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Atoms and molecules form the basis of everything around us. This chapter is very important in the Chemistry curriculum as it forms the basis of almost everything studied in chemistry. Have you ever wondered how things around us are made? Like the air we breathe, the phone we are currently using to read this article, or the food we eat daily to quench our appetite. The answer to all those questions is atoms and molecules. Everything in our environment, from the tiniest grain of sand to the vast universe, is made up of atoms and molecules. Atoms are the smallest unit of matter, while molecules are made up of a combination of two or more atoms, and they are so small that we cannot see them with the naked eye; to see them, we require a high-end microscope. Even a small drop of water contains countless molecules. Just as Bricks combine to make huge structures, atoms combine to form molecules, which make everything around us.
Science Class 9 NCERT solutions, created by experienced subject experts, in a very comprehensive and systematic way, covering all the in-text and exercise questions. These solution provides a clear and detailed understanding of these fundamental topics. The NCERT Solutions for Atoms and Molecules are aimed at simplifying complex questions and helping students grasp the concepts of atomic structure and molecular formation.
To get all the solved exercise questions, click below on the download PDF icon. In this PDF, you will get detailed solutions to all the questions that are given in the NCERT textbook.
All the in-text exercise questions with detailed solutions are given below:
Exercise-3.1 (Page: 27)
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
Answer:
Given the reaction
sodium carbonate + acetic acid → sodium acetate + carbon dioxide + water
5.3g 6g 8.2g 2.2g 0.9g
Now,
Total Mass on the Left Hand Side = 5.3g + 6g = 11.3g
Total Mass on the Right Hand Side = 8.2g + 2.2g + 0.9g = 11.3g
As Mass on the LHS is equal to the RHS, the observation is in agreement with the law of conservation of mass.
Answer:
Given:
Hydrogen and oxygen combine in the ratio of 1:8 by mass to form water.
Now,
For every 1 g of Hydrogen, 8 g of oxygen is needed for the reaction to take place.
Therefore, for 3 g of Hydrogen, the mass of oxygen needed = 8
Hence, 24 g of Oxygen is needed to complete a reaction with 3 g of Hydrogen.
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 is the result of the law of conservation of mass is
"Atoms can neither be created nor destroyed".
Question.4 Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Answer:
The postulate of Dalton’s atomic theory can explain the law of definite proportions is
"The relative number and kinds of atoms are equal in given compounds."
Exercise-3.2 (Page: 30)
Question 1. Define the atomic mass unit.
Answer:
An atomic mass unit is a unit of mass used to express the weight subatomic particles, where one unit is equal to exactly one-twelfth the mass of a carbon-12 atom.
Question 2. Why is it not possible to see an atom with naked eyes?
Answer:
We can't see the atom with the naked eye because they are minuscule in nature. they are measured in the nanometres. Also, except for noble gases, all-atom do not exist independently. they exist in the form of any compound.
Exercise 3.3 and 3.4 (Page: 34)
Question 1. Write down the formulae of
(i) sodium oxide
(ii) aluminium chloride
(iii) sodium sulphide
(iv) magnesium hydroxide
Answer:
The Formula of the given compounds are :
(i) sodium oxide :
(ii) aluminium chloride:
(iii) sodium sulphide:
(iv) magnesium hydroxide:
Question 2. Write down the names of compounds represented by the following formulae:
(i)
(ii)
(iii)
(iv)
(v)
Answer:
The Names of the following compounds are :
(i)
=Aluminium Sulphate
(ii)
=Calcium Chloride
(iii)
=Potessium Sulphate
(iv)
= Potassium Nitrate
(v)
= Calcium Carbonate.
Question 3. What is meant by the term chemical formula?
Answer:
The chemical formula of a compound is a symbolic representation of its composition. For example, The chemical formula for common salt is NaCl as it is made up of Sodium (Na) and Chlorine (Cl).
Q 4.(i) How many atoms are present in a
Answer:
and hence,
A total of 3 atoms are present in
Question 4.(ii) How many atoms are present in a
Answer:
And Hence,
A total of 5 atoms are present in
Exercise-3.5.1-3.5.2 (Page: 35)
Question 1. Calculate the molecular masses of
Answer:
The molecular mass of
= 2
= 2
=2u.
The molecular mass of
= 2
= 2
= 32u.
The molecular mass of
= 2
= 2
= 71u.
The molecular mass of
= Atomic mass of Carbon +2
= 12u + 2
= 44u.
The molecular mass of
= Atomic mass of Carbon +4
= 12u + 4
= 16u.
The molecular mass of
= 2
= 2
=24u+ 6u
= 30u.
The molecular mass of
= 2
= 2
=24u+ 4u
= 28u.
The molecular mass of
= Atomic mass of Nitrogen + 3
= 14u+ 3
= 17u.
The molecular mass of
= Atomic mass of Carbon +4
= 12u+ 4
= 32u.
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:
Given,
the atomic mass of Zn = 65 u,
the atomic mass of Na = 23 u,
the atomic mass of K = 39 u,
the atomic mass of C = 12 u,
and the atomic mass of O = 16 u
Now,
Formula unit mass of ZnO = Atomic mass of Zinc + Atomic mass of O
= 65 u + 16 u
= 81 u.
Formula unit mass of Na2O = 2
= 2
= 62 u.
Formula unit mass of K2CO3 = 2
= 2
= 138 u.
Answer:
The total mass of the compound = 0.24 g
Mass of boron in the compound = 0.096 g
Mass of oxygen in the compound = 0.144 g
Now, As we know,
The percentage of an element in the compound :
So,
The percentage of Boron in the compound by weight :
The percentage of Oxygen in the compound by weight :
Answer:
3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.
So, by the law of definite proportions,
When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will be used to produce 11.00 gram of carbon dioxide. The remaining 42.00 g of oxygen will remain unreacted. Law of constant proportion is Held.
Q 3. What are polyatomic ions? Give examples.
Answer:
Polyatomic ions are ions that contain more than one atom. These atoms can be of the same type or of a different type.
Some examples of polyatomic ions are NH4+ , OH- , SO42- , and SO32- .
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:
The chemical formula of Given compounds are :
(a) Magnesium chloride
(b) Calcium oxide
(c) Copper nitrate
(d) Aluminium chloride
(e) Calcium carbonate
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:
The names of the elements present in the given compounds are :
(a) Quick lime :
(b) Hydrogen bromide:
(c) Baking powder:
(d) Potassium sulphate:
Question 6.(a) Calculate the molar mass of the following substances.
Answer:
The molecular mass of ethyne,
= 2
= 24u + 2u
= 26u
Question 6. (b) Calculate the molar mass of the following substances.
Answer:
The molecular mass of Sulphur molecule, = 8
= 8
= 256u
Question 6.(c) Calculate the molar mass of the following substances
Answer:
The molecular formula of Baking soda is NaHCO3
So the molecular mass will be some of the individual atoms involved in the formula.
Molecular Mass of baking soda = Molecular mass (Na + H + C + 3xO)
= 23 + 1 + 12 + 3x16
= 84 g/mol
Question 6.(d) Calculate the molar mass of the following substances.
Answer:
The molecular mass of Hydrochloric acid, HCl = 1
= 35.5 u + 1u
= 36.5u
Question 6. (e) Calculate the molar mass of the following substances.
Answer:
The molecular mass of Nitric acid, HNO3 = 1
= 1
= 14u + 1u + 48u
= 63u
The chapter Atoms and Molecules will provide you with the basic concepts of atoms and molecules, the building blocks of matter. It begins with Dalton’s Atomic Theory, which explains the nature of atoms and their behavior. Students will learn about the laws of chemical combination: the Law of Conservation of Mass and the Law of Constant Proportions. The chapter will explain to you what atoms, molecules, and ions are and how they form compounds. It will also feature atomic and molecular masses, chemical formulae, and how to calculate the mole and molar mass using the concept of the Avogadro number (6.022 × 10²³). This chapter will help you understand chemical reactions and formulas.
To improve your performance, some important practice questions are given below:
Q1. State the Law of Conservation of Mass.
Answer:
The Law of Conservation of Mass states that mass can neither be created nor destroyed in a chemical reaction. The total mass of reactants is equal to the total mass of products.
Q2. What is the chemical formula of water and carbon dioxide? Name the elements present in them.
Answer:
Water (H₂O); Elements – Hydrogen and Oxygen
Carbon dioxide (CO₂); Elements – Carbon and Oxygen
Q3. Define atomicity. Give the atomicity of oxygen (O₂) and phosphorus (P₄).
Answer:
Atomicity is the number of atoms present in a molecule of an element.
Oxygen (O₂); Diatomic (Atomicity = 2)
Phosphorus (P₄); Tetratomic (Atomicity = 4)
Q4. Calculate the molecular mass of carbon dioxide (CO₂).
Answer:
Atomic mass of C = 12 u, O = 16 u
Molecular mass of CO₂ = 12 + (16 × 2) = 44 u
Q5. What is meant by a mole? How many particles are there in one mole of a substance?
Answer:
A mole is the amount of a substance that contains 6.022 × 10²³ particles (atoms, molecules, or ions). This number is called Avogadro’s constant.
To effectively solve questions of the chapter on Atoms and Molecules, follow a systematic and structured approach that helps understand the concepts and the theories:
All the topics and subtopics covered in the NCERT textbook are listed below:
3.1 Laws of Chemical Combination
3.1.1 Law of Conservation of Mass
3.1.2 Law of Constant Proportions.
3.2 What is an Atom?
3.2.1 What Are the Modern-Day Symbols of Atoms of Different Elements?
3.2.2 Atomic Mass
3.3 What is a Molecule?
3.3.1 Molecules of Elements
3.3.2 Molecules of Compounds
3.3.3 What is An Ion?
3.4 Writing Chemical Formulae
3.4.1 Formulae of Simple Compounds
3.5 Molecular Mass
3.5.1 Formula Unit Mass
1 nanometer (nm) is equal to 10-9 meters (m)
1 nm = 10-9 m
Number of moles (n) = Given quantity of substance (N) /Molar mass of substance (No)
Number of moles of atoms = Number of atoms Avogrdro's number [Avogrdro’s number=6.022 × 1023]
Mass = Molar mass × Number of moles
Formula unit mass = (Atomic mass of cation × Number of cations) + (Atomic mass of anion × Number of anions)
Class 9 NCERT chapter-wise solutions are given below:
Class 9 NCERT Subject-wise solutions are given below:
The NCERT books and syllabus links for class 9 are given below:
The three basic parts of an atom are:
Types of chemical bonds are:
Chemical bonds are attractive forces that hold atoms together in molecules and compounds. Atoms form chemical bonds to achieve a more stable electron configuration.
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