Science NCERT Grade 9, Chapter 3, Atoms and Molecules begins with a discussion on how the atoms were discovered and other historical aspects related to the discovery of the atom. The first important concept discussed in the chapter is the laws of chemical combination. The following two laws are explained under this:
-
Law of conservation of mass
-
Law of constant proportions
Dalton’s atomic theory and its postulates are also stated. The next concept discussed is based on the size of the atom. A table showing the relative size of different matters is presented in chapter, Atoms and Molecules. Symbols of atoms proposed by Dalton are also shown along with the IUPAC symbols of some basic elements. Thereafter, the concept of atomic mass along with examples is included in the chapter, Atoms and Molecules. Molecules of different elements- metals and non-metals are shown in a tabular format. Similarly, the chapter explains the concept of ion along with suitable examples. Subsequently, a detailed procedure of writing chemical formulas using symbols and combining capacity i.e valency of elements are thoroughly explained. Examples of some simple compounds and their chemical formula are mentioned in the chapter, Atoms and Molecules. Apart from this, the concept of molecular mass and mole concept is also discussed. A flowchart represents the relationship between Avogadro's number and mole followed by ample numericals based on this relationship.
The topics covered in the chapter, Atoms and Molecules are -
- Laws of Chemical Combination
- What is an Atom?
- What is a Molecule?
- Writing Chemical Formulae
- Molecular Mass and Mole Concept
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, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + ethanoic acid → sodium ethanoate + carbon dioxide + water
Answer:
In the given reaction, sodium carbonate reacts with ethanoic acid to produce sodium ethanoate, carbon dioxide, and water.
Mass of sodium carbonate = 5.3 g (Given)
Mass of ethanoic acid = 6 g (Given)
Mass of sodium ethanoate = 8.2 g (Given)
Mass of carbon dioxide = 2.2 g (Given)
Mass of water = 0.9 g (Given)
Now, total mass before the reaction = (5.3 + 6) g
= 11.3 g
And, total mass after the reaction = (8.2 + 2.2 + 0.9) g
= 11.3 g
∴Total mass before the reaction = Total mass after the reaction
Hence, the given observations are in agreement with the law of conservation of mass.
Video Solution for atoms and molecules (Page: 32 , Q.No.: 1)
NCERT Solution for Class 9 science - atoms and molecules 32 , Question 1
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 3 g 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 1 g of hydrogen gas is 8 g.
Therefore, the mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3 g = 24 g.
Video Solution for atoms and molecules (Page: 33 , Q.No.: 2)
NCERT Solution for Class 9 science - atoms and molecules 33 , Question 2
Page No 33:
Question 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.
Video Solution for atoms and molecules (Page: 33 , Q.No.: 3)
NCERT Solution for Class 9 science - atoms and molecules 33 , Question 3
Page No 33:
Question 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.
Video Solution for atoms and molecules (Page: 33 , Q.No.: 4)
NCERT Solution for Class 9 science - atoms and molecules 33 , Question 4
Page No 35:
Question 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’.
Video Solution for atoms and molecules (Page: 35 , Q.No.: 1)
NCERT Solution for Class 9 science - atoms and molecules 35 , Question 1
Page No 35:
Question 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, the atom of an element does not exist independently.
Video Solution for atoms and molecules (Page: 35 , Q.No.: 2)
NCERT Solution for Class 9 science - atoms and molecules 35 , Question 2
Page No 39:
Question 1:
Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer:
(i) Sodium oxide →Na2O
(ii) Aluminium chloride → AlCl3
(iii) Sodium sulphide → Na2S
(iv) Magnesium hydroxide → Mg(OH)2
Video Solution for atoms and molecules (Page: 39 , Q.No.: 1)
NCERT Solution for Class 9 science - atoms and molecules 39 , Question 1
Page No 39:
Question 2:
Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3
Answer:
(i) Al2(SO4)3 → Aluminium sulphate
(ii) CaCl2 → Calcium chloride
(iii) K2SO4 → Potassium sulphate
(iv) KNO3 → Potassium nitrate
(v) CaCO3 → Calcium carbonate
Video Solution for atoms and molecules (Page: 39 , Q.No.: 2)
NCERT Solution for Class 9 science - atoms and molecules 39 , Question 2
Page No 39:
Question 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.
Video Solution for atoms and molecules (Page: 39 , Q.No.: 3)
NCERT Solution for Class 9 science - atoms and molecules 39 , Question 3
Page No 39:
Question 4:
How many atoms are present in a
(i) H2S molecule and
(ii) PO43− ion?
Answer:
(i) In an H2S molecule, three atoms are present; two of hydrogen and one of sulphur.
(ii) In a PO43− ion, five atoms are present; one of phosphorus and four of oxygen.
Video Solution for atoms and molecules (Page: 39 , Q.No.: 4)
NCERT Solution for Class 9 science - atoms and molecules 39 , Question 4
Page No 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
= 2 u
Molecular mass of O2 = 2 × Atomic mass of O
= 2 × 16
= 32 u
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
= 30 u
Molecular mass of C2H4 = 2 × Atomic mass of C + 4 × Atomic mass of H
= 2 × 12 + 4 × 1
= 28 u
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
Video Solution for atoms and molecules (Page: 40 , Q.No.: 1)
NCERT Solution for Class 9 science - atoms and molecules 40 , Question 1
Page No 40:
Question 2:
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.
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
= 62 u
Formula unit mass of K2CO3 = 2 × Atomic mass of K + Atomic mass of C + 3 × Atomic mass of O
= 2 × 39 + 12 + 3 × 16
= 138 u
Video Solution for atoms and molecules (Page: 40 , Q.No.: 2)
NCERT Solution for Class 9 science - atoms and molecules 40 , Question 2
Page No 42:
Question 1:
If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Answer:
One mole of carbon atoms weighs 12 g (Given)
i.e., mass of 1 mole of carbon atoms = 12 g
Then, mass of number of carbon atoms = 12 g
Therefore, mass of 1 atom of carbon
Video Solution for atoms and molecules (Page: 42 , Q.No.: 1)
NCERT Solution for Class 9 science - atoms and molecules 42 , Question 1
Page No 42:
Question 2:
Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na = 23 u, Fe = 56 u)?
Answer:
Atomic mass of Na = 23 u (Given)
Then, gram atomic mass of Na = 23 g
Now, 23 g of Na contains = number of atoms
Thus, 100 g of Na contains number of atoms
= number of atoms
Again, atomic mass of Fe = 56 u(Given)
Then, gram atomic mass of Fe = 56 g
Now, 56 g of Fe contains = number of atoms
Thus, 100 g of Fe contains number of atoms
number of atoms
Therefore, 100 grams of sodium contain more number of atoms than 100 grams of iron.
Video Solution for atoms and molecules (Page: 42 , Q.No.: 2)
NCERT Solution for Class 9 science - atoms and molecules 42 , Question 2
Page No 43:
Question 1:
A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.
Answer:
Mass of boron = 0.096 g (Given)
Mass of oxygen = 0.144 g (Given)
Mass of sample = 0.24 g (Given)
Thus, percentage of boron by weight in the compound =
= 40%
And, percentage of oxygen by weight in the compound =
= 60%
Video Solution for atoms and molecules (Page: 43 , Q.No.: 1)
NCERT Solution for Class 9 science - atoms and molecules 43 , Question 1
Page No 43:
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:
Carbon + OxygenCarbon dioxide
3 g of carbon reacts with 8 g of oxygen to produce 11 g of carbon dioxide.
If 3 g of carbon is burnt in 50 g of oxygen, then 3 g 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 11 g of carbon dioxide will be formed.
The above answer is governed by the law of constant proportions.
Video Solution for atoms and molecules (Page: 43 , Q.No.: 2)
NCERT Solution for Class 9 science - atoms and molecules 43 , Question 2
Page No 44:
Question 3:
What are polyatomic ions? Give examples?
Answer:
A polyatomic ion is a group of atoms carrying a charge (positive or negative). For example, ammonium ion, hydroxide ion (OH−), carbonate ion, sulphate ion.
Video Solution for atoms and molecules (Page: 44 , Q.No.: 3)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 3
Page No 44:
Question 4:
Write the chemical formulae of the following:
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
Answer:
(a) Magnesium chloride → MgCl2
(b) Calcium oxide → CaO
(c) Copper nitrate → Cu (NO3)2
(d) Aluminium chloride → AlCl3
(e) Calcium carbonate → CaCO3
Video Solution for atoms and molecules (Page: 44 , Q.No.: 4)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 4
Page No 44:
Question 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 |
Video Solution for atoms and molecules (Page: 44 , Q.No.: 5)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 5
Page No 44:
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:
(a) Molar mass of ethyne, C2H2 = 2 × 12 + 2 × 1 = 26 g
(b) Molar mass of sulphur molecule, S8 = 8 × 32 = 256 g
(c) Molar mass of phosphorus molecule, P4 = 4 × 31 = 124 g
(d) Molar mass of hydrochloric acid, HCl = 1 + 35.5 = 36.5 g
(e) Molar mass of nitric acid, HNO3 = 1 + 14 + 3 × 16 = 63 g
Video Solution for atoms and molecules (Page: 44 , Q.No.: 6)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 6
Page No 44:
Question 7:
What is the mass of−−
(a) 1 mole of nitrogen atoms?
(b) 4 moles 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 14 g.
(b) The mass of 4 moles of aluminium atoms is (4 × 27) g = 108 g
(c) The mass of 10 moles of sodium sulphite (Na2SO3) is
10 × [2 × 23 + 32 + 3 × 16] g = 10 × 126 g = 1260 g
Video Solution for atoms and molecules (Page: 44 , Q.No.: 7)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 7
Page No 44:
Question 8:
Convert into mole.
(a) 12 g of oxygen gas
(b) 20 g of water
(c) 22 g of carbon dioxide
Answer:
(a) 32 g of oxygen gas = 1 mole
Then, 12 g of oxygen gas == 0.375 mole
(b) 18 g of water = 1 mole
Then, 20 g of water == 1.11 moles (approx)
(c) 44 g of carbon dioxide = 1 mole
Then, 22 g of carbon dioxide == 0.5 mole
Video Solution for atoms and molecules (Page: 44 , Q.No.: 8)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 8
Page No 44:
Question 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 = 16 g
Then, mass of 0.2 mole of oxygen atoms = 0.2 × 16g = 3.2 g
(b) Mass of one mole of water molecule = 18 g
Then, mass of 0.5 mole of water molecules = 0.5 × 18 g = 9 g
Video Solution for atoms and molecules (Page: 44 , Q.No.: 9)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 9
Page No 44:
Question 10:
Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.
Answer:
1 mole of solid sulphur (S8) = 8 × 32 g = 256 g
i.e., 256 g of solid sulphur contains = 6.022 × 1023 molecules
Then, 16 g of solid sulphur contains =
= 3.76 × 1022 molecules (approx)
Video Solution for atoms and molecules (Page: 44 , Q.No.: 10)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 10
Page No 44:
Question 11:
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)
Answer:
1 mole of aluminium oxide (Al2O3) = 2 × 27 + 3 × 16
= 102 g
i.e., 102 g of Al2O3 = 6.022 × 1023 molecules of Al2O3
Then, 0.051 g of Al2O3 contains =
= 3.011 × 1020 molecules of Al2O3
The number of aluminium ions (Al3+) present in one molecule of aluminium oxide is 2.
Therefore, the number of aluminium ions (Al3+) present in 3.011 × 1020 molecules (0.051 g ) of aluminium oxide (Al2O3) = 2 × 3.011 × 1020
= 6.022 × 1020
Video Solution for atoms and molecules (Page: 44 , Q.No.: 11)
NCERT Solution for Class 9 science - atoms and molecules 44 , Question 11
View NCERT Solutions for all chapters of Class 9