SOME BASIC CONCEPTS OF CHEMISTRY notes
Chemistry is the branch of
science that deals with the composition, structure and properties of matter.
Chemistry is called the science of atoms
and molecules.
IMPORTANCE OF CHEMISTRY-- Chemistry plays a central role
in science and is often intertwined with other branches of science like
physics, biology, geology etc. Chemistry also plays an important role in daily
life.
1.Chemical
principles are important in diverse areas, such as: weather patterns,
functioning of brain and operation of a computer.
2. Chemical industries manufacturing
fertilizers, alkalis, acids, salts, dyes,
polymers, drugs, soaps, detergents, metals,
alloys and other inorganic and organic chemicals,
including new materials, contribute in a big
way to the national economy.
3.Chemistry plays an important role in meeting human needs
for food, health care products and other materials aimed at improving the
quality of life. This is exemplified by the large scale production of a variety
of fertilizers, improved varieties of pesticides and insecticides. Similarly
many life saving drugs such as cisplatin and taxol, are effective
in
cancer therapy
and AZT (Azidothymidine) used for helping
AIDS victims, have been isolated from plant and animal sources or prepared by
synthetic methods.
4. With a better understanding of chemical principles it
has now become possible to design and synthesize new materials having specific
magnetic, electric and optical properties. This has lead to the production of
superconducting ceramics, conducting polymers, optical fibres and large scale
miniaturization of solid state devices.
5. In recent years chemistry has tackled with a fair degree
of success some of the pressing aspects of environmental degradation. Safer
alternatives to environmentally hazardous refrigerants like CFCs
(chlorofluorocarbons), responsible for ozone depletion in the stratosphere,
have been successfully synthesised.
Understanding
of bio-chemical processes, use of enzymes for large-scale production of
chemicals and
synthesis of new exotic materials are some of the intellectual challenges for
the future generation of chemists. A developing country like India needs
talented and creative chemists for accepting such challenges.
Anything which has
mass and occupies space is called matter.
example,
book, pen, pencil, water, air, all living beings etc.
Matter
can exist in three physical states viz. solid, liquid and gas.
The properties of these three physical states
can be understood with the help of following table---
A mixture contains two or more substances present in it (in any ratio) which are called its components. Many of the substances present around you are mixtures.
For example, sugar solution in water, air, tea etc.,
A mixture may be classify as homogeneous or
heterogeneous.
Homogeneous mixture, the components
completely mix with each other and its composition is uniform throughout. e.g.
Sugar solution, and air .
Heterogeneous mixtures, the composition
is not uniform throughout and sometimes the different components can be
observed. E.g. the mixtures of salt and sugar, grains and pulses along with
some dirt (often stone) pieces,
Components of a mixture can be separated by
using physical methods such as simple hand picking, filtration,
crystallisation, distillation etc.
Pure substance/Substances have
fixed composition e.g.Copper, silver, gold, water, glucose e.t.c. Pure
substances can be classified as elements and compounds.
An element consists of only one type
of particles. These particles may be atoms or molecules. E.g.
Sodium, copper, silver, hydrogen, oxygen etc.
However, the atoms of different elements are
different in nature.
Compound When two or more atoms of
different elements combine, the
molecule of a compound is obtained. E.g. water, ammonia, carbon dioxide, sugar etc.
Q. Write the
differences between--- (with suitable examples)
a) homogeneous
and heterogeneous mixtures
b) mixture and
compound
PROPERTIES OF MATTER AND THEIR MEASUREMENT--Every substance has unique or characteristic properties.
These properties can be classified into two categories – physical properties
and chemical properties.
Physical properties are
those properties which can be measured or observed without changing the
identity or the composition of the substance. E.g. colour, odour, melting
point, boiling point, density etc.
The measurement or observation of chemical properties require
a chemical change to occur. e.g.
Burning of Mg-ribbon in air
Chemical
properties are characteristic reactions of
different substances; these include acidity or basicity, combustibility etc.
Many properties of matter such as length, area, volume,
etc., are quantitative in nature.
Metric System were
being used in different parts of the world. The metric system which
originated in France in late eighteenth century, was more convenient
as it was based on the decimal system. The need of a common standard
system was being felt by the scientific community. Such a system was
established in 1960 so it is in detail.
The International System of Units (SI)
The International System of Units (in French Le Systeme International d’Unités – abbreviated as SI)
was established by the 11th General Conference on Weights and
Measures (CGPM from Conference Generale des Poids at Measures).
The CGPM is an inter governmental treaty organization created by a diplomatic
treaty known as Metre Convention which was signed in Paris in 1875.
Definitions of SI Base Units
Unit of length—metre-- The
metre is the length of the path traveled by light in vacuum during a
time interval of 1/299 792 458 of a second.
Unit of mass-- kilogram -- it is equal to the mass of the international prototype of
the kilogram.
Unit of time—second-- The
second is the duration of 9 192
631 770 periods of the radiation corresponding to the transition between the
two hyperfine levels of the ground state of the caesium-133 atom.
Unit of electric current--ampere --The ampere is that constant current which, if
maintained in two straight parallel conductors of infinite length, of negligible
circular cross-section, and placed 1
metre apart in vacuum, would produce between these conductors a force equal to
2 × 10–7 newton per metre of length.
Unit of thermodynamic temperature-- kelvin --The kelvin, unit of thermodynamic temperature, is the
fraction 1/273.16 of the thermodynamic temperature of the triple point of
water.
(Triple point- Temperature at which all three physical states of water can exist i.e. 00C)
Unit of amount of substance-- mole --1. The mole is the amount of substance of a system
which contains as many elementary entities as there are atoms in 0.012 kilogram
of carbon-12; its symbol is “mol.”
2. When the mole is used, the elementary entities must be
specified and may be atoms, molecules, ions, electrons, other particles, or
specified groups of such particles.
Unit of luminous intensity-- candela --The candela is the luminous intensity, in a given
direction, of a source that emits monochromatic radiation of frequency 540 × 1012
hertz and that has a radiant intensity in that direction of 1/683 watt per steradian.
Mass and Weight-- Mass
of a substance is the amount of matter
present in it while weight is the force
exerted by gravity on an object. The
mass of a substance is constant whereas
its weight may vary from one place to another
due to change in
gravity.
The mass of a
substance can be determined very accurately by using an analytical balance.
Maintaining the
National Standards of Measurement
Each modern
industrialized country including India has a National Metrology Institute (NMI) which maintains standards of
measurements. This responsibility has been given to the National Physical
Laboratory (NPL), New Delhi.
Derived units—Units
derived with the help of base units of measurement.
Volume-- Volume
has the units of (length)3. So volume has units of m3 or
cm3 or dm3.
A common unit, litre (L) is not an SI unit, is used for
measurement of volume of liquids.
1 L = 1000 mL , 1000 cm3 = 1 dm3
In the
laboratory, volume of liquids or solutions can be measured by graduated cylinder,
burette, pipette etc. A volumetric flask is used to prepare a known volume of a
solution.
Density Density
of a substance is its amount of mass per unit volume.
SI unit of density = SI unit of mass/SI unit of volume
= kg/m3
or kg m–3
This unit is quite large and a chemist often expresses
density in g cm–3.
Temperature--There are three common scales to measure temperature — °C (degree celsius), °F (degree fahrenheit) and K (kelvin). Here, K is the SI
unitGenerally, the thermometer with celsius scale are
calibrated from 0° to 100° where these two temperatures are the freezing point
and the boiling point of water respectively.
The fahrenheit scale is represented between 32° to 212°.
The temperatures on two scales are related to each other by
the following relationship:

The kelvin scale is related to celsius scale as follows :
K =
°C + 273.15
Note—Temperature
below 0 °C (i.e. negative values) are possible in Celsius scale but in Kelvin
scale, negative temperature is not possible.
UNCERTAINTY IN MEASUREMENT
The study of
chemistry, deals with experimental data as well as theoretical calculations.
There are meaningful ways to handle the numbers conveniently and present the
data realistically with certainty to the maximum possible extent. These ideas
are discussed below in detail.
Scientific Notation
As chemistry is the
study of atoms and molecules which have extremely low
masses and are present in extremely large numbers, a chemist has to deal with
numbers very large (Avogadro’s no.) as well as very small (mass of a H atom).
It is very difficult to count numbers involving so many zeros and it offers a
real challenge to do simple mathematical operations of addition, subtraction,
multiplication or division with such numbers.
This problem is solved by using scientific notation, i.e.,
exponential notation for such numbers.
In which any number
can be represented in the form N × 10n (Where n is an exponent
having positive or negative values and N can vary between 1 to 10).
e.g. We can write 232.508 as 2.32508 X102
in scientific notation. Similarly, 0.00016 can be written as 1.6 X 10–4.
Now, for performing mathematical operations on numbers
expressed in scientific notations, the following points are to be kept in mind.
Multiplication and Division
These two operations follow the same rules which are there
for exponential numbers, i.e.-----(cw)
Addition and Subtraction
For these two operations, first the numbers are written in
such a way that they have same exponent. After that, the coefficient are added
or subtracted.
Significant
Figures -- Every experimental measurement has
some amount of uncertainty associated
with it. Every one would like the
results to be precise and accurate.
Precision refers to
the closeness of various measurements for the same quantity.
Accuracy is the
agreement of a particular value to the true value of the result. e.g. if the
true value for a result is 2.00 g. Precision and accucacy clearly understood
from the data given in Table
1 2 Average (g) Remark
Student A
1.95 1.93 1.940 values are precise as they
are close to each
other but are
not accurate.
Student B
1.94 2.05 1.995 neither precise
nor accurate
Student C
2.01 1.99 2.000 These values are both
precise and accurate.
The uncertainty in the experimental or the calculated
values is indicated by mentioning the number of significant figures. Significant
figures are meaningful digits which
are known with certainty.
The uncertainty is indicated by writing the certain digits and the last
uncertain digit. E.g. If we write a result as 11.2 mL, we say the 11 is certain
and 2 is uncertain and the uncertainty would be +1 or -1 in the last digit.
There are certain
rules for determining the number of significant figures.
(1) All non-zero
digits are significant.e.g. in 285 cm-- three and in 0.25 mL -- two S.F.
(2) Zeros preceding to first non-zero digit are not
significant. E.g in 0.03-- one and in 0.0052 -- two significant figures.
(3) Zeros between two non-zero digits are significant.e.g.
in 2.005 -- four significant figures.
(4) Zeros at the end or right of a number are significant
provided they are on the right side of the decimal point. E.g. in 0.200 g
--three significant figures.
But, in 100 -- one significant figure, but 100. -- three
significant figures and 100.0 – four significant figures. Such numbers are
better represented in scientific notation. We can express the number 100 as
1×102 – one significant figure, 1.0×102 – two significant
figures and 1.00×102 – three significant figures.
(5) Counting numbers of objects, for example, 2 balls or 20
eggs, have infinite significant figures as these are exact numbers and can be
represented by writing infinite number of zeros after placing a decimal i.e., 2
= 2.000000 or 20 = 20.000000
In numbers written
in scientific notation, all digits are significant e.g., 4.01×102
has three significant figures, and 8.256 × 10–3 has four significant
figures.
Addition and Subtraction of Significant
Figures
The result cannot have more digits to the right of the
decimal point than either of the original numbers.
12.11 + 18.0 +
1.012 = 31.122
Here, 18.0 has only one digit after the decimal point so
the result should be reported as 31.1.
Multiplication and Division of Significant
Figures
2.5×1.25 = 3.125
Since 2.5 has two significant figures, the result should be 3.1.
The following
points should be keep in mind for rounding off the numbers to write the result
to the required number of significant figures
1. If the rightmost digit to be removed is more than 5, the
preceding number is increased by one.
E.g. 1.386 If we
have to remove 6, we have to round it to 1.39
2. If the rightmost digit to be removed is less than 5, the
preceding number is not changed. e.g.
4.334 if 4 is to
be removed, then the result is rounded up to 4.33.
3. If the rightmost digit to be removed is 5, then the
preceding number is not changed if it is an even number but it is increased by
one if it is an odd number. E.g. if 6.35 is to be rounded by removing 5, we
have to increase 3 to 4 giving 6.4 as the result. However, if 6.25 is to be
rounded off it is rounded off to 6.2.
Dimensional Analysis During calculations generally there is a need to convert
units from one system to other. This is called factor label method or
unit factor method or dimensional analysis. Examples----( C.W.)
LAWS OF CHEMICAL COMBINATIONS
The combination of
elements to form compounds is governed by the following five basic laws.
Law of Conservation of Mass (Given by
Antoine Lavoisier in 1789).
It states that matter (mass) can neither be
created nor destroyed.
C + O2 ---à
CO2
12g + 32g = 44g
Law of Definite Proportions (Given by, a
French chemist, Joseph Proust.)
He stated that a given compound always contains exactly the same proportion of elements by
weight.
(Proust worked with two samples of cupric carbonate one
of which was of natural origin and the other was synthetic one. He
found that the composition of elements present in it was same)
e.g. If we collect water from different
sources or prepare in lab, It always has H & O in fix ratio by mass or by
volume or by no. of atoms.
Law of Multiple Proportions (Given by Dalton in 1803.)
According to this
law, if two elements combine together and form two or more than two
compounds, the masses of one element that combine with a fixed mass of the
other element, are in the simple ratio of small whole numbers.
e.g. N and O combine together and form five oxides N2O, NO,
N2O3, NO2, and N2O5
Oxide
of N
|
reacting mass of N
|
reacting mass of O
|
fix mass of N
|
reacting mass of O with fix mass of N
|
Ratio of reacting mass of O with fix mass of N
|
N2O
|
28g
|
16g
|
14g
|
8g
|
1
|
NO
|
14g
|
16g
|
14g
|
16g
|
2
|
N2O3
|
28g
|
48g
|
14g
|
24g
|
3
|
NO2
|
14g
|
32g
|
14g
|
32g
|
4
|
N2O5
|
28g
|
80g
|
14g
|
40g
|
5
|
Gay Lussac’s Law
of Gaseous Volumes (Given by Gay Lussac in 1808.)
According to this law when gases combine or are
produced in a chemical reaction they do so in a simple ratio by volume provided
all gases are at same temperature and pressure.
E.g.
H2(g) + Cl2(g) ---à 2HCl(g)
1V 1V 2V
All reactants and products have simple
ratio 1:1:2.
Note--Gay-Lussac’s
discovery of integer ratio in volume relationship is actually the law of
definite proportions by volume.
Avogadro Law (In 1811, Given by Avogadro)
According to this law equal volumes of gases at the same temperature and pressure should contain equal number of molecules.
Avogadro made a
distinction between atoms and molecules Avogadro could explain
the result of chemical reactions by considering the molecules to be polyatomic. Two volumes of
hydrogen react with One volume of oxygen to give Two volumes of water vapour
DALTON’S ATOMIC THEORY--In 1808, Dalton published ‘A New System
of Chemical Philosophy’ in which he proposed the following
:
1. Matter consists of indivisible atoms.
2. All the atoms of a given element have identical
properties including identical mass. Atoms of different elements differ in
mass.
3. Compounds are formed when atoms of different elements
combine in a fixed ratio.
4. Chemical reactions involve reorganization of atoms.
These are neither created nor destroyed in a chemical reaction.
Dalton’s theory
could explain the laws of chemical combination.
ATOMIC AND
MOLECULAR MASSES
Today, we use mass
spectrometry technique for determining
the atomic masses.
Since 1961 C-12
(isotope of a carbon) taken as standard to calculate relative masses of other
elements.
In this system,
12C is assigned a mass of exactly 12 atomic mass unit (amu) and masses
of all other atoms are given relative to this standard.
One atomic
mass unit is defined as a
mass exactly equal to one twelfth the mass of one carbon - 12 atom. And 1 amu = 1.66056×10–24 g. Mass of an atom of
hydrogen = 1.6736×10–24 g.
Thus, in terms of
amu, the mass of hydrogen atom =1.6736×10–24g/1.66056×10–24g=1.0078
amu
=
1.0080 amu Today, ‘amu’ has been replaced by ‘u’ which is known
as unified mass.
Average Atomic Mass--Many naturally occurring elements exist as more than one isotope. When we take into account the existence of these isotopes and their relative abundance (per cent occurrence),the
average atomic mass of that element can be calculated. For example,
(C.W.)
Molecular Mass
Molecular mass is
the sum of atomic masses of the elements present in a molecule. E.g. (c.w.)
Q. Calculate molecular mass of glucose (C6H12O6)
molecule.
Formula Mass---In crystalline substances
e.g. sodium chloride do not contain discrete molecules as their constituent
units. In such compounds, positive (sodium) and negative (chloride)
entities are
arranged in a
three-dimensional structure, so for ease of calculations , simple ratio of
these entities taken as formula unit and its mass known as formula mass. e.g.
(c.w.)
Thus, formula mass
of sodium chloride = atomic mass of sodium + atomic mass of chlorine
= 23.0 u + 35.5 u = 58.5 u
MOLE CONCEPT AND MOLAR MASSES
One mole is the
amount of a substance that contains as many particles or entities (i.e. atoms,
molecules or ions) as there are atoms in exactly 12 g (or 0.012 kg) of the 12C
isotope.
It is very
important to note that the mole of a substance always contain the
same number of entities, no matter what the substance may be.
This number of entities in 1 mol is given a
separate name and symbol. It is known as ‘Avogadro constant’, denoted by
NA in honour of Amedeo Avogadro.
The mass of one mole of a substance in grams
is called its molar mass.
The molar mass is
numerically equal to atomic/molecular/ formula mass in u but
expressed in grams in place of u. e.g. Molecular mass of water=18.02u and Molar
mass of water = 18.02 g
(mass of 1- water molecule) (mass of 6.02x1023
- water molecules)
PERCENTAGE COMPOSITION—This determination is important to check
the purity of a given sample. Let us consider the example of water (H2O).
Water contains hydrogen and oxygen, the percentage composition of both these
elements can be calculated as :
Mass % of an element = mass of that
element in the compound × 100
molar mass of the compound
Molar mass of water = 18.02 g
Mass % of hydrogen = 2× 1.008 × 100
18.02
=
11.18
Mass % of oxygen = 16.00
× 100
18.02
= 88.79 (other examples C.W.)
Empirical Formula for Molecular Formula—
An empirical formula represents the
simplest whole number ratio of various atoms present in a compound. E.g. CH is
the empirical formula of benzene.
The molecular formula shows
the exact number of different types of atoms present in a molecule of a compound.
E.g. C6H6 is the molecular formula of benzene.
Determination of
the Empirical Formula and Molecular Formula---With the help of mass per cent of
various elements present in a compound, its empirical formula can be
determined. Molecular formula can further be obtained if the molar mass is
known. Example--
Q. A compound
contains 4.07 % hydrogen, 24.27 % carbon and 71.65 % chlorine. Its molar mass
is 98.96 g. What are its empirical and molecular formulas ?
Solution-----
Name of element
|
Percentage of elements
|
Step-1Conversion of mass per cent to grams.
|
Step 2. number moles of each element
|
Step 3. Divide the mole value by the
smallest number
|
C
|
24.27%
|
24.27g
|
24.27/12 = 2.0225
|
2.0225/2.018 = 1
|
H
|
4.07%
|
4.07g
|
4.07/1= 4.07
|
4.07/2.018 =2
|
Cl
|
71.65%
|
71.65g
|
71.65/35.5 = 2.018
|
2.018/2.018 =
1
|
In step- 3 we get the ratio of different
elements in the compound.
C:H:Cl
= 1:2:1
(Note-In
case the ratios are not whole numbers, then they may be converted into whole
number by multiplying by the suitable coefficient.)
Step 4. Write
empirical formula with the help of
ratio of elements.
C1H2Cl1
or CH2Cl, is the empirical formula of the above compound.
Step 5. Writing
molecular formula
Determine empirical
formula mass as— (For CH2Cl, empirical formula mass is)
(a)
12 + (1x2) +
35.5 = 49.5
(b)
Divide Molar mass by empirical formula mass
= Molar mass =
98.96 g = 2 = (n)
Empirical formula
mass 49.48 g
(c) Multiply empirical formula by n obtained
above to get the molecular formula
Empirical formula =
CH2Cl, n = 2. Hence molecular formula is C2H4Cl2.
STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS
We
know chemical equation is maximum informative, when it is written in the form
of balanced chemical equation. It gives quantitative relationship between the
various reactants and products in terms of moles, masses, molecules and
volumes. This is called stoichiometry. (The word ‘stoichiometry’ is
derived from two Greek words - stoicheion
(meaning element) and
metron (meaning measure) i.e. measurement of an element.). The
coefficients of the balanced chemical equation are known as stoichiometric
coefficients. (examples—C.W.)
Balancing a
chemical equation--According to the law of conservation of mass, a balanced
chemical equation has the same number of atoms of each element on both sides
of the equation.
Limiting Reagent—The
reactant which gets consumed first or limits the
amount of product formed is known as limiting
reagent. (Examples- c.w.)
Reactions in Solutions--A majority of
reactions in the laboratories are carried out in solutions. The
concentration of a solution or the amount of substance present in
its given volume can be expressed in any of the following ways.
1. Mass per cent or
weight per cent (w/w %)
2. Mole fraction
3. Molarity
4. Molality
1. Mass per cent--It is obtained
by using the following relation:
Mass per cent (w/w%) = Mass of solute x 100
Mass of solution
2. Mole Fraction (X)
It is the ratio of
number of moles of a particular component to the total number of moles of the
solution. If a
substance ‘A’ dissolves in substance ‘B’ and their number of moles are nA
and nB respectively; then the mole fractions of A and B are given as
Mole fraction of A (XA) = No.of moles of A =
nA
No.of
moles of solution nA + nB
Similarly we can calculate the mole fraction
of B (XB).
3. Molarity (M) -- It is defined
as the number of moles of the
solute in 1 litre of the solution. Thus,
Molarity (M) =
No. of moles of solute
Volume of solution in litres
Molarity on dilution can be calculated by
using the general formula
M1V1
= M2V2
4. Molality(m)-- It is defined as
the number of moles of solute present
in 1 kg of solvent.
Molality (m) = No.
of moles of solute
Mass of solvent in kg
Exercise-1
1.1 Calculate the
molecular mass of the following :
(i) H2O
(ii) CO2
(iii) CH4
1.2 Calculate the
mass per cent of different elements present in sodium sulphate (Na2SO4).
1.3 Determine the
empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen
by mass.
1.4 Calculate the
amount of carbon dioxide that could be produced when
(i) 1 mole of
carbon is burnt in air.
(ii) 1 mole of
carbon is burnt in 16 g of dioxygen.
(iii) 2 moles of
carbon are burnt in 16 g of dioxygen.
1.5 Calculate the
mass of sodium acetate (CH3COONa) required to make 500 mL of 0.375
molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1.
1.6 Calculate the
concentration of nitric acid in moles per litre in a sample which has a
density, 1.41 g mL–1 and the mass per cent of nitric acid in it
being 69%.
1.7 How much copper
can be obtained from 100 g of copper sulphate (CuSO4) ?
1.8 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.
1.9 Calculate the
atomic mass (average) of chlorine using the following data :
% Natural
Abundance Molar Mass
35Cl 75.77 34.9689
37Cl 24.23 36.9659
1.10 In three moles
of ethane (C2H6), calculate the following :
(i) Number of moles
of carbon atoms.
(ii) Number of
moles of hydrogen atoms.
(iii) Number of
molecules of ethane.
1.11 What is the
concentration of sugar (C12H22O11) in mol L–1
if its 20 g are dissolved in enough water to make a final volume up to 2L?
1.12 If the density
of methanol is 0.793 kg L–1, what is its volume needed for making
2.5 L of its 0.25 M solution?
1.13 Pressure is
determined as force per unit area of the surface. The SI unit of pressure,
pascal is as shown below :
1Pa = 1N m–2,
If mass of air at sea level is 1034 g cm–2, calculate the pressure
in pascal.
1.14 What is the SI
unit of mass? How is it defined?
1.15 Match the
following prefixes with their multiples:
Prefixes
Multiples
(i) micro
106
(ii) deca 109
(iii) mega
10–6
(iv) giga
10–15
(v) femto
10
1.16 What do you
mean by significant figures ?
1.17 A sample of
drinking water was found to be severely contaminated with chloroform, CHCl3,
supposed to be carcinogenic in nature. The level of contamination was 15 ppm
(by mass).
(i) Express this in
percent by mass.
(ii) Determine the
molality of chloroform in the water sample.
1.18 Express the
following in the scientific notation:
(i) 0.0048
(ii) 234,000
(iii) 8008
(iv) 500.0
(v) 6.0012
1.19 How many
significant figures are present in the following?
(i) 0.0025 ____ (ii) 208_____ (iii) 5005___ (iv)
126,000____ (v) 500.0____ (vi) 2.0034____
1.20 Round up the
following upto three significant figures:
(i)
34.216_______(ii) 10.4107_______(iii) 0.04597__________(iv) 2808__________
1.21 The following
data are obtained when dinitrogen and dioxygen react together to
form different
compounds :
Mass of dinitrogen and Mass of dioxygen are as respectively---
(i) 14 g 16 g, (ii)
14 g 32 g, (iii) 28 g 32 g, (iv) 28 g 80 g
(a) Which law of
chemical combination is obeyed by the above experimental data? Give its
statement.
(b) Fill in the
blanks in the following conversions:
(i) 1 km =
...................... mm = ...................... pm
(ii) 1 mg =
...................... kg = ...................... ng
(iii) 1 mL = ......................
L = ...................... dm3
1.22 If the speed
of light is 3.0 × 108 m s–1, calculate the distance
covered by light in 2.00 ns.
1.23 In a
reaction A + B2 ----àAB2
Identify the limiting reagent, if any, in
the following reaction mixtures.
(i) 300 atoms of A
+ 200 molecules of B
(ii) 2 mol A + 3
mol B
(iii) 100 atoms of
A + 100 molecules of B
(iv) 5 mol A + 2.5
mol B
(v) 2.5 mol A + 5
mol B
1.24 Dinitrogen and
dihydrogen react with each other to produce ammonia according to the following
chemical equation:
N2
(g) + H2 (g) -----à2NH3 (g)
(i) Calculate the
mass of ammonia produced if 2.00 x 103 g dinitrogen reacts with
1.00 x 103 g of dihydrogen.
(ii) Will any of
the two reactants remain unreacted?
(iii) If yes, which
one and what would be its mass?
1.25 How are 0.50
mol Na2CO3 and 0.50 M Na2CO3
different?
1.26 If ten volumes
of dihydrogen gas reacts with five volumes of dioxygen gas, how many volumes of
water vapour would be produced?
1.27 Convert the
following into basic units:
(i) 28.7 pm
(ii) 15.15 pm
(iii) 25365 mg
1.28 Which one of
the following will have largest number of atoms?
(i) 1 g Au (s)
(ii) 1 g Na (s)
(iii) 1 g Li (s)
(iv) 1 g of Cl2(g)
1.29 Calculate the
molarity of a solution of ethanol in water in which the mole fraction of
ethanol is 0.040 (assume the density of water to be one).
1.30 What will be
the mass of one 12C atom in g ?
1.31 How many
significant figures should be present in the answer of the following
calculations?
(i) 0.02856 x 298.15 x 0.112
=
0.5785
(ii) 5 x 5.364 =
(iii) 0.0125 + 0.7864 +
0.0215 =
1.32 Use the data
given in the following table to calculate the molar mass of naturally occuring argon
isotopes:
Isotope Isotopic molar
mass Abundance
36Ar 35.96755 g
mol–1
0.337%
38Ar 37.96272 g
mol–1 0.063%
40Ar 39.9624 g
mol–1 99.600%
1.33 Calculate the
number of atoms in each of the following
(i) 52 moles of Ar
(ii) 52 u of He
(iii) 52 g of He.
1.34 A welding fuel
gas contains carbon and hydrogen only. Burning a small sample of it in oxygen
gives 3.38 g carbon dioxide , 0.690 g of water and no other products. A volume
of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. Calculate
(i) empirical formula,
(ii) molar mass of the gas, and
(iii) molecular
formula.
1.35 Calcium
carbonate reacts with aqueous HCl to give CaCl2 and CO2 according
to the reaction,
CaCO3 (s) + 2 HCl (aq)
-------à CaCl2 (aq) + CO2(g)
+ H2O(l)
What mass of CaCO3
is required to react completely with 25 mL of 0.75 M HCl?
1.36 Chlorine is
prepared in the laboratory by treating manganese dioxide (MnO2) with aqueous
hydrochloric acid according to the reaction
4 HCl (aq) + MnO2(s) ------à 2H2O (l) + MnCl2(aq)
+ Cl2 (g)
How many grams of
HCl react with 5.0 g of manganese dioxide?
Chemistry for class 11.....
elements-of-group-13-p-block-elements
states-of-matter-liquids-and-solids
geometric-isomerism-different-geometries
chemical-thermodynamics
introducation-of-carbon-chemistry
electrons-in-atom-and-periodic-table
hybridisation
intermolecular-forces-liquid-and-solids
niels-bohr-atomic-model
iupac-nomenclature-of-organic-compounds
chemical-bonding-molecular-geometry
molecular-orbital-theory
heisenberg-uncertainty-principle
some-basic-concepts-of-chemistry
equilibrium
environmental-chemistry
hydrogen
structure-of-atom
classification-of-elements
Chemistry for class 11.....
elements-of-group-13-p-block-elements
states-of-matter-liquids-and-solids
geometric-isomerism-different-geometries
chemical-thermodynamics
introducation-of-carbon-chemistry
electrons-in-atom-and-periodic-table
hybridisation
intermolecular-forces-liquid-and-solids
niels-bohr-atomic-model
iupac-nomenclature-of-organic-compounds
chemical-bonding-molecular-geometry
molecular-orbital-theory
heisenberg-uncertainty-principle
some-basic-concepts-of-chemistry
equilibrium
environmental-chemistry
hydrogen
structure-of-atom
classification-of-elements
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