NCERT Class 9 Chapter 3 – Atoms and Molecules
This chapter deals with the introduction of atoms and molecules that constitute matter. The important topics touched upon have been summarized as follows: Introduction to Atoms and Molecules: Explanation of the basic units of matter and their interaction. Structure of Atoms: Studying the atomic structure – protons, neutrons and electrons. Composition of molecules : How atoms are combined into molecules and also studying molecular formulae. Chemical Reaction: Reactions involving atoms and molecules and laws governing these reactions.
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The NCERT Class 9 Chapter 3 – Atoms and Molecules are tailored to help the students master the concepts that are key to success in their classrooms. The solutions given in the PDF are developed by experts and correlate with the CBSE syllabus of 2023-2024. These solutions provide thorough explanations with a step-by-step approach to solving problems. Students can easily get a hold of the subject and learn the basics with a deeper understanding. Additionally, they can practice better, be confident, and perform well in their examinations with the support of this PDF.
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Students can access the NCERT Class 9 Chapter 3 – Atoms and Molecules. Curated by experts according to the CBSE syllabus for 2023–2024, these step-by-step solutions make Science much easier to understand and learn for the students. These solutions can be used in practice by students to attain skills in solving problems, reinforce important learning objectives, and be well-prepared for tests.
Exercise-3.1
In a reaction, 5.3g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.
Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water
Sodium carbonate + acetic acid → Sodium acetate + carbon dioxide + water
5.3g 6g 8.2g 2.2g 0.9g
As per the law of conservation of mass, the total mass of reactants must be equal to the total mass of
products.
As per the above reaction, L.H.S. = R.H.S. i.e., 5.3g + 6g = 2.2g + 0.9 g + 8.2 g = 11.3 g
Hence, the observations are in agreement with the law of conservation of mass.
Hydrogen and oxygen combine in a 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?
We know hydrogen and water mix in a ratio 1: 8.
For every 1g of hydrogen, it is 8g of oxygen.
Therefore, for 3g of hydrogen, the quantity of oxygen = 3 x 8 = 24g
Hence, 24g of oxygen would be required for the complete reaction with 3g of hydrogen gas.
Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
The relative number and types of atoms are constant in a given composition, says Dalton’s atomic theory, which is based on the rule of conservation of mass.
“Atoms cannot be created nor be destroyed in a chemical reaction.”
Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
The postulate of Dalton’s atomic theory that can explain the law of definite proportions is that the
relative number and kinds of atoms are equal in given compounds.
Exercise-3.2
Define the atomic mass unit.
An atomic mass unit is a unit of mass used to express the weights of atoms and molecules where one
atomic mass is equal to 1/12th the mass of one carbon-12 atom.
Why is it not possible to see an atom with the naked eyes?
Firstly, atoms are minuscule in nature, measured in nanometers. Secondly, except for atoms of noble
gases, they do not exist independently. Hence, an atom cannot be visible to the naked eyes.
Exercise-3.3
Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
The following are the formulae:
(i) sodium oxide – Na2O
(ii) aluminium chloride – AlCl3
(iii) sodium sulphide – Na2S
(iv) magnesium hydroxide – Mg (OH)2
Write down the names of compounds represented by the following formulae:
(i) Al2(SO4)3
(ii) CaCl2
(iii) K2SO4
(iv) KNO3
(v) CaCO3.
Listed below are the names of the compounds for each of the following formulae:
(i) Al2(SO4)3 – Aluminium sulphate
(ii) CaCl2 – Calcium chloride
(iii) K2SO4 – Potassium sulphate
(iv) KNO3 – Potassium nitrate
(v) CaCO3 – Calcium carbonate
What is meant by the term chemical formula?
Chemical formulas are used to describe the different types of atoms and their numbers in a compound or element. Each element’s atoms are symbolised by one or two letters. A collection of chemical symbols that depicts the elements that make up a compound and their quantities.
For example, the chemical formula of hydrochloric acid is HCl.
How many atoms are present in a
(i) H2S molecule and
(ii) PO43- ion?
The number of atoms present is as follows:
(i) H2S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total.
(ii) PO43- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.
Exercise 3.5
If one mole of carbon atoms weighs 12grams, what is the mass (in grams) of 1 atom of carbon?
Given: 1 mole of carbon weighs 12g
1 mole of carbon atoms = 6.022 x 1023
The molecular mass of carbon atoms = 12g = an atom of carbon mass
Hence, mass of 1 carbon atom = 12 / 6.022 x 1023 = 1.99 x 10-23g
Which has more number of atoms, 100 grams of sodium or 100 grams of iron
(given the atomic mass of Na = 23u, Fe = 56 u)?
(a) In 100 grams of Na:
m = 100g, Molar mass of Na atom = 23g, N0 = 6.022 x 1023, N = ?
N = (Given mass x N0)/Molar mass
N = (100 x 6.022 x 1023)/ 23
N = 26.18 x 1023 atoms
(b) In 100 grams of Fe:
m = 100 g, Molar mass of Fe atom = 56 g, N0 = 6.022 x 1023, N = ?
N = (Given mass x N0)/ Molar mass
N = (100 x 6.022 x 1023)/ 56
N = 10.75 x 1023 atoms
Therefore, the number of atoms is more in 100 g of Na than in 100 g of Fe.
Exercise 3.6
Give the names of the elements present in the following compounds.
(a) Quick lime
(b) Hydrogen bromide
(c) Baking powder
(d) Potassium sulphate
The following are the names of the elements present in the following compounds:
(a) Quick lime – Calcium and oxygen (CaO)
(b) Hydrogen bromide – Hydrogen and bromine (HBr)
(c) Baking powder – Sodium, Carbon, Hydrogen, Oxygen (NaHCO3)
(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K2SO4)
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
Listed below is the molar mass of the following substances:
(a) Molar mass of Ethyne C2H2= 2 x Mass of C+2 x Mass of H =
(2×12)+(2×1)=24+2=26g
(b) Molar mass of Sulphur molecule S8 = 8 x Mass of S = 8 x 32 = 256g
A 0.24g sample of a compound of oxygen and boron was found by analysis to contain 0.096g of boron and 0.144g of oxygen. Calculate the percentage composition of the compound by weight.
Given: Mass of the sample compound = 0.24g, mass of boron = 0.096g, mass of
oxygen = 0.144g
To calculate the percentage composition of the compound,
Percentage of boron = mass of boron / mass of the compound x 100
= 0.096g / 0.24g x 100 = 40%
Percentage of oxygen = 100 – percentage of boron
= 100 – 40 = 60%
What are polyatomic ions? Give examples.
Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.
Example: CO32-, H2PO4–
Write the chemical formula of the following.
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
The following are the chemical formula of the above-mentioned list:
(a) Magnesium chloride – MgCl2
(b) Calcium oxide – CaO
(c) Copper nitrate – Cu(NO3)2
(d) Aluminium chloride – AlCl3
(e) Calcium carbonate – CaCO3
When 3.0g 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 3.00g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?
When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is
produced.
Given that
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
Find out
We need to find out the mass of carbon dioxide that will be formed when 3.00 g of
carbon is burnt in 50.00 g of oxygen.
Solution
First, let us write the reaction taking place here.
C + O2 → CO2
As per the given condition, when 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g
of carbon dioxide is produced.
3g + 8g →11 g ( from the above reaction)
The total mass of reactants = mass of carbon + mass of oxygen
=3g+8g
=11g
The total mass of reactants = Total mass of products
Therefore, the law of conservation of mass is proved.
Then, it also depicts that carbon dioxide contains carbon and oxygen in a fixed ratio
by mass, which is 3:8.
Thus, it further proves the law of constant proportions.
3 g of carbon must also combine with 8 g of oxygen only.
This means that (50−8)=42g of oxygen will remain unreacted.
The remaining 42 g of oxygen will be left un-reactive. In this case, too, only 11 g of
carbon dioxide will be formed
The above answer is governed by the law of constant proportions.
Exercise 3.4
Calculate the formula unit masses of ZnO, Na2O, K2CO3, given atomic masses of
Zn = 65u,
Na = 23 u, K=39u, C = 12u, and O=16u.
The atomic mass of Zn = 65u
The atomic mass of Na = 23u
The atomic mass of K = 39u
The atomic mass of C = 12u
The atomic mass of O = 16u
The formula unit mass of ZnO= Atomic mass of Zn + Atomic mass of O = 65u + 16u
= 81u
The formula unit mass of Na2O = 2 x Atomic mass of Na + Atomic mass of O = (2 x
23)u + 16u = 46u + 16u = 62u
The formula unit mass of K2CO3 = 2 x Atomic mass of K + Atomic mass of C + 3 x
Atomic mass of O = (2 x 39)u + 12u + (3 x 16)u = 78u + 12u + 48u = 138u
Calculate the molecular masses of H2 , O2 , Cl2, CO2, CH4, C2H6,C2H4, NH3,CH3OH.
The following are the molecular masses:
The molecular mass of H2 – 2 x atoms atomic mass of H = 2 x 1u = 2u
The molecular mass of O2 – 2 x atoms atomic mass of O = 2 x 16u = 32u
The molecular mass of Cl2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u
The molecular mass of CO2 – atomic mass of C + 2 x atomic mass of O = 12 +
( 2×16)u = 44u
The molecular mass of CH4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x
1)u = 16u
The molecular mass of C2H6– 2 x atomic mass of C + 6 x atomic mass of H = (2 x
12) +
(6 x 1)u=24+6=30u
The molecular mass of C2H4– 2 x atomic mass of C + 4 x atomic mass of H = (2x 12)
+
(4 x 1)u=24+4=28u
The molecular mass of NH3– atomic mass of N + 3 x atomic mass of H = (14 +3 x
1)u= 17u
The molecular mass of CH3OH – atomic mass of C + 3x atomic mass of H + atomic
mass of O + atomic mass of H = (12 + 3×1+16+1)u=(12+3+17)u = 32u