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1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The
products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate.
Show that these observations are in agreement with the law of conservation of
mass.
sodium ethanoate + carbon dioxide +
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sodium carbonate + ethanoic acid
water
•
Solution:
Mass of the reactants = Mass of sodium carbonate + Mass of ethanoic acid
= 5.3+6.0=11.3 g
Mass of products = Mass of carbon dioxide + Mass of water + Mass of sodium ethonate
= 2.2 + 0.9 + 8.2 = 11.3 g
Since the mass of products is equal to the mass of reactants, mass is neither created nor lost
during the given chemical change and the observation made is in agreement with the law of
conservation of mass.
2.
Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water. What mass
of oxygen gas would be required to react completely with 3 g of hydrogen gas?
•
Solution:
X grams of hydrogen 8
x gram of oxygen will be required to form water.
Oxygen required to react with 3 g hydrogen to form water = 3 8 = 24 g.
3. Which postulate of Dalton’s atomic theory is the result of the law of
conservation of mass?
•
Solution:
Atoms are indivisible particles, which cannot be created nor destroyed in a chemical reaction.
4. Which postulate of Dalton’s atomic theory can explain the law of definite
proportions?
•
Solution:
The relative number and kinds of atoms are constant in a given compound.
5. Define the atomic mass unit.
in
•
Solution:
One atomic unit is a mass unit equal to exactly one twelfth (1/12th) the mass of one atom of
carbon-12.
6.
•
Why is not possible to see an atom with naked eye?
Solution:
The size of an atom is very small. So it is impossible to see an atom with naked eye.
7. Write down the formulae of
(i) sodium oxide
1
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide
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Solution:
The formulae are:
i) sodium oxide- Na2O
(ii) aluminium chloride-AlCl3
(iii) sodium suphide-Na2SO3
(iv) Magnesium hydroxide - Mg(OH)2
8. Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3.
Solution:
•
The names of compounds are as follows:
(i) Al2(SO4)3 – aluminium sulphate
(ii) CaCl2 – Calcium chloride
(iii) K2SO4 – potassium sulphate
(iv) KNO3 – potassium nitrate
(v) CaCO3- calcium carbonate
9.
•
What is meant by the term chemical formula?
Solution:
The chemical formula of a compound is a symbolic representation of its composition.
10. Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6,C2H4, NH3, CH3OH.
•
H2 = 1+ 1 = 2u
O2 = 16 + 16 = 32 u
Cl2 = 35.5 + 35.5 = 71 u
CO2 = 12 + 32 = 44u
CH4 = 12 + 1 4 = 16 u
C2H4 = 12 2 + 1 4 = 28 u
C2H6 = 12 2 + 1 6 = 30 u
NH3 = 14 + 1 3 = 17u
CH3OH. = 12 + 1 3 + 16 + 1 = 32 u
in
Solution:
Molecular mass of
Molecular mass of
Molecular mass of
Molecular mass of
Molecular mass of
Molecular mass of
Molecular mass of
Molecular mass of
Molecular mass of
11. Calculate the formula unit masses of ZnO, Na2O, K2CO3,
given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.
•
Solution:
Formula unit mass of ZnO = 65 + 16 = 81 u
Formula unit mass of Na2O = 23 2 + 16 = 62 u
Formula unit mass of K2CO3 = 39
2 + 12
1 + 16
3 = 138 u.
2
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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12. If one mole of carbon atom weighs 12 gram, what is the mass (in gram) of 1
atom of carbon?
•
Solution:
1 mole of carbon atom = 6.022
10 23 atoms
Now 6.022
10 23 atoms of carbon weigh 12 grams
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one atom of carbon weighs =
= 1.99
1023 g.
13. 0.24 g sample of compound of oxygen and boron was found to contain 0.096 g
of boron and 0.144 g of oxygen on analysis . Calculate the percentage composition of
the compound by weight.
•
Solution:
% of boron in the sample =
100 = 40%
% of oxygen in the sample =
100 = 60%
The sample of compound contains 40% boron and 60% oxygen by weight.
14. When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is
produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is
burnt in 50.00 g of oxygen? Which law of chemical combination will govern your
answer?
•
Solution:
The mass of the carbon dioxide formed will be only 11 grams. The remaining oxygen is not
used up. This indicates the law of definite proportions which says that in compounds, the
combining elements are present in definite proportions by mass.
15. What are polyatomic ions? Give examples.
•
Solution:
When two or more atoms combine together and behave like one entity with a net charge,
then it is called polyatomic ion. For example oxygen atom and hydrogen atom combine to
form hydroxide ion (OH-). One carbon atom and three oxygen atoms act as carbonate ion
(CO3-2).
•
in
16. Write the chemical formulae of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate.
Solution:
The chemical formulae for the following are
(a) Magnesium chloride – MgCl2
(b) Calcium oxide -CaO
(c) Copper nitrate –Cu(NO3)2
3
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CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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(d) Aluminium chloride –AlCl3
(e) Calcium carbonate. –CaCO3
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17. Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
•
Solution:
The names of the elements present in in the following compounds are
(a) Quick lime – calcium and oxygen
(b) Hydrogen bromide-Hydrogen and bromide
(c) Baking powder –Sodium, hydrogen , carbon and oxygen
(d) Potassium sulphate- Potassium, sulphur and oxygen.
18. Calculate the molar mass of the following substances.
(a) Ethyne, C2H2
(b) Sulphur molecule, S8
(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)
(d) Hydrochloric acid, HCl
(e) Nitric acid, HNO3
•
Solution:
The molar mass of the following substances are
(a) Ethyne, C2H2 – 2
12+ 1 2 = 26 u
(b) Sulphur molecule, S8 – 8
32 = 256 u
(c) Phosphorus molecule, P4 - 4
31 = 124 u
(d) Hydrochloric acid, HCl - 1 1 + 1 35.5 = 36.5 u
(e) Nitric acid, HNO3 - 1 1 + 1 14 + 3 16 = 63 u
19. What is the mass of—
(a) 1 mole of nitrogen atom
(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?
•
Solution:
a) 1 mole of nitrogen atom =14 u = 14 gm
(b) 4 moles of aluminium atoms = 4
(c) 10 moles of sodium sulphite = 10
27 = 108 u = 108 gram
(126 gram) = 1260 grams
•
Solution:
a)32 gram of oxygen gas
in
20. Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide.
= 1 mole
12 grams of oxygen gas
b) 18 grams of water
=
= 0.375 mole
= 1 mole
20 grams of water
=
= 1.1 mole
4
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MATHEMATIC CENTER D96 MUNIRKA VILLAGE NEW DELHI 110067 & VIKAS PURI NEW DELHI
CONTACT FOR COACHING MATHEMATICS FOR 11TH 12TH NDA DIPLOMA SSC CAT SAT CPT
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c) 44 grams of carbon dioxide = 1 mole
22 grams of carbon dioxide
= 0.5 mole
21. What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Solution:
1 mole of oxygen atoms = 16 grams
0.2 moles of oxygen atoms = 16 0.2 = 3.2 grams
b. 1 mole of water molecules = 18 grams
0.5 mole of water molecules = 18 0.5 = 9 grams
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22. Calculate the number of molecules of sulphur (S8) present in 16 g of solid
sulphur.
•
Solution:
1 mole of S8 = 32
8 = 256 grams
1 mole of S8 = 6.023
1023 molecules
256 grams S8 has = 6.023
16 grams S
= 3.76
8
10
23
has =
16
1022 molecules.
23. Calculate the number of aluminium ions present in 0.051 g of aluminium
oxide. (Hint: The mass of an ion is the same as that of an atom of the same element.
Atomic mass of Al = 27 u)
•
Solution:
1 mole of aluminium oxide = 2
27+ 3
102 grams aluminium oxide has =6.023
16 = 102 grams
10
23
aluminium oxide molecules
0.56 gram of aluminium oxide has =
Hence 0.56 grams aluminium oxide gives = 2
= 6.023
3.01 10
10
21
21
Al+++ ions.
aluminium ions
in
1 molecule of aluminium oxide gives
= 3.01
10 21aluminium oxide molecules
+++
= 2 Al
ions
5
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