Ncertt solution of class 9 chapter atom and molecules
Page no. 32
Question 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 1: 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 mass.
Page no. 33
Question 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 2: 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.
Question 3: Which postulate of Dalton's
atomic theory is the result of the law of conservation of mass?
Answer 3: 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”.
Question 4:
Which postulate of Dalton's
atomic theory can explain the law of definite proportions?
Answer 4: 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 no. 35
Question 1: Define atomic mass unit.
Answer 1: 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'.
Question 2: Why is it not possible to see
an atom with naked eyes?
Answer 2: 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 no. 39
Question 1:
Write down the formula of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium suphide
(iv) magnesium hydroxide
Answer 1:
(i) Sodium oxide → Na2O
(ii) Aluminium chloride → AlCl3
(iii) Sodium suphide → Na2S
(iv) Magnesium hydroxide → Mg(OH)2
Question 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 2:
(i) Al(SO4)3→ Aluminium sulphate
(ii) CaCl2→ Calcium chloride
(iii) K2SO4→ Potassium sulphate
(iv) CaCO3→ Calcium carbonate
Question 3:
What is meant by the term
chemical formula?
Answer 3: 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
Answer 4:
Page no.40
Question 1: Calculate the molecular masses
of H2, O2, Cl2, CO2, CH4, C2H6, C2H4, NH3, CH3OH.
Answer 1:
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
Question 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 2:
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 1:
Question 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
Therefore, 100 grams of sodium
contain more number of atoms than 100 grams of iron.
Exercise
Question 1: A 0.24 g sample of
compound of oxygen and boron was found by analysis to contain 0.096 g if boron
and 0.144 g of oxygen. Calculate the percentage composition of the compound by
weight.
Answer 1:
Mass of boron = 0.096g
(Given)
Mass of oxygen = 0.144g
(Given)
Mass of sample = 0.24g
(Given)
Question 2: When 3.0 g of carbon is
burnt in 8.00 g 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 combinations will govern your answer?
Answer 2:
Carbon + Oxygen
→ Carbon dioxide
3g of carbon reacts with 8
g of oxygen to produce 11g of carbon dioxide. If 3g of carbon is burnt in 50g
of oxygen, then 3g of carbon will react with 8 g of oxygen. The remaining 42 g
of oxygen will be left un-reactive. In this case also, only 11g of carbon dioxide
will be formed. The above answer is governed by the law of constant
proportions.
Question 3: What are polyatomic ions?
Give examples?
Answer 3:
Question 4: Write the chemical formula
of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer 4:
(a) Magnesium chloride
→MgCl2
(b) Calcium oxide →CaO
(c) Copper nitrate →Cu(NO3)2
(d) Aluminium
chloride →AlCl3
```(e) Calcium carbonate →CaCO3
Question 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 6:
(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
Question 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 7:
(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
Question 8: Convert into mole.
(a) 12g of oxygen gas
(b) 12g of water
(c) 22g of carbon dioxide
Answer 8:
(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
Question 9: What is the mass of: (a) 0.2
mole of oxygen atoms? (b) 0.5 mole of water molecules?
Answer 9: 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:
Question 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 11:
