Chemistry deals with the composition, structure and properties of matter.

The basic
constituents of matter are atoms and molecules so chemistry is called the science of
atoms and molecules.
In this chapter, we will discuss about matter and its properties alongwith discussion of
constituent particles of matter.

SOME BASIC
CONCEPTS OF
CHEMISTRY
|TOPIC | ‘
ine CHAPTER CHECKLIST
General Introduction, Nature of Matter Yition General Introduction| and
and Laws of Chemical Combination ___ Nature of Matter and Laws
of Chemical Combination
Chemistry is the branch of science that deals with structure, properties, e Atomic and Molecular
interaction, transformation of a matter and the energy changes accompanying Masses, Mole Concept and
these transformations. It plays a central role in science and is often intertwined Formulae of Compounds
with other branches of science. It also plays an important role in daily life. ¢ Chemical Reactions and
Importance and Scope of Chemistry Chemistry has helped us to meet with all their Equations and
our requirements that are necessary for the advancement of life such as food | _ Stoichiometric Calculations _|
additives, health care products, plastics, dyes, polymers, etc. fe
Some of its applications are listed below.
(i) Chemistry is important for the development of commercially important
polymers and understanding their impacts on environment.
(ii) Manufacture of fertilizers, drugs, soaps, alkalis, salts, dyes, metals, alloys and
other inorganic and organic chemicals including the new materials’ in
chemical industries made a big contribution towards national economy.
(iii) Chemistry helps in the synthesis of many life saving drugs such as cisplatin
and taxol (effective in cancer therapy), AZT or azidothymidine [effective
for AIDS (acquired immuno deficiency syndrome) victims].
(iv) Better understanding of the chemical principles leads to the production of
optical fibres, conducting polymers etc., and large scale miniaturisation of
solid state devices.
(v) Several advances in automotive and aerospace engineering rely on
improved materials such as ceramics, which are synthesised and
engineered in chemistry.
 |Allxnone | CHEMISTRY Class 11th
2
(vi) Chemistry was crucial for understanding the Physical Classification of Matter
formation of ozone hole and remains crucial to Depending upon their physical characteristics, matter is
understand the environment impact of many generally classified as solids, liquids and gases.
produced materials and how to render them safe.
|_Physical classification
(vii) Safer alternatives to environmentally hazardous
substances such as CFCs (chlorofluorocarbons,
responsible for ozone layer depletion), branched
Liquids
chain detergents (responsible for water pollution),
have been successfully synthesised with the (i) Solids The particles are held very close to each other
knowledge of chemistry. in an orderly fashion and there is no much freedom
However, advancement in chemistry also lead to of movement.
several problems such as, evolution of green house (ii) Liquids The particles are close to each other in an
gases like methane, carbon dioxide, oxides of orderly fashion but they can move around.
nitrogen, which are responsible for global warming;
(iii) Gases The particles are far apart as compared to
air and water pollution (because of the addition of
foreign substances into air and water from chemical
present in solids and liquids states and their
industries or due to human activity), etc. which are
movement is easy and fast.
the great challenges for future chemists. The three states of matter are interconvertible under
different conditions of temperature and pressure.
NATURE OF MATTER Sublimation

Matter is anything that has mass and occupies space

e.g. book, pen, pencil etc. Melting (heat Evaporation (heat)
Solid
Another term widely used in chemistry is material but it Freezing (cool) Condensation (cool)
has a limited meaning as it corresponds to matter having
specific uses, Deposition
e.g. glass, cement, paper, etc.
Note
Thus, all materials are matter but all matters are not material.
(i) Difference between gas and vapour is that vapours represent a
gaseous state of a substance which is liquid at room
Classification of Matter temperature. A substance which is in gaseous state of room
There are two ways to classify matter, viz temperature is called gas but not vapour. e.g. ammonia is a
gas (not vapour) but water on heating forms water vapours.
(i) Physical classification based upon the physical (iia= Besides above, the two other states of matters are also known.
properties These are named as plasma and Bose-Einstein

(i) Chemical classification based upon the chemical

composition.
SU NP ape ee
matter (i.e.
solid, liquid and gas) are considered.

Differentiating Properties of Solids, Liquids and Gases

Characteristic Solids Liquids Gases

Arrangement of Particles are held very close to each Particles are close to each Particles are far apart as compared
particles other in an ordered manner. other but not so ordered as solids. to that of solids and liquids.
Movement of Particles are not free to move. Particles can move around. The movement of particles is easy
particles and fast.
Compressibility Hard and incompressible due to More compressible due to more Most compressible due to large
and hardness close packed structure. empty space in structure. interparticle empty space.
Volume Definite y Definite Not definite
Shape Definite Not definite, take the shape of the Not definite, completely occupy the
2 Ny eae 2 ee container in which they are kept. container in which they are placed.
Diffusion They do not diffuse. They show slow diffusion. They show fast diffusion.
Some Basic Concepts of Chemistry

Chemical Classification of Matter

Depending upon the chemical composition, at the macroscopic or bulk level, matter is classified as

Matter
(have mass and occupy space)

Pure substance Mixture

(have fixed composition) (have variable composition)

Elements Compounds Homogeneous mixtures

Heterogeneous mixtures
(contain only one (contain more than (uniform composition
(non-uniform composition
kind of particles, one type of particles throughout)
throughout)
that can be atoms but in fixed ratio) e.g. salt solution,
e.g. colloids, alloys etc.
or molecules) sugar solution, air etc.
Metals Inorganic compounds
(mineral origin)
Non-metals
Metalloids
Organic compounds
(animal and plant origin)

A brief description for classification of matter is given below Depending upon the physical and chemical properties,
(i) Mixture the element are further divided into three categories.
Two or more substances mix together in any ratio to give I. Metals Metals are those elements which reflect light
and hence possess lustre. They are good conductor of
mixture. e.g. air, milk, tea etc.
heat and electricity, malleable and ductile (can be
A mixture can be homogeneous or heterogeneous. drawn in wire). At room temperature, metal exist as
(a) Homogeneous mixture A mixture having uniform solid e.g. silver, aluminium.
composition throughout is called homogeneous IT. Non-metals Non-metals are those elements
mixture. e.g. salt solution, air etc. which do not reflect light and hence do not
(b) Heterogeneous mixture A mixture in which the possess lustre. They are brittle, poor conductor of
composition is not uniform throughout and heat and electricity (except graphite).
different components can be observed is called II. Metalloids Metalloids are those elements which
heterogeneous mixture. e.g. mixture of salt and possess the characteristics of both metals as well
sugar, cereals and pulse etc. as non metals. e.g. bismuth, antimony etc.
(ii) Pure substance (b) Compounds A molecule of a compound is formed
They have fixed composition and their constituents by the combination of two or more atoms of
cannot be separated by physical methods. e.g. gold, different elements in a fixed ratio by mass.
silver, water etc. Compound may be categorised into two types.
A pure substance can be an element or a compound. Pure I. Organic Compound Those compound which
substances are always homogeneous. So, homogeneous contains carbon and a few other element like
mixture is also called ‘solution’. hydrogen, nitrogen, sulphur etc. These are
(a) Elements An element is defined as a pure substance originally obtained by plants and animals.
that contains only one kind of particle which may be II. Inorganic compound Those elements which
an atom or a molecule. contains any two or more element out of more
than 114 elements known so far.
4A |Allgzone | CHEMISTRY Class 11th

(i) Fundamental Units

Properties of Matter and
In this system there are 7 base units, called the fundamental units,
their Measurement which are independent and cannot be derived from any other units.
Each substance is associated with some unique set of These units along with their symbols are tabulated below.
characteristics, which are collectively called the Base Physical Quantities and their Units
properties of matter.
Base physical Symbol for Name of Symbol for SI
The properties of matter are categorised into two quantity quantity Slunit unit
types.
. Length ] metre St ee
(a) Physical Properties Mass ‘ola trhaed m si kilogram _kg
Those properties which can be measured or observed Time . ae i t %second 2
without changing the identity or the composition of
Electric current ; / ampere A
the substance are called physical properties. e.g. colour,
; Thermodynamic temperature T kelvin K
odour, melting point, boiling point, density etc.
Amount of substance n mole mol
(b) Chemical Properties
Luminous intensity ky candela cd
Those properties which require a chemical change
to occur for their measurement or observation are The definitions of the SI base units are given in the following
known as chemical properties. e.g. characteristic table .
reactions of different substances ; acidity, basicity,
Definitions of SI Base Units
ccombustibility etc.
Units of length metre The metre is the length of the path travelled
Unit of Measurement by light in vacuum during a time interval of
1/299 792 458 of a second.
Several properties of matter like length, area, volume
Unit of mass kilogram The kilogram is the unit of mass, it is equal
etc. are quantitative in nature and are called
to the mass of the international prototype of
physical quantities. In order to measure these the kilogram.
quantities some standard reference is chosen, which
Unit of time second The second is the duration of 9 192 631 770
is called the unit of measurement. Thus, periods of the radiation corresponding to the
“A unit is defined as the standard of reference chosen transition between the two hyperfine levels of
to measure a physical quantity”. the ground state of the caesium 133 atom.

Each physical quantity is represented by a number Unit of electric ampere The ampere is that constant current which, if
current maintained in two straight parallel conductors
followed by unit in which it is measured. e.g. the
of infinite length, of negligible circular
length of a pen is 10 cm, here 10 is a number and cm cross-section and placed 1 metre apart in
denotes centimetre in which, the unit length is vacuum, would produce between these
measured, conductors a force equal to2 x 107’ newton
per metre of length.
The simple numerical figure 10 does not convey any
meaningful information. Thus, it is essential to add Unit of kelvin The kelvin, unit of thermodynamic
thermodynamic temperature, is the fraction 1/273.16 of the
units with every experimental value.
temperature thermodynamic temperature of the triple
There were two different systems of measurement, point of water.
called the English system and the metric system. The Unit of amount mole (i) The mole is the amount of substance of a
metric system was originated in France and more of substance system which contains as many elementary
convenient but now-a-days SI (The International entities as there are atoms in 0.012 kilogram of
System of Units) system is widely accepted. carbon-12; its symbol is “mol”.
(ii) When the mole is used, the elementary
The international System of Units entities must be specified and may be atoms,
molecules, ions, electrons, other particles, or
(SI Units) specified groups of such particles.
Units of candela The candela is the luminous intensity, in a given
This system was established by 11th General Juminous direction, of a source that emits monochromatic
Conference on Weights and Measurements and in intensity radiation of frequency 540 x 10'* hertz and that
French is called “Le systeme International d’ unites. has a radiant intensity in that direction of 1/683
watt per steradian.
pone
Some Basic Concepts of Chemistry
5
Derived Units The various prefixes are listed in the following table.
The other physical quantities such as speed, volume, density »Prefixes Used in SI System
etc., can be derived from the above given base quantities and
are called derived quantities and their units are derived from Multiple Prefix Symbol Multiple Prefix Symbol

mass, ™
fundamental units. e.g. density, d = ———-—
10! deci d 10! deca da
volume, V dates eek ee a aR, ena
and volume = length x breadth x height = (length) > [for a cube] 10-3 milli m- 1 —— Kile * k
density = 7 105. micro m “408 mega MM
(length)?
1079) Sa haine n 10" giga G
On putting the units, we get density,
10-2 pico p ae
he ies 104 femto f ifthe peta P
Therefore, the unit of density is kgm = 10ne atto a 10'8 exa E
Some commonly used derived quantities and their units are 104! zepto z 407! zeta Z
tabulated below. 10°*4 ~—-yocto y faee yotta y
Physical
Quantity Formula Unit Symbol
Note
Area Length square Squaremetre mm? Though, base unit for the mass is kilogram, yet prefixes are used
with gram as in kilogram, kilo is already a prefix.
Volume Length cube Cubic metre mo :
Density Mass/volume Kilogram per
cubic metre
kgm Some Physical Quantities
Velocity Distance/time Metre per second ms7! (a) Mass and Weight
Acceleration Speed change/ Metre per ms~2 Mass (m) is the amount of matter present in a substance. It
time second per remains constant for a substance at all the places. Its unit
second (SI unit) is kilogram (kg) but in laboratories usually gram
Force Mass x Newton N =kgms> (1 kg= 1000 g) is used. Analytical balance or electrical
acceleration balance is used to measure the mass of a substance.
Pressure Force/area Pascal (Newton Pa =Nm~ Weight (w) is the force exerted by gravity on an
per square =kgm's@
metre) object. It varies from place to place due to change in
gravity. Its unit is Newton (N).
Work, energy Forcex distance Joule J=Nm=kgm? s?”
w=mXxg
Frequency Cycles/sec Hertz Hz =s"
where, m = mass
Electric charge Current x time Coulomb C=As
g = gravity.
Potential —— Volt V =kgm? s°A
difference =JAoS. = JG (b) Volume
Electric Potential ohm Q =VA" The space occupied by matter (usually by liquid or a
resistance difference/
current gas) is called its volume.
Volume = (length)? =m?
Supplementary Units Thus, SI unit of volume is m°. However, in laboratory,
These include radian (rad) for angle and steradian (Sr) for solid smaller volumes likecm? (or mL i.e. millilitre) ordm?
angle. This class of unit was created in 1960 to cover those units (or L ice. litre ) are used. These terms are correlated as
which are neither base units nor derived from base units. 1 L= 1000 mL = 1000 cm* =1 dm’
1 m? = (100 cm)? = 10° cm? = 10° dm? = 10° L
Prefixes
Some physical quantities are either too small or too large. To In the laboratory, volume of liquids or solutions can be
change the order of magnitude, these are expressed by using measured by graduated cylinder, burette, pipette etc.
prefixes before the name of the base units.
 |Allxnone |CHEMISTRY Class 11th
6
EXAMPLE |1] Convert the following into basic units. 373 K 100°C Boiling point 212°F
(ii) 15.15 ps of water
(i) 28.7 pm
(iii) 25365 mg
Sol. s-x- The basic units for length is meter (m), for time Human body
9 is second (s) and for mass is kilogram (kg). 310K 37°C} temperature 98.6°F
-12 298 K 25°C Room HA
(i) 287 pm x te = 287 x 107!'m temperature
1pm
a Freezing point oF
10 15 oe of water os
(ii) 1515us x =1515 X10 s sy
lus

iii) 25365mg x Beton. EN =o 5369107 ke

“) 3 1000 mg 1000 g
Kelvin Celsius Fahrenheit
(c) Density
Thermometers using different temperature scales
It is defined as the amount or mass per unit volume and
has units kg m~°> orgcm >. (Here kg or g represents “dl@ =2( — 32) or Fae C@)+32
mass and m? or cm” represents volume).
Density = Mass/Volume
EXAMPLE |3] Convert the following temperatures
EXAMPLE {2| A liquid has a volume of 49.0 cm? anda into degree Fahrenheit.
mass of 57.642 g. Find out the density of this liquid in SI (i) 25°C, physiological (human body) temperature.
unit. [HOTS] (ii) 35°C, the room temperature.
Mass (kg)
Sol. Density = Sol. (i) Given, C = 25°C
Volume (m?)
Mass = 57.642 g = 57.642x10° kg oF a2 +325 2x 25+32=45+32=77°F
5
Volume =49.0 cm? = 49.0 x(10*)? = 49.0 x 107° m? (ii) Given, C=35°C
57.642 x10° 9 9
“. Density = — = 1176 x 10° kg/m* Pe a eee
49.0 x 107

(d) Temperature EXAMPLE |4] At what temperature will both the

celsius and Fahrenheit scales read the same value?
It is defined as the degree of hotness or coldness. Several
scales are used to measure temperature, common three Sol. Suppose both read the same value as x.
scales are °C (degree celsius), K (kelvin) and °F (degree Then as "c=? (°F-32)
Fahrenheit).
x =5/9(x —32) or 9x =5 x -—160
These scales are shown in the adjacent figure.
or 4x =—160 or x= —40-
From the adjacent figure, it is clear that on the kelvin
scale freezing point of water is taken as 273.15 K and EXAMPLE |5) A measured temperature on Fahrenheit
boiling point of water as 373.15 K, at normal scale is 200°F. What will this reading be on celsius scale?
atmospheric pressure. [NCERT Exemplar]
Similarly, on the celsius scale, freezing point of water is Sol. There are three common scales to measure temperature
taken as 0°C and boiling point of water is taken as 100°C. °C (degree celsius), °F (degree Fahrenheit) and K (kelvin).
Thus, both the scales are divided into 100 equal parts. The K is the SI unit.
The temperatures on two scales are related to each other
The relationship between kelvin and celsius scale is
Oe 275 or he Cre 275.85 by the following relationship ° F = 2 iC 32
5
Practically speaking, the lowest temperature permitted in Putting the values in above equation
nature is — 273.15°C (0 K). This temperature is known as
200 ~32= =#°C => 2 CH 168
absolute zero. Similarly, the celsius and Fahrenheit scales 5
are related as
=> 1° = HBX «933°
Some Basic Concepts of Chemistry

(e) Length (v) (6.8 x 10~ = (1.4 x 107°)

Though the SI unit of length is metre, yet it is very common (vi) (4.56 x 10? + 2.62 x 10°)
to express length in Angstrom (A) or nanometre (nm) or
picometre (pm). These are related to SI unit as follows (vii) (9.87 x 10°? — 2.26 x 10°“)
Sol. (i) 0.000968=968 x 10* (ii) 157428 = 1.57428 x 10°
1A=10!° m, lnm = 10? m, 1 pm = 107? m
(iii) 90,000 = 9 x 104
(iv) (5.7 x 10°) x (4.2 x 107) = (57 x 4.2) (10°7)
UNCERTAINTY
= 2394 x 10* = 2394 x10
IN MEASUREMENT
In the study of chemistry we deal with (v) (68x10°) + (14 x10°°) = (*)x (10°)
the experimental 14
data as well as theoretical calculations. There are several = 4857 x10°
ways by which the numbers are handled conveniently and (vi) (4.56 x 10° + 2.62 x10”)
the data is presented realistically with certainty to the
= 45.6 X10 + 262 10° = (456 + 2.62) x 10°
possible extent. These ways are scientific notation,
significant figures and dimensional analysis. = 48.22 x 10° = 4.822 x 10°
(vii) (987 x 10° — 2.26 x10°*)
Scientific Notation = 987 X10-° — 0.226 x 10 °= (987 — 0.226) x 10>
It is the greatest way to represent very small or very large = 9644 x10°
numbers conveniently. In scientific notation, any number
can be represented in the form N X10” where 7 is an EXAMPLE |7| Calculate
exponent having positive or negative values and N is a
(i) (5.7 x 107%) x (4.2 x 107%)
number (called digit term) which varies between 1.000...
ang 9.999... % (ii) (5.7 x 107°) + (4.2 x 107)
(i) Calculations involving multiplication and division Sol. (i) Given, (5.7 x10 x (4.2 x 107°)
These two operations follow the same rule which are = (5.7 x 4.2) x (10-4 x10);
there for exponential numbers i.e. the number NV = 2394x104? = 2394 x10°
before the factor 10” in scientific notation are (ii) Given, (57x10) + (4.210)
multiplied and the exponents of 10 are added up.
(5.7 +4.2)x10°- ™) = 23.94 x 107
While, in division, the N are divided and the
exponents (7) are subtracted.
(ii) Calculations involving addition and subtraction For
Accuracy and Precision
these two operations, first the numbers are written in Accuracy is the agreement of a particular value to the true
such a way that they have some exponent. After that, value of the result or it is a measure of the difference
107 is taken out as common and the coefficients are between the experimental value or the mean value of a set of
added or substracted as the case may be. measurements and the true value.

Thus, in general number (JV) is written in scientific Accuracy = Mean value — True value
notation as number with a single non-zero digit to the left
Smaller the difference between the mean value and the true
of decimal and the number of places decimal point has
value, the larger is the accuracy. Accuracy also expresses the
moved is the exponent (7) of 10 with (+) sign if moved left
correctness of measurement. When the true value of a
and (—) sign if moved right. quantity is not known then it becomes very difficult to
e.g. 943.876 is written as 943876 x 107 in scientific calculate the accuracy. In such cases, precision of the
notation i.e. decimal is moved two places towards left so measurements is calculated.
that only one non-zero digit is left and the number of places Precision refers to the closeness of various measurements
moved (2) is the exponent of 10 in scientific notation. for the same quantity or it is expressed as the difference
between a measured value and the arithmetic mean value
EXAMPLE |6| Express the following in the scientific
for a series of measurements.
notation.
(i) 0.000968 (ii) 157428 |Precision = Individual value — Arithmetic mean value
(iii) 90,000 (iv) (5.7 x 10°) x (4.2 x10“)
 | Allxwone | CHEMISTRY Class 1th
8
Smaller the difference between the individual values of Thus, “the total number of digits in a number including the
repeated measurements, the greater is the precision. last digit whose value is uncertain is called the number of
* Accurate results are generally precise but precise results need not be significant figures.” e.g. there are 4 significant figures in
accurate. 3.200 but only two significant figures in 3200.
e.g. if true value of the mass of a given salt is 10.5 g and four Note
students reported the result of their three measurements as The number of certain digits depends upon the precision of the
follows. instrument used for the purpose.

Measurement (g) 1 2 3 Average (g) Rules for Counting the Number

~ StudentA 103 104 105 10.4 of Significant Figures
Student B 10.0 10.1 10.2 10.1 The following rules are followed to count the number of
~~ StudentC «104 106~=—«10.5 105 significant figures in a particular number.
Student D 10.8 10.7 ; 10.9 ey 10.8 _
(i) All non-zero digits are significant. e.g. in 285 cm,
there are three significant figures and in 0.35 mL,
(i) Measurement by student A has not so good accuracy there are two significant figures.
and good precision. : (ii) Zeros between two non-zero digits are significant.
(ii) Measurement by student B has poor accuracy but e.g. there are four significant figure in 1007.
good precision. (iii) Trailing zeros are significant only if the digit contains
(iii) Measurement by student C is both accurate and a decimal point. eg. 1.560 have 4 significant
precise. numbers whereas 450 has 2 significant figures. Such
(iv) Measurement by student D has poor accuracy and numbers are better represented in scientific notation.
poor precision. (iv) Leading zeros are not significant because their
EXAMPLE |{8| Two students performed the same purpose is simply to place the decimal point.
experiment separately and each one of them recorded e.g. In 0.05, there is only one significant figure
two readings of mass which are given below. Correct whereas 0.0060500 have five significant figures.
reading of mass is 3.0 g. On the basis of given data, (v) Defined values or counting numbers have infinite
what would you infer about the accuracy and precision number of significant figures, as these are exact
in the readings of students A x B ? [NCERT Exemplar]
numbers and can be represented by writing infinite
Students Readings number of zeros after placing a decimal.
(i) (ii) Ke. 2= 2.000000
A 3.01 2.99 or 20= 20.000000
B 3.05 2,95 (vi) Numbers written in scientific notation, all digits are
significant e.g. 4.01x 107 has three significant
Sol. Average of readings of student,A = EHS BEY figures, and 8.256 x 10> has four significant figures.
2

Average of readings of student,B =

3.05 + 2.95 _ 3.00 Significant Figures in Operational
Calculations
Correct reading = 3.00. For both the students, average
value is same as the correct value. Hence, readings of (1) In case of addition and subtraction
both are accurate. Readings of student A are close to When the two or more digits are added or substracted the
each other (differ only by 0.02) and also close to the result cannot have more digits to the right of the decimal
correct reading, hence, readings of A are precise also. point than either of the original numbers.
But readings of B are not close to each other (differ by
0.1) and hence are not precise. e.g.
(1)
Significant Figures 6.611 Each number has three
decimal places, so the
Significant figures are meaningful digits which are known 9.234 answer has to be reported
with certainty and the uncertainty is indicated by writing the +2.021 up to three decimal places.
Actualcurn 13.866 The significant figures in
certain digits and the last uncertain digit. e.g. if we write a
each of the terms is 4 but
measurement as 49.6 g, we say that 49 is certain and 6 is Reported sum 13.866 the result has 5 significant
uncertain and the uncertainty would be +1 in the last digit. figures.
Some Basic Concepts of Chemistry

The first number has 2 ‘ (ii) If the rightmost digit to be removed is less than 5,
decimal places and the
the preceeding number is not changed. e.g. 4.334 if
~11.02534 Second has 5 decimal
Actual eaacn
s e nce. SO Ih € answer is 4 is to be removed, the result is rounded off to 4.33.
difference ee reported up to 2 decimal (iii) If the rightmost digit to be removed is 5, then the
Reported
lace only. preceeding number is not changed if it is an even
8.23 ‘ Z
difference _ number but it is increased by one if it is an odd
number.
(11) In case of multiplication and division e.g. if 6.35 is to be rounded by removing 5, we
In these operations, the result must be reported with have to increase 3 to 4 giving 6.4 as the result.
no more significant figures as are there in the However, if 6.25 is to be rounded off, it is rounded
measurement with the least significant figures, e.g. off to 6.2.
41.012 The first term has 5 Note
ignificant figures,
gures, while In multiple step calculations, do not round off until the end, however
Actual product x 1.21 signi
49.62452 the second term has 3 keep significant figure in mind.
significant figures, so the
Reported product 49.6 result can have only 3 EXAMPLE |10| Round up the following upto three
significant figures.
significant figures.
The presence of exact numbers in an expression does
(i) 34.216 (ii) 10.4107
not affect the number of significant figures in the (iii) 0.04597 (iv) 2808
answer because exact numbers have infinite number Sol. (i) 34.216
of significant figures. The digit to be removed is less than 5, therefore the
preceeding digit is not changed.
4.28 0.146 3
= 44.84784. ”. 34.216 rounded upto three significant figures = 34.2
0.0418
(ii) 10.4107
Correct answer is 44.8.
The digit to be removed is less than 5, therefore the
preceding digit is not changed.
EXAMPLE |9| How many significant figures should be 10.4107 rounded up to three significant figure
present in the answer of the following calculations? =104
(i) 0.02856x 298.15x 0.112 (iii) 0.04597
0.5785 The digit to be removed is more than 5, therefore the
(ii) 5x 5.364 preceeding digit is increased by one.
(iii) 0.0125 + 0.7864+ 0.0215 [NCERT Textbook] . 0.04597 is rounded up to three significant figures
= 0.0460
Sol. (i) Least precise number of the calculation is 0.112.
Number of significant figures in the answer = Number (iv) 2808 .
of significant figures in the least precise number = 3. The digit to be removed is more than 5, therefore the
(ii) Least precise number of calculation = 5.364. preceeding digit is increased by one.
Number of significant figures in the answer 2808 rounded up to three significant figures
= Number of significant figures in 5.364= 4. = 2810

(iii) Since, the least number of decimal places in each term

is four, the number of significant figures in the EXAMPLE |11] Express the result of the given
answer is also 4. calculation to the appropriate number of significant
figures.
Rounding off the Numbers 3.24 x 0.08666
The numbers are rounded off to limit the result of a 5.006
mathematical operation to required number of significant Sol. cer ON 0.0560883 (Actual result)
figures. While rounding off the numbers, keep following 5.006
points in the mind. As 3.24 has least number of significant figures, viz 3, the
(i) If the rightmost digit to be removed is more than 5, result should contain 3 significant figures only.
the preceeding number is increased by one. e.g. Hence, the result will be reported as 0.561
1.386. If we have to remove 6, we have to round (after rounding off).
Ol 16 10 1.59.
 | Alléwone | CHEMISTRY Class 11th
lO
EXAMPLE |13| What is the mass (in grams) of a
Dimensional Analysis
copper block whose dimensions are 5.0 inch x 6.0 inch and
In calculations, many of the times it become necessary to convert
whose density is 8.96 g/cm 3? Given that 1 inch = 2.54cm.
units from one system to another. This is achieved by factor label 2.54cm_ linch
method or unit factor method or dimensional analysis. The Sol Here, unit conversion factors are1=
linch 2.54 cm
dimensions of a derived quantity are the powers to which the
basic quantities have to be raised in a product defining the Hence, required mass (in g) = 5.0 inch x 6.0 inch x 4.0 inch
quantity. Dimensional analysis involves calculations based on 2.54cm é 2.54 cm 3 2.54 cm y 8.96 E176x104 g
the fact that if two quantities have to be equated, they must 1 inch inch linch 1cm
have the same dimensions or the same units.
For the application of this method, conversion factors are LAWS OF CHEMICAL
required which are mentioned in the given table.
COMBINATIONS
Important Conversion Factors Elements combined to form molecules in accordance to the
1 ft = 12 inch TA = 10"Th following five basic laws, called the laws of chemical
1yd=3ft Pre = 10" b combinations.
1 mile = 5280 ft 1dm?=1L
1. Law of Conservation of Mass
1 inch = 2.54 cm tom? = 1079 L
1 m= 39.37 inch 1ft?
= 28.32 L This law was put forth by Antoine Lavoisier in 1789.
According to this law, “matter can neither be created nor
1 mile = 1.609 km 1kg=10° g
destroyed but one form of matter can be converted into
1 cal = 4.186 J 1 kHz
= 10° Hz other.” This law is also known as law of indestructibility of
1 MHz
= 10° Hz matter. The other statement for this law is “In all physical
1 minute = 60s terg=10°"J and chemical changes, the total mass of the reactants is
1 hr = 60 minute 1 eV= 1.6022x 1079 J equal to that of the products.”
1 day
= 24 x 60 x 60s 1 amu = 1.66053x 10°” kg i.e. sum of masses of reactants = sum of masses of products.
= 931.48 MeV
1mg=10°°g EXAMPLE |14| If 6.3 g of NaHCO, are added to
15.0 g of CH,COOH solution, the residue is found to
1 metric ton = 10° kg 1 atm= 760 mm Hg= 76 cm Hg
weigh 18.0 g. What is the mass of CO, released in the
1 Ib= 453.6 g = 760 torr reaction ?
1 HP= 746 W 1 bar = 10° Pa Sol. NaHCO, + CH,COOH —>
1 dyne =10°° N 1 atm = 101.325 kPa
1 kg = 2.2 pound (Ib) 1 Latm = 101.325 J H,0+ CH;COONa + CO,
1 g = 0.0353 ounce (0) Residue
Sum of the masses of reactants =6.3+ 15= 21.3 g
EXAMPLE |12| How many seconds are there in 2 Sum of the masses of products = x +18
days? 21.3= x +18 x = 213-18 = 3.3 g

Sol. Step I Here, we know 1 day = 24 hours (h) Thus, the mass of the CO, released is 3.3 g.
id
or een, | = sd ey yaml
24h 1 day
ie 2. Law of Definite Proportions
60 min This law was given by, a French chemist, Joseph Proust in
or =
|=——_
60 min 1h 1799. According to this law, “a given compound always
so, for converting 2 days to seconds. contains exactly the same proportion of elements by
i.e. 2 days Lae, = ,...seconds weight.” It is sometimes also referred to as law of definite
Step II and III The unit factors can be multiplied in composition. e.g. pure water obtained from any source
series in one step only as follows. (well, river, lake or sea) or any country (India, Russia,
24h 60min 60s America etc.) will always be made up of only hydrogen and
2 dayx Ne x
1 day lh 1 min oxygen elements combined together in the same fixed ratio
= 2X 24X 60X 60s = 172800 s. of 1 : 8 by mass.
Some Basic Concepts of Chemistry II
EXAMPLE |15| Copper oxide was prepared by the EXAMPLE |16| Three oxides of nitrogen contained
following methods. 63.6%, 46.7% and 30.4% nitrogen respectively. Show that
(i) In one case, 1.75 g of the metal was dissolved in nitric these figures illustrate the law of multiple proportions.
acid and igniting the residual copper nitrate yielded
Sol. In case first,
2.19 g of copper oxide.
Step 1 The oxide of nitrogen contains 63.6% N
(ii) In the second case, 1.14 g of metal dissolved in nitric
i.e. 63.6 g of N reacts with (100— 63.6) g
acid were precipitated as copper hydroxide by adding
of O= 36.4 g of O.
caustic alkali solution. The precipitated copper
hydroxide after washing, drying and heating yielded Step II .. 1g ofN will react with SATE
63.6
1.43 g of copper oxide.
of O=0.57g of O.
(iii) In the third case, 1.46 g of copper when strongly
heated in a current of air yielded 1.83 g of copper In case second,
oxide. Show that the given data illustrate the law of Step I The oxide of nitrogen contains 46.7% N
definite composition. ie. 46.7g of N reacts with (100 — 467) g
of O=53.3g of O.
Sol. Step 1 andII Inthe first experiment,
2.19 g of copper oxide contained 1.75 g of Cu. Step II .. 1g of N will react with aL:
- 100 g of copper oxide contained of O=1.14
gof O.
UNAS) ‘
Cu=— X 100 =79.91 g,ie.% of Cu=79.91 In case third,
219
Step 1 The oxide of nitrogen contains 30.4% N
In the second experiment,
i.e. 30.4 g ofN reacts with (100 — 30.4) g
1.43 g of copper oxide contained 1.14 g of copper. ofO = 69.6 g of O.
. 100 g of copper oxide contained
Step II ..1g of N will react with a g of
Cu=1* x 100= 79.72 g,
1.43 O= 2.26 g of O
Le. %of Cu= 79.72 Step III This means the ratio of the masses of oxygen
In the third experiment, which combine with 1 g of nitrogen is 0.57: 1.14:
1.83 g of copper oxide contained 1.46 g of copper 2.26, i.e. 1: 2: 4. is obviously in accordance with
the law of multiple proportions.
. 100 g of copper oxide contained

ene
1.83
100= 79.78 g, 4, Law of Reciprocal Proportions
ie. % of Cu= 79.78
This law was given by Richter in 1792. It states that, “the
Step III The percentage of copper in copper oxide derived
ratio of the masses of two elements A and B which combine
from all the three experiments is nearly the same. separately with a fixed mass of the third element C is either
Hence, the above data illustrate the law of the same or some simple multiple of the ratio of the masses
definite composition. in which A and B combine directly with each other” This
can be illustrated by the following examples.
3. Law of Multiple Proportions EXAMPLE |17]| The element H and O combine
This law was proposed by Dalton in 1803. It states that “if separately with the third element S to form H,S and SO,
two elements can combine to form more than one respectively, then show that they combine directly with
compound, the masses of one element that combine with a each other to from H,0.
fixed mass of the other element, are in the ratio of small
whole numbers”. e.g. two compounds, water and hydrogen
peroxide are formed when hydrogen combines with oxygen. H,S SO;
Hydrogen + Oxygen— Water
2¢g log 18 g
Hydrogen + Oxygen > Hydrogen peroxide
2¢g 32g 34 g Sol. As shown in figure, the masses of H and O which
combine with the fixed mass of S, i.e. 32 parts are 2 and
Here, the masses of oxygen (i.e. 16 g and 32 g) that 32 i.e. they are in the ratio 2 : 32 or 1:16. When H and O
combine with a fixed mass of hydrogen (i.e. 2 g) bear a combine directly to form H,O, the ratio of their
simple ratio, i.e. 16 : 32 or 1: 2. combining , masses are 2:16 or 1: 8.
 |AllZwone | CHEMISTRY Class 11th
[2
He thus put forward the hypothesis known as Avogadro’s
it al
These ratio are related to each other as F : F = OY)
hypothesis, which states that equal volumes of gases at
i.e. they are simple multiple of each other. the same temperature and pressure should contain equal
number of molecules.
EXAMPLE |18| The percentage composition of
Applications of Avogadro’s law are
elements in NH, H,0 and N,0; is given below.
NH, ————> 82.35% N and 17.65% H (a) Calculation of Atomicity of
H,0O ———> 88.90% 0 and 11.10% H Elementary Gases
N,0, ———— 63.15% 0 and 36.85% N Atomicity of an elementary substance is defined as the number
Show that these data are in accordance with the law of of atoms of the element present in one molecule of that
reciprocal proportion. eae substance,
Sol. (i) NH, 1 part of H reacts with = aoe 4.67 part of N e.g. atomicity of oxygen (O) is two while that of ozone
(O;) is three.
; 88.90
(ii) H,O 1 part of H reacts with = ae =8.01 part of O e.g. Hydrogen+Chlorine ——> Hydrogen chloride
1 vol 1 vol 2 vol
Thus, the!ratio Ne Oe: 4/67 = 8:01 1): 1:72
(iii) NO, (By experiment)
Here, N and O reacts with each other in the ratio n molecule n molecule 2n molecules
N: O:: 3685 :6315 =1:1.71. Thus, the two ratios are (By Avogadro’s law)
the same. Hence, it illustrates the law of reciprocal
proportions. x molecule i molecule 1 molecule
2 2
(On dividing throughout by 27)
5. Gay Lussac’s Law of Gaseous Volumes
It implies that one molecule of hydrogen chloride gas is
As the name implies this law was given by Gay Lussac in
1808. He stated that “when gases combine or are produced in made up of rymolecule of hydrogen and : molecule of
a chemical reaction they do so in a simple ratio by volume
chlorine.
provided all the gases are at same temperature and pressure”.
e.g. one volume of hydrogen and one volume of chlorine always Hence, 5 molecule of chlorine =1 atom of chlorine
combine to form two volumes of hydrogen chloride gas.
. 1 molecule of chlorine =2 atoms. So, its atomicity is 2.
lvol 1 vol 2 vol Note
Since, a molecule is made up of two or more atoms, so 1/2 molecule
The ratio between the volumes of the reactants and the
is possible and the molecule may contain one or more atoms.
product in this reaction is simple, i.e. 1: 1: 2
Hence, it illustrates the law of combining volumes. (b) Finding the Relationship between Mass
and Volume of a Gas
EXAMPLE |19| N, reacts with H, to form NH. If 10 L Molecular mass = 2 X vapour density
N, reacts with 30 L H, to form 20 L NH; under same = 9 ye MASS of certain volume of the gas at STP
conditions of pressure and temperature. Show that the
data is in accordance with law of gaseous volumes. mass of same volume of H, at STP
Sol. The reaction.taking place is N, + 3H,-——> 2NH,
ye mass of 1L of the gas at STP

_ The ratio of Ny :H,: NH, ::10:30: 20ie.1:3:2.

mass of 1L of H, at STP
Since, the ratio is simple and whole number, thus law of yz mass of 1L of the gas at STP
gaseous volume is proved. 0.089 ¢
= 22.4x mass of 1L of the gas at STP
6. Avogadro’s Law = mass of 22.4 L of the gas at STP
Avogadro found that the smallest particle of a gas which Thus, the weight of 22.4 L of any gas at STP is equal to the
can exist independently is the molecule, not the atom,
molecular mass of the gas expressed in gram.
therefore, the volume of a gas must be related to the
number of molecules (rather than atoms) present in it. This is called Gram-Molecular Volume (GMV) law.
Vapour density is also known as relative density of the gas.
Some Basic Concepts of Chemistry 13
DALTON’S ATOMIC THEORY But the discoveries in the beginning of 20th century by Sir J J
The concept that matter is composed of very tiny particles Thomson, Rutherford, Neils Bohr and others, resulted in a
was given by Indian and Greek philosophers. The Greek reviewed and modified theory.
philosopher Democritus (460 — 370 BC) suggested that The main drawbacks of Dalton’s atomic theory are
the ultimate particle of a matter is atom. The word ‘atom’ (i) It could explain the laws of chemical combination by
has been derived from the Greek word ‘atomos’ meaning mass but failed to explain the law of gaseous volumes.
indivisible. The old ideas were put on a scientific scale by (ii) It could not explain why atoms of different elements
John Dalton in the form of a theory, known as Dalton’s
have different masses, sizes, valencies etc.
atomic theory.
(iii) It failed to explain the nature of forces that bind
Main postulates of this theory are as follows. together atoms in a molecule.
(i) Matter consist of minute, indivisible, indestructible (iv) It could not explain how and why atoms of different
particles, called atoms. elements combine with each other to form compound
(ii) Atoms of the same element are identical to each atoms or molecules.
other. They have the same mass and size. (v) It did not make any distinction between ultimate
(iit) Atoms of different elements differ in properties and particle of an element or a compound.
have different masses and sizes.
(iv) Compounds are formed when atoms of different Modified Atomic Theory
elements combine with each other in simple The following are the modified views of Dalton’s atomic
numerical ratios and the relative numbers and kind theory.
of atoms are always the same in a given compound. (i) The atom is no longer supposed to be indivisible. It is
(v) Chemical reactions involve reorganisation of not a simple particle but a complex one.
atoms. These are neither created nor destroyed in a (ii) Atoms of the same element may not necessarily have the
chemical reaction. same mass but have the same atomic number and show
similar chemical properties (discovery of isotopes).
Explanation of the Laws of Chemical
(iii) Atoms of the different elements may have the same
Combination by Dalton’s Atomic Theory mass but they always have different atomic numbers
Dalton’s theory could explain the laws of chemical and differ in chemical properties (discovery of isobars).
combination. (iv) Atoms of one element can be transmuted into atoms
(i) Law of Conservation of Mass of other element (discovery of artificial
transmutation).
Matter is made up of atoms [postulate (i)] which can
(v) In certain organic compounds, like proteins, starch,
neither be created nor destroyed [postulate (v)]. Hence,
cellulose, etc., the ratio in which atoms of different
matter can neither be created nor destroyed.
elements combine cannot be regarded as simple. There
(ii) Law of Constant Composition are a number of compounds which do not follow the
law of constant proportions. Such compounds are called
law of multiple proportions and law of reciprocal
non-stoichiometric compounds.
proportions follows directly from postulate (v).
(vi) The mass of atom can be changed into energy
Limitations of Dalton’s Atomic according to Einstein’s equation, E = mc
Theory where, E = energy
m = mass
Dalton’s atomic theory gave a powerful initiative to scientists
in the beginning of the 19th century. and c = velocity of light.
 TOPIC PRACTICE 1|
OBJECTIVE Type Questions Readings of student A are close to each other (differ only
by 0.02) and also close to the correct reading, hence,
|1 Mark| readings of A are precise also. But readings of B are not
close to each other (differ by 0.1) and hence are not precise.
1. Which of the following statements about a
compound is incorrect? [NCERT Exemplar] The result reported in the following
(a) A molecule of a compound has atoms of different multiplication of significant figures,
elements 25% 1,25=3.125 should be
(b) A compound cannot be separated into its constituent (a) 3.125 (b) 3.1 (c) 3.12 (d) 3.10
elements by physical methods of separation
Sol (b) 2.5x 1.25= 3.125
(c) A compound retains the physical properties of its
constituent elements Since, 2.5 has two significant figures, the result should
(d) The ratio of atoms of different elements in a not have more than two significant figures thus, it is 3.1.
compound is fixed Which set of figures will be obtained after
Sol. (c) A compound is a pure substance containing two or rounding upto three significant figures 1.386,
more than two elements combined together in a fixed 4.334, 2.808?
proportion by mass and which can be decomposed into (a) 1.39, 4.34, 2.809 (b) 1.39, 4.33, 2.81
its constituent elements by suitable chemical methods. (d) 1.39, 4.34, 2.80
(c) 1.38, 4.34, 2.800
A measured temperature on Fahrenheit scale is Sol. (b) 1.39, 4.33 and 2.810
200°F. What will this reading be on celsius scale?
[NCERT Exemplar] If the density of a solution is 3.12 gm’, the mass
(a) 40 °C (b) 94°C (c) 93.3°C — (d) 30°C of 1.5 mL solution in significant figures is ......... :
[NCERT Exemplar]
Sol. (c) The temperatures on two scales are related to each
other by the following relationship. (a) 4.7 g (b) 4680 x 10° °g
9 (c) 4.680 g (d) 46.80 g
hime Hagea,

Sol. (a) For a solution, Mass = volume x density

9 168 x
a Le = | Cp= ca = 933°C =1.5 mLx 3.12g mL’ > 468g
The digit 1.5 has only two significant figures, so the
Two students performed the same experiment
answer must also be limited to two significant figures.
separately and each one of them recorded two So, it is rounded off to reduce the number of significant
readings of mass which are given below. Correct figures. Hence, the answer is reported as 4.7 g.
reading of mass is 3.0 g. On the basis of given
data, mark the correct option out of the Which of the following reactions is not correct
following statements [NCERT Exemplar] according to the law of conservation of mass?
(a) 2Mg(s) +O,(g)——> 2MgO(s) [NCERT Exemplar]
Students Readings
(b) CsHs(g)+ O2(g)—> CO,(g)+H,O(g)
iste seee Bel) {) sac tl) (c) P,(s) +50, (g¢) —> P,0,9(s)
A 3.01 age
(d) CH4(g)+ 202(g)—> CO,(g)+ 2H,O(g)
B 3.05 2.95 Sol. (b) In this equation,
(a) Results
precise
of both the students are neither accurate nor
.
C3H,(g)+ O2(g) —> CO,(g)}+ H,O(g)
44g 32g 44 ¢ 18g
(b) Results of student A are both precise and accurate ie. mass of reactants # mass of products.
(c) Results of student B are neither precise nor accurate Hence, law of conservation of mass is not followed.
(d) Results of student B are both precise and accurate
Which of the following statements indicates that
Sol. (b) Average of readings of student, A =
3.01 + 2.99 _ 3.00 law of multiple proportion is being followed?
[NCERT Exemplar]
Average of readings of student, B =
3.05 + 2.95 _ 3.00 (a) Sample of carbon dioxide taken from any source will
always have carbon and oxygen in the ratio 1: 2.
Correct reading = 3.00 (b) Carbon forms two oxides namely CO, and CO, where
For both the students, average value is close to the masses of oxygen which combine with fixed mass of
correct value. Hence, readings of both are accurate. carbon are in the simple ratio 2:1. °
Some Basic Concepts of Chemistry

(c) When magnesium burns in oxygen, the amount of

AgNO, + NaCl —-> AgCl+ NaNO,
magnesium taken for the reaction is equal to the
xg 5.85 g 14.35 g 8.5 g
amount of magnesium in magnesium oxide formed.
x+5.85= 14.35+ 85 or x=17.0¢g
(d) At constant temperature and pressure, 200 mL of
hydrogen will combine with 100 mL oxygen to Two substances X and Y combine to give a
produce 200 mL of water vapour. substance Z. The process is exothermic and Z has
Sol. (b) The masses of oxygen combine with a fixed mass of properties different from those of X and Y. Is the
carbon (12 parts) in CO, and CO are 32 and 16 substance Z, an element, a mixture or a compound?
respectively. These masses of oxygen bear a simple ratio Give explanation to support your answer.
of 32 : 16 or 2: 1 to each other.
Sol. The substance Z is a compound. This is because
This is an example of law of multiple proportion. I. heat is evolved during the formation of Z.
II. the properties of Z are different from those of X and Y.
VERY SHORT ANSWER Type Questions
How can you prove that red oxide of copper is
|1 Mark| not an element?
1. How is the term material different from matter? Sol. On heating red oxide of copper in the presence of
Sol. Anything which has mass and occupies space is called hydrogen, its mass decreases and it forms shining
matter. However, material corresponds to the matter metallic copper which is chemically different from the
which has specific uses e.g. glass, cement etc. original compound.
Cu,0--Hy >, 2Cu +50
Be What is the SI unit of mass? How is it defined?
Suppose the length of a cardboard has been
[NCERT Textbook]
reported to be 31.24 cm. What is the minimum
Sol. SI unit of mass is kilogram (kg). It is equal to the mass of
international prototype of the kilogram. uncertainty implied in this measurement?
. The minimum uncertainty implies in this measurement
What do you mean by significant figures? is + 0.01.cm.
Refer to page 8. [NCERT Textbook] What is the difference between 5.0 g and 5.00 g?
Note Sol. Though the two values seem to be equivalent but
All measured quantities are reported in such a way that only scientifically they are different. 5.0 g has two significant
the last digit is uncertain (usually by + 1). figures and hence, its precision is 0.1 part in 5, i.e. 20 ppt
5.00 has three significant figures and hence, its precision
How many significant figures should be present is 0.01 parts in 5 , i.e. 2 ppt. Hence, 5.00 g is more precise
in the answer of the following calculation? measurement than 5.0 g.
2.5 1.25~x 3.5
11. Why is the law of Gay Lussac’s not obeyed if any
2.01 [NCERT Exemplar] reactant or product is not a gas?
Sol. Least precise term i.e. 2.5 or 3.5 has two significant Sol. If any reactant or product is a liquid or solid, the volume
figures. Hence, the answer must have two significant occupied by them is extremely small as compared to the
figures. gas and hence, the law is not obeyed.
PMs h ey5a pol)
= 5.4415 =5.4
2.01 SHORT ANSWER Type I Questions
If the speed of light is 3.0x 10° ms", calculate |2 Marks
the distance covered by light in 2.00 ns. 1. Match the following with their multiples.
[NCERT Textbook] [NCERT Textbook]
Sol. Time 2.00 ns = 2.00 10°’s (.:1ns=10~” s) Prefixes Multiples
Distance covered = speed x time (i) micro 10°
= 30x 10° ms~ x 2.00X 10 s= 0.6m i) deca lS
What mass of silver nitrate will react with 5.85 g (ii) mega ; 10°
of sodium chloride to produce 14.35 g of silver (iv) giga ‘ 1002
chloride and 8.5 g of sodium nitrate, if the law of
conservation of mass is true? (vy) femto y (CRs Oats
Sol According to law of conservation of mass,
Sol. Micro =10~°, deca = 10, mega =10°, giga=10’ and
mass of reactants = mass of products.
femto =10”. [2]
16 |Allénvone | CHEMISTRY Class 11th

The digit 1.5 has only two significant figures, so the

2. The water level in a metric measuring cup is answer must also be limited to two significant figures.
0.75 L before the addition of a pebble weighing 150 So, it is rounded off to reduce the number of significant
g. The water level after submerging the pebble is figures. Hence, the answer is reported as 4.7 g. [1]
0.82 L. Determine the density of the pebble.
Sol. The volume displaced by the pebble 1.375 g of cupric oxide was reduced by heating in
= 0.82— 0.75= 0.07 L= 70 mL {1] a current of hydrogen and the weight of copper
that remained was 1.098 g. In another
Mass of the pebble = 150 g
experiment, 1.179 g of copper was dissolved in
Therefore, density of the pebble is
the nitric acid and the resulting copper nitrate
Mass _ 150
Density = = 2114
gmL {1] was converted into cupric oxide by ignition.
Volume 70
The weight of cupric oxide formed was 1.476 g.
3. Express each of the following in SI units. Show that these results illustrate the law of
(i) 14 pound per square inch ‘constant composition.
(atmospheric pressure)
Sol. Refer to example 15.
(ii) 100 mile per hour
(iii) 5 feet 2 inch 5.975 g of the higher oxide of metal gave 5.575 g
(iv) 150 pound
of lower oxide on heating. The quantity of the
lower oxide gave 5.175 g of metal on reduction.
Sol. (i) 1 pound per square inch = 6894.76 Nm ~~
Prove that these results are in accordance with
“. 14 pound per square inch
the law of multiple proportions.
2
= 14 6894.76= 96526.64 Nm ~
3 Sol. As 5.575 g of lower oxide on reduction gives 5.175 g of
6X 0.444 ms bi the metal, the mass of the oxygen is 5.575—5.175 =0.4g
(ii) 1 mile/h = Lox 10m
00s 0:4
For 1 ¢ of metal, mass of oxygen is meer: = 0.077 g
2 100 mile/h = 44.4 m/s
(iii) 5 feet 2 inch = 1.5748 m In case of higher oxides, mass of metal is 5.175 g
[.. 1 feet = 12 inch, 1m = 39.37 inch] Mass of oxygen is 5.975 —5.175= 0.8 g [1]
(iv) 1 pound = 0.454 kg 0.8
For 1 g of metal, mass of oxygen is = 0.155¢
*, 150 pound = 68.1 kg [2 x 4] 6 YS 5.175
For a given mass of metal, the ratio of oxygen is
4. For precious stone, carat is used for specifying 0.777 : 0.155 or 1: 2. Hence, the law of multiple
its mass. If 1 carat = 3.08647 grains (a unit of proportion is proved. [1]
mass) and 1 gram = 15.4324 grains. Find the total
mass in kilogram of a ring that contains 0.700 8. If ten volume of dihydrogen gas react with five
carat diamond and 5.00 gram gold. volume of dioxygen gas, how many volume of
Sol. Finding mass of diamond in kg. water vapour would be produced? [NCERT Textbook]
1 carat = 3.08647 grains Sol. 2H,(g)+O02 (g) —> 2H,0(g)
. 0.700 carat = 0.700 x 3.08647 grains = 216 grains 2V 1V 2V
Also, 1 gram = 15.4324 grains According to Gay Lussac’s law of gaseous volumes,
1 2 volume of dihydrogen react with 1 volume of dioxygen
lee rain = =0.064799 prams 1
P 15.4324 6 tH to produce 2 volume of water vapour. Therefore, 10
and 2.16 grains = 216 x 0.064799 = 01399 grams volume of dihydrogen on reaction with 5 volume of
dioxygen will produce 10 volume of water vapour.
1 gram =10 °/g 301399 g =1399 x 10° kg [2]

Total mass of the ring in kg One of the statements of Dalton’s atomic theory
=1399 x10 kg +5 x10 °kg =1899 x 10 kg (1) is given below “compounds are formed when
atoms of different elements combine in
5. Ifthe density of a solution is 3.12 g mL !,the a fixed ratio”.
mass of 1.5 mL solution in significant figures Which laws of chemical combination is not
will be [NCERT Exemplar] related to this statement? [NCERT Exemplar]
Sol. Given that, density of solution = 312 gmL"! Sol. Law of conservation of mass and Avogadro’s law
Volume of solution = 1.5 mL because law of conservation of mass is simply the law of
For a solution, Mass = volume x density indestructibility of matter during physical or chemical
=1.5mLx 3.12g
mL '=4.68¢ [1] changes.
Some Basic Concepts of Chemistry 17
Avogadro law states that equal volumes of different Sol. (i) (144.3 m*)+ (2.54 mx 8.4 m)
gases contain the same number of molecules under
similar conditions of temperature and pressure. 2.54 mX 8.4 m= 21.336 or 21 m*
{2]
10. (upto 2 significant figures)
Air contains 20% oxygen by volume. Calculate
144.3 m?
the theoretical volume of air which will be
+ 21 m?
required for burning completely 500 m? of
165.3 m?
acetylene gas. .
(As 144.3 contains (1) one digit (SF) after the decimal
All volumes are measured under the same
point.) (1]
conditions of temperature and pressure.
Sol. (ii) (4.05x 10° mL)— (0.0225x 10” mL)
2C,H, +50, — 4CO, +2H,0() =4.0275x 10° mL or 4.03 10° mL (1)
2 vol 5 vol 4 vol 2 vol (upto two decimal place as in 4.05)
(Gay-Lussac’s Law) (iii) (3.50 x 10” cm) x (4.00 x 10° cm)
or 1 vol 5/2 vol 4/2 vol 2/2 vol
= 14.0x 10° cm? (upto 3 significant figures) [1]

500 m? 5/2 x 500cm? 4/2 x 500m? 2/2 x 500ms 2. If two elements can combine to form more than
one compound, the masses of one element that
500m? =: 1250 m8 1000m? 500 m? combine with a fixed mass of the other element,
are in whole number ratio. [NCERT Exemplar]
[1] (i) Is this statement true?
Thus, 1250 m® oxygen is required for burning 500 m? of (ii) If yes, according to which law?
_ acetylene. But the percentage of oxygen in air is 20% (iii) Give one example related to this law.
“. Volume of air required =1250 x 100/20= 6250 m° [I] Sol. (i) Yes [1/2]
11. Pressure is determined as force per unit area of (ii) According to the law of multiple proportions. [1]
the surface. The SI unit of pressure, pascal is as (iii) H,+ O, —> H,0O
shown below 1 Pa=1Nm~ 2g 16g 18g {1/2]
If the mass of air at sea level is 1034 g cm™,
iO, > HO
Dry Syhis 34g {1/2]
calculate the pressure in pascal. Here, masses of oxygen, (i.e. 16 g in H,O and 32 g in
[NCERT Textbook]
H,0,) which combine with fixed mass of hydrogen (2 g)
‘Yy- ¢ The Sl unit of mass is kilogram and of length is are in the simple ratio i.e. 16:32 or 1: 2. [1/2]
? metre so convert gram to kilogram and cm? tom? as
3. Abox contains some identical red coloured
pressure is to be calculated in SI units.
balls, labelled as A, each weighing 2 g. Another
+ Write 1kg ms = 1Nand1 Nm = 1Pato obtain
box contains identical blue coloured balls,
pressure in pascal. labelled as B, each weighing 5 g. Consider the
Sol. Pressure is the force or weight per unit area. combinations AB, AB», A,B and A,B3 and shows
1034 gx 9.8 ms * that law of multiple proportion is applicable.
Pressure = 5 [-. F = ma] (1) [NCERT Exemplar]
1cm

_ 1034 kgx 100x 100x 9.8ms ” a: Combination Mass of A(Q) Mass of B(g)
1000 m? AB 2 re 5
(. IN=1kgms”)
=101332.0 Nm~
= 1,01332X 10° Pa [1]
Lateaig Wan > fe alt
AB a
SHORT ANSWER Type II Questions ABs 4 15
|3 Marks| (1)
Mass of B which is combined with fixed mass of A (say 1
1. Perform the following calculations and express the g) will be 2.5 g, 5 g, 1.25 g and 3.75 g in AB, AB,, A,B
results to proper number of significant figures. and A, B, respectively. They are in the ratio 2:4: 1:3
(i) 144.3 m* + (2.54 mx 8.4 m) which is simple whole number ratio. Hence, the law of
multiple proportion is applicable. [2]
(ii) (4.05 x 10? mL) — (0.0225x 107mL)
(iii) (3.50 x10? cm) (4.00x10° cm)
18 |Alléwone |CHEMISTRY Class 11th

= 4.35 atoms of oxygen. {1]

4. 45.4 L of dinitrogen reacted with 22.7 L of
Hence, ratio of M : Oin the second oxide
dioxygen and 45.4 Lnitrous oxide was formed. {1/2]
904. see OL Ais
The reaction is given below
. Formula of the other metal oxide is M,O3. [1/2]
2N2(g)+ O(g) —> 2N,0(g)
Which law is being obeyed in this experiment? 7. 10 mL of H, combine with 5 mL of O, to form
Write the statement of the law. [NCERT Exemplar] water. When 200 mL of H, at STP is passed over
Sol. 2N,(g) + On(g) —> 2N,0(g) heated CuO, the CuO loses 0.144 g of its weight.
2V 1V 2 V; Does the above data correspond to the law of
45.4L —> 22.7L —> 45.4L constant composition?
454 9 227 45.4 Sol. In the second experiment 0.144 g weight is lost from CuO.
2237) 22.7 22
This is due to the reduction of CuO into Cu. In other
Hence, the ratio between the volume of the reactants
words, 0.144 g oxygen combined with 200 mL H). [1]
and the product in the given question is simple
ie. 2:1: 2. It proves the Gay Lussac’s law of 32 g oxygen occupies 22400 mL volume at STP.
gaseous volumes. (2] 0.144
. 0.144 g oxygen will occupy = 22400 x ey
For Gay Lussac’s law of gaseous volumes : Refer page 12.
[1] = 100.8 mL O,
It means the ratio of H, and O, in water is
Describe what you need to do in the laboratory 200 : 100.8 = 2: 1. The same ratio is in first case
to test (i) the law of conservation of mass, (ii) the (10 : 5 or 2: 1). Thus, the law of constant
law of definite proportion and (iii) the law of composition is proved. [1]
multiple proportions.
Sol. (i) To test the law of conservation of mass, a reaction LONG ANSWER Type Questions
would have to be carried out in which the mass of the
reactants and the mass of the products are weighed |5 Marks|
and shown to be the same. {1]
l. The following data are obtained when
(ii) The law of definite proportions could be shown by
dinitrogen and dioxygen react together to form
demonstrating that no matter how a compound is
obtained, the reactants remain at the same proportions different compounds.
by mass. This can be done by decomposing a Mass of dinitrogen Mass of dioxygen
compound and showing that the masses of the
elements present are always in the same ratio. {1] (i) 14g 16g
(iii) To test the law of multiple proportions, two different (ii) 14g 32g
compounds made up of the same elements would have to
(iii) 28g 32g
be decomposed. Ifthe mass of one of the elements is kept
constant the masses of other elements combining with (iv) 28g 80g
that of the element in different samples would have to be
in the small whole number ratio. [1] (i) Which law of chemical combination is
Two oxides of a metal contain 27.6% and 30.0% obeyed by the above experimental data?
Give statement.
of oxygen respectively. If the formula of the first
oxide is M ,O,, find that of the second. (ii) Fill in the blanks in the following conversion
(a) Lkm=,..mm=...pm
Sol. In the first oxide, oxygen = 27.6
(b) 1 mg =... kg =... ng
Metal = 100— 27.6= 72.4 parts by mass.
As the formula of the oxide is M,O,, it means (ce) LmLecL=.cdm: [NCERT Textbook]
72.4 parts by mass of metal = 3 atoms of metal and Sol. (i) On fixing the mass of dinitrogen as 28 g, the masses of
4 atoms of oxygen = 27.6 parts by mass. dioxygen combined are 32, 64, 32 and 80 in the given
In the second oxide, oxygen = 30.0 parts by mass and four oxides. These are in the simple whole number
ratio ie. 2:4: 2:5. Hence, the given data obey the law
metal = 100 — 30 = 70 parts by mass.
of multiple proportions. [2]
But 72.4 parts by mass of metal = 3 atoms of metal.
Law of multiple proportions Refer to text on page 11.
. 70 parts by mass of metal = vi x 70 atoms of metal
4 (ii) (a) km =1 kmx ano x 100 cm 0 a = 10°mm
= 2.90 atoms of metal {1) 1 1m lcm
Also, 27.6 part by mass of oxygen = 4 atoms of oxygen. 1km = Lane Om me = 10” pm
4
. 30 part by mass of oxygen = ae x 30 atoms of oxygen
ikm 10m
“1km= 10° mm= 10% pm {1]
Some Basic Concepts of Chemistry
2
(b) 1 mg = 1 mgx —8 — x bE = 10 “kg Sol. Let x mole of oxalic acid and y mole offormic acid be
1000 mg 1000¢
heated with conc. H,SO, according to the following
equations.
1 mg = 1 mgx 1g pees cioing
1000 mg {0.7 COOH H,SO,/heat

“. 1mg = 107° kg= 10°ng (1]

| cop) Feo FHOw
COOH xmol x mol
x mol
=1 mixe—
areal = 19731,
| 1000 mL HCOOH
H,SO, /heat
==» CO (g)+H,O0 (1) [2]
1 dmx 1dmx 1 dm y mol y mol
1 mL= 1cm* = 1cm*> x
10 cmxX 10 cmx 10 cm Total moles of gaseous mixture = Moles of CO+ Moles
= 107° dm? .imL= 10°L = 10 dm? [1] of CO, = (x + y) mol+ x mol = (2x + y) mol
Now, KOH absorbs only CO, ie. x moles and the
2. Define the law of multiple proportions. Explain
volume of the solution decreases by 1/6th of its volume.
it with two examples. How does this law point to Since equal volume of gases have equal number of moles
the existence of atoms? [NCERT Exemplar] according to Avogadro’s law,
Sol. Refer to the text on page 11. moles of CO, x 1
= =— 2
moles of both the gases (2x+y) 6 Hi
A mixture of oxalic acid and formic acid is
heated with concentrated H,SO, and the gas or 6x=24x+y
evolved is collected. On treating the solution or 4x=y
with KOH, the volume of the solution decreases or Ya4
by 1/6th. Calculate the molar ratios of two acids 6
in the original mixture. *. Molar ratio of formic acid : oxalic acid = 4: 1 (1)

mews, YOUR TOPICAL UNDERSTANDING

OBJECTIVE Type Questions [1 Mark| (a) 3.47 g of BaCl, reacts with 2.36 g Na,SO, to give
3.88 g BaSO, and 1.95 g NaCl
1. Addition of 6.65x 10* and 8.95 10°, in terms of (b) Hydrogen sulphide contains 5.89% hydrogen, water
scientific notation will be contains 11.1% hydrogen and sulphur dioxide
(a) 7.545x 104 (b) 75.45x 10° contains 50% oxygen
(c) 754.5 107 (d) 75.45x 10° (c) An element forms two oxides, XO and XO,
containing 50% and 60% oxygen respectively. The
2. A student performs a titration with different burettes ratio of masses of oxygen which combines with 1 g
and finds titre values of 25.2 mL, 25.25 mL and 25.0 mL. of element X is 2:3
The number of significant figures in the average titre (d) 20 mL ammonia gives 10 volumes N, and 30
value is volumes H, at constant temperature and pressure
(a) 1 (b) 2 (Oi (d) 4 Ans. 1.(a) 2.(c) 3.(a) 4. (c)
Which of the following statements is correct about the
reaction given below? VERY SHORT ANSWER Type Questions
4Fe(s) + 30,(g) —> 2Fe,03(9) [1 Mark
[NCERT Exemplar]
1. Using the unit conversion factor, express 1.54
(a) Total mass of iron and oxygen in reactants = total mass of
iron and oxygen in product therefore it follows law of mms into pm is. [Ans. 1.54 x10° pmus ‘]
conservation of mass 2. Which one of the Dalton’s assumptions is now
(b) Total mass of reactants = total mass of product, therefore, known to be an error?
law of multiple proportions is followed
3. When 4.2 g of sodium bicarbonate is added to a
(c) Amount of FeO; can be increased by taking any one of
the reactants (iron or oxygen) in excess solution of acetic acid weighing 10.0 gq, it is
observed that 2.2 g of CO, is released into the
(d) Amount of Fe,O; produced will decrease if the amount of
any one of the reactants (iron or oxygen) is taken in excess atmosphere.
4. Which of the following statements illustrate the law of The residue left is found to weigh 12.0 g. Show
multiple proportions? that these observations are in agreement with
the law of conservation of mass.
 | Alléwone | CHEMISTRY Class 11th
20
SHORT ANSWER Type I Questions 3. Convert the following into metre.
|2 Marks| (i) 40 Em (thickness of milky way galaxy)
[Ans. 4.0 x10"”]
1. 0.7 g of iron combines directly with 0.4 g of S to
form FeS. If 2.8 g of Fe is dissolved in dil. HCl and (ii) 1.4 Gm (diameter of sun) [Ans.14 x10’m]
excess of Na,S solution is added, 4.4 g of FeS is (iii) 41 Pm (distance of nearest star)
precipitated. Prove that the law of definite (Ans. 41 x10'° m]
composition is true.
4. (i) Nitrogen forms five compounds with oxygen,
2. Convert the following into kg. in which 1.0 g of nitrogen combines with
(i) 0.91 10°*” g (mass of electron) 0.572 g, 1.14 g, 1.73 g, 2.28 g and 2.85 g of
oxygen, respectively. Show that these figures
[Ans. 91 x10°?”]
(ii) 700g (mass of human DNA molecule) agree with law of multiple proportions.
[Ans. 0.7 kg] (ii) Give the statement of law of definite
3. Hydrogen peroxide and water contain 5.93% and composition.
11.2% of hydrogen respectively. Show that the
data illustrates law of multiple proportions. LONG ANSWER Type Questions
4. Express the following number to three significant |5 Marks|
figures. 1. (i) A lady purchases a ring from a jeweller with
(i) 6.0263 [Ans. 5] diamonds embedded into it. The jeweller tells that
(ii) 2.3652 [Ans. 5] total diamond used in the ring is five carat. How
(iii) Sixty thousand [Ans. 1] much weight he should subtract from the weight
of the ring to get the weight of gold? [Ans. 1 g]
(iv) 2.861 10° [Ans. 4]
(ii) Perform the following calculation to proper
5. A certain element ‘X’ forms three different binary
number of significant figures.
compounds with chlorine, containing 59.68%,
68.95% and 74.75% chlorine, respectively. Show
(a) 108/7.2 [Ans. 15]
how these data illustrate the law of multiple (b) (1.6x 10°)? [Ans. 2.56 x10]
proportions. (c) (1.0042— 0.0034) (1.23) [Ans. 1.23]
(iii) Round up the following up to three significant
SHORT ANSWER Type IT Questions figures.
|3 Marks| (a) 34.216 [Ans. 34.2] (b) 10.4107 [Ans. 10.4]
1. How many significant figures are present in the (c) 0.04597 [Ans. 0.0460]
following?
2. (i) Calculate
(i) 0.0025 [Ans.2] (ii) 208 —[Ans. 3] (a) area of a square whose side is 1.2 m
(iii) 5005 [Ans.4] (iv) 126000 [Ans. 3] [Ans. 1.44 m7]
(v) 500.0 [Ans.] (vi) 2.0034 [Ans. 5] (b) volume of sphere with radius 1.6 cm
[Ans. 17.16 cm °]
[NCERT Textbook]
(c) length of a rectangle having area 10.25 m?
2. What is the weight in pounds of a gold bar 12.0 and breadth 2.5 m. [Ans. = 41m]
inch. long, 6.00 inch wide and 3.00 inch thick? The
(ii) Copper sulphate crystals contain 25.45% Cu and
density of gold is 19.3 g cm°. (Given linch = 2.54
36.07 % H,0. If the law of constant proportions is
cm, 1 lb = 453.6 g). [Ans. 150.6 Ib] true then calculate the weight of Cu required to
obtain 40 g of crystallineCuSO,. —[Ans. 10.18 g]
[TOPIC 2|
Atomic and Molecular Masses, Mole
Concept and Formulae of Compounds
ATOMIC MASS atom of carbon-12 taken as 12. It is determined with the
Atomic mass is defined as a number that expresses as to how help of mass spectrometer. If we have the percentage
many times an atom of the element is heavier than 1/12th abundances of all the isotopes, the average atomic mass is
of the mass of carbon atom ('*C). Therefore, calculated by the expression.
mass of an atom x A; ig piApe poy ee
Atomic mass = Average atomic mass = — = —____——— _ where,
(1/12) xXmass of acarbon atom ('*C) 100 100
p; 1s the percentage abundance of isotope having atomic
The scale of relative masses of atoms is termed as atomic mass A;. Fractional abundances, f; can also be used to
mass unit scale (amu). However, the new symbol used is u obtain the atomic mass of an element. Fractional
(known as unified mass) in place of amu. In this system 16 abundance is the total number of atoms that is comprised
is assigned a mass of exactly 12 atomic mass unit and masses of that particular isotope. Thus,
of all other atoms are given relative to this standard. Average atomic mass = 2 fA; = f,A, + fo A) +...
e.g. nitrogen has the following two isotopes with relative
abundance and atomic masses as shown against each ofthem.
Hence, one atomic mass unit is defined as the mass exactly : r Atomic
GaESI A set woeV EAT
fe}

equal to 1/12th of the mass of an atom of carbon ORGY

1 amu = 1.66056x 10774 g SNS iecetiie 99.33% 14,
15 0.67% 15
® Since, the atomic mass is the ratio of the masses, so It has no units.
However, we express it as amu, which only signifies that it is taken on From the above data, the average atomic mass of nitrogen
atomic mass unit scale in which 1/12th of carbon atom Is fixed as 1u.
can be calculated as
* C-12 is chosen as standard because in such case the masses of most (0.9933 x 14) = (0.0067 x 15)
of the elements are whole number or nearest to the whole number.
= 140067 u
Which corresponds closely to the mass of a proton or
neutron. More recently, these units have been called EXAMPLE |1| Boron occurs in nature in the form of
Daltons (Da). 1 Da=1u. two isotopes having atomic mass 10 and 11. What are the
percentage abundances of two isotopes in a sample of
Mass of an atom of hydrogen = 1.6736 107 74g boron having average atomic mass 10.8?
Thus, in terms of amu, the mass of hydrogen atom Sol. Let the % abundance of '°B isotope = x
_ 1.6736x 10° *g Then % abundance of ''B isotope = 100 — x
= 1.0078 u= 1.0080 u
1.66056x 10° *4g ; x X10+(100—- x) x11
The average atomic mass = FAD MOO Six) 2A
100
Similarly, the mass of oxygen 16 ('°O) atom would be
But, average atomic mass = 10.8
ISIS x X10
+ (100 — x)
—x)xX11 EsOR
100
Average Atomic Mass or 10x +1100
—11x = 10.8x 100
Many naturally occurring elements exist as mixture of or —x=-—1100+1080 or x= 20
several isotopes [Isotopes are the different atoms of the same Thus, percentage abundance,
element having same atomic number and different atomic 0B = 20%, ''B =80%
masses] and the atomic mass of an element is equal to the
average of atomic masses of all the isotopes. Thus, it is Note
termed as average atomic mass and is generally fractional. In the periodic table of elements, the atomic masses mentioned for
Average atomic mass is, thus, defined as the average different elements actually represented their average atomic masses.
relative mass of atoms of an element as compared with an
 |Allézone | CHEMISTRY Class 11th
22

Gram Atomic Mass EXAMPLE ]3|

(i) Calculate the gram molecular mass of sugar having
Atomic mass of an element expressed in gram is called its molecular formula C,,H,20,.
gram atomic mass or gram atom. Or, it may be defined as (ii) Calculate
that much quantity of the element whose mass in gram is (a) The mass of 0.5 g molecule of sugar and
~ numerically equal to its atomic mass. (b) Gram molecule of sugar in 547.2 g.
e.g. the atomic mass of hydrogen is 1.008 amu and thus, a Sol. (i) Molecular mass of sugar (C,,H2,0,;)
gram atom of hydrogen is 1.008 g. =12x atomic mass of C+ 22 x atomic mass of H
Gram atomic mass = Atomic mass expressed in gram +11 X atomic mass of O
= gram atom =12X12+ 22 xX1+11
x16 = 342
. Gram molecular mass of sugar = 342 g
Never get confused between the terms gram atom and mass
(ii) (a) 1 gram molecule of sugar = 342 g
of an atom of the element. “. 0.5 gram molecule of sugar = 342 0.5 =171 g
e.g. The mass of one atom ofoxygen is only 2.66 x erg g (b) 342 g of sugar = 1 gram molecule
whereas its gram atomic mass is 16 g. The mass of one 547.2 g of sugar = = x 547.2= 1.6 gram molecule
atom of an element is known as the actual mass of the
atom.
FORMULA MASS
MOLECULAR MASS Ionic compounds do not contain discrete molecules as their
Molecular mass is the sum of atomic masses of the constituent units. In these compounds, positive and negative
elements present in a molecule. Thus, it is obtained by ions are arranged in a three-dimensional structure. Thus, the
multiplying the atomic mass of each element by the formula of such compounds is used to calculate the formula
number of its atoms and then adding them together e.g. mass instead of the molecular mass.
Molecular mass of methane can be calculated as Formula mass of an ionic compound = number of cations
X its atomic mass + number of anions X its atomic mass
CH, = 1X atomic mass of C+ 4 X atomic mass of H
e.g. formula mass of sodium chloride
= (12.011u) + 4 (1.008 u). = 16.043 u
= atomic mass of sodium + atomic mass of chlorine
EXAMPLE |2| Calculate molecular mass of glucose = 23.0% 35510 = 585 0
(C3H,,0,) molecule. [NCERT] Formula mass expressed in grams is called gram formula mass.
Sol. Molecular mass of glucose (C,H,,0,)
= 6 (12.011u)+ 12 (1.008u)+ 6 (16.00u) MOLE CONCEPT AND
= (72.66u)+ (12.096u)+ (96.00u) = 180.162u MOLAR MASSES
The word mole (means heap or pile) was introduced by
Gram Molecular Mass Wilhelm Ostwald and is defined as “the amount of a
or Gram Molecule substance that contains as many particles or entities as
Molecular mass of a substance expressed in gram is called there are atoms in exactly 12 g (or 0.012 kg) of the ol
its gram molecular mass or gram molecule. isotope.
Gram molecular mass or gram molecule The mole of a substance always contain the same number of
= Molecular mass expressed in gram entities independent of the nature of the substance taken.
e.g. the molecular mass of oxygen is 32 and, therefore, its The number of atoms in 12 g of '*C is calculated as
gram molecular mass is 32 g. Gram molecular mass should Mass of a carbon atom = 1.992648x TOY?s
not be confused with the mass of one molecule of the (Determined by mass spectrometer)
substance in gram. 1 mol of carbon = 12 g
e.g. the mass of one molecule of oxygen is only “. The number of atoms in 12g C
53 2e0ree g, whereas the gram molecular mass, is 32
12 g/mol'*C
g. The mass of one molecule of a substance is known as
its actual molecular mass. ~ 1,992648x 1073 g/'?C atom
= 6.0221367 x 10°? atoms / mol
Some Basic Concepts of Chemistry
p28)
This number of entities in 1 mol is so important that it is (v) Number of molecules = Number of moles x N, ,
given a separate name and symbol.
where, N, = Avogadro number = 6.022x tap?
It is known as Avogadro constant, denoted by Ny in (vi) Number of atoms = Number of molecules x
honour of Amedeo Avogadro. atomicity (or number of atoms in the molecular
In other words, we can say a mole is a collection of formula or in 1 mole).
6.022x 107? particles. Note
Thus, 1 mole of atoms = 6.022 1079 The problems based on mole concept can be solved by unitary
atoms method or by directly applying formulae.
= Gram atomic mass of the elements
and 1 mole of molecules = 6.022 x 107? molecules EXAMPLE |4| Calculate the number of molecules
present in 44.8 cm? of oxygen gas at 273 K and
= gram molecular mass
2 atmosphere pressure.
The mass of one mole of a substance in grams is called its
Sol. At STP, volume of O, gas can be calculated as applying
molar mass. The molar mass in grams is numerically equal to
atomic/molecular/formula mass in ‘n’. gas equation : PVs _ PV
rir eer:
e.g. molar mass of water= 18.02 g mol”! and it contains 443 x 2 TV,
6.022 x 107° H,O molecules. We get [STP condition]
2783 ob 878
V, = 89.6 cm* = 0.0896 L
Moles in Case of Ionic Compounds Now, applying mole concept,
The mass of one mole formula units in gram is equal to volume of gas at STP (L)__ number of molecules of gas
formula mass expressed in grams or gram formula mass of 22.4L : avogadro’s number
the compound.
00896 number of molecules of O, gas
Thus, mass of 6.022 10”? formula units (or one mole 224 6.022 x10”
formula units) of any ionic substance in grams is equal to its . Number of molecules of O, gas
gram formula mass. _ 0.0896 x 6.022 x 10°°
= 240 x 107?
e.g. a mole of NaCl equals to 58.5 g (one gram formula « 22.4
mass) and contains 6.022 x 107° formula units of NaCl or
EXAMPLE |5| Calculate the number of moles in the
6.022 x 107? Na* ions and 6.022 107° CI7 ions. following.
(i) 7.85 g of iron (ii) 4.68 mg of silicon
Moles in Case of Gases (iii) 65.6 ug of carbon [NCERT]
In case of gases, a mole is defined as that amount of the gas
which occupies a volume of 22.4 L at STP, which is called Sol. (i) Moles of iron= a ass =0.141 mol
atomic mass 55.8
its molar volume. (iy Molesof silicon = mass of silicon a 4.68x 107°
The mass of one mole of a substance in gram. is called its atomic mass 28.1
molar mass. Its units are g mol” lor kg mol~!. The molar = 1.67x 10 *mol
mass in gram is numerically equal to Anatase ere
Gi) Moles cf carbone mass of carbon # 65.6x 10°
/formula mass
atomic mass iy?
e.g. molar mass of water = 18.02 g mol', = 5.47 x 10 ° mol

EXAMPLE |6| The cost of table salt (NaCl) and table

The above relations can be summarised as
- sugar (Cj2H,.0,,) is = 2 per kg and &6 per kg, respectively.
é mass of element
(i) Number of moles of atoms = —————_——— Calculate their costs per mol.
atomic mass
Sol. One mole of NaCl = 58.5 g
f molecule
(ii) Number of moles of molecule = ied lawendaaat Z
Cost of NaCl per mol = aan 58.5 =% 0.117
molecular mass
volume of the gas (STP) = 11.7 paise or 12 paise
(iii) Number of moles of gas = ERS:
One mole of sugar (C,,H,,0,,)= 342 g
gg
(iv) 1 mole = 6.022 10°? particles Cost of sugar per mol =—— x 342 = % 2.05
1000
= Gram atomic/ molecular mass.
 | Allénone | CHEMISTRY Class 11th
on

EXAMPLE |7| Calculate the number of atoms in each PERCENTAGE COMPOSITION

of the following. The percentage of any element or constituent in a
(i) 52 mole of Ar (ii) 52 u of He compound is the number of parts by mass of that element
[NCERT Textbook] or constituent present in 100 parts by mass of the
(iii) 52 g of He
6.022x 10° atoms
compound. Following two steps are involved to calculate
Sol. (i) «: 1 mole of Ar contains
the percentage composition of an element present in a
-. 52 moles of Ar will contain 6.022x 107° x 52 compound.
= 3.13x 10” atoms
Step I Calculate the molecular mass of the compound
(yes 4 u of He =1 atom from its formula by adding the atomic masses of
52 u of He == x52=13 atoms all the elements present.
(iii) 4 g of He contains 6.022 x10"? atoms
Step II Calculate the mass per cent of the element by
using the expression
_ 6.02210" x52
bo got We wilreontan Percentage or mass % of an element or constituent
= 7.8310" atoms
EXAMPLE |8| AeMass of f that
that element
el in
in ththe compoun d tae
(i) Assuming the density of water to be 1g/ cm?, calculate Molar mass of the compound
the volume occupied by one molecule of water.
Percentage composition helps in determining the
(ii) Assuming the water molecule to be spherical, calculate
molecular formula of a compound. Moreover, it is also
the diameter of the water molecule.
helpful to check the purity of a given sample.
(iii) Assuming that oxygen atom occupies half of the volume
occupied by the water molecule, calculate approximately
the diameter of the oxygen atom.
EXAMPLE |9| Calculate the mass percentage of
each element of water.
Sol. (i) 1 mole of H,O= 18g= 18cm*
Sol. Molar mass of water (H,O)= 2x atomic mass of H
(«density of H,O=1g/cm’) +1 x atomic mass of O= 2x 1.01 + 1X 16.00
= 6022 x10” molecules of H,O = 18.02 g
2X 1.008
Thus, 6.022 x 10°? molecules of H,O have volume Mass % of hydrogen = ares xX 100=11.18%
= 18cm> 16.
.“. 1 molecule of H,O will have volume and mass % of oxygen = x 100 = 88.79%
Le
cm? = 2.989x 10 cm?
6.022 x 10”
(ii) As water molecule is assumed to be spherical, if R is its
EMPIRICAL FORMULA
radius, then its volume will be An empirical formula of a compound represents the
4 - simplest whole number ratio of various atoms present in
- mR? = 2,989 x10 cm? or R® =7133 X10
one molecule of the compound.
or R=(7133)'? x10°° =1925 x10 °cm
Method to find Empirical Formula
.. Diameter of water molecule
of a Compound
= 21925 x10 ° cm = 385 X10° cm
Follow the following steps to find the empirical formula of
(iii) As oxygen atom occupies half of the volume occupied a compound.
by water molecule, hence if r is the radius of oxygen
4 Step Conversion of mass per cent into gram Convert
atom, then ; tr = i x 2.989 x 10°73 the mass per cent into gram by taking 100 g of the
1
compound as the starting material e.g. 407% means
or r? = 3566 x10 4 which gives
4.07 g if we take 100 g as the starting material.
r=1528 x10 ° cm Usually no direct method is used for finding the
.. Diameter of oxygen atom percentage of oxygen in a compound. It is calculated
= 2X1528x10° cm by finding the percentages of all elements except
oxygen. Thus, percentage of oxygen = 100 — (sum of
= 3.056 x 10° cm percentages of all other elements present in the
compound).
Some Basic Concepts of Chemistry 25
Step II Convert mass (in grams) into number of moles
for each element Convert mass obtained above
EXAMPLE |10| An organic substance containing
into number of moles by dividing them with atomic carbon, hydrogen and oxygen gave the percentage
masses of various elements. composition as C = 40.687%, H= 5.085%.
Step U1 Divide the mole value obtained above with the The vapour density of the compound is 59. Calculate the
smallest mole value to get the simplest ratio. molecular formula of the compound.
Step VI Write empirical formula by mentioning the numbers Sol. Step1 To calculate empirical formula of the compound.
after writing the symbols of respective elements. Here, % of O = 100 —(40.687 +5.085) = 54.228%
Calculate it in the tabular form as shown in the example 10. .. Empirical formula is C,H,0,.
(See table on Next Page)

MOLECULAR FORMULA Step II To. calculate the empirical formula mass. The
empirical formula of the compound is C,H,0,.
The molecular formula shows the exact number . Empirical formula mass
of different types of atoms present in a molecule of =(2x12)+(3X1)4+(2
x16) =59
a compound.
Step III To calculate the molecular mass of the salt. The
vapour density of the compound = 59 (Given)
Method to determint of the Molecular
Using the relation between vapour density and
Formula of a Compound
molecular mass.
Step 1 Determine the empirical formula as described We have molecular mass = 2X vapour density
above. = 59 =118
Step I Calculate the empirical formula mass by adding the Step WV To calculate the value of n
atomic masses of the atoms in the empirical _ molecular mass __
formula. empirical formula mass
Step III Determine the molecular mass by suitable method. Jee,
Step TV Determine the value of 7 as 3
7S molecular mass Ghansen t6 the Step V To calculate the molecular formula of the salt,
empirical formula mass. Molecular formula = 7 x empirical formula
nearest whole number. = 2xC,H3;0, =C,H,O,
=e , th lecular fi lai’ GG OFvie
RBG
Step V Multiply empirical formula by 7 to get the se A per ONT
molecular formula.
Molecular formula = x empirical formula

Percentage
EI Synibel Percentage Atomic mass Moles of the element See Simplest Simplest whole
ement y mbo of element of element (Relative number of moles) iy ti
ite ratio a mber molar ratio

© ere
40.687 5 12 40.687 _ 3 399 3,390
en ay 2
mea 12 3.389

5.0885 5,085 3
Hydrogen H 5.085 1: ; 5,085 ee
ea aS
54.228 = 3,389 3.389 2
54.228 16 tees 8
Peer = 16 3.389
 TOPIC PRACTICE 2|
OBJECTIVE Type Questions IV. 1x107'° mole of copper
(a) I< I< III< IV (b) I< I< IlI< IV
\1 Mark|
(c) I< I< IV<I1 (d) IV< I< I< I
1. An alkaloid contains 17.28% of nitrogen and Sol. (a) I. Mass of one atom of oxygen
its molecular mass is 162. The number of 16
nitrogen atoms present in one molecule of
=2.66x 10 g
~ 6,022 102
alkaloid is II. Mass of one atom of nitrogen
(a) 5 (b) 4 (c) 3 (d) 2 14
(d) 100 g alkaloid contains nitrogen= 17.28 g = 2.32x10 g
Sol.
~ 6.022% 1073
17.28 X 162
. 162 g alkaloid will contain nitrogen= ape Ill. Mass of 1x10"° mole of oxygen = 16x 107° g

= 27.9 g= 28g IV. Mass of 1x10-2° mole of copper = 63 X 10 °g

Atomic weight of nitrogen = 14 Hence, masses of atoms in increasing order :
So, number of atoms of nitrogen present in one molecule I<I<l<IV

eran 0.2429 g sample of potassium is heated in

14
oxygen, 0.440 g of acrystalline compound is
A gas is found to have the formula (CO),. Its obtained. What is the formula of this
compound?
vapour density is 70. The value of x will be
(a) 7 (b) 4 (c) 5 (d) 6 (a) KO (b) K,O
(c) KO, (d) KO,
Sol. (c) Vapour density = 70
Sol. (c) Mass of K= 0.242 g, mass of compound = 0.440 g,
Molecular mass = 2 x 70 = 140
Formula is [CO], mass ofO = 0.440
—0.242 = 0.198 g
Therefore molecular mass = (12 + 16), =140 0.2
Relative molar ratio of K= — = 0.0006,
= x X 28 =140
a Relative molar ratio of O = — = 0.012
1
The number of atoms present in one mole of an Simple ratio of K=1
element is equal to Avogadro number. Which of Simple ratio of O= 2
the following element contains the greatest
So, the formula is KO,.
number of atoms? [NCERT Exemplar]
(a) 4 g He (b)46gNa (c)040gCa (d)12¢He 6. The empirical formula and molecular mass of a
compound are CH,O and 180 g respectively.
Sol. (d) Moles of 4 g He =" =1 mol
What will be the molecular formula of the
compound? [NCERT Exemplar]
fae Naae = Jmol
23 (a)CyHygO, (b)CH,O = (c) C,H,,0, (d)C, H,O,
Sol. (c) Empirical formula mass = CHO
0.40 g Ca= 8" 01 mol
40 =12+ 2*1+16=30
12 Molecular mass = 180
PE a Aloo sa Be Molecular mass
“iEmpirical formula mass
Hence, 12 g He contains greatest number of atoms as it
180
possesses maximum number of moles. = —_ = 6
30
Arrange the following in the order of increasing Molecular formula =n x empirical formula
mass (Atomic mass of O = 16, Cu = 63 and = 6 X CH,0 = C,H,,0,
N = 14).
7. A compound contains 69.5% oxygen, 30.5%
I. One atom of oxygen nitrogen and its molecular weight is 92. The
II. One atom of nitrogen
formula of compound is
III. 1x 10° mole of oxygen (a) N,O (b) NO, (c) N20, d) N20;
Some Basic Concepts of Chemistry
oa|
Sol. (c) Element % % at. wt. Ratio The empirical formula is C,H,O,.
N 30.5 Empirical formula weight = 3 x12+5 1+ 2x16
30.5 : ——=2.1
14 8 1
= 36+5+4+32=73

69.5 Molecular weight of the compound = 2 x VD

O 69.5 16 = 4.34
—— 2 =2X73=146
mol. wt. ~ 140°"
Empirical formula= NO, 3 empirical formula wt. 73
Empirical formula weight = 46 Molecular formula = Empirical formula x 2
oy
n=—=2 =(C3H;0,)x 2= CoH 0,
46
= Molecular formula = (NO, ), = N,O, VERY SHORT ANSWER Type Questions
8. The empirical formula of a compound is CH,. One |1 Mark|
mole of this compound has a mass of 42 g. Its
I. Calculate the percentage of N in NH3 molecule.
molecular formula is
(a) C3H, (b) C3H, (c) CH, (d) C,H, Sol. Molar mass of NH, = 14+ 1x 3=17g mol |
Sol. (a) Weight of empirical formula, mass ofN in NH,
Percentage of N = x 100
CH, =12+(1 Xx 2)=12+2=14 molar mass of NH,
Mass of one mole of the compound = its molecular weight 14
=— X 100= 82.35%
= 42 17
pe Mol. wt. mi2<
2. Calculate the number of gram molecules of
S Empirical formula wt. ie
water in a beaker containing 576 g of water.
*. Mol. formula =(empirical formula) x n
Sol. Molecular mass of H,O= 2x 1+ 16= 18 g mol!
=(CH,)x 3= C,H,
18 g of water = 1 gram molecule
9. An organic compound containing C and H has 1
92.3% of carbon, its empirical formula is “. 576 g of water = 6 x 576 = 32 gram molecule
(a) CH (b) CH, (c) CH, (d) CH,
3. What is the symbol for SI unit of mole ? How is
Sol. (a) Element % % atomic weight Simplest ratio
the mole defined? [NCERT Exemplar]
Sol. Refer to text page 22.
92.3 7.69
=f 69 ——— a
Cray 92: 12 7.69 4. One mole of oxygen gas at STP is equal to?
[NCERT Exemplar]
a Tf7h0)
et t=7, 70 | Sol. 1 mole of O, gas at STP = 6.022 x10” molecules of O,
H bide 1 7.69
(Avogadro number) = 32 g of O,
*, Empirical formula = CH Hence, 1 mole of oxygen gas is equal to molecular
weight of oxygen as well as Avogadro number.
10. An organic compound containing C, H and O has
49.3% carbon, 6.84% hydrogen and its vapour 5. What will be the mass of one atom of C-12 in
density is 73. Molecular formula of the gram? [NCERT Exemplar; Textbook]
compound is Sol. Mass of 1atom of *C = asionnonssco
Avogadro’s number
(a) C3H50, (b) C4H i992

(c) C6HyoO4 (d) C 3H 90, 12g

~ 6.022x 10
Sol. (c) Element % % eclani
atomic ;
Simplest :
ratio | =1.9927x 10° g
49.3 4.1
6. Give an example of molecule in which the ratio
C 49.3 12 274 of the molecular formula is six times the
6.84 6.84
—~* = 6.84 eS ea ed 40) empirical formula.
H 6.84 1 274 Sol. The compound is glucose. Its molecular formula is
43.86 (ade Zs C,H,.0, while empirical formula is CH,0.
O 43.86 SS ae ett das
 | Allgwone | CHEMISTRY Class 11th
28
{(Natural abundance of 1 41x molar mass of ‘H)
7. Why are the atomic mass of most of the
elements is fractional? ~ +(Natural abundance of 2 Hx molar mass of 7H)}
Sol. It is because most of the elements occur in nature as a 100
mixture of isotopes and their atomic masses are the _ 99.985 x 1+ 0.015 2
average relative atomic masses of the isotopes ~ 100
depending on their abundance. _ 99.985 + 0.030 _ 100.015 _ | nnoi5 uy
[2]
8, Ablack dot used as a full stop at the end of a 100
sentence has a mass of about one attogram.
Assuming that the dot is made up of carbon,
3. Use the data given in the following table to
calculate the molar mass of naturally occurring
calculate the approximate number of carbon [NCERT]
argon.
atoms present in the dot ?
Isotope Isotopic molar mass Abundance
Sol. Mass of carbon in the dot =1 attogram =10°'*g
Gram atomic mass of carbon = 12 g, 36 35.96755 g mol 0.337%
38a 37.96272 g mol” 0.063%
i.e. 12 g of carbon contains 6.022 x10” atoms of carbon.
—4pr 39.9624 g mol" 99.600%
“. 107'8g of carbon will contain carbon atoms.
6.022 x 107° Sol. Average molar mass of Ar = & f; x A;
= ————_ X 10!8 = 5.02x 10’ atoms
12 = (0.00337X 35.96755) + (0.00063xX37.96272)
+ (0.99600x 39.9624)
SHORT ANSWER Type I Questions
= 0.121 + 0.024 + 39.803 = 39.948 g mol’ [2]
|2 Marks|
Calculate the number of moles in the
1, Calculate the atomic mass (average) of chlorine
following masses
using the following data.
(i) 1.46 metric ton of Al (1 metric ton =10° kg)
% natural
Isotope abundance Molar mass (ii) 7.9 mg of Ca
Sol. (i) 1.46 metric ton of Al =1.46x 10° x 10° g of Al
8c 75.77 34.9689
= 146x 10° ¢
ota 24,23 36.9659
Atomic mass of Al = 27
[NCERT Textbook] mass ofAl _ 1.46X 10°
Moles of Al = (1)
Sol. Average atomic mass is the sum of the products of atomic mass _ 27
fractional abundances (f;) of the isotopes and their = 5.41 10* mol
corresponding mass number (A; ). Average atomic mass, (ii) 7.9 mg of Ca = 7.9x10™° g of Ca
A= 2 fA, =f; X Apt fo X Aact nun [1] Atomic mass of Ca = 40.1
Average atomic mass,
of Ca _7.9 X 10 i
Moles of Ca =
_ A= 0.7577 34,9689+ 0.2423 x 36.9659 atomic mass 40.1
= 26.4959 + 8.9568 = 35.4527 {1]
= 1.97x 107 mol [1]
2. Calculate the average atomic mass of hydrogen
One mole of any substance contains 6.022x 107°
using the following data. [NCERT Exemplar]
atoms/molecules. Number of molecules of
Seton
; % Natural
eirindance
BDA
Molar mass
H,SO, present in 100 mL of 0.02 MH,SO,. What
will be the solution? [NCERT Exemplar]
ahs eS Sol. One mole of any substance contains 6.022 x 107%
*H 0.015 2 atoms/molecules.
Sol. Many naturally occurring elements exist as more than Hence, number of millimoles of H,SO,
one isotope. When we take into account the existence of = molarityx volume in mL
these isotopes and their relative abundance (per cent = 0.02 x 100 = 2 millimoles = 2 x 10” ? mol
occurrence), the average atomic mass of the element can
Number of molecules = number of moles x N ,
be calculated as. Average atomic mass
= 2x10 ° x 6.022 x 107? =12.044 x 102° molecules [2]
Some Basic Concepts of Chemistry pe)
6. A flask P contains 0.5 mole of oxygen gas. 9. What is the mass per cent of carbon in carbon
Another flask Q contains 0.4 mole of ozone gas. dioxide? [NCERT Exemplar]
Which of the two flasks contain greater number
Sol. Molecular mass of CO, =1x12+ 2X16 = 44g
of oxygen atoms?
1 g molecule of CO, contains 1g atoms of carbon
Sol. 1 molecule of oxygen (O,)= 2 atoms of oxygen
44 g of CO, contain C = 12 g atoms of carbon
1 molecule of ozone (O,)= 3 atoms of oxygen 12
In flask P, 1 mole of oxygen gas = 6.022 107° molecules % ofC in CO, = pg 100 27.27%
*. 0.5 mole of oxygen gas = 6.022 10”? x 0.5 molecules Hence, the mass per cent of carbon in CO) is 27.27%. [2]
= 6.022x 10” x 0.5x 2 atoms = 6.022x 1073 atoms
10. Calculate the mass per cent of different elements
In flask Q, 1 mole of ozone gas = 6.022 10” molecules
present in sodium sulphate, Na SO,. [NCERT]
0.4 mole of ozone gas = 6.022 x 10”° x 0.4 molecules
Sol. Mass per cent of an element
= 6.022 10” x 0.4x 3 atoms = 7.23x 10” atoms _ Mass of that element in the compound x 100
.. Flask Q has greater number of oxygen atoms as Molar mass of the compound
compared to the flask P. [2]
Molar mass of Na,SO,
7. How much copper can be obtained from 100 g = (2X 22.99) + 32.06 + (4X 16.00) = 142.04 g
of copper sulphate (CuSO,) ?
F 45.98 x 100
Sol. Molar mass of CuSO, = 63.54+32.06+(4 X16) = 159.6 g mol” ; Mass per cent of sodium = —————— = 32.37 [1]
142.04
159.6 g CuSO, contains = 63.54 g Cu {1} 32.06 X 100
Mass per cent of sulphur = —————— = 22.57
63.54 42.04
1 gCuSO, contains = —— g Cu
E f 159.6 © Mass per cent of oxygen =
64x 100
= 45.06 {1]
63.54 100 42.04
100 g CuSO, contains = = 39.81gCu [Il]
159.6 11. The empirical formula and molecular mass of a
8. Which one of the following will have the largest compound are CH.O and 180 g respectively.
number of atoms? [NCERT Textbook] What will be the molecular formula of the
(i) 1g Au (s) (ii) 1g Na (s) compound? [NCERT Exemplar]
(iii) 1g Li (s) (iv) 1g of CL,(g) Sol. Empirical formula mass = CH,O =12 + 2x1+16=30
Molecular mass = 180
~~. @ We know that number of atoms = moles x N, x Molecular mass _ 180
ae atomicity, so first calculate the number of moles by ne = — =6[l]
Empirical formula mass 30
using the formula,
mass of a substance (g) “. Molecular formula = n x empirical formula
« Moles of a substance = = 6X CH,O = C,H,,0, (1)
molar mass
¢ Then, find number of atoms from moles of the substance
12. Acompound on analysis was found to contain
and compare them.
(Atomic masses : Au = 197, Na = 23, Li = 7,Cl = 35.5 u) C = 34.6%, H= 3.85%, and 0= 61.55%. Calculate its
empirical formula.
1
Sol. (i) 1g Au= =e mole of Au = — x 6.022x 10” atoms Sol. Calculation of the empirical formula.
197 197
Simplest
of Au = 3.057 107! atoms Atomic Gram atoms ‘ whole
Element Percentage eee (Moles) Molar ratio aargecy
(ii) 1g Na ib mole of Na= S x 6.022 107 atoms of : ratio
23
Na = 26 x10” atoms
s 34.6 12 5 98 2.88_,
S46 3
oe 4 = Ee ei 12 ae oe'2.88
(iii) 1 g Li= 2 mole of Li= = X 6.022 x 107° molecules of Li
J i H 3.85 Jie eeIDE gees) Bs ly es
= 860 x 22”" atoms 8
4
Or =
1
(iv) 1g Cl, = ae mole of Cl, toys | at. roe ee =
O 61.55 16 61.55_
3 96 3.85_ 1 396 4
1
=— x 6,022 10”? atoms Cl, = 8.48 x10 atoms 16
71 On
Hence, 1 g Li have the largest number of atoms. :
[4x]2
 | Allxwvone| CHEMISTRY Class 11th
30
, The simplest whole number ratios of the different 2. Inthree moles of ethane (C,Hg), calculate the
elements are: C: H: O:: 3:4: 4 and the empirical
following.
formula of the compound = C3,H,0O,. [2]
(i) Number of moles of carbon atoms.
13. What is the molecular mass of a substance each (ii) Number of moles of hydrogen atoms.
molecule of which contains 9 atoms of carbon, (iii) Number of molecules of ethane.
13 atoms of hydrogen and 2.33x 10-8 g other [NCERT Textbook]
component? Sol. (i) 1 mole of C,H, contains 2 moles of carbon atoms.
Sol. Mass of 9 atoms of carbon = 9X 12 amu= 108 u «. Number of moles of carbon atoms in 3 moles of
Mass of 13 atoms of hydrogen = 13x 1 amu = 13 u C,H, = 2x 3=6. [1]
Mass of 2. 33x 10” g of other component (ii) 1 mole of C,H, contains 6 moles of hydrogen atoms.
(2.33x 10°” g) .. Number of moles of hydrogen atoms in 3 moles of
=(1u)x 7 fa 14.04 u {1] C,H, = 3X 6= 18 [1]
1.66x10-4 g
(iii) 1 mole of C,H, contains 6.022x 10° molecules of
Molecular mass of the substance
ethane.
= (108 + 13+ 14.04) u = 135.04u {1]
. 3 moles of C,H,
14. Chlorophyll, the green colouring matter of = 3 x 6.022 x 107? =18066 x 10** molecules
plants contains 2.68% magnesium by weight. Number of ethane molecules in 3 moles of
Calculate the number of magnesium atoms in C,H, = 3 6.023x 107 = 1.8069x 10% [1]
2.00 g of chlorophyll (Atomic mass of Mg = 24).
Sol. Mass of chlorophyll = 2.0 g 3. Calculate the mass per cent of calcium,
Percentage of Mg = 2.68 g phosphorus and oxygen in calcium phosphate
2.68 X 2.0 Ca3(PO4)>.
Mass of Mg in 2.0 g of chlorophyll = Be ERE IS
100 “(Ps To calculate the mass per cent of atom, using the formula.
6.022x 10”atoms of magnesium= 24 g
Sol. Mass per cent of an element
.”. 24 g of Mg contains 6.022 x 10” atoms __ atomic mass of the element present in the compound ane
6.022 107° molar mass of the compound
. 0.054 g of Mg contains x 0.054
Mass per cent of calcium
= 1.3x 107! atoms [2] % 3x (atomic
(atomi mass of calcium)
i a
molecular mass of Ca,(PO,),
SHORT ANSWER Type II Questions _ 120u
X100 = 38.71% {1]
|3 Marks| Ou
Mass per cent of phosphorus
1. Calculate the molar mass of the following.
ae 2X |(atomic mass of ofphosp
phosph orus) i0G
(i) HO (ii) CO, ~— (iii) CH, +[NCERT Textbook]
molecular mass of Ca,(PO,),
Sol. (i) Molar mass of H,O= 2x atomic mass of hydrogen
2 ax ol
+1 atomic mass of oxygen — x 100= 20% (1)
= 2X 1.0079 u+ 1X 16.00 u= 18.0158u {1] 310 u
(ii) Molar mass of CO, = 1 x atomic mass of carbon Mass per cent of oxygen
+ 2x atomic mass of oxygen _ 8x (atomic mass of oxygen) ey
= 1X 12.01 u+ 2X16.00 u= 44.01 u [1] molecular mass of Ca,(PO,),
(iii) Molar mass of CH, =1 X atomic mass of carbon _ 8xl6u
+ 4 Xatomic mass of hydrogen x100 = 41. 29 % (1)
310 u
=1x12.01 u+4X1. 0079
u =16.0416 {1]
Some Basic Concepts of Chemistry
31
4. Determine the molecular formula of an oxide of iron in which the mass per cent of iron and oxygen are
69.9 and 30.1 respectively. Given that molar mass of the oxide is 159. 898 mol~!
[NCERT Textbook]
Element Symbole cs by mace ehtom| Relative number Olean: ; ; Simple whole number
Sol.
bf Cire Bose eae te Atsig Pete eee moles of element S!mple molarratio a: ratio
Iron Fe 69.9 55.85 69.9 _ 495 1.25 _, inp =2
Se a Ol A 8 __ 55.85 {wee 1.25)
Oxygen O 30.1 16.00 30.1 _ 4. 188_45 1.5x2=3
: 16.00 Poany
.. Empirical formula = Fe,O,. Empirical formula mass of Fe,O3 = (2X 55.85)+ (3X 16.00) = 159.7 g mol *
molar mass _ 159.8 _
[3]
- empirical formula mass 71597
Hence, molecular formula of the compound is Fe,03.

A crystalline salt when heated becomes anhydrous _ aX16+bx28

= 20 g mol | (Given) [1]
and loses 51.2% of its weight. The anhydrous salt a+b
on analysis gave the percentage composition as ie. 16a + 28b = 20(a+b) or 4a+7b=5(a+b)
Mg= 20.0%,S= 26.66% and O = 53.33%.
or a= 20 Qe SSS Sve {1]
Sol. part
% of Atomic Relative no. of If the ratio is reversed, now the ratioa:b =1: 2
Element
mass mass moles element °!mple molar ratio ile x
Average molar mass = ee eee 24 g mol ‘[1]
Mg 20 24 a= 0.8333 1+ 2
24
So» 26166. 32 SOP = 0.8125 7. Calculate the total number of electrons present
in 1.6 g of methane.
O,).' 53:43 *° 76 ene i 3.333125
Sol. (i) Molar mass of methane (CH,)= 12+ 4x 1= 16g
16 g of methane contains = 6.022x 10”* molecules
1.6 g of methane will contain
The empirical formula of the anhydrous salt comes out fe7.022x 10"
x (1.6 g) = 6.022 10” molecules _—[2]
to be MgSO,. Empirical formula mass = 120. [2] (16 g)
Molecular mass = 120. (ii) Number of electrons in 6.022x 10° molecules of
Hence, molecular formula = MgSO,. methane. 1 molecule of methane contains = 6 + 4 =10
electrons 6.022 107” molecules of methane contain
As crystalline salt on becoming anhydrous loses 51.2% by
mass, this means 48.8 g of anhydrous salt contains
electrons= 6.022 10” x 10 = 6.022 10” [1]
H,O= 51.2 g. Therefore, 120 g of anhydrous salt contains 8. A 0.005 cm thick coating of copper is deposited
51.2 on a plate of 0.5 m’ total area. Calculate the
x 120 g= 126 g= = molecules = 7 H,O molecules
number of copper atoms deposited on the plate
Hence, molecular formula of crystalline salt (density of copper = 7.2 g cm~?, atomic
= MgSO, :7H,0 [1] mass = 63.5).
The average molar mass of a mixture of Sol. Area of plate = 0.5m? = 0.5 x 10° cm’?
methane (CH,) and ethane (C;H,) present in the Thickness of coating = 0.005 cm
ratio of a: b is found to be 20.0 g mol". If the Volume of copper deposited = 0.5 x 10* x 0.005 = 25 cm®
ratio were reversed, what would be the molar Mass of copper deposited = 25 x 7.2 =180 g
mass of the mixture? Now,63.5 g of copper contains atoms = 6.022 x 10°
Sol. Molar mass of CH, = 16g mol? 6.022 x 10°
.”. 180 g of copper will contain atoms = x 180
63.5
Molar mass of C,H, = 28 g mol*
= 1.71x 10” atoms
When they are present in the a: b, their average molar mass [3]
 |AllZwone | CHEMISTRY Class 11th
32
0.9218
LONG ANSWER Type Questions %Z of C in
as, ry
the compound =
e 0.9985
x 100 = 92.32

|5 Marks| Zenda 4 22.0787 x 100 = 7.68

0 compound =

i (i) The density of the water at room temperature eee aeein Kage 0.9985
is 0.1 g/mL. How many molecules are there in Calculation for Empirical Formula
a drop of water if its volume is 0.05 mL? Relative number
Percent Atomic Simplest
(ii) An alloy ofiron (53.6%), nickel (45.8%) and of moles of .
Rigpeut by mass mass ploniants molar ee
manganese (0.6%) has a density of 8.17 g em’,
Calculate the number of Ni atoms present in OS 0g:80 ta SS ee eee
the alloy of dimensions 12 : 7.68
10.0 cm x 20.0 cm x 15.0 cm. H 7.68 1 7.68 _ 768 Mabe
Sol. (i) Volume of a drop of water = 0.05 mL ile : ros
Mass of a drop of water = volume x density Hence, empirical formula = CH [3]
= (0.05 mL) x (1.0 g/mL) = 0.05 g [1/2]
(ii) Calculation for molar mass of the gas
Gram molecular mass of water
10.0 L of the given gas at STP weigh = 11.6 g
(H,0)= 2x 1+ 16= 18 g; 18 g of water=1mol [1/2]
1 mol . 22.4 L of the given gas at STP will weigh
0.05 g of water = x (0.05 g) = 0.0028 mol % 11.6X 22.4 = 25.984 g
(18 ey) [1/2]
10
‘+ 1 mole of water contains molecules = 6.022x 107°
Molar mass = 25.984 ~ 26 g mol’. [1]
0.0028 mole of water will contain molecules
(iii) Empirical formula mass (CH)= 12+ 1= 13
= 6.022x 107 x 0.0028 = 1.68x 107’ molecules [1]
molecular mass + 26 ‘
(ii) Volume of the alloy — — — 2
empirical formula mass 13
= (10.0 cm)x (20.0 cm)x (15.0 cm)= 3000 cm? [1/2]
Hence, molecular formula
Mass of the alloy = density x volume =nx CH= 2x CH=C,H, [1]
= (8.17 g cm °)x (3000 cm®) = 24510 g
3. Arrange the following in order of their
8 increasing masses in gram (i) One atom of silver,
Mass ofNi in the alloy = (24510 g ise = 11225.6 g
[1/2] (ii) one gram-atom of nitrogen, (iii) one mole of
59 g Ni have atoms = 6.022 x10” calcium, (iv) one mole of oxygen molecules, (v)
Gram atomic mass of Ni =59 g {1] 1073 atoms of carbon and (vi) one gram of iron.
11225.6 g of Ni have atoms Sol. (i) 1 mole of Ag atom = 108 g= 6.022 107° atoms.

= 6.022% 10° 41225-68)= 9 15x10" atoms Mass of 6.022 10° atoms of Ag = 108 g.
(59.0 g) [1/2]
Mass of 1 atom of Ag = ( a = 1,793X 10“ g
2. Awelding fuel gas contains carbon and 6.022x 10° (1
hydrogen only. Burning a small sample of it in (ii) Mass of gram atom of N = atomic mass ofN in gram
oxygen gives 3.38 g carbon dioxide, 0.690 g of = 14.0 g.
water and no other products. A volume 10.0 L (iii) Mass of a mole of Ca= atomic mass of Ca in gram
(measured at STP) of this welding gas is found to = 40.0 g. [1]
weigh 11.6 g. Calculate (i) empirical formula (ii) (iv) Mass of mole of oxygen molecules = molar mass of
molar mass of the gas and (iii) molecular oxygen in gram = 32.0 g. [1]
formula. [NCERT Textbook] (v) Mass of mole of C-atom = 12 g= 6.023 107° atoms.
Sol. (i) 44g CO, = 12g carbon Mass of6.023 x 10*° atoms of C = 12 g.
12
3.38 g CO, = ie x 3.38 g = 0.9218 g carbon Mass of 1 atom of C = Geer g
6.023 10”
18 g H,O= 2g hydrogen Mass of 10”° atoms of
0.690 g H,O = =x 0.690 g 12
C=| ————_ |x 10% = 1.992 [I]
Ges os) °
= 0.0767 g hydrogen (vi) Mass of iron =1.0 g. Hence, the required order of
Total mass of compound = 0.9218+ 0.0767 = 0.9985 g increasing masses is one atom of silver < one gram of
(because compound contains only carbon and hydrogen) iron <10*? atoms of C < one-gram atom of nitrogen
< one mole of oxygen < one mole of calcium. [lig
Some Basic Concepts of Chemistry
33
4. A compound made up of two elements A and B Calculation of Empirical Formula
has A= 70%, B = 30%. Their relative number of
Relative se a Simplest whole
moles in the compound are 1.25 and 1.88. Calculate Simpl , t
Element ©number of : number molar
(i) atomic masses of the elements A and B Ae moles molar Kgtio ratio
i
(ii) molecular formula of the compouna, if its A 1.25 12bq 4 2
molecular mass is found to be 160. ant W246)

Sol. Relative number of moles of an element B 1,88 1.88 _ 4. 3

_ % of the element Wes)

atomic mass . Empirical formula= A,B, | [2]

% of element Calculation of molecular formula
Qi’ Aitopaelton mks —e——
relative number of moles Empirical formula mass = 2 x 56 +3 X16 = 160
. Atomic mass of A = tls =156 molecular mass 160
1e25
n= — =—=1 [2]
empirical formula mass 160
: 30
and Atomic mass of B =—— = 16 {1] . Molecular formula = A,B,
1.88

SSE YOUR TOPICAL UNDERSTANDING

OBJECTIVE Type Questions 4. Which of these is not an empirical formula?
|1 Mark| N,0,, CCl, Ce H i 0, C,H,0.

1. Given that, the abundances of isotopes **Fe, °°Fe SHORT ANSWER Type I Questions
and °’Fe are 5%, 90% and 5% respectively, the |2 Marks|
atomic mass of Fe is
(a) 55.85 (b) 55.95 (c) 55.75 (d) 56.05 1. Fe,(SO,)3 is used in water and sewage treatment to
2. How many number of aluminium ions are present in aid the removal of suspended impurities. Calculate
0.051 g of aluminium oxide? the mass percentage of iron and sulphur in this
(a) 6.023x 10” ions (b) 3 ions compound. [Ans. Fe = 28%; s = 24%]

(c) 6.023x 107” ions (d) 9 ions 2. Compute the mass of one molecule and the
molecular mass of C,H, (benzene)
3. An organic compound on analysis was found to
contain 10.06% carbon, 0.84% hydrogen and
(atomic mass of C = 12u, H = 1u). [Ans. 13x10-*’g]
89.10% chlorine. What will be the empirical formula 3. An organometallic compound on analysis was found
of the substance? to contain, C = 64.4%, H=5.5% and Fe = 29.9%.
(a) CH,Cl, (b) CHCl, (©) CCl, (d) CH,Cl Determine its empirical formula (atomic mass of
4. 1.020 g of metallic oxide contains 0.540 g of the Fe = 56 uw). [Ans. C,,H,,Fe]
metal. If the specific heat of the metal, M is 0.216 4. Calculate the number of gram of oxygen in 0.10
cal deg-1g~. The molecular formula of its oxide is mole of Na,CO, -10H,0. ; [Ans. 20.8 g]
(a) MO (b)M,0, (c)M,0, — (a) M,0 5. Calculate the percentage composition of the various
Ans. 1.(b) 2.(c) 3.(b) 4.(b) elements in MgSO,. A
[Ans. Mg = 20%; S = 26.67; O = 53.33]
VERY SHORT ANSWER Type Questions
[1 Mark| SHORT ANSWER Type II Questions
1. How many moles of atoms are present in 9.0 g of |3 Marks|
aluminium? [Ans 0.33 mol] 1. To account for atomic mass of nitrogen as 14.0067 u,
2. Calculate the mass of a sample of iron metal that what should be the ratio of °N and ‘N atoms in
contains 0.250 mole of iron atoms. [Ans. 14g] natural nitrogen? (atomic mass of “N= 14.00307 u
3. Describe the difference between the mass of a mole and °N= 15.001 wv).
of oxygen atom (0) and the mass of a mole of [Ans = 0.364 : 99.636]
oxygen molecule (0,). [Ans A = 14g]
 |Allézone | CHEMISTRY Class 1ith
34
2. A compound contains 4.07% hydrogen, 24.27% LONG ANSWER Type Questions
carbon and 71.65% chlorine. Its molar mass is 98.96 g. |5 Marks|
Determine its empirical and molecular formulae.
1. (i) The volume of a drop of rain was found to be
[Ans. EF = CH,Cl; MF = C,H,Cl,] 0.448 mL at NTP. How many molecules of water
3. The mass of a litre of oxygen at standard conditions and number of atoms of hydrogen are present
of temperature and pressure is 1.43 g and that of a inthis drop? [Ans15 x10” molecules of water
litre of SO, is 2.857 g. 30 x 10” atoms of Hydrozen]
(i) How many molecules of each gas are there in
this volume? (ii) How many atoms of He are present in 52 u
of He? [Ans 13 atoms]
[Ans 2.6910” molcules of each]
(iii) Insulin contains 4.5% sulphur. Calculate the
(ii) What is the mass in gram of a single molecule
minimum molecular mass of sulphur.
of each gas?
[Ans 711u]
[Ans mass of O, molecule = 5.3210 “’g;
mass of SO, molecule = 1.06 x10 “g] 2. Distinguish between the following.
(i) Atomic and molecular mass.
(iii) What are the molecular masses of SO, and 0, (ii) Atomic mass and atomic weight
respectively?
(iii) Empirical and molecular formulae.
[Ans 64u and 32u]
(iv) Gram atom and mass of one atom
4. (i) How many molecules approximately do you (v) Moles and molecules.
expect to be present in (a) a small sugar crystal
3. (i) How empirical formula and molecular formula are
which weighs 10 mg (b) one drop of water with
related to each other? Explain with an example.
0.05 cc volume?
(ii) A compound on analysis gave the following
[Ans (a) = 1.76 x10"” molecules of sugar
percentage composition Na = 14.31%,
(b) =1.67 x10”! molecules of water]
S = 9.97% ,H = 6.22% and 0 = 69.50%.
(ii) 9.7x 10’” atoms of iron weigh as much as 1 cc Calculate the molecular formula of the
of H, at STP. What is the atomic mass of iron? compound assuming that all the hydrogen in
[Ans A = 55.41u] the compound is present in combination with
5. What is molecular mass? How is it calculated? oxygen as water of crystallisation. The
Explain with two examples. molecular mass of the compound is 322.
[Ans Na,SH90,4]

ITOPIC 3|
Chemical Reactions and their Equations and Stoichiometric Calculations
CHEMICAL REACTIONS whereas, the substances which are produced as a result of
A chemical reaction is a change in which one or more the chemical change, are known as products. Reactants and
substance(s) react(s) to form new substance(s) with entirely products of a chemical equation are separated by arrow
different properties. pointing towards the products.
Characteristics of a Chemical Equation
Chemical Equations A chemical equation must fulfill the following conditions.
A chemical equation is a brief representation of a chemical
(i) It must be consistent with the experimental facts.
reaction in terms of symbols and formulae of substances
involved in it, e.g. the reaction of silver nitrate with sodium (ii) It should have molecular species.
chloride to give silver chloride and sodium nitrate can be (iii) It should be balanced i.e. it follows law of
represented as conservation of mass and have equal number of
atoms of each elements on both the sides.
AgNO, + NaCl ——> AgCl+ NaNO,
Se eee) oS erie tOi ee tee
Reactants Products
Information Conveyed by a Chemical Equation
The substances which react among themselves to bring A chemical equation gives qualitative as well as quantitative
about the chemical changes are known as reactants informations.
Some Basic Concepts of Chemistry
35
(a) Qualitative Information * Old STP conditions, 273.15 K, 1 atm, volume occupied by 1mole of a
gas =22.4L.
It tells about the names of various reactants and products
involved in a reaction. It gives an idea for the number of * New STP conditions, 273.15 K, 1 bar, volume occupied by 1 mole of a
molecules taking part in the reaction or formed in the gas =22.7 L
reaction.
Balancing a Chemical Equation
The chemical equation can be made more information by
A chemical equation must be in accordance with the law of
incorporating the following changes. conservation of mass which states that the total mass of the
(i) The physical states of reactants and products can be reactants must be equal to the total mass of the products.
indicated by using the abbreviations (s) for solid, (J for
In other words, the number of atoms of each kind in the
liquid, (g) for gas and (aq) for aqueous solution. e.g. reactant side must be equal to the number of atoms of same
Cu(s) + 2HCl (aq) —> CuCl, (aq) +H, (g) kind in the product side.
(ii) In order to indicate the strength of acid or base, dil. Thus, a chemical equation having an equal number of
for dilute or conc. for concentrated is written before atoms of each element in the reactant and the product sides
the formula of acid or base. is called the balanced chemical equation.
Cu(s) +2 dil. HCl (aq) —> CuCl, (aq) +H, (g) e.g. 4 Fe(s)
+ 30) (g) —> 2 Fe,O3(s)
(iii) The reaction conditions, such as presence of catalyst, is a balanced equation.
temperatue, pressure, etc., may be written above the
arrow between the reactants and products. whereas
Fe/Mo,723K is an unbalanced equation.
N,(g) +3 H, (g) aaa eee p) NH; (g)
There are several methods to balance a chemical equation.
These methods are
(iv) Heat change taking place during the reaction may be
expressed in any one of the following two ways. (a) hit and trial method or trial and error method.
C(g) +O (g) ——> CO, (g) + 93.6 kJ ; (b) partial equation method.
AH =—936 kJ (c) oxidation number method.
(b) Quantitative Information (d) ion-electron method.
Quantitative information conveyed by a chemical equation The first two methods are discussed below while the
is as follows. other methods will be taken up in chapter 8 (Redox
reactions).
(i) the relative number of reactant and product species
(atoms or molecules) taking part in the reaction. (a) Hit and Trial or Trial and Error Method
(ii) the relative number of moles of the reactants and products. It is the simplest method used to balance a chemical
(iii) the relative masses of the reactants and products. equation. Though there is no definite rules for balancing an
(iv) the relative volumes of gaseous reactants and products.
equation by this method, yet the following steps are
generally used to balance an equation.
Consider the following chemical equation.
Algorithm for Hit and Trial Method
CH,4(g) + 20,(g) —> CO,(g) + 220 (g)
The following information is obtained from the chemical Step 1 Write down the skeletal equation using symbols
equation. and formulae of the reactants and products.
(i) One molecule of CH4(g) reacts with 2 molecules of Step Il Change elementary gases (like hydrogen, oxygen
O, (g) to give one molecule of CO,(g) and and nitrogen), if present, to their atomic states.
2 molecules of H,O (g). Step III Start balancing the equation by selecting the
(ii) One mole of CH4(g) reacts with 2 moles of O2(g) formula containing the maximum number of atoms
and balance the number of atoms of each of its
to give 1 mole of CO,(g) and 2 moles of HO (4g).
constituents on both sides of the equation by
(iii) 16 g of CH4(g) reacts with 2 x 32g of O2(g) to give multiplying with suitable numbers. Then proceed to
44 g of CO, and 2x 18 g of Hj,O (g). balance the other atoms, if they are not balanced
(iv) 22.7 L of CH4(g) reacts with 45.4 L of O,(g) to already.
give 22.7 L of CO, and 45.4 L of H,O(g) at STP.
 |Alléwone | CHEMISTRY Class 11th
36
Or EXAMPLE |2| Balance the following skeleton
In case, the above method fails, then start balancing equation by the method of partial equations.
the atoms which appear minimum number of times KMnO, + H,SO, + (COOH), —~K,SO, +MnSO, + CO, + H,0
and atoms of elementary gases are balanced last of all. Sol. The oxidation of oxalic acid, (COOH), by potassium
permanganate, KMnO, takes place in the following steps
Step TV Once all the atoms are balanced, change the
(i) KMnO, reacts with dil. H,SO, to produce nascent
equation into the molecular form. oxygen.
Step V Verify that the number of atoms of each element KMnO, + H,SO, —> K,SO, + MnSO,
is balanced in the final equation. + 3H,O+ [O]
On balancing this skeleton equation by hit and trial
Note :
method, we get
Remember that subscripts in formulae of reactants and products
cannot be changed to balance an equation. 2KMnO, + 3H,SO, —> K,SO, + 2MnSO,
+ 3H,O + 5[O] ...(i)
(ii) Oxalic acid is oxidised to CO, and H,O by the nascent
EXAMPLE |1] Balance the following equation.
oxygen produced in equation (i). The balanced partial
Fe+ H,0 —— Fe,0, + H, equation for this reaction is
Sol. Step 1 As the question is already in skeletal form of (COOH), + [0] —> 2 CO, + H,O ..-(ii)
equation, so, step is completed. To cancel the intermediate product ie. nascent
Step Il Changing the elementary substance hydrogen to oxygen, multiply equation (ii) by 5 and adding to (i),
atomic form. we have
Fe+ H7O=—>\Fe,0, 42h 2KMnO, + 3H,SO, + 5 (COOH), —>
Step Ill Fe,0, has the largest number of atoms. To K,SO, + 2 MnSO, + 10 CO, + 8H,O
balance this, multiply H,Oby 4 to balance oxygen This represents the balanced chemical equation for
atoms. In 4 molecules of H,O, there are 8 atoms of the above reaction.
H which are balanced by multiplying H on RHS
by 8.
3 Fe+ 4H,0—> Fe,0, + 8H
STOICHIOMETRY AND
Step IV Converting H-atom to molecular form. STOICHIOMETRIC
3Fe+ 4H,0O—— Fe,0, + 4H, CALCULATIONS
Step V The number of Fe, H and O atoms on both the
sides of the equation are respectively 3, 8 and 4.
The word ‘stoichiometry’ is derived from the combination
The equation obtained is a balanced chemical of two Greek words i.e. stoicheion (means element) and
equation. metron (means measure).

(b) Partial Equation Method Stoichiometry, thus, deals with the calculation of masses
(sometimes volumes also) of the reactants and the products
The above method is helpful only for simple reactions but involved in a chemical reaction or in other words, the
in case of complex reactions, where the same element is relationship between the amounts of reactants and products
repeated in a number of compounds, partial equation is called the stoichiometry.
method is used.
The number before the formula unit or molecules used to
It involves the following steps.
balance the equation are called stoichiometric coefficients.
(i) The given chemical reaction is written in various These shows the number of moles/molecules of that
probable steps which are called as partial equation. particular substance. Calculations based on chemical
(ii) Each partial equation is separately balanced by hit equations are termed as stoichiometric calculations.
and trial method as discussed earlier.
(iii) In order to cancel out the intermediate species (not Algorithm for Stoichiometric
involed in the final equation), the partial equations Calculations
are multiplied by suitable integers.
Step 1 Write the balanced chemical equation.
(iv) Finally, the partial equations are added to get the
final equation. Step II Write the stoichiometric coefficient that shows the
moles of the respective substance.
Some Basic Concepts of Chemistry
ay
Step III Convert moles into required quantity (if needed). StepIV Write the given mass under the respective
Step TV Write the given data. substance.
Step V Calculate the required information by unitary CH, (g)+ 20,(g)
—> CO,(g)+ 2H,0 (g)
method. 16g 64 g 44g 36 g
These are the general steps involved in all types of StepV Calculate the mass of unknown substance by
stoichiometric calculations. unitary method.
These calculations are of the following types. From the equation, it is already clear that 16 g of
methane produces 36 g of water.
(a) Calculations Involving
Mole-Mole Relationship (c) Calculations Involving Mass-Mole
In such problems, the moles of one of the reactant/product is or Mole-Mass Relationship
to be calculated, if the moles of others are given. In such problems, the mass or mole of one of the
reactant/product is to be calculated if we have the mole or
EXAMPLE |3| How many moles of CO, will be mass of others.
obtained when 0.274 mole of C,H.OH is burnt?
Follow the following steps to solve such problems EXAMPLE |5| Oxygen is prepared by the catalytic
Sol. Step Write down the balanced equation for the given decomposition of potassium chlorate (KC10;).
reaction and then write the stoichiometric Decomposition of potassium chlorate gives potassium
coefficient and after that given moles. chloride (KCl) and oxygen (0,). If 2.4 moles of oxygen is
C,H,OH + 30,— > 2CO, + 3H,O needed for an experiment, how many grams of
potassium chlorate must be decomposed?
v2
Stoichiometric coefficient 1 mol
Given moles 0.274 mol
Sol. The balanced equation is
Step II Calculate the amount of the CO, by unitary
method. 2 KCIO,(s)
—> 2 KCl(s) + 30,(g)
2 mol 3 mol
1 mole of C,H,OH produces 2 moles of CO, 2x (39+ 355+3%X16)=245¢g 2.4 moles
. 0.274 mole of C,H,OH will produce CO, ?

= ‘3xX 0.274 = 2X 0.274 = 0.548 mol 3 moles of O, is produced by decomposition of

1 KCIO, = 245 g
Thus, 0.548 moles of CO, is obtained from 0.274 ‘. 2.4 moles of O, will be produced by the decomposition of
moles of C,H,OH.
KCIOg SrtA p96.
(b) Calculations Involving Mass-Mass
Relationship EXAMPLE |6| How many moles of methane are
In such problems, the mass of one of the reactant/product is required to produce 22 g CO, (g) after combustion?
to be calculated if we have the mass of others. [NCERT Textbook]
Sol. According to the chemical equation,
EXAMPLE |4| Calculate the amount of water (g) CH,(g) + 20,(g)
—> CO,(g) + 2 H,0(g)
produced by the combustion of 16 g of methane. 44 g CO,(g) is obtained from 16 g CH,(g).
[NCERT Textbook] [.. 1 mol CO, (g) is obtained from 1 mol of CH, (g)]
Solution of such problems involve following steps 1 mol CO,(g)
Moles of CO,(g) = 22g CO,(g) x
Sol. Step I Write the balanced chemical equation. 44¢ CO,(g)

CH,(g) + 20,(g) —> CO,(g) + 2 H,0(g) = 0.5 mol CO, (g)

Step II Write the stoichiometric coefficients of each Hence, 0.5 mol CO, (g) would be obtained from 0.5 mol
substance. These represent mole of that CH, (g) or 0.5 mol of CH, (g) would be required to
particular substance. produce 22 g CO, (g).

CH,(g) +20,(g)—> CO,(g)+2H,0 (s) (d) Calculations Involving

1 mol 2 mol 1 mol 2 mol
Mass-Volume Relationship
Step III Convert moles into mass by multiplying them
with molar mass. In these problems, mass or volume of one of the reactants
CH, (g) + 20,(g)——> CO, (g) + 2H,0 (g) or products is calculated from the mass or volume of
1 mol 2x 32 1x 44 2x18 other substances.
= 16g = 64g =44¢ = 36g
 | AllZwone | CHEMISTRY Class 11th
38
Sol. The balanced chemical equation is
EXAMPLE |7| Calculate the amount of KCIO, needed to
2C,Hy + 5 Opee—y4COpe 2H,0
supply sufficient oxygen for burning 112 L of CO gas at NTP. 2 mol 5 mol 4 mol :

5x 22400 cm 4x 22400 cm
Sol. Calculation of O, gas required to burn 112 L of CO. 2x 22400cm?

COG 502 — CO, 2x 22400 cm? of acetylene require O, for complete

1mol 0.5 mol 1 mol combustion = 5x 22400 cm®
=22.4L .
200 cm? of acetylene will require O, for complete
22.4 L of CO at NTP require O, = 0.5 mol 5X 22400
112 L of CO at NTP will require combustion = x 200 = 500 cm? at STP
2 X22400
O, _ aay 112= 2.5 mol Further 2 22400 cm? of acetylene produce CO,
22.4
This O, is to be obtained by heating KCIO3. = 4x 22400cm*
.. 200 cm? of acetylene will produce CO,
2 mol 3 mol _ 4x 22400 x 200= 400 cm? at STP
=
2 (39+ 35.5+ 3X 16)= 245 g
2 x 22400
3 moles of O, are produced from KCIO, = 245g
2.5 moles of O, will be produced from KCIO,
xX 2.
Limiting and Excess Reagents
The reactant which is present in the lesser amount and gets
consumed after sometime i.e. which limits the amount of
EXAMPLE |8| Hydrogen gas is prepared in the product formed is called the limiting reagent.
laboratory by reacting dilute HCl with granulated zinc. The reactant other than the limiting reagent, which is in
Following reaction takes place. somewhat excess is called the excess reagent. The
Zn+ 2HCL—> ZnCl, + H, remaining amount of excess reagent is calculated by
Calculate the volume of hydrogen gas liberated at STP when subtracting the available amount of limiting reagent from
32.65 g of zinc reacts with HCl. 1 mole of a gas occupies 22.7 L the amount of the excess reagent. Remember that in
volume at STP; atomic mass of Zn= 65.3u. stoichiometric calculations, it is very important to choose
[NCERT Exemplar] the limiting reagent as it decides the amount of the
Sol. Given that, mass of Zn = 32.65 g product obtained.
1mole of gas occupies = 22.7 L volume at STP
Atomic mass of Zn = 65.3 u EXAMPLE |10| 50.0 kg of N,(g) and 10.0 kg of
The given equation is H,(g) are mixed to produce NH;(g). Calculate the
Zn + 2HCI—=> ZnCl,+ H, NH;3(g) formed. Identify the limiting reagent in the
65.3.g 1 mol= 22.7 L at STP production of NH; in this situation. [NCERT Textbook]
From the above equation, it is clear that Sol. Step1 A balanced equation for the above reaction is
written as follows.
65.3 g Zn, when reacts with HCl, produces
= 22.7 L of H, at STP N,(g)+ 3H2(g) == 2 NH; (g)
Step II Calculation of moles.
. 32.65 g Zn, when reacts with HCl, will produce
Moles of N,
£122),7%32', 65
= 11.35 L of H, at STP. =50.0kgN, x 1000 gN, sed molN,
65.3 lkgN, 280gN,
(e) Calculations Involving =17.86x10° mol
Volume-Volume Relationship Moles of H,
In these problems, the volume of one of the reactant/ product = 10.00 kg H, x
1000 g H, lImol H,
is given and that of the other is to be calculated. 1kg H, 2.016 g H,

EXAMPLE |9| What volume of oxygen at STP is = 4.96 x10° mol

required to effect complete combustion of 200 cm? of Step III Then, find out the limiting reagent.
acetylene and what would be the volume of carbon dioxide 17.86 x 10° molN, x 3molH, (g) =5.36 10° mol H,
formed? 1 mol N, (g)
Some Basic Concepts of Chemistry
oY
But we have only 4.96x10° mol H,. Hence, The concentration of the solution is usually expressed in the
dihydrogen is the limiting reagent in this case. So, following ways.
NH, (g) would be formed only from that amount
of available dihydrogen ive. 4 -96 x 10° mol. (a) Mass Per cent
StepIV After that, find out the amount of ammonia. It is the mass of acomponent per 100 g ofthe solution. Ina
Since, 3 mol H,(g) gives 2 mol NH, (g) solution of the two components
2mol NH, (g) f sol
4.96 x 10° mol H, (g) x Mass per cent = nical av [yy
3 mol H, (g) mass of solution
= 3.30 x 10° mol NH; (g) e.g. 10 % (w/W) solution of sodium chloride means that 10
3.30 x 10° mol NH,,(g) is obtained from g of sodium chloride is present in 90 g of water so that the
total mass of the solution is 100 g.
4.96 X10° mol H,(g). |
If they are to be converted to grams, it is done as follows.
EXAMPLE |12] A solution is prepared by adding 2 g of
1 mol NH, (g) = 17.0gNH,(g) 3.30 10° mol
substance A to 18 g of water. Calculate the mass per cent of
NH, (g)x
17.0 g NH; (g) the solute. [NCERT Textbook]
1 mol NH, (g)
Sol. Mass per cent of A = vies che x 100
3.30 x 10° x 17g NH, (g)= 561 x10°g NH, Mass of solution
= 561kg NH, pe
oe ge ee N10
2g of A+ 18 g of water 20g
EXAMPLE |11| Ina reaction, A + B, ——> AB,
identify the limiting reagent if any in the following (b) Volume per cent
reaction mixtures. It is defined as the volume of the component per 100 parts
(i) 300 atoms of A + 200 molecules of B by volume of the solution. e.g. if V4 and Vg are the volumes
(ii) 2 moles of A+3 moles of B of two components A and B respectively in a solution, then
(iii) 100 atoms of A+100 molecules of B
Volume per cent of A= iss ORLA aee x 100
(iv) 5 moles of A + 2.5 moles of B Volume of A+ volume of B
(v) 2.5 moles of A+5 moles of B [NCERT Textbook] This may be expressed as v/V.
Sol. The given equation shows that 1 mole of A reacts Sometimes, we express the concentration as weight/volume.
with 1 mole of B, and 1 atom of A reacts with 1
e.g. a 10% solution of sodium chloride (w/V) means that 10
molecule of B,
g of sodium chloride is dissolved in 100 mL of solution.
(i) Bis the limiting reagent because 200 molecules of B,
will react with 200 atoms of A and 100 atoms of A will
(c) Parts per million
be left in excess.
(ii) A is the limiting reagent because 2 moles of A will react When a solute is present in very minute amounts (trace
with 2 moles of B and 1 mole of B will be left in excess. quantities), the concentration is expressed in parts per
(iii) Both will react completely because it is stoichiometric million, abbreviated as ppm. It is defined as the parts of a
mixture. No limiting reagent. component per million parts of the solution. It is expressed as
(iv) 2.5 moles of B will react with 2.5 moles of A, hence Bis mass of component A
the limiting reagent. ppm of A= x 10°
total mass of solution
(v) 2.5 moles of A will react with 2.5 moles of B. Hence A
is the limiting reagent. For example, suppose a litre of public supply water contains
about 310° g of chlorine. The parts per million of
Reactions in Solutions chlorine is
Most of the reactions occurring in the laboratories are
axdow x 10°>
carried out in solutions. In solutions, generally two . ppm of chlorine =
1000
components are present.
* Atmospheric pollution in cities due to harmful gases is generally
The one which is in lesser amount is called the solute and expressed in ppm though in this case the values refer to volumes
the other one which is in higher amount is called the rather than masses. e.g. the concentration of SO, in Delhi has been
solvent. The amount of solute present in a given quantity of found to be as high as 10 ppm. This means that 10 crf of SO, are

solvent or solution is expressed in terms of concentration. present in 10° cnt (or 10° L) ofair.
 |AllZwone | CHEMISTRY Class 11th
40

(d) Mole Fraction (x) If we have % by mass and specific gravity or density, the
molarity is calculated by the expression,
It is the ratio of number of moles of a particular component Hy, % by weight x specific gravity x 10
to the total number of moles of the solution.
al molar mass
In case ofa solution of two components A and B,
In case of dilution, M,V, =M,V,, where, M,V, are
Mole fraction of A, volume before dilution and
ny respectively the molarity and
number of moles of A dilution.
Qo M,V, are molarity and volume after
4 number of moles of solution 4 +78
© Molarity of asolution depends upon temperature because volume of
(where n 4 and mp are the moles of A and B respectively). a solution is temperature dependent.
Mole fraction of B,
number of moles of B np EXAMPLE |14| Calculate the molarity of NaOH in the
xg =
number of moles of solution n,4 +n, solution prepared by dissolving its 4 g in enough water to
form 250 mL of the solution.
Also remember that x 4 + x, =1 number of moles of solute
ie j it M) = ——
Mole fraction is independent of temperature. Sokesinee ae volume of solution in litres
_ Mass of NaOH / Molar mass of NaOH
EXAMPLE |13| A solution contains 25% water, 25% 0.250 L
ethanol and 50% acetic acid by mass. Calculate the mole _4g/40g __ 0.1 mol
fraction of each component. 0.250L 0.250 L
Sol. Let the total mass of solution = 100 g
=0.4mol L'=0.4M
Mass of water = 25 g, Mass of ethanol = 25 g
Note
Mass of acetic acid = 50g
Molarity of a solution depends upon temperature because volume of
Moles of water = = = 1.388 ( Molar mass of H,O = 18) a solution is temperature dependent.

Moles of ethanol = 25/46 = 0.543 EXAMPLE |15| What volume of 10 M HCl and 3 M HCl
(.“ Molar mass of C,H,OH = 46) should be mixed to get 1L of 6 M HCl solution?
Moles of acetic acid = 50/60 = 0.833 Sol. Suppose volume of 10 M HCI required to prepare 1L of 6
(.“ Molar mass of CH,COOH = 60) MHCl=xL
Total number of moles = 1.388+ 0.543+ 0.833 = 2.764 Volume of 3 M HCI required = (1— x) L

Mole fraction of water = = 0.502 Applying molarity equation M,V, + M,V, = M,V,
— e—— ee
4 10MHCl 3MHClL 6MHCI

Mole fraction of ethanol = Gees = 0.196 10x x +30—= x)=6x1

2.764
— be cas taySn eye ey i x =5=0428 L
, ; ; 0.83
Mole fraction of acetic acid = Dern=10'302
.76 “. Volume of 10 M HCI required = 0.428 L= 428 mL
(e) Molarity (M) Volume of 3 M HCI required = 1 — 0.428 = 0.572 L = 572 mL

It is defined as the number of moles of the solute in 1 L of

(f) Molality (m)
the solution. It is denoted by M. Thus, It is defined as the number of moles of solute present in
number of moles of solute 1 kg of solvent. It is denoted by m.
Molarity (M) = number of moles of solute
volume of solution in litres Thus, molality (7) =
mass of solvent in kg
As we know, moles = ae
molar mass mass of solute x 1000
or m =
Mena ae mass of solute (W) molar mass of solute X mass of solvent (in g)
molar mass of solute (M/) * Molality is considered better for expressing the concentration as
x volume (V) ofsolution (Z) compared to molarity because molarity changes with temperature
due to the expansion or contraction of the liquid but, molality does
is i es mass of solute x 1000 not change with temperature as the mass of the solvent does not
molar mass of solute X V (in mL) change with change in temperature.
* However, molarity is widely used because it is easier to measure the
The units of molarity are mol L™! or mol dm~, volume of a solution.
Some Basic Concepts of Chemistry
ad
EXAMPLE |16| Calculate the molality of a solution = 1074.5 g = 1.0745 kg.
containing 20.7 g potassium carbonate dissolved in 500 Mobuiye number of moles of solute _ 3mol
mL of solution (assume density of solution = 1 mle = = 2.79 m
mass of solvent in kg 1.0745 kg
Sol. Mass of K,CO, = 20.7 g
Molar mass of K,CO, = 2x 39+ 12+ 3X 16 = 138 mol!
(g) Normality (JV)
It is defined as the number of gram equivalents of solute
Moles of K,CO, = eB ee 0.15 present per litre of solution. It is denoted by ‘N’.
138
Mass of solution = (500 mL) x (1 g mL”')= 500 g number of gram equivalents of solute
Normality (V) =
Amount of water = 500— 20.7= 479.3 g volume of solution in litres
Moliity moles of solute eee mass of solute
mass of solvent in gram Number of gram equivalent of solute =
equivalent mass
0.15
=
479.3
1000
= 0.313 m The unit of normality is g equiv L"'. Like molarity,
normality of a solution also changes with temperature.
* Often in a chemical laboratory, a solution of a desired
concentration is prepared by diluting a known solution of higher
concentration. The solution of higher concentration is also known EXAMPLE |18} Calculate the normality of solution
as Stock solution. containing 62.3 g of hydrated copper sulphate
(CuSO, -5H,0) in 500 mL of solution.
EXAMPLE |17| The density of 3 M solution of NaCl is Sol. Mass of solute = 62.3 g
1.25 g mL’. Calculate molality of the solution.
. 5 F 249.5
[NCERT Textbook]
Equivalent mass of oxalic acid = ae = 124.75 g
Sol. M=3 molL 62.3
Mass of NaCl in 1 L solution = 3 x 58.5= 175.5 g Gram equivalents of oxalic acid = — = 0.
124.75
Mass of 1 L solution = 1000 x 1.25= 1250 g
Volume of solution = 500mL
(since density = 1.25 g mL‘)
0.
Mass of water solution = 1250— 175.5 Normality = sii x 1000=1N
500

TOPIC PRACTICE 3 |
OBJECTIVE Type Questions 12
% of CinCO»o= wT x 100 = 27.27%
{1 Mark|
Hence, the mass per cent of carbon in CO,» is 27.27%.
1. Which of the following equations is 3. Air contains 20% O, by volume. How much
unbalanced?
(a) 4Fe (s) + 30,(g) —— 2Fe,0,(s)
volume of air will be required for combustion
of 100 cc of acetylene?
(b) 2Mg(s) + O2(g) —> 2MgO(s ) (a) 500 cc (b) 1064 cc
(c) Py(s) + O2(g) —> PO4o(s) (c) 212.8 cc (d) 1250 cc
(d) CH,(g) + 20,(g)
—> CO,(g) + 2H,0(g) Sol. (d) 2C,H, + 50, —— 4CO, + 2H,0
Sol. (c) The unbalanced equation is 2 cc Dice
100 cc 250 cc
P,(s) + O2(g) ——> P4Ojo(s)
100
It can be balanced as follows : Hence, air will be needed = Pr PAVE IVA we,
P,(s) + 502(g) —— > P,Oj9(s)
2. What is the mass per cent of carbon in carbon 4. One mole of any substance contains 6.022x Oe
dioxide? [NCERT Exemplar] atoms/molecules. Number of molecules of
(a) 0.034% (b) 27.27% — (c) 3.4% (d) 28.7% H,SO, present in 100 mL of 0.02MH,SO,
Sol (b) Molecular mass of CO, =1x12+ 2x16 = 44g solution is ......... : [NCERT Exemplar]
1 g molecule of CO, contains 1g atoms of carbon (a) 12.044 x10” molecules _(b) 6.022 x 10° molecules

- 44g of CO, contain C = 12g atoms of carbon (c)1 x 10”? molecules (d) 12.044 x 10”? molecules
42 |Allzwone | CHEMISTRY Class 11th

Sol. (a) Number of moles of H,SO, 8. Ifthe concentration of glucose (CgH;0¢) in

= molarity x volume in mL blood is 0.9 gL *, what will be the molarity of
= 0.02 x 100 = 2 millimoles glucose in blood? [NCERT Exemplar]
= 2x10 * mol (a)5M (b) 50 M (c)0.005M (d)0.5M
Number of molecules = number of moles x N 4 Sol. (c) In the given question, 0.9 g L-' means that 1000 mL
(or 1L) solution contains 0.9 g of glucose.
== Dsl Qi ax 6102 21e 1088
0.9
.. Number of moles = 0.9 g glucose = a mol glucose
= 12.044 x 107° molecules

5. How many grams of concentrated nitric acid ‘. (molecular mass of glucose

(C,H,,O,)= 12x 6+ 12x 1+ 6X 16= 180u)
solution should be used to prepare 250 mL of
2.0 MHNO,? The concentrated acid is 70% =5x10 °* mol glucose
HNOs3. i.e. 1L solution contains 0.005 mol glucose or the molarity
(a) 45.0 g conc. HNO, of glucose is 0.005 M.
(b) 90.0 g conc. HNO;
9. What will be the molality of the solution
(c) 70.0 g conc. HNO,
containing 18.25 g of HCl gas in 500 g of water?
(d) 54.0 g conc. HNO,
[NCERT Exemplar]
Sol. (a): Molarity
(a) 0.1m (b) 1M (c) 0.5m (d)1m
Weight of HNO, Moles of solute ;
* Molecular mass of HNO, x Volume of solution ( in L) Sol. (d) Molalit = i
@) pee hig Mass of solvent (in kg) @
“. Weight of HNO,
Given that, Mass of solvent (H,O) = 500 g = 05 kg
= Molarity x Molecular mass x Volume (in L) Weight of HCl = 18.25 g
Molecular weight of HC] =1x1+1X35.5 = 36.5g/mol
= 2x63x = 315¢
Moles of HCl = si = (5
It is the weight of 100% HNO, 36

But the given acid is 70% HNO, 0.5

100
ae
=—=l1m [fromEq.q. (i)]
(i
Its weight = 31.5 x age =45¢
10. What will be the molality of the solution made
6. What will be the molarity of a solution, which by dissolving 10 g of NaOH in 100 g of water?
contains 5.85 g of NaCl(s) per 500 mL? (a) 25m (b) 5m (c) 10m (d) 1.25 m
[NCERT Exemplar] Sol 'Malality Mass of solute : 1000
(a) 4 mol L"! (b) 20 mol L™! Mol. mass of solute Mass of solvent
(c) 0.2 mol L”! (d) 2 mol L™! 10 1000
m=— X —=2.5m
Sol. (c) Molarity =
weight x 1000 40 100

molecular weight x volume (mL)

585 x 1000 VERY SHORT ANSWER Type Questions
585 x 500
=0.2molL”
|] Mark|
7. 1f£500 mL of a 5M solution is diluted to 1500 I. What do you understand by stoichiometric
mL, what will be the molarity of the solution coefficients in a chemical equation?
obtained? [NCERT ¥ <emplar] Sol. The coefficients of reactants and products involved in a
(a) 1.5M (b) 1.66 M chemical equation represented by the balanced form, are
(c) 0.017 M (d) 1.59M known as stoichiometric coefficients.
Sol. (b) For dilution, a general formula is e.g. N2(g)+ 3H2(g) —> 2NH,(g)
MV, = MV,
The stoichiometric coefficients are 1, 3 and 2 respectively.
(Before dilution) (After dilution)
2. Volume of a solution changes with change in
500 x 5M =1500 x M, temperature, then, will the molality of the
5 solution be affected by temperature? Give
M, = 3 =1.66M
reason for your answer. [NCERT Exemplar]
Some Basic Concepts of Chemistry 3)
Sol. No, molality of solution does not change with Sol. Fe,0,+3CO—-+ 2Fe +3CO,
temperature since mass remains unaffected with 16 mol ?
temperature. 3 moles of CO are used to make 2 moles of Fe.
2;
What is the difference between molality and Hence, 16 moles of CO are used to make —x16=10.67 mol
molarity? 3
[NCERT Exemplar]
Refer to text page 40. If 2 Lof Ny is mixed with 2 L of H, at a constant
Calculate the mass of sodium acetate, temperature and pressure, then what will be the
volume of NH; formed ?
CH3COONa required to make 500 mL of 0.375
molar aqueous solution. Molar mass of sodium Sol. N,(g)+ 3H, (g) —> 2NH;(g)
acetate is 82.0245 g molt. [NCERT] 1L of N, reacts with 3 L of H,.
w X 1000 Therefore, 2L of N, will react with 6 L of H, but we
Sol. Molarity = have only 2 L of H,, therefore, H, is the limiting
m X volume of solution (mL)
reactant. 3 L of H, gives 2 L of NH3.
where, w = mass of solute and m = molar mass of solute.
Given, molarity of the solution = 0.375 M .. 2LofH, gives = 2x pc
3 3
Molar mass of solute, m = 82.0245 g mol” !
= 1.33 L of NH;
Volume of solution = 500 mL
Mass of solute = ?
0.375 X 82.0245 x 500
SHORT ANSWER Type I Questions
. Mass of solute, w = = 15.3799
1000 |2 Marks|
~ 15.38 g
1. If the concentration of glucose (CgH,,0g) in
What is the concentration of sugar (Cj9H»70),) in blood is 0.9 gL !,what will be the molarity of
mol Lif its 20g are dissolved in enough water glucose in blood? [NCERT Exemplar]
to make a final volume up to 2 L? Sol. In the given question, 0.9 g L’ ‘means that 1000 mL
(or 1L) solution contains 0.9 g of glucose.
‘Ox To find molar concentration, calculate the molar mass
of the sugar, by adding atomic masses of different .. Number of moles =0.9g glucose = = mol glucose
elements as molarity = i
mxV =5x10 * mol glucose
Sol. Molar mass of the sugar, C,,H..0,,, (where, molecular mass of glucose

m = (12x 12.01) + (22x 1.0079) + (11x 16.00)

(C,H,,0,)
=12 X 6+12 X1+6 X16 =180u)
i.e. 1L solution contains 0.05 mole glucose or the molarity
= 342.2938 g mol | = 342 of glucose is 0.005 M. [2]
Given, w = 20g,V=2L
What will be the molality of the solution
Molarity ep ee ee og209 mol Le! containing 18.25 g of HCl gas in 500 g of water?
THY (Lie B42 a2
[NCERT Exemplar]
= 0.0292 M Sol. Molality is defined as the number of moles of solute
What will be the molarity of a solution, which present in 1 kg of solvent. It is denoted by m.
moles of solute
‘contains 5.85 g of NaCl(s) per 500 mL? Thus, Molality (m) = ...(i) [1]
mass of solvent (in kg)
[NCERT Examplar]
Given that, Mass of solvent
Sol. Since, molarity (M) is calculated by following formula.
(H,O)= 500 g =05kg
weight x 1000
Molarity = Weight of HCl = 18.25 g
molecular weight x volume (mL)
Molecular weight of HCl
[Molecular weight of NaCl = 585g] = OG xi3 350.
5.85 x 1000 =0.2molL”! = 36.5g
~ 585X500 5
Moles of HC] (solute) = cS =105
How many moles of iron can be made from
Fe,O3 by the use of 16 moles of carbon @ OY -= i [from Eq. (i)] [1]
monoxide in the following reaction? 0.5
Fe,03 + 3CO—> 2Fe+3CO,
 |Allévone | CHEMISTRY Class 11th
a4
For dilution, a general formula is
3. The reactant which is entirely consumed in
M,V, = M2V,
reaction is known as limiting reagent. In the Before diution After dilution
reaction2A + 4B —> 3C+4D, when5 moles of A
react with 6 moles of B, then 500 X5M =1500x M => M=—=1.66M (1]
wluo
(i) which is the limiting reagent?
(ii) calculate the amount of C formed? Sulphuric acid reacts with sodium hydroxide as
[NCERT Exemplar] follows.
Sol 2A+4B——> 3C+4D H.SO,+ 2NaOH —> Na,SO, + 2H,0
According to the given reaction, 2 moles of A react with When 1L of 0.1M sulphuric acid solution is
4 moles of B. allowed to react with 1L of 0.1M sodium
Hence, 5 moles of A will react with 10 moles of hydroxide solution, the amount of sodium
sulphate formed and its molarity in the solution
B ee = 10 mots|
obtained is [NCERT Exemplar]
(i) It indicates that reactant B is limiting reagent as it will Sol. For the reaction,
consume first in the reaction because we have only 6 H,SO, + 2NaOH ——> Na,SO, + 2H,0
moles of B. {1] 1L of 0.1 MH,SO, contains = 0.1 mole of H,SO,
(ii) Limiting reagent decide the amount of product 1L of 0.1 M NaOH contains = 0.1 mole of NaOH
produced. According to the reaction, 4 moles of B
produces 3 moles of C. According to the reaction, 1 mole of H,SO, reacts with 2
moles of NaOH. Hence, 0.1 mole of NaOH will react with
.. 6 moles of B will produce SENG =45 moles of C. [1] 0.05 mole of H,SO, (and 0.05 mole of H,SO, will be left
unreacted), i.e. NaOH is the limiting reactant. Since, 2
4. How are 0.50 mole Na,CO, and 0.50 MNa,CO, moles of NaOH produces 1 mole of Na,SOq.
Hence, 0.1 mole of NaOH will produces 0.05 mole of Na,SO, .
different? [NCERT Textbook]
Mass of Na,SO,= moles xX molar mass
Sol. Molar mass of Na,CO, = (2X 22.99) + 12.01+ (3x 16)
=05x(46+32+64)g=710g [I]
= 105.99= 106 g mol! (1)
Volume of solution after mixing = 2 L
0.50 mole Na,CO, = 0.50 106 = 53g Na,CO, Since, only 0.05 mole of H,SO, is left behind, as NaOH is
0.50 M Na,CO, means 53 g Na,CO,; is present in 1 L completely used in the reaction.
of the solution. [1]

5. How many gram of Na,CO; should be dissolved Therefore, molarity of the given solution is calculated
from moles of H,SO,.
to make 100 cm? of 0.15 M Na,COg solution?
H,SO, left unreacted in the solution = 0.05 mole
Sol. 1000 cm*of 0.15 M Na,CO, contains Na,CO, (1]
.. Molarity of the solution = = = 0.025 mol L”! [1]
= 0.15 mol
100 cm? of 0.15 M Na,CO, will contain Na,CO,
0.15
= — X 100= 0.015 mol
SHORT ANSWER Type IT Questions
1000 |3 Marks|
(Molar mass of Na,CO, = 106 g mol') [1]
L Calculate the amount of carbon dioxide that
Mass of Na,CO, = 0.015x 106= 1.59 g could be produced when
6. If 500 mL of a 5M solution is diluted to 1500 (i) 1 mole of carbon is burnt in air.
mL, what will be the molarity of the solution (ii) 1 mole of carbon is burnt in 16 g of dioxygen.
obtained? [NCERT Examplar] (iii) 2 moles of carbon are burnt in 16 g of
dioxygen. [NCERT Textbook]
Y In case of solution, molarity is calculated by using
molarity equation, M,V, =M Vo, we have, V, (before “o ¢ In order to find the moles of different elements/
dilution) and V, (after dilution), so calculate molarity of compounds taking part in the reaction, write a
the given solution from this equation. balanced chemical equation for combustion of carbon
Sol. Given that, M,=5M > V, =500 mL in dioxygen (air).
* Calculate the amount of CO, produced with the help
V, =1500
mL > M,=M (1) of information available from the balanced chemical
equation,
Some Basic Concepts of Chemistry 305)
Sol. C(s)+0,(g) —> CO,(g)
1 mol 1 mol 1 mol
Se Dinitrogen and dihydrogen react with each
32g 44g other to produce ammonia according to the
(i) According to the equation when 1 mole of carbon is following chemical equation,
burnt completely, CO, produced is 44 g. {1] N,(g)+ 3H2(g) —> 2NH;(g).
(ii) 1 mole of carbon requires 32 g dioxygen from the (i) Calculate the mass of ammonia produced if
reaction. But we have only 16 g dioxygen. Hence, 2.00x 10°g dinitrogen reacts with 1.00 x 10° g
dioxygen is the limiting reagent. So, the amount of of dihydrogen.
CO, produced by 16 g dioxygen is 22 g. [1] (ii) Will any of the two reactants remain unreacted?
(iii) In this case again, dioxygen is the limiting reagent. 16 (iii) If yes, which one and what would be its mass?
g dioxygen can react only with 0.5 mole of carbon and [NCERT Textbook]
produce 22 g CO). [1]
Sol. (i)N2(g) + 3H,(g) —> 2NH,(g)
1 mol 3 mol 2 mol
2. 1f4 g of NaOH dissolves in 36 g of H,O, 28 g 6g 34g
calculate the mole fraction of each component 28 g N, reacts with 6 g H,.
in the solution. Also, determine the molarity of 1g N, will react with = g H..
solution (specific gravity of solution is 1 g mL”).
20 00
X6
[NCERT Exemplar] “. 2000 g N, will react with fnaa = 428.57 g H,
“9 ¢ To proceed the calculation, first calculate the number
Hence, N, is the limiting reagent and H, is in excess. N,
of moles of NaOH and H,0.
limits the amount of ammonia produced.
¢ Then, find mole fraction of NaOH and H,0 by using the
28 g N, produces 34 g NH,and1 g N, produces = g NH,
formula,
X \,04 = ANaOH or Xy0 = Ny,0
Mao + 9H,0 Anan
+ IH,0 34
., 2000 SNe
gN, will P produce —38 x 2000 = 2428.57 8gNH ay)
¢ Then, calculate molarity Ae so in order to
mx (ii) H, is in excess so it will remain unreacted. {1/2]
~ calculate molarity we require volume of solution which * (iii) Amount of H, remains unreacted
v= es 2 ee =1000— 428.57 = 571.43 g. {1]
specific gravity
4. Commercially available concentrated
Sol Number of moles of NaOH, hydrochloric acid contains 38% HCl by mass.
Mass (g) (i) What is the molarity of the solution (density
NaOH = a
40
=) 1 mol n= 1h Ll ee
Molar mass (g mol
Cs
") of solution = 1.19 g mL” ')?
(ii) What volume of the above concentrated HCl
36 -
Similarly,Y, nNy ,0 =18 NO) is required to make 1.0 Lof 0.10 M HCl?
Sol. (i) 38% HCl by mass means 38 g of HCl is present in 100 g
Mole fraction of NaOH, of solution.
x moles of NaOH mass _ 100
Volume of solution = : — = 84.03 mL
NaOH “" moles of NaOH + moles of H,O density 1.19 [1/2]
0.1
= = 0.0476 Moles of HCl = eh 8 1.04
ANOS Oa [1] 36.5 [1/2]
ny,0 04 1000
Sinulatly. yi o= 4. 2 Molarity = eRe OR 12.38 M
[1/2]
NyaoH + H,0
(ii) From the molarity equation, M,V, = M2V,
ae —
= = = 0.9524
acid acid>
0.1+ 2
12.38 MxV, = 0.10 Mx 1.0L [1/2]
Total mass of solution = mass of solute + mass of solvent
=4+36=40g (1) = 0.1% +2 — 0.00808 L = 8.08 cm? (1)
Volume of solution 12.38
_massofsolution 40g 5. Ifthe density of methanol is 0.793 kg L, what is
= 40 mL
specific gravity 1g mL” its volume needed for making 2.5 L of its 0.25 M
moles of solute x 1000 solution? [NCERT Textbook]
Molarity =
volume of solution (mL) Sol. Given, d = 0.793 kg L | = 0.793x 10° gL!
_ 0.11000 =2.5M [1] Mass of methanol (m) = 0.793 x 10° g [if volume =1L]
40
 | AllZwone | CHEMISTRY Class 1ith
46
Final volume, V, = 2.5 L Sol. 4HCl(aq) +MnO,(s) —> 2H,0(/)
Final molarity, M, =0.25M ie
D 87
: + MnCl, (aq)+Cly(g)
Molarity of initial solution M, =?
According to the balanced chemical equation,
Initial volume V, = ?. Molar mass of methanol,
87 g of MnO, reacts with 4x 36.5 g HCl
CH,OH= (1x 12.01)+(4 1.0079)+ 16.00 4x 36.5 5
= 32.0416~ 32g mol * [1/2] 5 g of MnO, will react with = 8.39gHCl [3]
87
3 = 1
Mil ee : = 24.781 mol L”!
32 g mol LONG ANSWER Type Questions
v=12 => M,V,=M,V, (1] |5 Marks|
24.781 X V, = 0.25% 2.5 [1/2]
4. (i) Asample of drinking water was found to be
Viz 0.252:9_ 0.02522 L = 25.22 mL severely contaminated with chloroform CHCl,
24.781 [1] which is carcinogenic in nature. The level of
Calculate the concentration of nitric acid in mol contamination was 15 ppm (by mass).
per litre in a sample which has a density (a) Express this in per cent by mass.
1.41g mL” and the mass per cent of nitric acid (b) Determine the molarity of chloroform in
in it being 69%. the water sample. [NCERT Textbook]
vty w x 1000
ats ¢ We know that molarity = so in order to (ii) Calculate the molarity of a solution of
ethanol in water in which the mole fraction of
calculate molarity,we require mass and molar mass
ethanol is 0.040. [NCERT Textbook]
of solute and volume of the solution. To calculate
mass, convert mass per cent into gram. Sol. (i) (a) 15 ppm means 15 parts in one million (10)° parts.
* Calculate, molar mass of nitric acid by adding atomic 15 X100
Therefore, % by mass = =1.5x10°% [Il
masses of different atoms and volume of solution by
using the formula, d = 7 (b) Molar mass of CHCl, = 119 g mol [1/2]
¢ Now put these values in the formula of molarity. 1.5X 10°% means1.5x 10° g chloroform is present
in 100 g sample.
w X 1000
Sol. Molarity = w X 1000
m X volume of solution (mL) Molarity, M = ——————_____ [1/2]
m X volume of sample
Given, d =1.4lg mL!, mass % of HNO, = 69%
(For water, density = 1 g cm °, so mass = volume)
69% HNO; means 100 g of its solution contains 69 g
HNO, (nitric acid). [1/2] 1.5x10 > x1000 &
Mie ee 1 6
Hence, mass of HNO, (solute) = 69 g 119100 [1]
Molar mass of nitric acid, (ii) Molarity is defined as the moles of solute (ethanol) in
1L of the solution.
HNO, = 1.0079 +14.0067 + (3 x 16.00) = 63.0146 g mol!
1L of ethanol solution (as it is diluted) = 1L of water
Density, d = igi aL 1000 g
Number of moles of H,O= = 55.55 mol
V d 1.41 je mL’ [1/2] [1/2]
Molarity = w X1000 *H,0 =l—Xc Hon
m X volume of solution (mL) [1] Xy,0 =1-0.040= 0.96 [1/2]
69 x 1000 x1.4 mH 40
nap ESE =15.44M
63.0146 x 100 (1)
Xy 20 a :
™y,0 +"c,H.0H
Note
Concentration of a substance in mol per litre is known as —s 0.96 = ee eee
molarity. 55.55-+ Nc .H.0H (1/2)
Chlorine is prepared in the laboratory by treating or n CoN ONe=y pines
3, vases2.3145 mo mol
le. 0.96 {1/2]
manganese dioxide (MnO,) with aqueous
hydrochloric acid according to the reaction, 2. Calcium carbonate reacts with aqueous HCl to
4 HCl(aq) + MnO,(s) —> 2H.0 (1) give CaCl, and CO, according to the reaction,
+ MnCl,(aq) + Cl.,(g) CaCO3(s) + 2 HCl(aq) —>
How many gram of HCl reacts with 5.0 g of
manganese dioxide? [NCERT Textbook] CaCl, (aq) + CO, (g)+ H,O (I)
 Some Basic Concepts of Chemistry Aap
What mass of CaCO, is required to react
3. Calcium carbonate reacts with aqueous HCl to
completely with 25 mL of 0.75 M HCl?
give CaCl, and CO, according to the reaction
[NCERT Textbook] given below CaCO3(s) + 2HCI (aq) ——> CaCl,(aq)
'
“9 + To calculate the mass of CaCO, required to react + CO,(g)+ H,0 (J)
completely with 25 mL of 0.75 M HCI first we will
What mass of CaCl, will be formed when 250 mL
calculate the mass of HCl in 25 mL of 0.75 M HCl,
of 0.76 M HCl reacts with 1000 g of CaCO? Name
¢ Now, calculate the mass of CaCO,(g) by using the
the limiting reagent. Calculate the number of
information available from a balanced chemical
moles of CaCl, formed in the reaction.
equation.
[NCERT Exemplar]
Sol. (i) Calculation for mass of HCI in 25 mL of 0.75 M HCl
Sol. Number of moles of HCl
w X 1000 a _ wX 1000
Molarity =
m X vol (mL) 36.5 X 25 = 250 mLx ovo 0.19 mol;
1000
(Molar mass of HCl = 1+ 35.5= 36.5 g/mol) Number of moles of
25
w = 0.75X 36.5 X—— = 0.6844
1000
pe
2! CaCO, = 8 __= 10 mol [2]
100 g mol”
(ii) Calculation for required mass of CaCO, to react
completely with 0.6844 g HCl. For the 10 moles of CaCO,(s) number of moles of HCl
CaCO,(s)+ 2HCl(aq) ——> required would be 10 x -= 20 mol HCl (aq) {1}
100 g 2x 36.5
= 73g
CaCl,(aq)+ CO,(g)+ H,O(1) But we have only 0.19 mole HCl (aq), hence, HCl (aq) is
According to balanced chemical equation, the limiting reagent. Since, 2 moles HCl (aq) forms
1 mole of CaCl,, therefore, 0.19 moles of HCl(aq) would
73 g HCl completely reacts with 100 g CaCO,
give
1 g HCl completely reacts with = g CaCO,
0.19 x := 0.095 mol
“. 0.6844 g HCI will completely reacts with
Mass of CaCl, = 0.095 x 111= 10.54 g [2]
TNO o0-
x 0.
CBA odie
5 g (24)
73 :

— ESSSSS YOUR TOPICAL UNDERSTANDING

OBJECTIVE Type Questions 4. xg of Ag was dissolved in HNO, and the solution was
treated with excess of NaCl, when 2.87g of AgCl was
|1 Mark|
prec pitated. The value of x is
1. In the following reaction, (a) 1.08, (b)216g (c)270g (d) 1.62 g
MnO, + 4HCl ——> MnCl, +2H,0+Cl, 5. The weight of iron which will be converted into its
2 moles of MnO, react with 4 moles of HCl to form oxide (Fe,0,) by the action of 18 g of steam on it
tie Pele at STP. will be (Atomic weight of Fe = 56)
Thus, per cent yield of Cl, is (a) 168 g (b) 84¢ (c) 42 ¢ (d) 21¢
(a) 25% (b) 50% (c) 100% (d) 75% Ans. 1.(b) 2.(d) 3.(c) 4.(b) 5.(c)
2. The concentration of a solution or the amount of
substance present in its given volume can be VERY SHORT ANSWER Type Questions
expressed in which of the following ways? |1 Mark|
(a) Mass per cent or weight per cent (w/w%)
(b) Mole fraction or molarity 1. If 6.022 x 10°? molecules of N, react completely
(c) Molality with H, according to the equation
(d) All of the above N,(g)+ 3H.(g) ——> 2NH;(g) then calculate the
3. Asolution is prepared by adding 2 g of asubstance A ~ number of molecules of NH; formed.
to 18 g of water. Calculate the mass per cent of the [Ans. 1. 2 x1074]
solute. 2. What do you mean by mole fraction?
(a) 8% (b) 9% (c) 10% (d) 11%
48 | Alléwone | CHEMISTRY Class 11th

3. Calculate the mass of ferric oxide that will be with water upto the mark on the neck. What is the
obtained by complete oxidation of 2 g of Fe. [Atomic molarity of the solution? [Ans. A = 0.09M]
weights of Fe=56u, 0= 16u] [Ans. A = 286g] 3. How many gram of barium chloride (BaCl,) is needed
4. Calculate the mass percentage of C inC,H,. to prepare 100 cm? of 0.250 MBaCl, solution?
[Ans. A = 85 71%] [Ans. A = 5.2g]
5. What do you mean by excess reagent? 4. What volume of 0.250 M HCl (aq) is required to
react completely with 22.6 g of sodium carbonate
SHORT ANSWER Type I Questions according to the reaction?
|2 Marks| Na,CO, (s)+ 2HCl (aq) —>
1. A solution is prepared by dissolving 18.25 g of NaOH 2NaCl (aq) + H,0(1) + CO, (g) [Ans.1.7 L]
in distilled water to give 200 mL of solution.
Calculate the molarity of the solution. LONG ANSWER Type Questions
[Ans. 3.65 m] |5 Marks|
2. What mass of solid AgCl is obtained when 25 mL of
0.068 M AgNO, reacts with excess of aqueous HCl? 1. (i) Calculate the amount of KCLO; needed to supply
[Ans. 0.24 g] sufficient oxygen for burning 112 L of CO gas at
3. What volume of 0.34 M KOH is sufficient to react NTP. [Ans. 204.16 g]
with 20 mL of 0.15 MH,SO, solution? (ii) In a reaction vessel, 0.184 g of NaOH is required
[Ans.17.65 mL] to be added for completing the reaction. How
4. How many moles and how many gram of sodium many millilitre of 0.150 M NaOH should be added
chloride are present in 250 mL of 0.50 M NaCl solution? for this requirement? [Ans. 30.67]
[Ans. No. of mole 0125 mol and No. of grams = 7.31 g]
2. (i) Commercially available sulphuric acid contains
SHORT ANSWER Type II Questions 93% acid by mass and has density of 1.84g m_?.
Calculate [Ans.18.71 m]
|3 Marks|
(a) the molarity of the solution
1. For the reaction, 4 Fe+ 30, ——> 2Fe,03, 4.80 g of (b) volume of concentrated acid required to
oxygen is used to burn 0.150 mole of iron. What prepare 2.5Lof0.50MH,SO,. [Ans.93 mL]
mass of Fe,0; will be produced ? What mass of Fe (ii) 250 mL of 0.5 M sodium sulphate (Na,SO,)
will be left over at the end of the reaction ? What solution are added to an aqueous solution
mass of 0, will be left over at the end of the containing 10.0 g of BaCl, resulting in the
reaction? formation of white precipitate of BaSO,. How
[Ans. mass of Fe,O, produced = 12.0 g, No mass of Fe will many moles and how many gram of barium
be left over, mass of O, left = 1.2 g] sulphate will be obtained?
2. A sample of NaNO, weighing 0.38 g is placed in a [Ans. Number of mole = 0.048 and mass = 1118 g]
50.0 mL measuring flask. The flask is then filled
SUMMARY
Any thing having mass and occupying space is called matter.
Matter can be a pure substance or a mixture. Pure substance with only one kind of atoms are known as elements.
Pure substance having fixed composition are called compounds.
When composition of various components are not fixed is called mixture.
Compositions of components of a mixture is uniform through out is called homogeneous mixture. If it is not uniform
through out, it is known as heterogeneous mixture.
A matter possess properties like mass, volume, density, temperature, etc.
Representation of number as N x 10” (where, N = number, n =power over 10) is called scientific notation
(n is (-) ve if decimal moves to right and is (+) ve if it moves towards left).
All certain digits with last digit uncertain are called significant figures. Dimensional analysis can be done by using the
concept. Required unit = given value x conversion factor aws of combinations :
(i) Law of conservation of mass, (il) Law of definite proportion.
(iii) Law of multiple proportion. (iv) Gay Lussac’s law of gaseous volume. (v) Avogadro's law (V < n).
Dalton’s atomic theory state that matter is made up of tiny indivisible particles called atoms.
Atoms of same elements are like. Compounds have different atoms but in fixed proportion.
Masses can be expressed in various ways:
(i) Mass of an atoms relative to the mass of C,, is called atomic mass.
(ii) AMU (atomic mass unit) is the mass exactly equal to ath mass of C-12 atom (1 amu or 1 — u = 1.66056 x 10 -24
°“" g).

(ili) Average atomic mass Average of all isotopic atomic mass is called average atomic mass. It is given as
} xXxXA
Average atomic mass = oer (where, xX = % abundance, A =atomic mass).
x
(iv) Molecular mass is the sum of atomic mass of all the elements present in a compound.
(v) Formula mass in case of ions, molecular mass is called formula mass.
Formula showing exact number of atoms, present in a molecules is called molecular formula.
Molecular formula = n x empirical formula.
Mass of constituents in 100 parts of a compound is known as percentage composition.
The simplest whole number ratio of different atoms present in a compound is known as empirical formula.
Values (number of moles) used for calculation based on balanced chemical equations are called stoichiometric calculations.
The reagent which limits the amount of product formed is known as limiting reagent.
Reaction occurs in aqueous medium are called reactions in solutions.
Relation between solute and solvent can be expressed as:
(i) Mass per cent is the mass of substance (solute/solvent) per 100 g of solution.
(ii) Mole fraction is ratio of moles of substance (solute/solvent) to the total number of moles present in the solution.
; W x 1000
(iii) Molarity (M) is the number of moles of solute present per litre of solution. M = ann ca (m =molecular mass of
solute)
w x 1000
(iv) Molality (m) is the number of moles of solute present per kg of solvent. m = a ViSr Va(W =mass of solvent)

Mole concept is the relation in terms of moles and other properties like number of particles, volume of gas, mass etc.
(i) 1 mole of particles = 6.023 x 10”? particles. (ii) 1 mole = 22 .4L (for gases at NTP).

(iii) Number of moles = —imaspopelpstace, (W)_ (iv) Number of atoms = Number of molecules x atomicity.
atomic or moler mass (m)
 7. Abivalent metal has an equivalent mass of 32.

CrLAP LES
The molecular mass of the metal nitrate is
(a) 182 (b) 168 (c) 192 (d) 188

PRACTICE
. If1 mL of water contains 20 drops then number
of molecules in a drop of water is
(a) 6.023 x10” molecules (b) 1376 x 107° molecules
(c) 1.62 x 1071 molecules _—_(d) 4.346 x 107° molecules

OBJECTIVE Type Questions 9, What will be the molarity of pure water?

(a) 18M (b) 50.0 M (c) 55.6M (d) 100 M
|1 Mark|
1. Which of the following statements is/are 10. How many number of molecules and atoms
respectively are present in 2.8 L of a diatomic
incorrect?
(a) The weight of a substance can be determined gas at STP?
very accurately by using an analytical balance (a) 6.023 x 107°, 7.5 x 10” (b) 6.023 x 107°, 15 x 107
(b) Volume is denoted in dm? units (c) 7.5 X10", 15 x 107” (d)15 x 107", 7.5 x10”
(c) Density of a substance is its amount present per ANS: (Tsay acta) ae (a) 4.(b) 5.(c) 6.(b)
unit volume
(d) Candela is the luminous intensity, that emits
To (dr eetey Fake) 10. (c)
monochromatic radiation of frequency, 540 x 10? Hz
VERY SHORT ANSWER Type Questions
2. Which of the following options is not correct? |1 Mark|
(a) 8008 = 8.008 x10° (b) 208 =3
(c) 5000 = 5.0 x10° (d) 2.0034 = 4 il. What is the mass in gram of one molecule of
caffeine (CgH;jN 40>)? [Ans. A = 3. 22 x107"g]
3. Which of the following statements is/are
correct regarding significant figures? 12. How many moles of sulphur atoms does 45.5 g
(a) All non-zero digits are significant of sulphur contain? [Ans. A =1. 42]
(b) Significant figures are meaningful digits which are
known with certainty
13. Explain, why diamond is considered as an
(c) Zero between two non-zero digits are significant
element, not as a compound?
(d) All of the above 14. Round up the following upto three significant
figures.
18.72 g of a substance X occupies 1.81 cm*. What
will be its density measured in correct (i) 38.216 [Ans. 38.2] (ii) 10.4107 [Ans. 10.4]
significant figures? 15. 32 g of sulphur reacts with 32 g of oxygen even
(a) 10.3 g/cem?® (b) 10.34 g/em?* if more than 32 g of sulphur are available.
(c) 10.4 g/em* (d) 10.3425 g/em® Which law of chemical combination does this
illusti ate?
5. Which law states that if two elements can
combine to form more than one compound, the » Give the einpirical formulae of the following
masses of one element that combine with a substances.
fixed mass of other element, are in the ratio of (i) CgHg (ii) Graphite (iii) Magnesium chloride
small whole numbers? 17. Write balanced chemical equation for the
(a) Avogadro’s law following.
(b) Law of definite composition
(c) Law of multiple proportions Elemental phosphorus (P,) reacts with
concentrated nitric acid to give nitrogen
(d) Gay Lussac’s law of gaseous volumes
dioxide and phosphoric acid.
6. Zinc sulphate contains 22.65% zinc and 43.9%
18. 3 L of water is added to 2 Lof 5 M HCl. What is
water of crystallisation. If the law of constant
the molarity of HCl in the resultant solution?
proportions is true then the weight of zinc
required to produce 20 g of the zinc sulphate [Ans. 2M]
crystals will be Or Calculate the mass of a molecule of carbon
(a) 453g (b) 4.53¢ (c) 0.453g (d) 453g dioxide (“*CO,)? [Ans. 7.64 x10%g]
Some Basic Concepts of Chemistry
5!
SHORT ANSWER Type I Questions 28. Lithium oxide is used to remove water from air
|2 Marks| according to the reaction
19. The molality and molarity of a solution of Li,O(s)+ H,O (g)— 2 LiOH (s). If 72 kg of water is
to be removed and 35 kg of Li,O is available, then
sulphuric acid are 4.13 mol/kg and 11.12 mol/L
respectively. Calculate density of the solution. (i) which reactant is limiting? [Ans. Li,O]
(ii) how many kg of excess reactant is left?
[Ans. 3.809 g/L]
[Ans. 43.36 Kg]
20. The “star of India” sapphire weighs 563 carats.
29. Weighing 3104 carats (1 carat = 200 mg), the
If one carat is equal to 200 mg, what is the mass
Cullinan diamond was the largest natural
of the gemstone in gram? [Ans. 1126] diamond ever found. How many carbon atoms
21. Give one experiment involving a chemical were present in the stone? [Ans. 31210” atoms]
reaction to prove that the law of conservation
30. Discuss Avogadro's hypothesis and its
of mass is true. [Ans. C,)H¢]
important applications. What is the significance
22. Naphthalene (compound of C and H) contain of terms: Avogadro’s constant and mole?
93.71% carbon. If its molar mass is 128 g mol, 31. Commercially available concentrated
calculate its molecular formula. hydrochloric acid contains 38% HCl by mass.
(i) What is the molarity of this solution? The
23. What volume of oxygen at STP can be produced
density is 1.19 gcm®. [Ans. 12.42 m]
by 6.125 g of potassium chlorate according to
the reaction, 2KClO;> 2KCl + 30,? (ii) What volume of concentrated hydrochloric
acid is required to make 1.00 L of 0.10 M HCl?
[Ans. 1.68 L]
[Ans. 81 mL]
24. Calculate the number of gram of SO, which can
32. Calculate the percentage by mass of chromium
be prepared by the treatment of 100 g of Na,SO3
in each of the following oxides .
with HCI. [Ans. 50.82]
(i) CrO [Ans. 76.47%] (ii) Cr,Oz [Ans. 68.42% ]
25. Agold coloured metal object has a mass of 365 (iii) CrOz [Ans. 52.00%]
. g and a volume of 22.12 cm®. Is the object
33. Seema and Rano were doing experiment in the
composed of pure gold? laboratory. Seema suggested that if we keep
[Ans. No] 100 g biscuit of gold and 100 g aluminium bar
separately in two measuring cylinders having
SHORT ANSWER Type II Questions equal amount of water then water will rise more
|3 Marks| in case of aluminium but Rano was not agreed
with this. Her assumption was opposite to it.
26. (i) 0.5 mole each of H,S and SO, mixed together
in a reaction flask, react according to (i) According to you whose assumption was
equation
right and why?
(ii) Calculate the moles of both gold and
2H,S + SO, —— 2H,0+ 3S
aluminium.
Calculate the number of moles of S formed. [Ans. A: moles of gold =0.51 mol and moles of
[Ans. 0.75 mol] aluminium = 3.7 mol]
(ii) Calculate the volume of 0.015 M HCI solution (iii) What values are associated with Seema?
required to prepare 250 mL of a 8.25x 10° M (iv) Mention the values exhibited by Rano.
HCI] solution. [Ans. 87.5 mL]

27. Calculate the volume of 1.00 mol L™ aqueous LONG ANSWER Type Questions
solution of sodium hydroxide that is neutralised |5 Marks|
by 200 mL of 2.00 mol L aqueous hydrochloric 34. (i) An element X forms four oxide having
acid and the mass of sodium chloride obtained. percentages ofX equal to 77.4%, 63.2%, 69.9%
Neutralisation reaction is and 72.0%. Verify law of multiple proportions.
NaOH (aq) + HCl (aq) —> NaCl (aq)+ H,0 (J) (ii) A solution of glucose in water is labelled as
[Ans. volume of 1.00 m NaOH = 400 mL 10% (w/W). The density of the solution is
and mass and of NaCl produced = 23.4g] 1.20 g mL". Calculate (a) molarity (b)
molality [Ans. (a) 0.67 m (b) 0.62 m]