Page 32
Q.No. 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, 0.9g water and 8.2 g of sodium ethanoate. Show that these observations are in
agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Answer:
In a reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate,
carbondioxide, and water.
Sodium + Ethanoic → Sodium + Carbon + Water
Carbonate acid ethanoate dioxide
Mass of sodium carbonate = 5.3g (Given)
Mass of ethanoic acid = 6g (Given)
Mass of sodium ethanoate = 8.2g (Given)
Mass of carbon dioxide = 2.2 (Given)
Mass of water = 0.9g (Given)
Now, total mass before the reaction = (5.3 + 6)g
= 11. 3g
and total mass after the reaction = (8.2 + 2.2 + 0.9)g
= 11.3g
Therefore, Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of ma
Page 33
Q.No. 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 3g of hydrogen gas?
Answer:
It is given that the ratio of hydrogen and oxygen by mass to form water is 1:8. Then, the mass of
oxygen gas required to react completely with 1g of hydrogen gas is 8g. Therefore, the mass of
oxygen gas required to react completely with 3g of hydrogen gas is 8 × 3g = 24 g.
Q.No. 3:
Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?
Answer:
The postulate of Dalton's atomic theory which is a result of the law of conservation of mass is
“Atoms are indivisible particles, which can neither be created nor destroyed in a chemical
reaction”.
Q.No. 4:
Which postulate of Dalton's atomic theory can explain the law of definite proportions?
Answer:
The postulate of Dalton's atomic theory which can explain the law of definite proportion is “The
relative number and kind of atoms in a given compound remains constant”.
Page 35
Q.No. 1:
Define atomic mass unit.
Answer:
Mass unit equal to exactly one- twelfth the mass of one atom of carbon - 12 is called one atomic
mass unit. It is written as 'u'.
Q.No. 2:
Why is it not possible to see an atom with naked eyes?
Answer:
The size of an atom is so small that it is not possible to see it with naked eyes. Also, atom of an
element does not exist independently.
Page 39
Q.No. 1:
Write down the formula of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide
Answer:
(i) Sodium oxide → Na2O
(ii) Aluminium chloride → AlCl3
(iii) Sodium suphide → Na2S
(iv) Magnesium hydroxide → Mg(OH)2
Q.No. 2:
Write down the names of compounds represented by the following formula:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3
Answer:
(i) Al(SO4)3→ Aluminium sulphate
(ii) CaCl2→ Calcium chloride
(iii) K2SO4→ Potassium sulphate
(iv) CaCO3→ Calcium carbonate
Q.No. 3:
What is meant by the term chemical formula?
Answer:
The chemical formula of a compound means the symbolic representation of the composition of a
compound. From the chemical formula of a compound, we can know the number and kinds of
atoms of different elements that constitute the compound. For example, from the
chemical formula CO2 of carbon dioxide, we come to know that one carbon atom and two oxygen
atoms are chemically bonded together to form one molecule of the compound, carbon dioxide.
Question 4:
How many atoms are present in a
(i) H2S molecule and
(ii) PO4
3-
ion?
Answer :
(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO4
3-
ion, five atoms are present; one of phosphorus and four of oxygen.
Page 40
Question 1:
Calculate the molecular masses of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Answer:
Molecular mass of H2 = 2 × Atomic mass of H
= 2 × 1 = 2u
Molecular mass of O2 = 2 × Atomic mass of O
= 2 × 16 = 32u
Molecular mass of Cl2 = 2 × Atomic mass of Cl
= 2 × 35.5 = 71 u
Molecular mass of CO2 = Atomic mass of C + 2 × Atomic mass of O
= 12 + 2 × 16 = 44 u
Molecular mass of CH4 = Atomic mass of C + 4 × Atomic mass of H
= 12 + 4 × 1 = 16 u
Molecular mass of C2H6 = 2 × Atomic mass of C + 6 × Atomic mass of H
= 2 × 12 + 6 × 1 = 30u
Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1 = 28u
Molecular mass of NH3 = Atomic mass of N + 3 × Atomic mass of H
= 14 + 3 × 1 =17 u
Molecular mass of CH3OH Atomic mass of C+4 ×Atomic mass of H+Atomic mass of O
= 12 + 4 × 1 + 16 = 32 u
Q.No. 2:
Calculate the formula unit masses of ZnO, Na2O, K2CO3, given masses of Zn = 65u, Na = 23u, K
= 39u, C = 12u, and O = 16u.
Answer :
Formula unit mass of ZnO = Atomic mass of Zn + Atomic mass of O
= 65 + 16 = 81 u
Formula unit mass of Na2O = 2 × Atomic mass of Na + Atomic mass of O
= 2 × 23 + 16 = 62u
Formula unit mass of K2CO3
= 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 3 × 16 = 138u
Page 42
Question 1:
If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?
Answer:
One mole of carbon atoms weighs 12g (Given)
i.e., mass of 1 mole of carbon atoms = 12g
Then, mass of 6.022× 1023 number of carbon atoms = 12g
Therefore, mass of 1 atom of carbon =
12
6.022× 10– 23 g g
= 1.9926 × 10– 23 g
Q.No. 2:
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass
of Na = 23u, Fe =56 u)?
Answer 2:
Atomic mass of Na = 23u (Given)
Then, gram atomic mass of Na = 23g
Now, 23g of Na contains = 6.022×1023 number of atoms
Thus, 100g of Na contains =
6.022×1023 ×10⁻²³
number of atoms
= 2.6182 × 1024 number of atoms
Again, atomic mass of Fe = 56u (Given)
Then, gram atomic mass of Fe = 56g
Now, 56 g of Fe contains = 6.022×1023 number of atoms
Thus, 100 g of Fe 6.022×1023 ×100
56
number of atoms
= 1.0753 × 1024 number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.
Q.No. 5:
Give the names of the elements present in the following compounds:
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate.
Answer:
Compound Chemical formula Elements present
Quick lime CaO Calcium, oxygen
Hydrogen bromide HBr Hydrogen, bromine
Baking powder NaHCO3 Sodium, hydrogen, carbon,
oxygen
Potassium sulphate K2SO4 Potassium, sulphur, oxygen
Q.No. 6:
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
Answer:
(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 28g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256g
(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63g
Q.No. 7:
What is the mass of
(a) 1 mole of nitrogen atoms?
(b) 4 mole of aluminium atoms (Atomic mass of aluminium = 27)?
(c) 10 moles of sodium sulphite (Na2SO3)?
Answer:
(a) The mass of 1 mole of nitrogen atoms is 14g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27)g = 108g
(c) The mass of 10 moles of sodium sulphite (Na2SO3) is 10 × [2 × 23 + 32 + 3 × 16]g
= 10 × 126g = 1260g
Q.No. 8:
Convert into mole.
(a) 12g of oxygen gas
(b) 12g of water
(c) 22g of carbon dioxide
Answer:
(a) 32 g of oxygen gas = 1 mole
Then, 12g of oxygen gas = 12/32 mole = 0.375 mole
(b) 18g of water = 1 mole
Then, 20 g of water = 20/18 mole = 1.11 moles (approx.)
(c) 44g of carbon dioxide = 1 mole
Then, 22g of carbon dioxide = 22/44 mole = 0.5 mole
Q.No. 9:
What is the mass of:
(a) 0.2 mole of oxygen atoms?
(b) 0.5 mole of water molecules?
Answer:
(a) Mass of one mole of oxygen atoms = 16g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2g
(b) Mass of one mole of water molecule = 18g
Then, mass of 0.5 mole of water molecules = 0.5 × 18g = 9g
Question 10:
Calculate the number of molecules of sulphur (S8) present in 16g of solid sulphur.
Answer 10:
1 mole of solid sulphur (S8) = 8 × 32g = 256g
i.e., 256g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16g of solid sulpur contains 6.022×1023
256
× 16 molecules
= 3.76 × 1022 molecules (approx)
Q.No. 11:
Calculate the number of aluminium ions present in 0.051g 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
= 27u)
Answer:
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16 = 102g
i.e., 102g of Al2O3 = 6.022 × 10⁻²³ molecules of Al2O3
Then, 0.051 g of Al2O3 contains =
6.022 × 10⁻²³
102
× 0.051 molecules
= 3.011 × 1020
molecules of Al2O3
The number of aluminium ions (Al3+
) present in one molecules of aluminium oxide is 2.
Therefore, The number of aluminium ions (Al3+
) present in
3.11 × 1020 molecules (0.051g) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020
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