ml b) 40 ml c) 66 ml d) 88 ml
6] 10 millimoles of a diacidic base exactly neutralises 100 ml of an acid. Then the Normality of that acid is
a) 0.2 N b) 0.1 N c) 0.4 N d) 0.5 N
7] The molarity and normality of the solution obtained by mixing
100 ml of 0.2 N H2SO4 with 50 ml of 0.1N HCl is
a) 0.1 M, 0.167 N b) 0.2 M, 0.4 M c) 0.1 M, 0.1 N d) 0.167 M, 0.167 N
8] The molarity and normality of the solution obtained by mixing 50 ml of 0.1N with 100 ml of
0.1N NaOH is
a) 0.01 M, 0.01 M b) 0.033 M, 0.066 N c) 0.1 M, 0.1 N d) 0.033 M, 0.033 N

41
SESSION – 10
AIM
 To discuss different methods of expressing concentration of solution [Contd.]
THEORY
6. Molality: It is defined as the number of moles of the solute present in 1 kg of the solvent. It is denoted
by ‘m’. The units of molality are moles/kg. When one mole of a solute is dissolved in 1 kg of the
solvent, the solution is called 1 molal or 1m solution.
Molality, m =
.
∴ m= ×
7. Mole Fraction: It is defined as the ratio of number of moles of one component to the total number of
moles of the solution. Mole fraction is denoted by ‘x’
Consider a solution containing two components A and B. If xA and xB are mole fractions of A and B
respectively and nA and nB are the number of moles of A and B respectively. Then
xA= =
xB = =
Where, wA and wB are the weights of A and B in grams respectively and MA and MB are the
molecular masses of A and B respectively.
The sum of the mole fractions of the components of a solution is equal to unity. In a binary solution,
mole fraction of solute + mole fraction of solvent = 1
8. Formality: It is the number of formula masses in grams of the solute dissolved per litre of the solution.
It is denoted by ‘F’. It is used for ionic compounds in which there is no existence of molecule.

42
CLASS EXERCISE
1] The molality of 10% (W/W) NaOH solution is
a) 2.77 m b) 5.54 m c) 0.0025 m d) 2.5 m
2] 16 g. of methanol is present in 100 ml of the solution. If the density of the solution is 0.96gml-1, the
molality of the solution is
a) 6.75 m b) 6.25 m c) 5.75 m d) 5 m
3] 100 ml of ethyl alcohol [d = 0.92 g/ml] and 900 ml of water [d = 1 g/ml] are mixed to form
1 lit solution. The Molarity and molality of the resulting solution are
a) 2M and 2m b) 2M and 2.22m c) 2.2 M and 1.1m d) 2M and 1 m
4] 6 g. of Urea is present in 100 gm. of water, the concentration of the solution is
a) 1M b) 1m c) 1N d) 1F
5] A solution contains 410.3 g of H2SO4 per litre of solution at 20o
C. If the density is 1.243 g/ml what will
be its molarity and molality.
6] The density of a 2.03 M solution of acetic acid in water is 1.017 g/ml calculate the molality of the
solution.
7] NaOH aqueous solution is labelled as 10% (w/v). Density of the solution is 1.02 g/ml. Then the mole
fraction of the solute in the solution is
a) 0.05 b) 0.0466 c) 0.53 d) 0.053
HOME EXERCISE
1] The molalilty of 2% (W/W) NaCI solution nearly
a) 0.02 m b) 0.35 m c) 0.25 m d) 0.45 m
2] The molality of the solution prepared by dissolving 18 g of glucose in 500g of water
a) 01.2 m b) 0.4 m c) 0.1m d) 0.2 m
3] The densities of a 3M sodium thio–sulphate (Na2S2O3) is 1.25 g/ml, the % mass of
a) 12.5 b) 65.84 c) 37.92 d) 25
4] The mole fraction of ethyl alcohol in a solution of total volume 95 ml prepared by adding 50 ml ethyl
alcohol(den = 0.789 g ml-1
) to 50 ml water (den=1.0 g ml-1
), is
a) 0.21 b) 0.24 c) 0.23 d) 0.25
5] 6 g. of Urea is dissolved in 90 g. of water: The mole fraction of solute is
a) 0.764 b) 0.0196 c) 0.534 d) 0.466
6] A gaseous mixture contains fours gases A, B, C and D. The mole fraction of “B” is 0.5 The mole fraction
of “A” is
a) 0.525 b) 0.375 c) 0.625 d) 0.732
NCERT Text Book Questions: 1.17, 1.29

43
MAIN
LEVEL – I
1] The number of significant figures in π are
a) One b) Two c) Three d) Infinite
2] Given the numbers 786, 0.786 and 0.0786. The number of significant figures for the three numbers is
a) 3, 4 and 5 respectively b) 3, 3 and 3 respectively
c) 3, 3 and 4 respectively d) 3, 4 and 4 respectively
3] In which of the following numbers all zeros are significant?
a) 0.0005 b) 0.0500 c) 50.000 d) 0.0050
4] 1087.2 = 14.583. The correct answer to this problem in proper number of significant digits is
a) 15 b) 14.58 c) 14.5 d) None of these
5] 14.90 + 0.0070 + 1.0 + 0.081 = 15.9880. The sum to proper number of significant digit is
a) 15.9 b) 16.0 c) 15.99 d) 16
6] Which is larger quantity?
a) Mega b) Femto c) Milli d) Giga
7] The scientific notation of 0.0000000540 is
a) 5.40 x 10-7
b) 5.40 x 10-8
c) 54.0 x 10-7
d) 54.0 x 10-9
8] 3g of a hydrocarbon on combustion with 11.2 g of oxygen produces 8.8g CO2 and 5.4g H2O. The data
illustrates the law of
a) Conservation of mass b) Multiple proportions
c) Definite proportions d) Reciprocal proportions
9] The gram mole of a gas at N.T. P. occupies 22.4 L. This fact was derived from
a) Law of gaseous volumes b) Avogadro’s hypothesis
c) Berzelius hypothesis d) Dalton’s atomic theory
10] Among the following pairs of compound the one that illustrates the law of multiple proportion is
a) NH3 and NF3 b) CO2 and CS2 c) CS2 and FeSO4 d) SnCl2, SnCl4
11] The percentage of carbon and oxygen in samples of CO2 obtained by different methods were found to
be the same. This illustrates the law of:
a) conservation of mass b) constant proportions
c) multiple proportions d) reciprocal proportions
12] One of the following combinations which illustrates the law of reciprocal proportions?
a) N2O3. N2O4, N2O5 b) NaCl, NaBr, NaI c) CS2, CO2, SO2 d) MgO, Mg (OH)2
13] Two elements X and Y combine in gaseous state to form XY in the ratio 1: 35 .5 by mass. The mass of
Y that will be required to react with 2g of X is
a) 7.1g b) 3.55g c) 71g d) 35.5g
14] Calculate the number of gram atoms in 2.3 g of sodium.
a) 23 b) 10 c) 2.24 d) 0.1
15] Calculate the mass of 2.5 gram atoms of oxygen
a) 40g b) 80.0g c) 8g d) 11.2g

44
16] Calculate the mass of 1.5 gram molecule of sulphuric acid
a) 151g b) 129.0 c) 147.0 g d) 200gm
17] 19.7 kg gold was recovered from a smuggler. The atoms of gold recovered are: (Au = 197)
a) 100 b) 6.02 x 1023 c) 6.02 x 1024 d) 6.02 x 1025
18] The molecular mass of CO2 is 44 amu and Avogadro’s number is 6.02 x 1023. Therefore, the mass of
one molecule of CO2 is:
a) 7.31 x 10-23 b) 3.65 x 10-23 c) 1.01 x 10-23 d) 2.01 x 10-23
19] What is the mass of 3.01 x 1022 molecules of ammonia?
a) 1 kg. b) 0.85g c) 2gm d) 5mg
20] The largest number of molecules is in:
a) 28 g of CO b) 46 g of C2H5OH c) 36 g of H2O d) 54 g of N2O5
21] The number of molecules in 89.6 litre of a gas at NTP are:
a) 6.02 x 1023 b) 2 x 6.02 x 1023 c) 3 x 6.02 x 1023 d) 4 x 6.02 x 1023
22] How many electrons are present in 1.6 g of methane?
a) 16 b) 6.02 x 1023 electrons c) 1.6 x 1023 d) 6.023 x 1022
23] The density of a gaseous element is 5 times that of oxygen under similar conditions. If the molecule of
the element is diatomic, what will be its atomic mass?
a) 49.0 b) 53.33 c) 80 d) 41.7
24] The gram atoms present in 5 gram of Calcium are
a) 0.125 b) 0.21 c) 0.117 d) 0.512
25] Number of motes present in 100 Kg of lime stone is
a) 104 b) 103 c) 105 d) 106
26] The number of molecules of CO2 in 4.4g of the gas at STP
a) 6.02 × 1023 b) 5.02× 1023 c) 6.02 × 1024 d) 6.02 × 1022
27] Weight of 6.02 ×1020 molecules of hydrogen is
a) 0.002g b) 0.02g c) 2g d) 0.01g
28] Which of the following pairs of gases contains the same number of molecules?
a) 11g of CO2 and 7g of N2 b) 44g of CO2 and 14g of N2
c) 22g of CO2 and 28g of N2 d) All the above pairs of gases
29] 2.0 × 1022 atoms of an element weights 6g. The atomic weight of the element is approximately
a) 290 b) 180 c) 34.4 d) 104
30] A mixture of 2 mole of H2 and 1 mole of He occupies litres at NTP.
a) 22.4 b) 44.8 c) 67.2 d) 22400
31] Which of the following will contain the same number of atoms as 20 gram of Calcium?
a) 24 gram of Mg b) 12 gram of carbon
c) 24 gram of carbon d) 12 gram of Mg

45
32] Which of the following has the smallest number of molecules?
a) 11.2 litre of SO2 gas at STP b) 1 mole of SO2 gas
c) 1× 1023 molecules of SO2 gas d) 3.2 gram of SO2 gas
33] The weight of a single molecule of a substance is 8.5  10-23 gram. The molecular weight of the
substance is
a) 51.2 b) 30.1 c) 60 × 1023 d) 14.5
34] The number of molecules present in 35.5 gram of chlorine is
a) 3.0115×1015 b) 3.0115 × 1023 c) 2.0115 × 1023 d) 6.023× 1023
35] 16 grams of a gas at STP occupies 11.2 litres. The molecular weight of the gas is
a) 23 b) 25 c) 30 d) 32
36] The total number of electrons present in 18 ml of water is
a) 6.024 10-24 b) 7.240  10-23 c) 6.023  1024 d) 6.023  1023
37] The ratio between the number of molecules in equal masses of nitrogen and oxygen is
a) 7 : 8 b) 1 : 9 c) 9 : 1 d) 8: 7
38] Which of the following pairs of gases contains the same number of molecules
a) 11g of CO2 and 7 g of N2 b) 44g of CO2 and 14 g of N2
c) 22 g of CO2 and 28 g of N2 d) All the above pairs of gases
39] Equal masses of oxygen, hydrogen and methane are kept under identical conditions. The ratio of the
volumes of the gases will be
a) 2 : 16 : 2 b) 2 : 16 : 1 c) 1 : 16 : 2 d) 1 : 1 : 1
40] 4.4 gram of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of molecules
present in the container will be
a) 6.022 × 1023 b) 1.2044 × 1023 c) 2 moles d) 6.023 × 1024
41] Out of the following the largest number of atoms are contained in
a) 11g of CO2 b) 4g of H2 c) 5g of NH3 d) 8g of SO2
42] The percentage of nitrogen in urea, (NH2CONH2) is:
a) 38.4 b) 46.6 c) 59.1 d) 61.3
43] The empirical formula of a compound is CH2 O, if vapour density is 90. Then the molecular formula is
a) CH2O b) C2H4O2 c) C3H6O3 d) C6H12O6
44] An organic compound having carbon and hydrogen has 80% carbon. The empirical formula of the
hydrocarbon is
a) CH4 b) CH3 c) CH2 d) CH
45] A compound contains 90% C and 10% H. The empirical formula of the compound is
a) C8H10 b) C15 H30 c) C3H4 d) C15H32
46] An organic compound contains C = 50% and H = 9.25%. Its empirical formula is
a) C3H6 b) C3H7O2 c) C2H4O d) C4H8O

46
47] A metal M having an atomic weight of 197 yields a chloride containing 35.1% chlorine. The empirical
formula of the compound is
a) MCl3 b) MCl c) MCl2 d) MCl4
48] When 1 gram CaCO3 is dissolved in excess dilute acid the volume of CO2 evolved at STP will be
a) One litre b) 224 ml c) 22.4 litre d) 2.24 litre
49] 8 gram of sulphur is completely burnt in a large excess of oxygen, the volume in litres of SO2 formed as
reduced to STP is
a) 5.6 b) 8.0 c) 11.2 d) 16.0
50] 0.01 mole of iododorm (CHI3) reacts with Ag powder to produce a gas whose volume at NTP is
a) 224 ml b) 112 ml c) 336 ml d) 1120 ml
51] Molarity of liquid HCl with density equal to 1.17 g/ cc is
a) 36.5 b) 18.25 c) 32.05 d) 4.65
52] How many milliliters (mL) of 1 M H2SO4 solution are required to neutralize 10 mL of 1 M NaOH
solution?
a) 2.5 mL b) 5.0 mL c) 10.0 mL d) 20.0 mL
53] How much water should be added to 200 cc of seminormal solution of NaOH to make it exactly
decinormal?
a) 200 cc b) 400cc c) 800 cc d) 600 cc
54] 100 mL of 1.0 M HCl is mixed with 75 mL of 1.0 M Na2CO3. The resulting solution will be
a) Acidic b) Basic c) Neutral d) Amphoteric.
55] The molarity of concentrated sulphuric acid (density = 1.834 g cm-3
) containing 90 % of H2SO4 by mass
is
a) 16.84 b) 1.68 c) 9.18 d) 0.918
56] Density of a 2.05 M solution of acetic acid in water is 1.02 g/ mL, the molality of the solution is
a) 0.44 mol kg-1
b) 1.14 mol kg-1
c) 3.28 mol kg-1
d) 2.28 mol kg-1
57] The volume of 0.5 M KMnO4 solution which can oxidise 20 ml of 0.2 M Mohr salt solution in acidic
medium is
a) 1.6 ml b) 3.2 ml c) 4.8 ml d) 5 ml
58] The concentration unit which changes with temperature is
a) Molarity b) Molality
c) Mole fraction d) both molality and mole fraction
59] 7 g. of N2, 8 g. of O2 and 22 g. of CO2 are present in a gaseous mixture. The mole fraction of Nitrogen
in the mixture is
a) 0.33 b) 0.63 c) 0.25 d) 0.75

47
LEVEL – II
1] The No. of moles of barium carbonate which contain 1.5 moles of oxygen atoms is
2] A molecule of Haemoglobin contains 0.33% of iron by weight. The molecular weight of Haemoglobin
is 67200. The number of iron atoms (At. wt. = 56) present in one molecule of Haemoglobin is
a) 1 b) 2 c) 3 d) 4
3] 48 gram of Mg contains the same number of atoms as 160 gram of another element. The atomic mass
of the element is
a) 24 b) 320 c) 80 d) 40
4] The number of moles of oxygen in one litre of air (21% oxygen by volume) at STP would be
5] What is correct for 10 g of CaCO3?
a) It contains 1 g atom of carbon b) It contains 0.3 g atoms of oxygen
c) It contains 12 g of calcium d) It refers to 0.1 g equivalent of CaCO3.
6] If isotopic distribution of C-12 and C-14 is 98% and 2% respectively, then the number of
C-14 atoms in 12 g of carbon is
a) 1.032×1022 b) 3.01 × 1022 c) 5.88 × 1023 d) 6.02 × 1023
7] The number of moles of water present in 100g of water of 90% purity is
a) 3 b) 18 c) 5 d) 5.5
8] Which of the gases contains the same number of molecules as that of 16 grams of oxygen
a) 16g of O3 b) 32g of SO2 c) 16g of SO2 d) All
9] The correct arrangement of the following in order of increasing mass is
I) N2 molecule II) oxygen atom III) I Avogram VI) 1×10-10 gram atom of copper
a) I > II > III > IV b) III < II < I < IV c) I > III > II > IV d) IV > I > II > III
10] 0.5 mole of a gas (Mol. Wt. 80) occupies 11.2 litres at STP. The volume occupied by 0.25 mole of a
lighter gas (Mol. Wt. 20) at STP will be
a) 11.2 lit b) 5.6 lit c) 8 lit d) 22.4 litre
11] The molecular weight of an unknown substance is found to 24000. If it contains 0.2% Magnesium,
then the number of Magnesium atoms that can be present in a molecule of it is
a) 1 b) 2 c) 4 d) 10
12] The percentage of magnesium in chlorophyll is 2.68%. The number of magnesium atoms in 2 gram of
chlorophyll is
a) 1.34 × 1021 b) 1.34 × 10–21 c) 1.35 × 10–24 d) 1.35 × 1024
13] Ordinary water contains one part of heavy water per 6000 parts by weight. The number of heavy water
molecules present in a drop of water of volume 0.01 ml is
a) 2.5 × 1016 b) 5 × 1017 c) 5 × 1016 d) 7.5 × 1016
14] The number of molecules present in a drop of water. If its volume is 0.05 ml are
a) 1.66 × 1021 b) 1.60 × 1022 c) 1.66 × 1023 d) 1.60 × 1024

48
15] Which of the following will not have a mass of 10 g?
a) 0.1 mol CaCO3 b) 1.51 × 10
23
Ca
2+
ions
c) 0.16 mol of C ions d) 7.525 × 1022 Br atom.
16] x L of N2 at STP contains 3 × 1022 molecules. The number of molecules in x/2 L of ozone at STP will
be
a) 3 × 1022 b) 1.5 × 1022 c) 1.5 × 1021 d) 1.5 × 1011
17] 10 ml of a gaseous hydrocarbon combustion gives 40 ml of CO2 and 50 ml of H2O vapour under the
same condition. The hydrocarbon is
a) C4H6 b) C6H10 c) C4H8 d) C4H10
18] 15 ml hydrocarbon requires 45 ml of O2 for complete combustion and 30 ml of CO2 is formed. The
formula of the hydrocarbon is
a) C3H6 b) C2H6 c) C4H10 d) C2H4
19] Complete combustion of a sample of a hydrocarbon gives 0.66 2 grams of CO2 and 0.362 grams of
H2O the formula of a compound is
a) C3H8 b) CH4 c) C2H6 d) C2H4
20] The simplest formula of a compound containing 50% of element x (atomic weight = 10) and 50% of
element y (atomic weight = 20) is
a) xy b) x2y c) xy2 d) x2y3
21] The empirical formula of an organic compound is CH. 6.023 1022 molecules of same organic
compound weigh 7.8 g. The molecular formula is
a) C2H2 b) C6H6 c) C2H4 d) None
22] An organic compound contains C = 21.56%, H = 4.56% and Br = 73.36%. Its molecular weight is 109.
Its molecular formula is
a) C2H5Br b) C3H7Br c) C4H8Br d) C6H6Br
23] 0.078 gram of hydrocarbon occupies 22.4 ml volume at STP. The empirical formula of hydrocarbon is
CH. The molecular formula of hydrocarbon is
a) C5H5 b) C6H6 c) C2H2 d) C8H8
24] Empirical formula of a compound is CH2. The mass of one litre of this organic gas is exactly equal to
that of one litre of nitrogen. Therefore the molecular formula of the organic gas is
a) C3H8 b) C2H6 c) C2H4 d) C3H6
25] When 1.2 g of carbon is completely burnt in 6 litres of oxygen at STP, the remaining volume of oxygen
is
a) 3.76 lit b) 2.6 lit c) 5.8 lit d) 37.6 lit
26] 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca (OH)2. The maximum number of moles of CaSO4
formed is
a) 0.5 b) 0.2 c) 0.4 d) 0.25

49
27] 70 gram of a sample of magnesite on treatment with excess of HCl gave 11.2 litre of CO2 at STP. The
percentage purify of the sample is
a) 80 b) 70 c) 60 d) 50
28] When 100 gram of ethylene polymerises to polythene according to the equation
nCH2 = CH2 - (CH2 - CH2)n - the weight of polythene produced will be
a) gram b) 100 gram c)
100
gram d) 100n gram
29] Air contains 20% by volume of oxygen. The volume of air required for the complete combustion of one
litre of methane under the same conditions is
a) 2 litre b) 4 litre c) 10 litre d) 0.4 litre
30] The hydrated Na2SO4nH2O undergoes 56% loss in weight on heating and become anhydrous. The
value of n will be
a) 5 b) 3 c) 7 d) 10
31] 1.25 g of a solid dibasic acid is completely neutralized by 25 mL of 0.25 molar Ba (OH)2 solution.
Molecular mass of the acid is
a) 100 b) 150 c) 120 d) 200
32] 0.126 g of acid requires 20 ml of 0.1 N NaOH for complete neutralization. The equivalent mass of the
acid is
a) 45 b) 53 c) 40 d) 63
33] The mole fraction of the solute in one mole aqueous solution is
a) 0.009 b) 0.018 c) 0.027 d) 0.036
34] How many grams of phosphoric acid would be needed to neutralize 100 g of magnesium hydroxide?
a) 66.7 g b) 252 g c) 112 g d) 168 g
35] Normality of solution of FeSO4. 7H2O containing 5.56 g / 200 mL which converts to ferric from in a
reaction is (Fe = 56, s = 32, O = 16, H = 1)
a) 1 b) 0.1 c) 0.01 d) 10
36] 100 g of a sample of HCl solution of relative density 1.17 contains 31.2 g of HCl. What volume of this
HCl solution will be required to neutralize exactly 5 litres of
20
N
KOH solution?
a) 25 ml b) 29.2 ml c) 34.2 ml d) 250 ml
37] 300 ml of 1 M HCI and 100 ml of 1 M NaOH are mixed. The chloride ion concentration in the resulting
solution is
a) 1 M b) 0.5 M c) 0.75 M d) 0.25 M
38] 200 ml of 1 M H2SO4, 300 ml 3 M HCI and 100 ml of 2 M HCI is mixed and made up to 1 litre. The
proton concentration in the resulting solution is
a) 1.25 M b) 1.5 M c) 2.5 M d) 0.75 M
39] 20 ml of decinormal solution of NaOH neutralises 25 ml of a solution of dibasic acid containing 3g. of
the acid per 500 ml. The Molecular weight of the acid is
a)150 b) 75 c) 225 d) 300

50
40] In a compound C, H and N are present in the ratio 9: 1: 3.5 by weight. Molecular mass of the
compound is 108. Molecular formula of the compound is: [AIEEE 2002]
a) C2H6N2 b) C3H4N c) C6H8N2 d) C9H12 N3
41] Number of atoms is 560 g of Fe (atomic mass = 56 g mol -1
) [AIEEE 2002]
a) Is twice that of 70 gN b) Is half that of 20 gH
c) Both are correct d) none is correct
42] What volume of hydrogen gas at 273 K and 1 atm pressure will be consumed in obtaining 21.6 g
elemental boron (atomic mass = 10.8) from the reduction of boron trichloride by hydrogen?
[AIEEE 2003]
a) 67.2L b) 44.8L c) 22.4 L d) 89.6 L
43] One mole of magnesium nitride on reaction with an excess of water gives [AIEEE 2004]
a) One mole of ammonia b) Two moles of nitric acid
c) Two moles of ammonia d) One moles of nitric acid
44] How many moles of magnesium phosphate, Mg3 (PO4)2 will contain 0.25 mole of oxygen atoms?
[AIEEE 2006]
a) 3.125 × 10-2
b) 1.25 × 10-2
c) 2.5 × 10-2
d) 0.02
45] In the reaction, 2Al (s) + 6HCl (aq) →2Al3+
(aq) + 6Cl-
(aq) + 3H2 (g) [AIEEE 2007]
a) 33.36 L H2 (g) is produced regardless of temperature & pressure per mole Al that reacts
b) 67.2 L H2 (g) at STP is produced for every mole of Al that reacts
c) 11.2 L H2 (g) at STP is produced for every mole HCl (aq) consumed
d) 6 L HCl (aq) is consumed for every 3 L H2 (g) is produced.
46] Volume occupied by one molecule of water (density = 1 g cm-3
) is [AIPMT 2008]
a) 9.0× 10 b) 6.023 × 10 c) 3.0× 10 d) 5.5× 10
47] How many moles of lead (II) chloride will be formed from a reaction between 6.5 g of Pbo and 3.2 g of
HCl? [AIPMT 2008]
a) 0.044 b) 0.333 c) 0.011 d) 0.029
48] 100 ml of PH3 on heating forms P and H2. The volume change in the reaction is
[DPMT 2009]
a) an increase of 50 ml. b) an increase of 100 ml.
c) an increase of 150 ml. d) an decrease of 50 ml.
49] Excess of carbon dioxide is passed through 50 mL of 0.5 M calcium hydroxide solution. After the
composition of the reaction, the solutionwas evaporated to dryness. The solid calcium carbonate was
completely neutralized with 0.1 N hydrochloric acid. The volume of the hydrochloric acid required is
(At. mass of carbon = 40) [KCET 2009]
a) 200ml b) 500ml c) 400ml d) 300ml
50] 25.3 g of sodium carbonate, Na2CO3 is dissolved in enough water to make 250 mL of solution. If
sodium carbonate dissociates completely, molar concentration of sodium ions, Na+
and carbonate ions,
are respectively. [AIPMT Prelim 2010]
a) 0.477 M and 0.477 M b) 0.955 M and 1.910 M
c) 1.910 M and 0.955 M d) 1.90 M and 1.910 M

51
51] The number of atoms in 0.1 mol of a triatomic gas is [AIPMT Prelim 2010]
a) 1.800× 10 b) 6.026× 10 c) 1.806× 10 d) 3.600× 10
52] How much time (in hours) would it take to distribute one Avogadro number of wheat grains if
10 grains are distributed each second? [Kerala 2010]
a) 0.1673 b) 1.673 c) 16.73 d) 167.3
53] For areaction A + 2B → C, the amount of C formed by starting the reaction with 5 moles of A and 8
54] A mixture of CaCl2 and NaCl weighing 4.44 g is treated with sodium carbonate solution to preciopitate
all the Ca2+
ions as calcium carbonate. The CaCO3 so obtained is heated strongly to get 0.56 g of CaO.
The percentage of NaCl in the mixture is
a) 75 b) 30.6 c) 25 d) 69.4
55] If 1 mL of water contains 20 drops, then number oif moilecules in a drop of water is [AFMC 2010]
a) 6.023× 10 b) 1.376× 10 c) 1.673× 10 d) 4.346× 10
56] In an experiment, 4 g of M2Ox oxide was reduced to 2.8g of the metal. Tf the atomic mass of the metal
is 56 g mol-1, the number of O atoms in the oxide is [AFMC 2010]
a) 1 b) 2 c) 3 d) 4
57] Which has the maximum number of molecules among the following [AIPMT Mains 2011]
a) 44g CO2 b) 48g O3 c) 8g H2 d) 64g SO2
58] The density of a solution prepared by dissolving 120 g of urea in 1000 g of water is 1.15 g/ml. The
molarity of this solution is: [AIEEE 2012]
a) 0.50 M b) 1.78 M c) 1.02 M d) 2.05 M
moles of B is [Kerala 2010]
a) 5 moles b) 8 moles c) 16 moles d) 4 moles

52
KEY
SESSION –1
CLASS EXERCISE
2. d 3. c 4. d 5. d 6. a
HOME EXERCISE
1. c 2. d 3. c 4. a 5. c
SESSION –2
CLASS EXERCISE
1. c 2. b 3. b 4. a 5. c
HOME EXCERSICE
1. a 2. d 3. d 4. c 5. a
SESSION – 3
CLASS EXERCISE
1.d 2. b 3. a 4. c 5. a 6. B
HOME EXERSICE
1. b 2. b 3. b 4. a 5. c
SESSION – 4
CLASS EXERCISE
1. b 2. a 3. c 4. c 5. c 6. b 7. B
HOME EXCERSICE
1. b 2. d 3. b 4. d 5. b 6. d
7.d 8. c 9. c
SESSION – 5
CLASS EXERCISE
1. d 2. b 3. b 4. b 5. a 6. d
7. CH2Cl 8.C4H6O4
HOME EXERSICE
1. d 2. d 3. a 4. c 5. d 6. b
SESSION – 6
CLASS EXERCISE
1. b 2. b 3. c 4. a 5. a 6. b
7. c 8. c 9. d 10. a
HOME EXERCISE
1.b 2. d 3. d 4. a 5. d 6. b

53
SESSION – 7
CLASS EXERCISE
1.b 2. 36.21,3, 108.63 3. 3,114.72 4. a
HOME EXERCISE
1. a 2. c 3. c 4. a
SESSION – 8
CLASS EXERCISE
1. 10% 2. a 3. c 4. a 5. b 6. b
7. b 8. c 9. b 10. 136 ml
HOME EXERCISE
1.c 2. a 3. d 4. c 5. b 6. c 7. d
SESSION –9
CLASS EXERCISE
1. c 2. d 3. c 4. b 5. b
6. i) 0.167 M, 0.3 N ii) 0.116N, 0.233 N
HOME EXERCISE
1. a 2. b 3. c 4. b 5. c 6. a
7. a 8. d
SESSION- 10
CLASS EXERCISE
1. a 2. b 3. b 4. b 5. 4.18 M, 5.027 m 6. 2.267 7. b
HOME EXERCISE
1. b 2. d 3. c 4. b 5.b 6. b
MAIN
LEVEL – I
1) d 2) b 3) c 4) a 5) b 6) d 7) b 8) a
9) b 10) d 11) b 12) c 13) c 14) d 15) a 16) c
17) d 18) a 19) b 20) c 21) d 22) b 23) c 24) a
25) b 26) d 27) a 28) a 29) b 30) c 31) d 32) d
33) a 34) b 35) d 36) c 37) d 38) a 39) c 40) b
41) b 42) b 43) d 44) b 45) c 46) b 47) a 48) b
49) a 50) b 51) c 52) b 53) c 54) b 55) a 56) d
57) a 58) a 59) c

54
LEVEL – II
1) c 2) d 3) c 4) d 5) b 6) a 7) c 8) b
9) b 10) b 11) b 12) a 13) c 14) a 15) c 16) b
17) d 18) d 19) a 20) b 21) b 22) a 23) b 24) c
25) a 26) b 27) c 28) b 29) c 30) d 31) d 32) d
33) b 34) c 35) b 36) a 37) c 38) b 39) a 40) c
41) c 42) a 43) c 44) a 45) c 46) c 47) d 48) a
49) b 50) c 51) c 52) b 53) d 54) a 55) c 56) c
57) c 58) d
********

55
HOME EXERCISE
SESSION – 01
1] c
Sol: Diamond Graphite – Allotropies of carbon only ozone – 18 molecules also of oxygen but silica -
not comes under elemental group
2] d
Sol: Homogenous mixtures which we cannot differentiate by naked eye i.e, mixture of gas (or) mixture of
compounds have same physical properties
3] c
Sol: 294.406
280.208
------------
398.614≅398.6 rounding of value for 0.614
Because here one less than 5 so we write it as 398.6
4] a
Sol: In given solution mixture except + all are solution
Here is prespitate
5] c
Sol: 7.00g i.e, Y is better representation than X
SESSION – 02
1] c
Sol: Law of multiple proportions: when two elements combine to form two or more compounds is
simple numerical ratio in given reactions. Oxygen will combine with 12gr of Carbon CO and in
the ratio 1:2
2] d
Sol: + → +
Here , combines to form , which reflects Law of multiple proportions
3] d
Sol: The isotopes of Carbon , doesnot follows Laws because in mixure ratio of , is not
fixed
4] c
Sol: At S.T.P condition any one mole gas will occupies 22.4L.60gr Nitrogen Oxide will give 22.4L but
in those compounds Oxygen will have fixes proportions i.e, 1:2 ratio
i.e, Law of multiple proportions.
not by proportion from another Oxide, it follows Law of Constant Proportions
5] a
Sol: CO,

56
SESSION – 03
1] b
Sol: molecular weight of is greater(double) than , P container has more molecular than Q of
constant condition(1 liter,300K,1atm)
2] b
Sol: At NTP 1 mole has gas occupies 22.4L
? gas occupies 1 =1000L
No. of mole of gas at NTP in 1 vessel
=
.
≅44.6mole
3] b
Sol: 80gr of has 2.5 × 6.023 × 10 atoms
= = 2.5mole
Similarly = = 2.5 have same number of atoms
4] a
Sol: 18gr of water has 1N molecules
18mg of water has=
×
=10
5] c
Sol: Mwt of Sulphur =32
MWt of Carbon Sulphur and Carbon= =
SESSON – 04
1] b
Sol: 98 mg of have =
×
× 6.023 × 10
=6.023 × 10 molecules
If 3.01 × 10 molecular are left then
=(6.023 − 3.01) × 10 molecules
=3.22 × 10
=0.5 × 10
2] d
Sol: 5L of gas at STP=
.
=0.22 mole of
0.5 gr of =
.
=0.25moles
10gr od = =0.31 moles
15L of gas of STP=
.
= 0.67

57
3] b
Sol: 1 mole of Ozone( ) it has 3 moles of oxygen atoms
4] d
Sol: Each Ethanol contains ( − ) 6 atoms
0.04gr of Ethanol has=
.
× 6 × 6.023 × 10 H-atoms
=0.031 × 10 H- atoms
5] d
Sol: For any element atomic number less than mass number
X<Y
6] d
Sol: one mole contaions=6.023 × 10 atoms of C
=2 × 6.023 × 10 atoms of O
7] d
Sol: molecular weight of ( ) = 27 × 2=5
32 × 3 = 96
64 × 3 = 192
-------
342
( ) =342
50gr of substance has= =0.14mole
8] c
Sol: per 1 of alloy which has weight8.17g has
. × .
=4.4gr of Iron
For (210 × 20 × 15) of alloy contains
=4.47 × 210 × 20 × 15 of iron
=281610 gr of Iron
Atoms of Iron in given volume is × 6.03 × 10
=3.03 × 10 atoms of Iron
9] c
Sol: Constituents of Pyrex glass
. . . . .
Dividing the % composition of each element by their respective mass
. . . . .
0.18 0.021 0.0061 0.004 1.345
Dividing with small proportions
47.25 5.25 15.32 1 336.25
Ratio of Boron and Silicon atom=
× .
.
≅

58
SESSION – 05
1] d
Sol: .
∆
→ +
It loses 55.9% of weight in the form of
So remaining 44.1% of substance contains
44.1% contains 142gr
55.9% contains =
.
× 55.9
=179.9≅180gr of water molecules
∴ . = =10
2] d
Sol: Molecularity Nitrogen is 2 only
3] a
Sol: for every 100gr of compound 63.70 of Mg is present
Remaining substance is 36.3gr oxygen is present
That present is
.
≅2.27 proportions
4] c
Sol: 1.625gr of anhydrous means=0.01mole
After hydration with water is has (2.857-1.625)=1.23gr
of weight in form of molecules
that is =
.
≅ 0.07 mole which 7 times than that of (0.1 moles)
so hydrated as 7
5] d
Sol: In given composition 6.2% of hydrogen made molecules with Oxygen as molecules
So, weight of hydrogen in given substance for 322gr of complex is
. ×
= 19.96
19.96 gr of hydrogen can approximately forms 10 moles of water molecules
Our crystalized compound 10
6] b
Sol: properties of C H O
40 6.67 53.3
Divided with 6.67 .
.
.
.
.
And divides with
Corresponding M.wts
1
= ( ) Mwt=30
Total weight of compound is 180
n= =6
( ) ≅

59
SESSION – 06
1] b
Sol:
∆
→ + ↑
1 mole 1 mole 1 mole
By heating , escaped as a gas, so lost of weight due to remaining only
.
mole of loss
.
× 44gr of weight by evolves gas
Loss of weight=
.
× 44=5.5gr
2] d
+ → 2
1 mole 1 mole 2 mole
10ml(12.5-10)=2.5ml 2 × 10
Here will remain 2.5ml with 20ml of Hcl
3] d
Sol: ( ) + 2 → + ( )
At S.T.P conditions 20ml of reacts with40ml of , but here 20ml of oxygen is available which
can form 10ml of , 10ml of ( )
( ) + 2 ( )
→ ( ) + ( )
20ml 20ml
(20-10) ml 0 10ml
=10ml
So, total volume of gas is 20ml
4] a
Sol: 2 + → 3 + 2
2 mole 1mole 3 mole
But 11.2L of of NTP is represents mole of . Then it gives only 1.5 moles of sulphur.
5] d
Sol:
In a given reaction 1 mole CaO will form 1 mole weight of residence(CaO) is 28
Moles of CaO will form 0.5mole of weight of =0.5 × 100=50gr
6] b
Sol: → +
1mole 1 mole 1.5mole
.
.
=0.2mole 0.2mole 0.3mole
So 24.5gr of will give 0.3 moles of .
0.6 mole of required to completely react with to form water
2 + → 2
To evolve 0.6 mole of we need to take 0.6mole of Mg
+ 2 → +
Weight 0.6mole of Mg=0.6 × 24=14.4gram

60
SESSION – 07
1] a
Sol: → +
1 gr of salt of chloride have 0.835gr of chlorine vapour density od salt=170
Molecular weight of salt=170
170gr of salt of chloride will give=170 × 0.835gr of
1 mole of salt of chloride will give=
× .
= 2 moles of chlorine
∴ = 2 ⟹n=4
=
2] c
Sol: General formula
M-32.33% O-67.66%
.
= 2 n=
.
≅ 4
≅
M+(35.5)4=158
M=158-148
=10
3] c
Sol: Number of equivalents of metal = no of equivalents of oxygen
=
=
∴Equivalent weight of metal=
_
× 8
4] a
Sol: Molecular formula=
Valency of metal=+3
Atomic mass=valency×equivalent mass
=3 × 9
=27
SESSION – 08
1] c
Sol: 10ml of decimal solutions means
It has 10 × 0.1=1mole of solute
[∴ = =10 moles of solute
2] a
Sol: M= ×
.
× =0.1M

61
3] b
Sol: =
=decamolar=10M = = 0.1
=10ml =?
= =
×
.
=1000ml
4] c
Sol: =
=
( )( )( )( ) ( )( )
= =
=2.33M
5] b
Sol: Lets take x lit of 1MHCl and remaining (2-x)lit of 5M HCl is required to get 2 liters of 2M HCl
=
( )( ) ( )
=2
X+10-5x=4
6=4x
x=1.5liters
volume of 1M HCl is (x)=1.5liters
6] c
Sol: from the given data
weight = ( )( )
=
×
= 10.6
x=10.6y
a] y=0.05 x=10.6 × 0.05=0.53≠2.12
b] y=0.2 x=10.6 × 0.2=2.12≠1.06
c] y=0.1 x=10.6 × 1=1.06
d] y=0.1, x=1.06≠2.12
7] d
Sol: to complete reaction 0.10.18gr of NaOH we have to add
To get 0.18gr of NaOH the volume 0.15M solutions is
0.15=
.
×
V=
.
× .
= =30.6ml
SESSION – 09
1] a
Sol: No. of milli equivalents= × ( )
=0.1 × 10 = 1
2 1
Volume to be added = V - V = 1000 - 10 = 990 mL

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