Transcript Document

Chemistry
Class Exercise
Class Exercise - 1
Express the following numbers to
three significant figures.
(i) 6.022 × 1023
(iii) 0.0652 g
Solution
(i) 6.02 × 1023
(ii) 5.36 g
(iii)0.0652 g
(iv)13.2
(ii) 5.356 g
(iv) 13.230
Class Exercise - 2
What is the sum of 2.368 g and
1.02 g?
Solution
2 .3 6 8 g
 1 .0 2 g
3 .3 8 8
= 3.39 g
Class Exercise - 3
Express the result of the following
calculation to the appropriate number
of significant figures
816 × 0.02456 + 215.67
Solution
816 × 0.02456 = 20.0
Product rounded off to 3 significant figures because
the least number of significant figure in this
multiplication is three.
20 . 0
 215 . 67
235 . 67
Rounded off to 235.7
Class Exercise - 4
Solve the following calculations and
express the results to appropriate
number of significant figures.
(i) 1.6 × 103 + 2.4 × 102 – 2.16 × 102
(ii)
23
6.02  10
 5.00
20
4.0  10
Solution
(i) 1.6 × 103 + .24 × 103
1 .6  1 0
3
Rounded off to 1.8 × 103
1 .8  1 0
3
 .2 4  1 0
3
 .2 1 6  1 0
1 .8 4  1 0
3
1 .5 8 4  1 0
3
3
Class Exercise - 4
Rounded off to 1.6 × 103 or 16 × 102
(ii)
6 .0 2  1 0
23
 5 .0 0
4 .0  1 0
20

3 0 .1 0  1 0
4 .0  1 0
23
20
= 7.525 × 103 (rounded off to 7.5 × 103)
Class Exercise - 5
Convert 10 feet 5 inches into SI unit.
Solution
10 feet 5 inches = 125 inches
1 inch = 2.54 × 10-2 m
 125 inches = 2.54 × 10-2 × 125 m
= 317.5 × 10-2 m
Rounded off to 317 × 10–2 m
Class Exercise - 6
A football was observed to travel at a speed
of 100 miles per hour. Express the speed
in SI units.
Solution
1 mile = 1.60 × 103 m
100 miles per hour

1 0 0  1 .6 0  1 0
3
60  60
= 4.4 × 10-4 × 105 m/s
= 4.4 × 10 m/s
= 44 m/s
Class Exercise - 7
What do the following abbreviations
stand for?
(i) O
(ii) 2O
(iii) O2
Solution
(i) Oxygen atom
(ii) 2 moles of oxygen atom
(iii)Oxygen molecule
(iv)3 moles of oxygen molecule
(iv) 3O2
Class Exercise - 8
Among the substances given below
choose the elements, mixtures and
compounds
(i) Air
(iii) Diamond
(ii) Sand
(iv) Brass
Solution
(i)
(ii)
(iii)
(iv)
Air
Sand (SiO2)
Diamond (Carbon)
Brass (Alloy of metal)
- Mixture
- Compound
- Element
- Mixture
Class Exercise - 9
Classify the following into elements
and compounds.
(i) H2O
(ii) He
(iii)Cl2
(iv)CO
(v) Co
Solution
Element: He, Cl2, Co
Compound: H2O and CO
Class Exercise – 10
Explain the significance of the
symbol H.
Solution
(i) Symbol H represents hydrogen element
(ii) Symbol H represents one atom of hydrogen atom
(iii)Symbol H also represents one mole of atoms, that is,
6.023 × 1023 atoms of hydrogen.
(iv)Symbol H represents one gm of hydrogen.
BASIC CONCEPTS OF CHEMISTRY
Session - 2
Session Opener
Session Objectives
Session Objectives
1. The law of conservation of mass
2. The law of definite proportions
3. The law of multiple proportions
4. The law of reciprocal proportions
5. Gay Lussac’s law of gaseous volumes
6. Dalton’s atomic theory
7. Modern atomic theory
Law of conservation of mass
Total mass of the products remains
equal to the total mass of the
reactants.
H2 + Cl2
2g
71g
2 HCl
73g
Question
Illustrative Problem
8.4 g of sodium bicarbonate on
reaction with 20.0 g of acetic acid
(CH3COOH) liberated 4.4 g of carbon
dioxide gas into atmosphere. What is
the mass of residue left?
Solution:
8.4 + 20.0 = m + 4.4
m = 24 g
It proves the the law of conservation of
mass.
Law of definite proportions
Ice water
H 2O
1:8
River water H2O
1:8
Sea water
1:8
H 2O
A chemical compound always contains same elements
combined together in same proportion of mass.
Question
Illustrative Problem
Two gaseous samples were analyzed.
One contained 1.2g of carbon and
3.2 g of oxygen. The other contained
27.3 % carbon and 72.7% oxygen. The
experimental data are in accordance
with
(a) Law
(b) Law
(c) Law
(d) Law
of
of
of
of
conservation of mass
definite proportions
multiple proportions
reciprocal proportions
Solution
% of C in the 1st sample

1 .2
1 .2  3 .2
 100  27 . 3 %
Which is same as in the second sample.
Hence law of definite proportion is obeyed.
Law of multiple proportions
14:8
14:16
14:24
14:32
14:40
Ratio of oxygen combining with 14 parts of nitrogen
8:16:24:32:40
=
Two elements combine
1:2:3:4:5
two or more compounds
the mass of one of the elements which combines with
fixed mass of the others, bear a simple whole number
ratio to one another.
Statement of law of reciprocal
Proportions
The ratio of the weights of two elements
A and B which combine separately with
a fixed weight of the third element ‘C’ is
either the same or some simple multiple
of the ratio of the weights in which A and
B combine directly with each other.
WA : WC
WB : WC
 k ( W A :WB )
k may be 1
Law of reciprocal proportions
H 2S
2 : 32, 1 : 16
SO2
32 : 32, 1 : 1
H2S : SO2 
1
:
1
 1 : 16
S
16 1
H2O  2 : 16, 1: 8
H2S : SO2  : H2O 
H 2S
1
16

8
1

SO 2
1
2
H
H 2O
O
Gay Lussac’s law of gaseous
volumes
Gases reacts in volume which bear a
simple ratio to one another and to the
volume of the products under similar
conditions of temperature and pressure.
Gay Lussac’s Law of gaseous
volumes
1 volume
nitogen
gas
+
N2
1
2 volume
hydrogen
gas
O2
:
+
1
2NO
:
1 volume
oxygen
gas
:
1
2
2 volume
steam
O2
2H2
2
2 volume
nitrogen
oxide
1 volume
oxygen
gas
2H2O
:1
Ask yourself
Why Gay–Lussac’s law is not applicable to
solids and liquids ?
because they have negligible volumes as compared
to gases.
Do you know
The laws of chemical combinations are based
on quantitative results of chemical reactions.
Questions
Illustrative Problem
Carbon is found to form two oxides,
which contains 42.8% and 27.27% of
carbon respectively. Find out which of
the laws of chemical combination is
proved correct by this data?
Solution:
In the first oxide,
Carbon :Oxygen = 0.428 : 0.572
= 0.748 : 1
In the second oxide,
Carbon :Oxygen = 0.2727 : 0.7273
= 0.374 : 1
= 0.748 : 0.374
= 2:1
Illustrative Problem
Ammonia contains 82.35% of nitrogen
and 17.65% of hydrogen. Water
contains 88.90% of oxygen and
11.10% of hydrogen. Nitrogen trioxide
contains 63.15% of oxygen and
36.85% of nitrogen. Find out which of
the laws of chemical combination is
proved correct by this data?
Solution:
82 . 35
17 . 65

11 . 10
88 . 90
K
( 36 . 85 )
H
NH
H 2O
3
63 . 15
Hence k=1 which proves
law of reciprocal proportion.
N
N 2O 3
O
Ask your self
The balanced chemical reaction is an
expression of
Law of
multiple
proportion
Law of
conservatio
n of mass
Law of
constant
proportion
None of
the above
Illustrative example
Zinc sulphate crystals contain 22.6%
of zinc and 43.6% of water. How much
Zinc should be used to produce 13.7 gm
of zinc sulphate crystals and how much
water they will contain?
(Hint: Law of constant composition)
Solution:
100 gm of ZnSO4 will have 43.9 gm water and 22.6 gm
zinc
 Zn required to produce 100gm of ZnSO 4 crystals  22 .6 gm
Solution
 Z n re q u ire d to p ro d u ce 1 3 .7g m o f Z n S O 4 cry sta ls 
2 2 .6
100
 1 3 .7
 3 .0 9 6 2 g m
1 0 0 g m o f cry sta l co n ta in w a te r  4 3 .9g m
 1 3 .7 g m o f cry sta l co n ta in w a te r 
4 3 .9
1 3 .7
 100
 6 .0 1 4 3 g m
Dalton’s Atomic Theory
1. Matter is made up of indivisible
particles called atoms.
2. Atoms of the same element are
similar with respect to shape, size
and mass.
3. Atoms combine in simple whole
number ratio to form molecule.
4. An atom can neither be created nor
destroyed.
5. Atoms of two elements may
combine in different ratios to form
more than one compound.
SO2
1:2
S+O2
SO3
1:3
Limitations of Dalton’s Atomic Theory
It could not explain:
Why atoms of different
elements
have different masses,
sizes, etc.
Nature of binding force
between atoms and
molecules.
Limitations of Dalton’s Atomic
Theory
1. Can not explain causes of chemical combination
2. Can not explain law of combining volume
3. It does not give idea about structure of atom
4. It does not distinguish between the ultimate particle
of an element and a compound
5. It does not give idea about isotopes and isobars.
Modern Atomic Theory
1. Atom is divisible
2. Same atom may have different
atomic masses like 1H, 2H and 3H.
3. Different atoms may have same atomic mass like
40Ca and 40Ar.
4. Atom is the smallest particle that takes part in a
chemical reaction.
5. The mass of an atom can be changed into energy.
Class Test
Class Exercise - 1
Percentage of copper and oxygen in
samples of CuO obtained by different
methods were found to be the same.
This proves the law of
(a) constant proportions
(c) multiple proportions
Solution
Definition
(b) reciprocal proportions
(d) none of these
Class Exercise - 2
A balanced chemical equation is in
accordance with
(a) Boyle’s law
(b) Avogadro’s law
(c) Gay Lussac’s law
(d) law of conservation of mass
Solution
Definition
Class Exercise - 3
Different samples of water were found
to contain hydrogen and oxygen in the
approximate ratio of 1 : 8.
This shows the law of
(a) multiple proportion
(b) constant proportion
(c) reciprocal proportion
(d) none of these
Solution
Definition
Class Exercise - 4
Law of multiple proportion is illustrated
by the following pair of compounds.
(a) HCl and HNO3
(b) KOH and KCl
(c) N2O and NO
(d) H2S and SO2
Solution
Definition
Class Exercise - 5
The oxides of nitrogen contain
63.65%, 46.69% and 30.46%
nitrogen respectively. These data
proves the law of
(a) definite proportions (b) multiple proportions
(c) reciprocal proportions (d) conservation of mass
Solution
Definition
Class Exercise - 6
12 g carbon combine with 64 g sulphur
to form CS2. 12 g of carbon also
combine with 32 g oxygen to form CO2.
10 g sulphur combine with 10 g oxygen
to form SO2. These data proves the
(a) law of multiple proportions
(b) law of definite proportions
(c) law of reciprocal proportions
(d) Cray Lussac’s Law of gaseous volume
Solution
12 g
CS2
64 g S
10 g
C
CO2
O 32 g
S O2
10 g
Ratio of the weights of S and O combining with fixed
weight of C is 64 : 32 = 2 : 1. Ratio of weights of S
and O combining directly = 10 : 10 = 1 : 1. The two
ratios are simple multiple of each other. This proves
the law of reciprocal proportions.
Class Exercise - 7
Nitrogen forms five stable oxides with
oxygen of the formula, N2O, NO,
N2O3, N2O4, N2O5. The formation of
these oxides explain the
(a) law of definite proportions
(b) law of partial pressure
(c) law of multiple proportion
(d) law of reciprocal proportions
Solution
The mass of oxygen which combine with the fixed mass
of nitrogen (= 14 g) is N2O, NO, N2O3, N2O4, N2O5 are 8,
16, 24, 32, 40 g respectively. They are in the ratio of
1 : 2: 3 : 4 : 5. This proves the law of multiple
proportions.
Class Exercise - 8
Two metallic oxides contains 27.6%
and 30.0% oxygen respectively.
If the formula of the first oxide is
M3O4, then that of the second will be
(a) MO
(b) MO2
(c) M2O5
Solution
7 2 .4
M  3
2 7 .6
4
16
7 2 .4
M

16
2 7 .6

3
4
M = 56
(d) M2O3
Class Exercise - 8
Two metallic oxides contains 27.6%
and 30.0% oxygen respectively.
If the formula of the first oxide is
M3O4, then that of the second will be
(a) MO
(b) MO2
Solution:
70
56
x
y

16

2
30

x
y
3
Hence oxide is M2O3.
(c) M2O5
(d) M2O3
Class Exercise - 9
One litre of nitrogen combines with
three litre of hydrogen to form two
litre of ammonia under the same
conditions of temperature and pressure.
This explain the
(a) law of constant composition
(b) law of multiple proportion
(c) law of reciprocal proportions
(d) Gay Lussac’s law of gaseous volumes
Solution
The ratio of volumes of N2, H2 and NH3 is 1 : 3 : 2, which
is a simple ratio. This proves Gay Lussac’s law of gaseous
volumes.
Class Exercise - 10
2.16 g of copper metal when treated
with nitric acid followed by ignition
of the nitrate gave 2.70 g of
copper oxide. In another
experiment 1.15 g of copper
oxide upon reduction with hydrogen gave
0.92 g of copper. Show that the above
data illustrate the law of definite proportions.
Solution
2 .1 6
st
 100  80%
% of Cu in copper oxide in 1 case 
2 .7 0
% of oxygen = 20% % of Cu in copper oxide in 2nd case

0 .9 2
1 .1 5
 100  80%
% of oxygen = 20%
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