Chemistry. Class Exercise Class Exercise - 1 Express the following numbers to three significant...
-
Upload
buck-harper -
Category
Documents
-
view
221 -
download
1
Transcript of Chemistry. Class Exercise Class Exercise - 1 Express the following numbers to three significant...
Chemistry
Class Exercise
Class Exercise - 1
Express the following numbers tothree significant figures.
(i) 6.022 × 1023 (ii) 5.356 g(iii) 0.0652 g (iv) 13.230
Solution
(i) 6.02 × 1023
(ii) 5.36 g(iii) 0.0652 g(iv)13.2
Class Exercise - 2
What is the sum of 2.368 g and1.02 g?
Solution
2.368 g
1.02 g
3.388
= 3.39 g
Class Exercise - 3
Express the result of the followingcalculation to the appropriate numberof significant figures816 × 0.02456 + 215.67
Solution
816 × 0.02456 = 20.0
Product rounded off to 3 significant figures becausethe least number of significant figure in thismultiplication is three.
67.235
67.215
0.20
Rounded off to 235.7
Class Exercise - 4
Solve the following calculations andexpress the results to appropriatenumber of significant figures.
(i) 1.6 × 103 + 2.4 × 102 – 2.16 × 102
(ii)23
20
6.02 10 5.00
4.0 10
Solution
(i) 1.6 × 103 + .24 × 103
3
3
3
1.6 10
.24 10
1.84 10
Rounded off to 1.8 × 103
3
3
3
1.8 10
.216 10
1.584 10
Class Exercise - 4
Rounded off to 1.6 × 103 or 16 × 102
(ii)23 23
20 20
6.02 10 5.00 30.10 10
4.0 10 4.0 10
= 7.525 × 103 (rounded off to 7.5 × 103)
Class Exercise - 5
Convert 10 feet 5 inches into SI unit.
10 feet 5 inches = 125 inches1 inch = 2.54 × 10-2 m
Solution
Rounded off to 317 × 10–2 m
125 inches = 2.54 × 10-2 × 125 m
= 317.5 × 10-2 m
Class Exercise - 6
A football was observed to travel at a speedof 100 miles per hour. Express the speedin SI units.
Solution
1 mile = 1.60 × 103 m100 miles per hour
3100 1.60 1060 60
= 4.4 × 10-4 × 105 m/s= 4.4 × 10 m/s= 44 m/s
Class Exercise - 7
What do the following abbreviationsstand for?
(i) O (ii) 2O (iii) O2 (iv) 3O2
Solution
(i) Oxygen atom(ii) 2 moles of oxygen atom(iii) Oxygen molecule(iv)3 moles of oxygen molecule
Class Exercise - 8
Among the substances given belowchoose the elements, mixtures andcompounds
(i) Air (ii) Sand(iii) Diamond(iv) Brass
Solution
(i) Air - Mixture(ii) Sand (SiO2) - Compound(iii) Diamond (Carbon) - Element(iv) Brass (Alloy of metal) - Mixture
Class Exercise - 9
Classify the following into elementsand 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 CHEMISTRYSession - 2
Session Opener
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
Session Objectives
Law of conservation of mass
Total mass of the products remains equal to the total mass of the reactants.
H2 + Cl2 2 HCl
2g 71g 73g
Question
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?
Illustrative Problem
8.4 + 20.0 = m + 4.4 m = 24 g
It proves the the law of conservation of mass.
Solution:
A chemical compound always contains same elements combined together in same proportion of mass.
Law of definite proportions
Ice water H2O 1 : 8
River water H2O 1 : 8
Sea water H2O 1 : 8
Question
Two gaseous samples were analyzed.One contained 1.2g of carbon and3.2 g of oxygen. The other contained 27.3 % carbon and 72.7% oxygen. The experimental data are in accordancewith
(a) Law of conservation of mass(b) Law of definite proportions(c) Law of multiple proportions(d) Law of reciprocal proportions
Illustrative Problem
% of C in the 1st sample
%3.271002.32.1
2.1
Which is same as in the second sample.Hence law of definite proportion is obeyed.
Solution
the mass of one of the elements which combines with fixed mass of the others, bear a simple whole number ratio to one another.
Law of multiple proportions
8:16:24:32:40 1:2:3:4:5=
14:8 14:16 14:24 14:32 14:40
Two elements combine two or more compounds
Ratio of oxygen combining with 14 parts of nitrogen
The ratio of the weights of two elementsA and B which combine separately witha fixed weight of the third element ‘C’ iseither the same or some simple multipleof the ratio of the weights in which A andB combine directly with each other.
Statement of law of reciprocalProportions
)W:W(kW:W
W:WBA
CB
CA
k may be 1
Law of reciprocal proportions
SO2
OH
S
H2S
H2O
H2S 2 : 32, 1 : 16
SO2 32 : 32, 1 : 1
16:11
1:
16
1SO:SH 22
8:1,16:2OH2
2
1
1
8
16
1OH:SO:SH 222
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
Gay Lussac’s Law of gaseous volumes
N2 O2 2NO
2H2O2 2H2O
1 volumenitogen
gas
1 volumeoxygen
gas
2 volumenitrogen
oxide+
2 volumehydrogen
gas
1 volumeoxygen
gas
2 volumesteam
+
1 : 1 : 2
2 : 1 : 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 basedon quantitative results of chemical reactions.
Questions
Carbon is found to form two oxides,which contains 42.8% and 27.27% ofcarbon respectively. Find out which of the laws of chemical combination is proved correct by this data?
Illustrative Problem
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
= 2:1
= 0.748 : 0.374
Solution:
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?
Illustrative Problem
N
H
O
NH3
N2O3
H2O
15.63
)85.36(K
90.88
10.11
65.17
35.82
Hence k=1 which proves law of reciprocal proportion.
Solution:
Ask your self
The balanced chemical reaction is an expression of
Law of multiple proportion Law of
conservation 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 muchZinc should be used to produce 13.7 gm of zinc sulphate crystals and how muchwater they will contain?
Solution:
100 gm of ZnSO4 will have 43.9 gm water and 22.6 gm zinc
4Zn required to produce 100gm of ZnSO crystals 22.6 gm
(Hint: Law of constant composition)
Solution
43.913.7gm of crystal contain water 100
13.76.0143gm
100gm of crystal contain water 43.9gm
4
22.6Zn required to produce 13.7gm of ZnSO crystals 13.7
100
3.0962gm
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.
Dalton’s Atomic Theory
5. Atoms of two elements may combine in different ratios to form more than one compound.
SO2 1:2
SO3 1:3
S+O2
Limitations of Dalton’s Atomic Theory
It could not explain:
Why atoms of different elementshave 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 particleof 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 differentatomic 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 insamples of CuO obtained by differentmethods were found to be the same.This proves the law of
(a) constant proportions (b) reciprocal proportions(c) multiple proportions (d) none of these
Solution
Definition
Class Exercise - 2
A balanced chemical equation is inaccordance 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 foundto contain hydrogen and oxygen in theapproximate 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 illustratedby 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 contain63.65%, 46.69% and 30.46%nitrogen respectively. These dataproves 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 sulphurto form CS2. 12 g of carbon alsocombine with 32 g oxygen to form CO2.10 g sulphur combine with 10 g oxygento 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
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.
C
S O
CS2 CO2
SO2
12 g
10 g 10 g
64 g 32 g
Class Exercise - 7
Nitrogen forms five stable oxides withoxygen of the formula, N2O, NO,N2O3, N2O4, N2O5. The formation ofthese oxides explain the
(a) law of definite proportions(b) law of partial pressure(c) law of multiple proportion(d) law of reciprocal proportions
The mass of oxygen which combine with the fixed massof 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.
Solution
Class Exercise - 8
Two metallic oxides contains 27.6%and 30.0% oxygen respectively.If the formula of the first oxide isM3O4, then that of the second will be
(a) MO (b) MO2 (c) M2O5 (d) M2O3
Solution
72.43M
27.6 416
72.4 16 3
M 27.6 4M = 56
70 16 x56 30 y
x 2y 3
Hence oxide is M2O3.
Class Exercise - 8
Two metallic oxides contains 27.6%and 30.0% oxygen respectively.If the formula of the first oxide isM3O4, then that of the second will be
(a) MO (b) MO2 (c) M2O5 (d) M2O3
Solution:
Class Exercise - 9One litre of nitrogen combines withthree litre of hydrogen to form twolitre of ammonia under the sameconditions 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, whichis a simple ratio. This proves Gay Lussac’s law of gaseousvolumes.
Class Exercise - 102.16 g of copper metal when treatedwith nitric acid followed by ignitionof the nitrate gave 2.70 g ofcopper oxide. In anotherexperiment 1.15 g of copperoxide upon reduction with hydrogen gave0.92 g of copper. Show that the abovedata illustrate the law of definite proportions.
Solution
% of Cu in copper oxide in 1st case 2.16
100 80%2.70
% of oxygen = 20% % of Cu in copper oxide in 2nd case
0.92
100 80%1.15
% of oxygen = 20%
Thank you