case study based questions class 9 science chapter 3

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Class 9 Science Case Study Questions

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If you are wondering how to solve class 9 science case study questions, then myCBSEguide is the best platform to choose. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added Class 9 Science case study questions.

You can find a wide range of solved case studies on myCBSEguide, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

The rationale behind Science

Science is crucial for Class 9 students’ cognitive, emotional, and psychomotor development. It encourages curiosity, inventiveness, objectivity, and aesthetic sense.

In the upper primary stage, students should be given a variety of opportunities to engage with scientific processes such as observing, recording observations, drawing, tabulating, plotting graphs, and so on, whereas in the secondary stage, abstraction and quantitative reasoning should take a more prominent role in science teaching and learning. As a result, the concept of atoms and molecules as matter’s building units, as well as Newton’s law of gravitation, emerges.

Science is important because it allows Class 9 Science students to understand the world around us. It helps to find out how things work and to find solutions to problems at the Class 9 Science level. Science is also a source of enjoyment for many people. It can be a hobby, a career, or a source of intellectual stimulation.

Case study questions in Class 9 Science

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Examples of Class 9 science class case study questions

Class 9 science case study questions have been prepared by myCBSEguide’s qualified teachers. Class 9 case study questions are meant to evaluate students’ knowledge and comprehension of the material. They are not intended to be difficult, but they will require you to think critically about the material. We hope you find Class 9 science case study questions beneficial and that they assist you in your exam preparation.

The following are a few examples of Class 9 science case study questions.

Class 9 science case study question 1

• due to its high compressibility
• large volumes of a gas can be compressed into a small cylinder
• transported easily
• all of these
• shape, volume
• volume, shape
• shape, size
• size, shape
• the presence of dissolved carbon dioxide in water
• the presence of dissolved oxygen in the water
• the presence of dissolved Nitrogen in the water
• liquid particles move freely
• liquid have greater space between each other
• both (a) and (b)
• none of these
• Only gases behave like fluids
• Gases and solids behave like fluids
• Gases and liquids behave like fluids
• Only liquids are fluids

Answer Key:

• (d) all of these
• (a) shape, volume
• (b) the presence of dissolved oxygen in the water
• (c) both (a) and (b)
• (c) Gases and liquids behave like fluids

Class 9 science case study question 2

• 12/32 times
• 18 g of O 2
• 18 g of CO 2
• 18 g of CH 4
• 1 g of CO 2
• 1 g of CH 4 CH 4
• 2 moles of H2O
• 20 moles of water
• 6.022  ×  1023 molecules of water
• 1.2044  ×  1025 molecules of water
• (I) and (IV)
• (II) and (III)
• (II) and (IV)
• Sulphate molecule
• Ozone molecule
• Phosphorus molecule
• Methane molecule
• (c) 8/3 times
• (d) 18g of CH ​​​​​4
• (c) 1g of H ​​​​​​2
• (d) (II) and (IV)
• (c) phosphorus molecule

Class 9 science case study question 3

• collenchyma
• chlorenchyma
• It performs photosynthesis
• It helps the aquatic plant to float
• It provides mechanical support
• Sclerenchyma
• Collenchyma
• Epithelial tissue
• Parenchyma tissues have intercellular spaces.
• Collenchymatous tissues are irregularly thickened at corners.
• Apical and intercalary meristems are permanent tissues.
• Meristematic tissues, in its early stage, lack vacuoles, muscles
• (I) and (II)
• (III) and (I)
• Transpiration
• Provides mechanical support
• Provides strength to the plant parts
• None of these
• (a) Collenchyma
• (b) help aquatic plant to float
• (b) Sclerenchyma
• (d) Only (III)
• (c) provide strength to plant parts

Cracking Class 9 Science Case Study Questions

There is no one definitive answer to Class 9 Science case study questions. Every case study is unique and will necessitate a unique strategy. There are, nevertheless, certain general guidelines to follow while answering case study questions.

• To begin, double-check that you understand the Class 9 science case study questions. Make sure you understand what is being asked by reading it carefully. If you’re unclear, seek clarification from your teacher or tutor.
• It’s critical to read the Class 9 Science case study material thoroughly once you’ve grasped the question. This will provide you with a thorough understanding of the problem as well as the various potential solutions.
• Brainstorming potential solutions with classmates or other students might also be beneficial. This might provide you with multiple viewpoints on the situation and assist you in determining the best solution.
• Finally, make sure your answer is presented simply and concisely. Make sure you clarify your rationale and back up your claim with evidence.

A look at the Class 9 Science Syllabus

The CBSE class 9 science syllabus provides a strong foundation for students who want to pursue a career in science. The topics are chosen in such a way that they build on the concepts learned in the previous classes and provide a strong foundation for further studies in science. The table below lists the topics covered in the Class 9 Science syllabus of the Central Board of Secondary Education (CBSE). As can be seen, the Class 9 science syllabus is divided into three sections: Physics, Chemistry and Biology. Each section contains a number of topics that Class 9 science students must study during the course.

CBSE Class 9 Science (Code No. 086)

Theme: Materials Unit I: Matter-Nature and Behaviour Definition of matter; solid, liquid and gas; characteristics – shape, volume, density; change of state-melting (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation. Nature of matter:  Elements, compounds and mixtures. Heterogeneous and homogenous mixtures, colloids and suspensions. Particle nature and their basic units:  Atoms and molecules, Law of constant proportions, Atomic and molecular masses. Mole concept: Relationship of mole to mass of the particles and numbers. Structure of atoms:  Electrons, protons and neutrons, valency, the chemical formula of common compounds. Isotopes and Isobars.

Theme: The World of the Living Unit II: Organization in the Living World Cell – Basic Unit of life:  Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number. Tissues, Organs, Organ System, Organism: Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Theme: Moving Things, People and Ideas Unit III: Motion, Force and Work Motion:  Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, derivation of equations of motion by graphical method; elementary idea of uniform circular motion. Force and Newton’s laws:  Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration. Elementary idea of conservation of Momentum. Gravitation:  Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation:  Thrust and Pressure. Archimedes’ Principle; Buoyancy. Work, energy and power:  Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy. Sound:  Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Theme: Food Unit IV: Food Production Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

PRESCRIBED BOOKS:

• Science-Textbook for class IX-NCERT Publication
• Assessment of Practical Skills in Science-Class IX – CBSE Publication
• Laboratory Manual-Science-Class IX, NCERT Publication
• Exemplar Problems Class IX – NCERT Publication

myCBSEguide: A true helper

There are numerous advantages to using myCBSEguide to achieve the highest results in Class 9 Science.

• myCBSEguide offers high-quality study materials that cover all of the topics in the Class 9 Science curriculum.
• myCBSEguide provides practice questions and mock examinations to assist students in the best possible preparation for their exams.
• On our myCBSEguide app, you’ll find a variety of solved Class 9 Science case study questions covering a variety of topics and concepts. These case studies are intended to help you understand how certain principles are applied in real-world settings
• myCBSEguide is that the study material and practice problems are developed by a team of specialists who are always accessible to assist students with any questions they may have. As a result, students may be confident that they will receive the finest possible assistance and support when studying for their exams.

So, if you’re seeking the most effective strategy to study for your Class 9 Science examinations, myCBSEguide is the place to go!

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Case Study Questions of Class 9 Science PDF Download

Download PDF Case Study Questions of Class 9 Science to prepare for the upcoming CBSE Class 9 Exams Exam 2023-24. With the help of our well-trained and experienced faculty, we provide solved examples and detailed explanations for the recently added  Class 9 Science case study questions .

Case study questions are based on real or hypothetical scenarios that require students to analyze, evaluate, and apply scientific concepts to solve problems or make informed decisions. They often present a detailed context, providing students with the opportunity to demonstrate their understanding of the subject matter beyond basic recall.

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Class 9 Science: Case Study Questions

The inclusion of case study questions in Class 9 science CBSE is a great way to engage students in critical thinking and problem-solving. By working through real-world scenarios, Class 9 Science students will be better prepared to tackle challenges they may face in their future studies and careers. Class 9 Science Case study questions also promote higher-order thinking skills, such as analysis and synthesis. In addition, case study questions can help to foster creativity and innovation in students. As per the recent pattern of the Class 9 Science examination, a few questions based on case studies/passages will be included in the CBSE Class 9 Science Paper. There will be a paragraph presented, followed by questions based on it.

Chapterwise Case Study Questions of Class 9 Science

• Case Study Questions for Chapter 1 Matter in Our Surroundings
• Case Study Questions for Chapter 2 Is Matter Around Us Pure?
• Case Study Questions for Chapter 3 Atoms and Molecules
• Case Study Questions for Chapter 4 Structure of Atom
• Case Study Questions for Chapter 5 The Fundamental Unit of Life
• Case Study Questions for Chapter 6 Tissues
• Case Study Questions for Chapter 7 Diversity in Living Organisms
• Case Study Questions for Chapter 8 Motion
• Case Study Questions for Chapter 9 Force and Laws of Motion
• Case Study Questions for Chapter 10 Gravitation
• Case Study Questions for Chapter 11 Work and Energy
• Case Study Questions for Chapter 12 Sound
• Case Study Questions for Chapter 13 Why do we Fall ill
• Case Study Questions for Chapter 14 Natural Resources
• Case Study Questions for Chapter 15 Improvement in Food Resources

You can find a wide range of solved case studies on cbseexperts, covering various topics and concepts. Class 9 Science case studies are designed to help you understand the application of various concepts in real-life situations.

Class 9 Science Syllabus

Unit I: Matter-Nature and Behaviour

 Definition of matter; solid, liquid, and gas; characteristics – shape, volume, density; change of statementing (absorption of heat), freezing, evaporation (cooling by evaporation), condensation, sublimation.

Nature of matter:  Elements, compounds, and mixtures. Heterogeneous and homogenous mixtures, colloids, and suspensions. Physical and chemical changes (excluding separating the components of a mixture).

Particle nature and their basic units:  Atoms and molecules, Law of Chemical Combination, Chemical formula of common compounds, Atomic and molecular masses.

Structure of atoms:  Electrons, protons and neutrons, Valency, Atomic Number and Mass Number, Isotopes and Isobars.

Unit II: Organization in the Living World

Cell – Basic Unit of life:  Cell as a basic unit of life; prokaryotic and eukaryotic cells, multicellular organisms; cell membrane and cell wall, cell organelles and cell inclusions; chloroplast, mitochondria, vacuoles, endoplasmic reticulum, Golgi apparatus; nucleus, chromosomes – basic structure, number.

Tissues, Organs, Organ System, Organism:  Structure and functions of animal and plant tissues (only four types of tissues in animals; Meristematic and Permanent tissues in plants).

Unit III: Motio n, Force,  and Work

Motion:  Distance and displacement, velocity; uniform and non-uniform motion along a straight line; acceleration, distance-time and velocity-time graphs for uniform motion and uniformly accelerated motion, elementary idea of uniform circular motion.

Force and Newton’s laws:  Force and Motion, Newton’s Laws of Motion, Action and Reaction forces, Inertia of a body, Inertia and mass, Momentum, Force and Acceleration.

Gravitation:  Gravitation; Universal Law of Gravitation, Force of Gravitation of the earth (gravity), Acceleration due to Gravity; Mass and Weight; Free fall. Floatation: Thrust and Pressure. Archimedes’ Principle; Buoyancy.

Work, Energy and Power:  Work done by a Force, Energy, power; Kinetic and Potential energy; Law of conservation of energy (excluding commercial unit of Energy).

Sound: Nature of sound and its propagation in various media, speed of sound, range of hearing in humans; ultrasound; reflection of sound; echo.

Unit IV: Food Production

Plant and animal breeding and selection for quality improvement and management; Use of fertilizers and manures; Protection from pests and diseases; Organic farming.

Books for Class 9 Science Exams

Benefits of Case Study Questions

• Enhancing Analytical Skills : Case study questions challenge students to analyze complex scenarios, identify relevant information, and derive meaningful insights. By engaging with these questions, students develop critical analytical skills that are essential for scientific thinking and problem-solving.
• Promoting Critical Thinking : Case study questions encourage students to think critically and evaluate different perspectives. They require students to reason, make logical deductions, and justify their answers with supporting evidence. This process helps in honing their critical thinking abilities, enabling them to approach problems from multiple angles.
• Encouraging Practical Application of Concepts : By presenting real-world or hypothetical situations, case study questions promote the application of scientific concepts in practical scenarios. This application-based approach fosters a deeper understanding of the subject matter and helps students see the relevance of what they learn in the classroom to everyday life.

Case study questions of Class 9 Science provide students with an opportunity to apply their knowledge, enhance analytical skills, and think critically. By understanding the format, benefits, and effective strategies for answering case study questions, students can excel in this form of assessment. While challenges may arise, practicing time management, improving information extraction skills, and enhancing observation abilities will enable students to overcome these obstacles and perform well. Embracing case study questions as a valuable learning tool can contribute to a holistic understanding of scientific concepts and foster problem-solving abilities.

1. What is the purpose of case study questions in Class 9 Science?

Case study questions serve the purpose of evaluating a student’s understanding of scientific concepts, their ability to apply knowledge in real-life situations, and their analytical and critical thinking skills.

2. How can case study questions help improve analytical skills?

Case study questions require students to analyze complex scenarios, identify relevant information, and derive meaningful insights. Regular practice with such questions can significantly enhance analytical skills.

3. Are case study questions difficult to answer?

Case study questions can be challenging due to their comprehensive nature and the need for critical thinking. However, with practice and effective strategies, students can develop the skills necessary to answer them effectively.

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Class 9 Science Case Study Questions PDF Download

• Post author: studyrate
• Post published:
• Post category: class 9th
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Class 9 Science Case Study Questions play a crucial role in the field of science education as they provide real-life scenarios for students to analyze, apply their knowledge, and develop problem-solving skills. This article aims to present a comprehensive collection of case study questions for Class 9 Science , covering various topics and concepts.

Join our Telegram Channel, there you will get various e-books for CBSE 2024 Boards exams for Class 9th, 10th, 11th, and 12th.

CBSE Class 9 Science Exam will have a set of questions based on case studies in the form of MCQs. The CBSE Class 9 Science Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

If you want to want to prepare all the tough, tricky & difficult questions for your upcoming exams, this is where you should hang out.  CBSE Case Study Questions for Class 9  will provide you with detailed, latest, comprehensive & confidence-inspiring solutions to the maximum number of Case Study Questions covering all the topics from your  NCERT Text Books !

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CBSE Class 9th SCIENCE Chapterwise Case Study Question & Solution

Case study questions provide students with real-life scenarios that require critical thinking and application of scientific concepts. They help students understand the practical application of scientific principles and develop problem-solving skills in various scientific disciplines.

Chapterwise Case Study Questions for Class 9 Science

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked. For Science subjects, there would be 5 case-based sub-part questions, wherein a student has to attempt 4 sub-part questions.

• Case Study Questions for Chapter 1 Matter in Our Surroundings
• Case Study Questions for Chapter 2 Is Matter Around Us Pure?
• Case Study Questions for Chapter 3 Atoms and Molecules
• Case Study Questions for Chapter 4 Structure of Atom
• Case Study Questions for Chapter 5 The Fundamental Unit of Life
• Case Study Questions for Chapter 6 Tissues
• Case Study Questions for Chapter 7 Diversity in Living Organisms
• Case Study Questions for Chapter 8 Motion
• Case Study Questions for Chapter 9 Force and Laws of Motion
• Case Study Questions for Chapter 10 Gravitation
• Case Study Questions for Chapter 11 Work and Energy
• Case Study Questions for Chapter 12 Sound
• Case Study Questions for Chapter 13 Why do we Fall ill
• Case Study Questions for Chapter 14 Natural Resources
• Case Study Questions for Chapter 15 Improvement in Food Resources

The above  Case studies for Class 9 Science  will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 9 Science Case Studies have been developed by experienced teachers of schools.studyrate.in for the benefit of Class 10 students.

Class 9 Maths Case Study Questions

Benefits of Case Studies in Science Education

Case studies offer several advantages over traditional teaching methods. Here are some key benefits:

• Real-World Application : Case studies present authentic scenarios, enabling students to understand how scientific concepts are applied in real-life situations.
• Critical Thinking : Analyzing case studies requires students to think critically, make connections, and apply scientific knowledge to solve problems.
• Interdisciplinary Approach : Case studies often involve multiple scientific disciplines, fostering an interdisciplinary understanding of complex issues.
• Engagement and Active Learning : Case studies actively engage students in the learning process, promoting active participation, discussion, and collaboration.
• Skill Development : Case studies develop essential skills such as analytical thinking, problem-solving, and effective communication of scientific concepts.

Importance of Practicing Case Study Questions

Practicing case study questions is crucial for Class 9 Science students to enhance their understanding and application of scientific concepts. Here’s why it is important:

• Application of Knowledge : Case studies allow students to apply their theoretical knowledge to practical situations, bridging the gap between theory and real-world scenarios.
• Developing Analytical Skills : Analyzing case studies improves students’ ability to identify relevant information, make connections, and draw logical conclusions.
• Problem-Solving Skills : Case studies present complex problems that require students to think critically and develop effective problem-solving strategies.
• Enhanced Exam Performance : Practicing case study questions familiarizes students with the format and types of questions they may encounter in exams, leading to improved performance.

Subjects Covered in the Case Study Questions for Class 9 Science

The case study questions for Class 9 Science cover the following subjects:

• Motion and Forces
• Light and Reflection
• Electricity
• Matter and Its Properties
• Atoms and Molecules
• Structure of the Atom
• Chemical Reactions
• Cell: The Fundamental Unit of Life
• Diversity in Living Organisms
• Natural Resources

Tips for Approaching Case Study Questions

To tackle case study questions effectively, consider the following tips:

• Read Carefully : Pay close attention to the details provided in the case study, as they hold crucial information for solving the problem.
• Analyze Methodically : Break down the problem into smaller components and analyze each part systematically.
• Apply Relevant Concepts : Identify the scientific principles relevant to the case study and apply them appropriately.
• Consider Multiple Perspectives : Explore different angles and viewpoints while proposing solutions, taking into account various scientific factors.
• Provide Justifications : Support your answers with scientific explanations and logical reasoning to strengthen your responses.

The Class 9 Science Case Study Questions provided in this article serve as a valuable resource for students seeking to enhance their scientific knowledge and problem-solving skills. By practicing these case studies, students can develop a deeper understanding of scientific concepts and their practical applications. Embrace this opportunity to engage with real-world scenarios and strengthen your scientific acumen.

Q1: Are the Class 9 Science Case Study Questions aligned with the official curriculum?

Yes, the Class 9 Science Case Study Questions presented in this article are aligned with the official curriculum. They cover relevant topics and concepts that students need to study for their exams.

Q2: Can practicing case study questions alone guarantee success in Class 9 Science exams?

Practicing case study questions is an important part of exam preparation, but it should be complemented with a thorough understanding of the subject matter. It is advisable to study the concepts in detail, refer to textbooks, and engage in other learning activities to achieve success in exams.

Q3: Where I Can get Class 9 Science Case Study Questions ?

You can practice Class 9 Science Case Study Questions on schools.studyrate.in for free.

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Class 9 Science Chapter 3 Case Based Questions - Atoms and Molecules

Q3: 0.25 mole of an element ‘X’ is 9.75 g. What is X ? Ans:  0.25 mole of X = 9.75 g 1 mole of X = 9.75 ÷ 0.25 = 39.0 g mol –1  The element is Potassium.

Q3: Write name of (NH 4 ) 2 SO 4 Ans: Ammonium sulphate.

Q4: Give one example of polyatomic anion. Ans: CO 3 2- (Carbonate).

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CBSE Class 9 Science Important Case Study Questions with Answers for Term 2 Exam 2022 (PDF)

Check important case study questions of cbse class 9 science to prepare for the cbse term 2 exam 2022. all these questions have been put together by subject experts..

CBSE Class 9 Term 2 Exam 2022:  Important case based questions for CBSE Class 9 Science are provided here students to prepare for the upcoming Term 2 Exam 2022. All the questions provided below are curated by the subject experts. These questions are really helpful to revise important concepts and prepare the case study questions for the exam. Answers to all questions have been provided for reference. So, students should practice the chapter-wise questions to clearly understand the right way to attempt the case based questions. Download the chapter-wise questions in PDF.

Check some of the important case study questions below:

Q. Read the following and answer the questions :

A student was asked by his teacher to verify the law of conservation of mass in the laboratory. He prepared 5% aqueous solutions of NaCl and Na 2 SO 4 . He mixed 10 mL of both these solutions in a conical flask. He weighed the flask on a balance. He then stirred the flask with a rod and weighed it after sometime. There was no change in mass.

• Was the student able to verify the law of conservation of mass?
• If not, what was the mistake committed by him?
• In your opinion, what he should have done?
• What is the molar mass of Na 2 SO 4 ?
• No, he could not verify the law of conservation of mass in-spite of the fact that there was no change in mass.
• No chemical reaction takes place between NaCl and Na 2 SO 4 . This means that no reaction actually took place in the flask.
• He should have performed the experiment by using aqueous solutions of BaCl 2 and Na 2 SO 4 . A chemical reaction takes place in this case and a white precipitate of BaSO 4 is formed.
• Will the weight of the precipitate be the same as that of the reactants before mixing?
• If not, what she should have done?
• Which law of chemical combination does this support?
• State the law of conservation of mass.
• No, it will not be the same.
• She should have weighed the total contents of the beaker after the reaction and not the precipitate alone.
• It supports the law of conservation of mass.
• Mass can neither be created nor destroyed during a chemical reaction.

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Extra Questions for Class 9 Science Chapter 3 Atoms and Molecules

Extra questions for Class 9 Science Chapter 3 Atoms and Molecules with answers is given below. Our subject expert prepared these solutions as per the latest NCERT textbook. These questions will be helpful to revise the all topics and concepts. CBSE Class 9 extra questions are the most simple and conceptual questions that are prepared by subject experts for the students to study well for the final exams. By solving these extra questions, students can be very efficient in their exam preparations.

Atoms and Molecules Class 9 Science Extra Questions and Answers

Very short answer questions.

1: Define law of conservation of mass.  Answer:  In a chemical reaction mass can neither be created nor destroyed.

2: Explain law of constant proportion. Answer:  In a chemical substance the elements are always present in definite proportions by mass. E.g., In water, the ratio of the mass of hydrogen to the mass of oxygen H : O is always 1:8

3: Who coined the term atom? Answer:  John Dalton coined the term atom.

4: Define atom. Answer:  The smallest particle of matter, which can take part in a chemical reaction is called atom.

5: Define molecule. Answer:  The smallest particle of an element or compound which can exist independently is called molecule.

6: Define atomicity. Answer:  The number of atoms constituting a molecule is known as its atomicity.

7: What is atomic mass unit? Answer:  The sum of the atomic masses of all the atoms in a molecule of the substance is expressed.in atomic mass unit. E.g., H 2 0 = 1 × 2 + 16 = 18 amu

8: How do atoms exist? Answer:  Atoms exist in the form of atom, molecule or ions.

9: Give the atomicity of phosphorous and nitrogen. Answer:  The atomicity of phosphorus is P 4  i.e., 4. The atomicity of nitrogen is N 2  i.e., 2. 

10: What is an ion? Answer:  Charged atom is called as an ion. The ion can be positively charged called cation or negatively charged called anion. 

11: Give one example of cation and anion. Answer:  Cation = Na + Anion =  Cl –

12: Give one difference between cation and anion. Answer:  Cations are positively charged ion. Anions are negatively charged ion.

13: Give the chemical formula for ammonium sulphate. Answer:  Ammonium sulphate – NH 4 +  SO 4 2- Chemical formula – (NH 4 ) 2 S0 4 .

14: What is Avogadro’s constant? Answer:  The Avogadro’s constant (6.022 x 10 23 ) is defined as the number of atoms that are present in exactly 12 g of carbon-12.

15: Find the molecular mass of H 2 O. Answer:  Molecular mass of H 2 O = (2 × 1) + (16) = 2 + 16 = 18 u

Short Answer Type Questions

1: Give the unit to measure size of atom and give size of hydrogen atom. Answer:  The unit to measure size of atom, is nanometer, size of hydrogen atom is 10 -10 m.

2: What is IUPAC, give its one function? Answer:  IUPAC is International Union for Pure and Applied Chemistry. It approves the names of elements.

3: Give the Latin name for sodium, potassium, gold and mercury. Answer:  Sodium → Natrium, Gold → Aurum Potassium → Kalium, Mercury → Hydrargyrum

4: What is the ratio by mass of combining elements in H 2 O, CO 2  and NH 3 ?

Answer:  H 2 O ratio by mass of combining elements 2 : 16 →1 : 8 (H : O) CO 2  ratio by mass of combining elements 12 : 32 → 3 : 08 (C : O) NH 3  ratio by mass of combining elements 14 : 3 → 14 : 3 (N : H)

5: Define valency and give the valency for the following elements: Magnesium, Aluminium, Chlorine and Copper. 

Answer:  Valency: The combining capacity of an element is called its valency. Valency of the following elements: Magnesium – 2 Aluminium – 3 Chlorine – 1 Copper – 2 

6: What is polyatomic ton? Give one example.

Answer:  A group of atoms carrying a charge is known as a polyatomic ion. E.g., Ammonium – NH 4 + Nitrate – NO 3 –  

7: Write down the formula for: Copper nitrate, calcium sulphate and aluminium hydroxide.

Answer:  Chemical formula: Copper nitrate → Cu(NO 3 ) Calcium sulphate → CaSO 4  Aluminium hydroxide Al(OH) 3

8: What is formula unit mass? How is it different from molecular mass?

Answer:  The formula unit mass of a substance is a sum of the atomic masses of all atoms in a formula unit of a compound. The constituent particles of formula unit mass are ions and the constituent particles of molecular mass are atoms.

9: Find the number of moles in the following: (i) 50 g of H 2 O (ii) 7 g of Na

Answer:   (i) Molar mass of H 2 O = 18 g Given mass of H 2 O = 50 g ∴ No. of moles in 50g of H 2 O = 58/18 = 2.78 moles.

(ii) Molar mass of Na = 23 g Given mass of Na = 7 g ∴ No. of moles in 50g of H 2 O = 7/23 = 0.304 moles.

10: Find the number of atoms in the following: (i) 0.5 mole of C atom (ii) 2 mole of N atom

Answer:  (i) 0.5 mole of C atom: Number of atoms in 1 mole of C atom = 6.022 × 10 23  atoms Number of atoms in 0.5 mole of C atom = 6.022 × 10 23  × 0.5  = 3.011 × 10 23  atoms

(ii) 2 mole of N atom: Number of atoms in 1 mole of N atom = 6.022 × 10 23  atoms Number of atoms in 2 mole of N atom = 6.022 × 2 × 10 23 = 1.2044 × 10 24  atoms

11: Find the mass of the following: (i) 6.022 × 10 23  number of O 2  molecules (ii) 1.5 mole of CO 2  molecule

Answer:  (i) 6.022 × 10 23  number of 02 molecules: Mass of 1 mole of O 2  molecule = 6.022 × 10 23  molecules = 32 g

(ii) 1.5 mole of CO 2  molecule: Mass of 1 mole of CO 2  molecule = 6.022 × 10 23  molecules = 44 g Mass of 1.5 mole CO 2  molecule = 44 × 1.5 = 66 g

12: Show the relationship between mole, Avogadro number and mass. Answer:

13: What are the rules for writing the symbol of an element?

Answer:  IUPAC → International Union of Pure and Applied Chemistry approves name of elements. Symbols are the first one or two letters of the element’s name in English. The first letter of a symbol is always written as a capital letter (upper case) and the second letter as a small letter (lower case). e.g., Hydrogen → H Helium → He Some symbols are taken from the names of elements in Latin, German or Greek. e.g., Symbol of iron is Fe, its Latin name is Ferrum. Symbol of sodium is Na, its Latin name is Natrium.  14: Explain relative atomic mass and relative molecular mass.

Answer:  Relative atomic mass: It can be defined as the number of times one atom of given element is heavier than 1/12 th of the mass of an atom of carbon-12. Relative Molecular Mass: It is defined as the number of times one molecule of a substance or given element is heavier than 1/12 th of the mass of one atom of carbon-12.  15: The formula of carbon-dioxide is CO 2 . What information do you get from this formula? 

Answer:  (i) CO 2  represents carbon-dioxide. (ii) CO 2  is one molecule of carbon-dioxide. (iii) CO 2  is one mole of carbon-dioxide i.e., it contains 6.022 × 10 23  molecules of carbon dioxide. (iv) CO 2  contains 1 atom of carbon and two atoms of oxygen. (v) CO 2  represents 44 g of molar mass. 

16: State 3 points of difference between an atom and an ion. Answer:

17: Calculate the formula unit mass of NaCl and CaCl 2 . (Na = 23, Cl = 35.5, Ca = 40)

Answer:  Formula unit mass of NaCl = 23 + 35.5 = 58.5 u

Formula unit mass of CaCl 2  = 40 + (2 × 35.5) = 40 + 71 = 111 u 

18: The ratio by mass for hydrogen and oxygen in water is given as 1 : 8 respectively. Calculate the ratio by number of atoms for a water molecule. Answer:  The ratio by number of atoms for a water molecule are:

Thus, the ratio by number of atoms for water is H : O = 2 : 1.

19: Write down the chemical formula for the following compounds: (a) Aluminium carbonate (b) Calcium sulphide (c) Zinc carbonate (d) Copper phosphate (e) Magnesium bicarbonate (f) Aluminium hydroxide.

Answer:  The chemical formula are:

20: Give the atomicity of the following compounds: (a) Ca(OH) 2   (b) Mg(HCO 3 ) 2   (c) Cu 2 O. (d) H 2 SO 4 (e) Al 2 (SO 4 ) 3 (f) MgCl 2

Answer:  The atomicity of the molecules are: (a) Ca(OH) 2   → 05 (b) Mg(HCO 3 ) 2  → 11 (c) Cu 2 O → 03 (d) H 2 SO 4 → 07 (e) Al 2 (SO 4 ) 3 → 17 (f) MgCl 2 → 03

21: Explain the difference between 2O, O 2  and O 3 .

Answer:  2O → It represents 2 atoms of oxygen (cannot exist independently). O 2  → It represents one molecule of oxygen (made up of 2 atom) can exist freely. O 3  → It represents one molecule of ozone (made up of 3 atoms) it can exist independently.

Long Answer Type Questions

1: (a) How do atoms exist? (b) What is atomicity? (c) What are polyatomic ions?

Answer:  (a) Atoms of some elements are not able to exist independently. For such elements atoms form molecules and ions. In case of metals and inert gases atoms can exist independently.

(b) The number of atoms constituting a molecule is known as its atomicity. E.g.,O 3  → atomicity is 3 O 2  → atomicity is 2

(c) Polyatomic ions: When more than two atoms combine together and act like an atom with a charge on it is called polyatomic ion. E.g., OH – , N0 3 – , NH 4 +

2: Calculate (a) the mass of one atom of oxygen (b) the mass of one molecule of oxygen (c) the mass of one mole of oxygen gas (d) the mass of one ion of oxygen (e) the number of atoms in 1 mole of oxygen molecule

Answer:  (a) Mass of one atom of oxygen 1 mole of oxygen atom = 16 gm = 6.022 × 10 23  atoms. ∴ Mass of one atom of oxygen = 16/6.022 × 10 23   = 2.65 × 10 23

(b) Mass of one molecule of oxygen 1 molecule of oxygen = O 2 = 2 × 16 = 32 u

(c) Mass of one oxygen gas 1 molecule of oxygen gas is O 2  = 32 u

(d) Mass of one ion of oxygen One mole of oxygen = 6.022 × 10 23  atoms = 16g. 16   Mass of one ion of oxygen = 16/6.022 × 10 23   = 2.65 × 10 23

(e) Number of atoms in one mole of oxygen molecule 1 mole of oxygen molecule i.e. 0 2  = 6.022 × 10 23  molecules. 1 molecule of O 2  = 2 atoms.

∴ Number of atoms in 1 mole of oxygen molecule = 6.022 × 10 23  × 2 atoms = 1.2044 × 10 24  atoms

3: What is meant by atomic mass, gram atomic mass of an element? Why is the mass have different expressions i.e., ‘u’ and ‘g’?

Answer:   The atoms are very tiny and their individual mass cannot be calculated as it is negligible. Hence the mass of atoms is expressed in units with respect to a fixed standard. Initially hydrogen atom with mass 1 was taken as standard unit by Dalton. Later, it was replaced by oxygen atom (0=16). But due to the isotopes the masses were found in fractions instead of whole number. Hence, carbon (C=12) isotope was taken as standard unit and was universally accepted. The atomic mass unit is equal to one twelfth (1/12) the mass of an atom of carbon-12, its unit is u.

Gram atomic mass:  When the atomic mass of an element is expressed in grams, it is called the gram atomic mass of the element. The mass of atoms, molecules is expressed in ‘u’ and the mass of moles i.e., molar mass is expressed in g.

4: Define a mole. Give the significance of the mole.

Answer:  Mole-One mole of any species (atoms, molecules, ions or particles) is that quantity or number having a mass equal to its atomic or molecular mass in grams. 1 mole = 6.022 × 10 23  in number (atoms, molecules, ions or particles)

Significance of the mole 1. A mole gives the number of entities present i.e, 6.022 × 10 23  particles of the substance. 2. Mass of 1 mole is expressed as M grams. 3. Mass of 1 mole = mass of 6.022 × 10 23  atoms of the element.

• NCERT Solutions
• NCERT Class 9
• NCERT 9 Science
• Chapter 3: Atoms And Molecules

NCERT Solutions for Class 9 Science Chapter 3 - Atoms and Molecules

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NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules is a detailed study material prepared by experts. It provides answers to the questions given in the textbook. NCERT Solutions are very helpful for a better understanding of the concepts and self-analysis.

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Questions from all the topics are covered in the solutions. Everything is presented in a way that students can easily understand. It will help them score well in the CBSE examination. The answers to all kinds of long and short questions, MCQs, tricks and tips are provided in the NCERT Solutions for Class 9 Science . Try solving the questions after completing the entire syllabus and overcoming the shortcomings before the CBSE exams arrive.

The chapter Atoms and Molecules form the basis of the upcoming chapters; therefore, it should be dealt with thoroughly. The questions and solutions will help the students clarify all their doubts related to the topic.

The NCERT Solutions for Class 9 carries all the important questions and answers for all the subjects and chapters. The students can refer to these solutions to excel in the CBSE examinations.

• Chapter 1 Matter in Our Surroundings
• Chapter 2 Is Matter Around Us Pure
• Chapter 4 Structure of the Atom
• Chapter 5 The Fundamental Unit of Life
• Chapter 6 Tissues
• Chapter 7 Diversity in Living Organisms
• Chapter 8 Motion
• Chapter 9 Force and Laws of Motion
• Chapter 10 Gravitation
• Chapter 11 Work and Energy
• Chapter 12 Sound
• Chapter 13 Why Do We Fall Ill
• Chapter 14 Natural Resources
• Chapter 15 Improvement in Food Resources

NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules

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Class 9 science chapter 3 exercise-3.1 questions with answers, exercise-3.1 page: 32.

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

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

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.

2. 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.

3. 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.”

4. 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 Page: 35

1. 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.

2. 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-3.3-3.4 Page: 39

1. 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 – Na 2 O

(ii) aluminium chloride – AlCl 3

(iii) sodium sulphide – Na 2 S

(iv) magnesium hydroxide – Mg (OH) 2

2. Write down the names of compounds represented by the following formulae:

(i) Al 2 (SO 4 ) 3

(ii) CaCl 2

(iii) K 2 SO 4

(v) CaCO 3 .

Listed below are the names of the compounds for each of the following formulae:

(i) Al 2 (SO 4 ) 3 – Aluminium sulphate

(ii) CaCl 2 – Calcium chloride

(iii) K 2 SO 4 – Potassium sulphate

(iv) KNO 3 – Potassium nitrate

(v) CaCO 3 – Calcium carbonate

3. 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.

4. How many atoms are present in a

(i) H 2 S molecule and

(ii) PO 4 3- ion?

The number of atoms present is as follows:

(i) H 2 S molecule has 2 atoms of hydrogen and 1 atom of sulphur hence 3 atoms in total.

(ii) PO 4 3- ion has 1 atom of phosphorus and 4 atoms of oxygen hence 5 atoms in total.

Exercise-3.5.1-3.5.2 Page: 40

1. Calculate the molecular masses of H 2 , O 2 , Cl 2 , CO 2 , CH 4 , C 2 H 6 , C 2 H 4 , NH 3 , CH 3 OH.

The following are the molecular masses:

The molecular mass of H 2 – 2 x atoms atomic mass of H = 2 x 1u = 2u

The molecular mass of O 2 – 2 x atoms atomic mass of O = 2 x 16u = 32u

The molecular mass of Cl 2 – 2 x atoms atomic mass of Cl = 2 x 35.5u = 71u

The molecular mass of CO 2 – atomic mass of C + 2 x atomic mass of O = 12 + ( 2×16)u = 44u

The molecular mass of CH 4 – atomic mass of C + 4 x atomic mass of H = 12 + ( 4 x 1)u = 16u

The molecular mass of C 2 H 6 – 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 C 2 H 4 – 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 NH 3 – atomic mass of N + 3 x atomic mass of H = (14 +3 x 1)u= 17u

The molecular mass of CH 3 OH – 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

2. Calculate the formula unit masses of ZnO, Na 2 O, K 2 CO 3 , 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 Na 2 O = 2 x Atomic mass of Na + Atomic mass of O = (2 x 23)u + 16u = 46u + 16u = 62u

The formula unit mass of K 2 CO 3  = 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

Exercise-3.5.3 Page: 42

1. 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 10 23

The molecular mass of carbon atoms = 12g = an atom of carbon mass

Hence, mass of 1 carbon atom = 12 / 6.022 x 10 23  = 1.99 x 10 -23 g

2. 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, N 0 = 6.022 x 10 23 , N = ?

N = (Given mass x N 0 )/Molar mass

N = (100 x 6.022 x 10 23 )/ 23

N = 26.18 x 10 23 atoms

(b) In 100 grams of Fe:

m = 100 g, Molar mass of Fe atom = 56 g, N 0 = 6.022 x 10 23 , N = ?

N = (Given mass x N 0 )/ Molar mass

N = (100 x 6.022 x 10 23 )/ 56

N = 10.75 x 10 23 atoms

Therefore, the number of atoms is more in 100 g of Na than in 100 g of Fe.

Exercise Page: 43

1. 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%

2. 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.

3.0 g of carbon combines with 8.0 g of oxygen to give 11.0 of carbon dioxide.

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.

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

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.

3. What are polyatomic ions? Give examples.

Polyatomic ions are ions that contain more than one atom, but they behave as a single unit.

Example: CO 3 2- , H 2 PO 4 –

4. 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 – MgCl 2

(b) Calcium oxide – CaO

(c) Copper nitrate – Cu(NO 3 ) 2

(d) Aluminium chloride – AlCl 3

(e) Calcium carbonate – CaCO 3

5. 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 (NaHCO 3 )

(d) Potassium sulphate – Sulphur, Oxygen, Potassium (K 2 SO 4 )

6. Calculate the molar mass of the following substances.

(a) Ethyne, C 2 H 2

(b) Sulphur molecule, S 8

(c) Phosphorus molecule, P 4 (Atomic mass of phosphorus =31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO 3

Listed below is the molar mass of the following substances:

(a) Molar mass of Ethyne C 2 H 2 = 2 x Mass of C+2 x Mass of H = (2×12)+(2×1)=24+2=26g

(b) Molar mass of Sulphur molecule S 8 = 8 x Mass of S = 8  x 32 = 256g

(c) Molar mass of  Phosphorus molecule, P 4 = 4 x Mass of P = 4 x 31 = 124g

(d) Molar mass of Hydrochloric acid, HCl = Mass of H+ Mass of Cl = 1+35.5 = 36.5g

(e) Molar mass of Nitric acid, HNO 3 =Mass of H+ Mass of Nitrogen + 3 x Mass of O = 1 + 14+

3×16 = 63g

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 (Na 2 SO 3 )?

The mass of the above-mentioned list is as follows:

(a) Atomic mass of nitrogen atoms = 14u

Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen atoms

Therefore, the mass of 1 mole of nitrogen atom is 14g.

(b) Atomic mass of aluminium =27u

Mass of 1 mole of aluminium atoms = 27g

1 mole of aluminium atoms = 27g, 4 moles of aluminium atoms = 4 x 27 = 108g

(c) Mass of 1 mole of sodium sulphite Na 2 SO 3 = Molecular mass of sodium sulphite = 2 x Mass of Na + Mass of S + 3 x Mass of O =  (2 x 23) + 32 +(3x 16) = 46+32+48 = 126g

Therefore, mass of 10 moles of Na 2 SO 3   = 10 x 126 = 1260g

8. Convert into a mole.

(a) 12g of oxygen gas

(b) 20g of water

(c) 22g of carbon dioxide

Conversion of the above-mentioned molecules into moles is as follows:

(a) Given: Mass of oxygen gas=12g

Molar mass of oxygen gas = 2 Mass of Oxygen =  2 x 16 = 32g

Number of moles = Mass given / molar mass of oxygen gas = 12/32 =  0.375 moles

(b) Given: Mass of water = 20g

Molar mass of water = 2 x Mass of Hydrogen + Mass of Oxygen = 2 x 1 + 16 = 18g

Number of moles = Mass given / molar mass of water

= 20/18 = 1.11 moles

(c) Given: Mass of carbon dioxide = 22g

Molar mass of carbon dioxide = Mass of C + 2 x Mass of Oxygen = 12 + 2x 16 = 12+32=44g

Number of moles = Mass given/ molar mass of carbon dioxide = 22/44 = 0.5 moles

9. What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

The mass is as follows:

(a) Mass of 1 mole of oxygen atoms = 16u; hence, it weighs 16g.

Mass of 0.2 moles of oxygen atoms = 0.2 x 16 = 3.2g

(b) Mass of 1 mole of water molecules = 18u; hence, it weighs 18g.

Mass of 0.5 moles of water molecules = 0.5 x 18 = 9g

10. Calculate the number of molecules of sulphur (S 8 ) present in 16g of solid sulphur.

To calculate the molecular mass of sulphur,

Molecular mass of Sulphur (S 8 ) = 8xMass of Sulphur = 8×32 = 256g

Mass given = 16g

Number of moles = mass given/ molar mass of sulphur

= 16/256 = 0.0625 moles

To calculate the number of molecules of sulphur in 16g of solid sulphur,

Number of molecules = Number of moles x Avogadro number

= 0.0625 x 6.022 x 10²³ molecules

= 3.763 x 10 22 molecules

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)

To calculate the number of aluminium ions in 0.051g of aluminium oxide,

1 mole of aluminium oxide = 6.022 x 10 23 molecules of aluminium oxide

1 mole of aluminium oxide (Al 2 O 3 ) = 2 x Mass of aluminium + 3 x Mass of oxygen

= (2x 27) + (3 x16) = 54 +48 = 102g

1 mole of aluminium oxide = 102g = 6.022 x 10 23 molecules of aluminium oxide

Therefore, 0.051g of aluminium oxide has = 6.022 x 10 23 / 102 x 0.051

      = 3.011 x 10 20 molecules of aluminium oxide

One molecule of aluminium oxide has 2 aluminium ions; hence, the number of aluminium ions present in 0.051g of aluminium oxide = 2 x 3.011x 10 20 molecules of aluminium oxide.

= 6.022 x 10 20

NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules

The chapter holds a weightage of 23 marks in the examinations. The questions are not specific and can be asked from any topic. The students are therefore advised to be thorough with the entire chapter. The chemical formulae and the numerical should be practised well.

The important topics provided in this chapter include

3.1 Laws of Chemical Combination

3.2 What Is an Atom?

3.3 What Is a Molecule?

3.4 Writing Chemical Formulae

3.5 Molecular Mass and Mole Concept

Exercise Solutions 11 Questions (8 numerical, 3 short)

Molecular mass and Mole concept – 8 numerical

Chemical Formula – 2 Questions

What Is an Atom – 1 Question

NCERT Solutions for Class 9 Science Chapter 3

The smallest unit of matter is an atom. It has the properties of an element. An atom comprises a dense core called a nucleus, surrounded by a series of outer shells. The electrons are present in these shells. The nucleus contains protons and neutrons. Protons have a positive charge, while neutrons are neutral.

Two or more atoms tightly bound together form a molecule. The molecules made up of two atoms are known as diatomic. Oxygen, nitrogen, hydrogen, and iodine are diatomic molecules. Earth’s atmosphere is comprised mainly of diatomic molecules. A molecule is the smallest part of a compound.

Key Features of NCERT Solutions for Class 9 Science Chapter 3 – Atoms and Molecules

• The answers are provided by Science experts.
• The answers are non-erroneous.
• The Solutions have questions from every important topic.
• It gives a thorough understanding of the concepts.

Disclaimer:

Dropped Topics –  Mole concept.

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NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules

Chapter 3 Atoms and Molecules NCERT Solutions for Class 9 Science

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Important Questions for CBSE Class 9 Science Atoms and Molecules 2024-25

• Class 9 Important Question
• Chapter 3: Atoms And Molecules

CBSE Class 9 Science Chapter-3 Important Questions - Free PDF Download

Class 9th is one of the most challenging years in a student's life as there are so many things they need to learn in a single year. As a subject science has always been on a stricter side of things and thus we at Vedantu came up with a solution to provide students with important questions chapter-wise. In this article, we will talk about the fundamentals of atoms and molecules and how they are important while providing students with extra questions that will help them understand the topic in a better way. We will be discussing laws of chemical combination, atoms, their symbols, the gram atomic mass, ions, molecules, and many more.

Vedantu is a platform that provides free CBSE Solutions (NCERT) and other study materials for students. You can download Class 9 Maths and Class 9 Science NCERT Solutions to help you to revise complete syllabus and score more marks in your examinations. 

Download CBSE Class 9 Science Important Questions 2024-25 PDF

Also, check CBSE Class 9 Science Important Questions for other chapters:

Study Important Questions for Class 9 Science Chapter 3 – Atoms and Molecules

Very Short Answer Questions (1 Mark)

Atomic radius is measured in nanometers and

$1nm={{10}^{-10}}m$

$1m={{10}^{-10}}nm$

$1nm={{10}^{-9}}m$

$1m={{10}^{-9}}nm$

Ans: (c) $1nm={{10}^{-9}}m$

Symbol of Iron is –

None of these

Ans:   (c) $Fe$

Atomicity of Chlorine and Argon is

Diatomic and Monoatomic

Monoatomic and Diatomic

Monoatomic and Monoatomic

Diatomic and Diatomic

Ans: (a) Diatomic and Monoatomic

Molecular mass of water $({{H}_{2}}O)$ is

Ans:   (a) $18g$

It is said that 1 mole of a compound contains –

$6.023\times {{10}^{23}}atoms$

$6.023\times {{10}^{24}}atoms$

$60.23\times {{10}^{23}}atoms$

$6.023\times {{10}^{25}}atoms$

Ans: (a) $6.023\times {{10}^{23}}atoms$

Oxygen is –

Tetravalent

Ans: (b) Bivalent

What is the molecular formula for Calcium Hydroxide?

$CaO{{H}_{2}}$

$Ca{{(OH)}_{2}}$

$C{{a}_{2}}OH$

$Ca{{H}_{2}}$

Ans: (b) $Ca{{(OH)}_{2}}$

Chargeless and Massless

Chargeless and has Mass

Has charge and Mass

Has charge and Massless.

Ans: (b) Chargeless and has Mass

Which of the following statements is correct?

Cathode rays travel in a straight line and have momentum.

Cathode rays travel in a straight line and have no momentum

Cathode rays do not travel in a straight line but have Momentum.

Cathode rays do not travel in a straight line and have no momentum.

Ans: (a) Cathode rays travel in a straight line and have momentum.

 How are \[\beta \]–particles represented?

$e_{-1}^{0}$

${{e}_{+1}}$

$e_{-1}^{1}$

$e_{0}^{1}$

Ans: (a) $e_{-1}^{0}$

 Elements $Ar_{18}^{40}$ and $Ca_{20}^{40}$ are

Both b and c

Ans:   (b) Isobars

The maximum number of electrons in L shell is

Ans:   (a) $8$

Short Answer Questions (3 Marks)

Define the atomic mass unit.

Ans: One atomic mass unit is a mass unit equal to exactly one-twelfth (${1}/{12}\;th$) the mass of one atom of $carbon-12$. The relative atomic masses of all elements have been found with respect to an atom of $carbon-12$.

According to the latest IUPAC (International Union of Pure and Applied Chemistry) recommendations, the atomic mass unit (written as ‘u’ – unified mass) is equal to the mass of one-twelfth (${1}/{12}\;th$) of$carbon-12$ atom.

$1\text{ }amu={1}/{12}\;th\text{ }Mass\text{ }Of\text{ }C_{6}^{12}$

Write down the formulae of

Sodium oxide

Ans:   Sodium oxide – $N{{a}_{2}}O$

Aluminium chloride

Ans:   Aluminium chloride – $AlC{{l}_{3}}$

Sodium sulphide

Ans:   Sodium sulphide – $N{{a}_{2}}S$

Magnesium hydroxide

Ans:   Magnesium hydroxide – $Mg{{(OH)}_{2}}$

Write down the names of compounds represented the following formulae:

$A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$

Ans: $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$  - Aluminium sulphate

$CaC{{l}_{2}}$

Ans: $CaC{{l}_{2}}$ - Calcium chloride

\[{{K}_{2}}S{{O}_{4}}\]

Ans:   \[{{K}_{2}}S{{O}_{4}}\] - Potassium sulphate

\[KN{{O}_{3}}\]

Ans:   \[KN{{O}_{3}}\] - Potassium nitrate

$CaC{{O}_{3}}$

Ans:   $CaC{{O}_{3}}$ - Calcium carbonate

What is meant by the term chemical formula?

Ans: The term chemical formula of a compound is said to be the symbolic representation of its composition or it is a notation that shows the type and number of atoms in a molecule of a compound with the help of atomic symbols and numbers.

They provide information on the elements that constitute the molecules of a compound and the ratio in which the atoms of those elements combine to form the molecules.

Example: A molecule of water, which is a compound, contains two molecules of hydrogen and one molecule of oxygen. Its chemical formula is \[{{H}_{2}}O\] .

What are polyatomic ions? Give examples.

Ans:   Polyatomic ions are a group of atoms carrying a charge. They are typically clusters of atoms that act as an ion, which carry a fixed charge on them.

Ammonium – \[N{{H}_{4}}^{+}\]

Hydroxide – \[O{{H}^{-}}\]

Nitrate – \[N{{O}_{3}}^{-}\]

Hydrogen carbonate – \[HC{{O}_{3}}^{-}\]

Write the chemical formulae of the following.

Magnesium chloride

Ans:   Magnesium chloride – $MgC{{l}_{2}}$

Calcium oxide

Ans: Calcium oxide –$CaO$

Copper nitrate

Ans:   Copper nitrate –\[CuN{{O}_{3}}\]

Ans:   Aluminium chloride –$AlC{{l}_{3}}$

Calcium carbonate

Ans:   Calcium carbonate – $CaC{{O}_{3}}$

Give the names of the elements present in the following compounds.

Ans: Quick lime –$CaO$

Elements present – Calcium, Oxygen

Hydrogen bromide

Ans: Hydrogen bromide –$HBr$

Elements present – Hydrogen, Bromine

Baking powder

Ans: Baking powder –$NaHC{{O}_{3}}$

Elements present – Sodium, Hydrogen, Carbon, Oxygen

Potassium sulphate

Ans:   Potassium sulphate –\[{{K}_{2}}S{{O}_{4}}\]

Elements present – Potassium, Sulphur, Oxygen

Calculate the molar mass of the following substances.

Atomic mass of –

$C=12u,\text{ }H=1u,\text{ }S=32u,\text{ }P=31u,\text{ }Cl=35.5u,\text{ }N=14u,\text{ }O=16u$

Ethyne – ${{C}_{2}}{{H}_{2}}$

Ans: ${{C}_{2}}{{H}_{2}}=(12\times 2)+(1\times 2)=24+2=26u=26{g}/{mole}\;$

Sulphur molecule –${{S}_{8}}$

Ans: ${{S}_{8}}=32\times 88=256u=256{g}/{mole}\;$

Phosphorus molecule – ${{P}_{4}}$ (Atomic mass of phosphorus is $31$)

Ans: ${{P}_{4}}=31\times 4=124u=124{g}/{mole}\;$ 

Hydrochloric acid – $HCl$

Ans: $HCl=(1\times 1)+(35.5\times 1)=1+35.5=36.5u=36.5{g}/{mole}\;$

Nitric acid – \[HN{{O}_{3}}\]

Ans: \[HN{{O}_{3}}=(1\times 1)+(14\times 1)+(16\times 3)=1+14+48=63u=63{g}/{mole}\;\]

What is the mass of –

$S=32u,\text{ Al}=27u,\text{ Na}=23u,\text{ }N=14u,\text{ }O=16u$

1 mole of nitrogen atoms?

Ans:   Given its atomic mass, the mass of 1 mole of nitrogen atoms is $14g$

4 moles of aluminium atoms (Atomic mass of aluminium is $27$)?

Ans:   Given its atomic mass, the mass of 1 mole of aluminium atoms is $27g$

Thus, the mass of 4 moles of aluminium atoms is $27\times 4=108g$

10 moles of sodium sulphite (\[N{{a}_{2}}S{{O}_{3}}\] )?

Ans:   Given its atomic mass, the mass of 1 mole of sodium sulphite (\[N{{a}_{2}}S{{O}_{3}}\]) is $(23\times 2)+(32\times 1)+(16\times 3)=46+32+48=126u=126{g}/{mole}\;$

Thus, the mass of 10 moles of sodium sulphite (\[N{{a}_{2}}S{{O}_{3}}\]) is $126\times 10=1260g$

Convert into mole.

Atomic mass of – $C=12u,\text{ }H=1u,\text{ }O=16u$

12 g of oxygen gas

Ans: Molar mass of ${{O}_{2}}=(16\times 2)=32{g}/{mole}\;$

$\Rightarrow 1\text{ }mole\text{ }of\text{ }{{O}_{2}}=32g$

$\Rightarrow 1g\text{ }of\text{ }{{O}_{2}}=\dfrac{1}{32}\text{ }moles\text{ }$

$\Rightarrow 12g\text{ }of\text{ }{{O}_{2}}=12\times \dfrac{1}{32}\text{ =0}\text{.375}moles\text{ }$ 

20 g of water

Ans: Molar mass of ${{H}_{2}}O=(1\times 2)+(16\times 1)=18{g}/{mole}\;$

$\Rightarrow 1\text{ }mole\text{ }of\text{ }{{H}_{2}}O=18g$

$\Rightarrow 1g\text{ }of\text{ }{{H}_{2}}O=\dfrac{1}{18}\text{ }moles\text{ }$

$\Rightarrow 20g\text{ }of\text{ }{{H}_{2}}O=20\times \dfrac{1}{18}\text{ =1}\text{.11}moles\text{ }$

22 g of carbon dioxide

Ans: Molar mass of $C{{O}_{2}}=(12\times 1)+(16\times 2)=44{g}/{mole}\;$

$\Rightarrow 1\text{ }mole\text{ }of\text{ C}{{O}_{2}}=44g$

$\Rightarrow 1g\text{ }of\text{ C}{{O}_{2}}=\dfrac{1}{44}\text{ }moles\text{ }$

$\Rightarrow 12g\text{ }of\text{ }{{O}_{2}}=22\times \dfrac{1}{44}\text{ =0}\text{.5}moles\text{ }$

State the Postulates of Dalton Theory?

Ans: Dalton’s atomic theory states that all matter, be it an element, a compound, or a mixture is composed of small particles called atoms. 

The postulates of the theory are:

All matter is made of very tiny particles called atoms, which participate in chemical reactions. 

Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction. 

Atoms of a given element are identical in mass and chemical properties. 

Atoms of different elements have different masses and chemical properties. 

Atoms combine in the ratio of small whole numbers to form compounds.

The relative number and kinds of atoms are constant in a given compound.

Find the percentage of water of crystallization in \[FeS{{O}_{4}}.7{{H}_{2}}O\].

Ans: Atomic mass of –

$Fe=55.9u,\text{ S}=32u,\text{ H}=1u,\text{ }O=16u$

Molar mass of$FeS{{O}_{4}}.7{{H}_{2}}O=(55.9\times 1)+(32\times 1)+(16\times 4)+7\times \left[ (1\times 2)+(16\times 1) \right]$

$=55.9+32+64+7\times \left[ 18 \right]{=151.9+126=227.9g}/{mole}\;$

Thus, we can say that \[227.9{g}/{mole}\;\]of $FeS{{O}_{4}}$contains $126g$water of crystallization.

So, \[1g\] of \[FeS{{O}_{4}}\]contains $\dfrac{126}{277.6}g$ water of crystallization.

Converting this fraction into percentage –

$\dfrac{126}{277.6}=0.4534g$ water of crystallization

Thus, we get $\dfrac{126}{277.6}\times 100=0.4534\times 100=45.34\%$

The percentage of water of crystallization in \[FeS{{O}_{4}}.7{{H}_{2}}O\] is $45.34\%$.

 \[2.42g\] of copper gave \[3.025g\] of a black oxide of copper, \[6.49g\] of a black oxide, on reduction with hydrogen, gave \[5.192g\] of copper. Show that these figures are in accordance with the law of constant proportion?

Ans: Given:

Mass of copper: \[2.42g\]

Mass of copper oxide: \[3.025g\]

Mass of black copper oxide: \[6.49g\]

Mass of copper obtained after reduction: \[5.192g\]

Verification: To prove the law of constant proportions, we need to find out the percentage of copper in copper oxide in both cases A and B.

Percentage of copper in Case A $=\dfrac{Mass\text{ }of\text{ }Copper}{Mass\text{ }of\text{ }Copper\text{ }Oxide}\times 100\%$

$=\dfrac{2.42}{3.025}\times 100\%=0.8\times 100\%=80\%$

Percentage of copper in Case B $=\dfrac{Mass\text{ }of\text{ }Copper}{Mass\text{ }of\text{ }Copper\text{ }Oxide}\times 100\%$

$=\dfrac{5.192}{6.49}\times 100\%=0.8\times 100\%=80\%$

It is clear from the above two calculations that the percentage of copper in copper oxide in both cases A and B is the same. This proves the law of constant proportions – copper always combines with oxygen in the same proportion.

A compound was found to have the following percentage composition by mass \[Zn=22.65\%\], \[S=11.15\%\], \[H=4.88\%\], \[O=61.32\%\]. The relative molecular mass is \[287{g}/{mole}\;\]. Find the molecular formula of the compound, assuming that all the hydrogen in the compound is present in water of crystallization.

\[Zn=22.65\%\]

\[S=11.15\%\]

\[H=4.88\%\]

\[O=61.32\%\]

Relative molecular mass: \[287{g}/{mole}\;\]

To find: Molecular formula of the compound.

$Zn=65.4u,\text{ S}=32u,\text{ H}=1u,\text{ }O=16u$

To find the formula, we need to find the proportion in which these atoms have combined.

It is known that – $Percentage\text{ }of\text{ }element\text{ present }in\text{ }a\text{ }compound=\dfrac{number\text{ }of\text{ }atoms\times atomic\text{ }mass}{mass\text{ }of\text{ }compound}\times 100\%$$\Rightarrow Number\text{ }of\text{ }atoms=\dfrac{percentage\text{ }of\text{ }element\text{ present }in\text{ }a\text{ }compound\times mass\text{ }of\text{ }compound}{atomic\text{ }mass\times 100}$Using the formula above,

\[Number\text{ }of\text{ }Zn\text{ }atoms=\dfrac{22.65\times 287}{65.4\times 100}=0.99=1\]

\[Number\text{ }of\text{ S }atoms=\dfrac{11.15\times 287}{32\times 100}=1.00=1\]

\[Number\text{ }of\text{ H }atoms=\dfrac{4.88\times 287}{1\times 100}=14\]

\[Number\text{ }of\text{ O }atoms=\dfrac{61.32\times 287}{16\times 100}=10.99=11\]

Here, all the Hydrogen atoms belong to the water of crystallization.

Water has the molecular formula – \[{{H}_{2}}O\]with two molecules of Hydrogen and one molecule of oxygen. 

Since we have $14$ atoms of Hydrogen, we can say that there are 7 molecules of water in this compound.

That leaves one atom of zinc, one atom of sulphur, and 4 atoms of oxygen (out of $11$, $7$ atoms of oxygen are in the water of crystallization). It is clear that the compound is Zinc Sulphate with the formula – \[ZnS{{O}_{4}}\]

Thus, the formula of the compound is \[ZnS{{O}_{4}}.7{{H}_{2}}O\]

Which element will be more reactive and why – the element whose atomic number is 10 or the one whose atomic number is 11?

Ans:   The element with atomic number $11$ is more reactive than the element with atomic number $10$. This is because of the electronic configuration of the atoms. 

The element with the atomic number $11$, has the configuration of $(2,8,1)$, which means it can easily lose an electron to attain stability. Thus, before losing the electron, it is not stable and is said to be more reactive.

While the element with the atomic number $10$, has the configuration of $(2,8)$, which means it is already stable with a completely filled L shell and does not have to gain or lose electrons to attain stability. Thus, it is said to be less reactive.

What are the failures of Dalton's Atomic theory?

Ans:    

It does not account for subatomic particles: It stated that atoms were the smallest unit of matter. But, the discovery of subatomic particles namely, protons, electrons, and neutrons disproved this postulate.

It does not account for isotopes: For example hydrogen $H_{1}^{1}$, deuterium $H_{1}^{2}$, and tritium$H_{1}^{3}$, have the same atomic number, but different mass numbers.

It does not account for isobars. Example: $Ar_{18}^{40}$ and$Ca_{20}^{40}$, they have different atomic numbers, but the same mass number.

Elements need not combine in simple, whole-number ratios to form compounds: There are complex organic compounds that do not combine in simple ratios of constituent atoms. Example: sugar/sucrose (${{C}_{11}}{{H}_{22}}{{O}_{11}}$).

It does not account for allotropes: The differences in the properties of diamond and graphite, even though they contain only carbon, cannot be explained by Dalton’s atomic theory.

Calculate the Molecular Mass of

Atomic mass of – $S=32u,\text{ H}=1u,\text{ C}=12u,\text{ }N=14u,\text{ }O=16u$

Ammonium sulphate ${{(N{{H}_{4}})}_{2}}S{{O}_{4}}$

Ans: Molar mass of \[{{(N{{H}_{4}})}_{2}}S{{O}_{4}}=2\times \left[ (14\times 1)+(1\times 4) \right]+(32\times 1)+(16\times 4)\]

\[=2\times \left[ (14)+(4) \right]+(32)+(64)=(2\times 18)+96=36+96=132{g}/{mole}\;\]

Penicillin ${{C}_{16}}{{H}_{18}}{{N}_{2}}S{{O}_{4}}$

Ans: Molar mass of \[{{C}_{16}}{{H}_{18}}{{N}_{2}}S{{O}_{4}}=(12\times 16)+(1\times 18)+(14\times 2)+(32\times 1)+(16\times 4)\]

\[=(192)+(18)+(28)+(32)+(64)=334{g}/{mole}\;\]

Paracetamol ${{C}_{8}}{{H}_{9}}NO$

Ans: Molar mass of \[{{C}_{8}}{{H}_{9}}NO=(12\times 8)+(1\times 9)+(14\times 1)+(16\times 1)\]

\[=(96)+(9)+(14)+(16)=135{g}/{mole}\;\]

 Answer the following questions are about one mole of sulphuric acid ${{H}_{2}}S{{O}_{4}}$

Atomic mass of – \[S=32u,\text{ H}=1u,\text{ }O=16u\]

Find the number of gram atoms of hydrogen in it?

Ans: Mass of $1$ mole of ${{H}_{2}}S{{O}_{4}}=(1\times 2)+(32\times 1)+(16\times 4)=2+32+64=98{g}/{mole}\;$

It is known that $1$ mole of any substance contains $6.023\times {{10}^{23}}\text{ atoms/molecules}$.

Thus, $1$ mole of ${{H}_{2}}S{{O}_{4}}=98{g}/{mole}\;=6.023\times {{10}^{23}}\text{ molecules}$

From the molecular formula, we can say that ${{H}_{2}}S{{O}_{4}}$ has two atoms of hydrogen.

i.e. $(2\times atomic\text{ mass of H)}=2\times 1=2g$

Thus, the number of gram atoms of hydrogen in ${{H}_{2}}S{{O}_{4}}$is, $2g$

How many atoms of hydrogen does it have?

Ans: Number of atoms of H $=number\text{ of atoms of H in }{{H}_{2}}S{{O}_{4}}\times AvogadroNumber$

$=2\times 6.023\times {{10}^{23}}\text{ =12}\text{.046}\times {{10}^{23}}atoms$

How many atoms (in grams) of hydrogen are present for every gram atom of oxygen in it?

Ans: From the molecular formula, we can say that ${{H}_{2}}S{{O}_{4}}$

has two atoms of hydrogen for every four atoms of oxygen.

i.e. $2H:4O$

$\dfrac{2}{4}H:1O\Rightarrow \dfrac{1}{2}H:1O\Rightarrow 0.5H:1O$

Thus, for one atom of oxygen we get $0.5$hydrogen atoms (in grams).  

Calculate the number of atoms in ${{H}_{2}}S{{O}_{4}}$?

Ans: $1$ mole of ${{H}_{2}}S{{O}_{4}}$ contains $6.023\times {{10}^{23}}\text{ molecules}$

Write an experiment to show that cathode rays travel in a straight line?

Ans: An experiment to show that cathode rays travel in a straight line can be performed using a fluorescent coated discharge tube and a source of cathode rays, an opaque object, and a high voltage source.

Set-up for the experiment:

(Image will be uploaded soon)

In a discharge tube coated with a fluorescent substance initiate the production of cathode rays using a high voltage source.

In the path of the cathode rays, place an opaque object and observe the fluorescence phenomena.

When cathode rays strike against the screen, they produce fluorescence. But due to the placement of the opaque object, we will observe a sharp shadow being formed on the screen in the shape of the object. 

This shadow of the object can be formed if and only if the cathode rays travel in a straight line and do not bend around the edges of the object.

This experiment shows that cathode rays travel in a straight line.

What is radioactivity? What are the applications of radioisotopes?

Ans: Radioactivity is defined as the spontaneous emission of radiation in the form of particles or high-energy photons that are a result of a nuclear reaction. It is the release of energy from the decay of the nucleus of atoms and/or isotopes.

Applications of radioisotopes:

The isotope of $Co-60$ emits $\gamma $-radiation that is used to treat cancer.

$I-131$ is used in the diagnosis and treatment of thyroid gland diseases.

$P-32$ is used in the treatment of leukemia and the identification of malignant tumors.

$C-14$ is used to study biochemical processes.

There are two elements C and B. C emits an $\alpha $ – particle and B emits a $\beta $ – particle. How will the resultant elements charge?

Ans: When an element emits $\alpha $ particle, its atomic number decreases by $2$ , and its mass number decreases by $4$. This is because alpha particles are positively charged nuclei of Helium with two protons and two neutrons.

Thus, in the case of element C that emits $\alpha $particle, its atomic number decreases by $2$ and its mass number decreases by $4$.

When an element emits $\beta $ particle, its atomic number increases by $1$ and its mass number remains the same. This is because a beta particle is essentially an electron.

Thus, in the case of element B that emits $\beta $ particle, its atomic number increases by $1$ and its mass number remains the same.

What are isotopes? Name the isotopes of hydrogen and draw the structure of their atoms.

Ans: Isotopes are defined as the atoms of the same element that have different mass numbers; i.e. elements having the same atomic number but different mass numbers.

Example – Isotopes of Hydrogen: 

Hydrogen $H_{1}^{1}$

Deuterium $H_{1}^{2}$

Tritium$H_{1}^{3}$

Structure of Isotopes of Hydrogen:

Long Answer Questions (5 Marks)

In a reaction, $5.3g$ of sodium carbonate reacted with $6g$ of ethanoic acid. The products were $2.2g$ of carbon dioxide, $0.9g$ water and $8.2g$ of sodium ethanoate. Show that these observations are in agreement with the Law of Conservation of Mass.

\[\mathbf{Sodium}\text{ }\mathbf{carbonate}\text{ + }\mathbf{Ethanoic}\text{ }\mathbf{acid}\to \mathbf{Sodium}\text{ }\mathbf{ethanoate}\text{ }+\text{ }\mathbf{Carbondioxide}\text{ }+\text{ }\mathbf{Water}\]

Ans: The law of conservation of mass states that mass can neither be created nor destroyed in a chemical reaction. This means that the mass of the constituents of a closed chemical reaction will remain the same before and after the reaction.

Mathematically - $\text{Mass of reactants = Mass of products}$

Here, the reactants are Sodium carbonate and Ethanoic acid. 

The products are Sodium ethanoate, carbon dioxide and water.

To prove the law of conservation of mass, we need to prove the mass of reactants is equal to the mass of the products.

Mass of Sodium carbonate: $5.3g$

Mass of Ethanoic acid: $6g$

Mass of Sodium ethanoate: $8.2g$

Mass of Carbon Dioxide: $2.2g$

Mass of Water: $0.9g$

The reaction – 

\[Mass\text{ of reactants = Mass of }\mathbf{Sodium}\text{ }\mathbf{carbonate}\text{ + Mass of }\mathbf{Ethanoic}\text{ }\mathbf{acid}\]$=5.3+6=11.3g$

\[Mass\text{ of products = Mass of }\mathbf{Sodium}\text{ ethanoate + Mass of carbondioxide + Mass of Water}\]$=8.2+2.2+0.9=11.3g$

It is clear from the above calculations that – 

$\text{Mass of reactants = Mass of products=11}\text{.3g}$

Thus this proves the law of conservation of mass.

Calculate the molecular masses of

Atomic mass of – $\text{H}=1u,\text{ C}=12u,\text{ }N=14u,\text{ }O=16u,\text{ }Cl=35.5u$

${{H}_{2}}$

Ans: Molar mass of ${{H}_{2}}=(1\times 2)=2u$

${{O}_{2}}$

Ans: Molar mass of ${{O}_{2}}=(16\times 2)=32u$

$C{{l}_{2}}$

Ans: Molar mass of $C{{l}_{2}}=(35.5\times 2)=71u$

$C{{O}_{2}}$

Ans: Molar mass of $C{{O}_{2}}=(12\times 1)+(16\times 2)=12+32=44u$

$C{{H}_{4}}$

Ans: Molar mass of $C{{H}_{4}}=(12\times 1)+(1\times 4)=12+4=16u$

${{C}_{2}}{{H}_{6}}$

Ans: Molar mass of \[{{C}_{2}}{{H}_{6}}=(12\times 2)+(1\times 6)=24+6=30u\]

${{C}_{2}}{{H}_{4}}$

Ans: Molar mass of ${{C}_{2}}{{H}_{4}}=(12\times 2)+(1\times 4)=24+4=28u$

$N{{H}_{3}}$

Ans: Molar mass of $N{{H}_{3}}=(14\times 1)+(1\times 3)=14+3=17u$

$C{{H}_{3}}OH$

Ans: Molar mass of $C{{H}_{3}}OH=(12\times 1)+(1\times 3)+(16\times 1)+(1\times 1)=12+3+16+1=32u$

If one mole of carbon atoms weighs $12$ grams, what is the mass (in grams) of one atom of carbon?

It is known that $1$ mole of any substance contains$6.023\times {{10}^{23}}\text{ atoms/molecules}$.

Thus, $1$ mole of $C=6.023\times {{10}^{23}}\text{ C-atoms}$

It is given that one mole of carbon atoms weighs $12$ grams

Combining these two observations,

$1\text{ }mole\text{ }of\text{ }C=12g=6.023\times {{10}^{23}}\text{ C-atoms}$

We need to find the mass of one carbon atom.

Since – $12g\text{ }of\text{ }C=6.023\times {{10}^{23}}\text{ C-atoms}$ i.e. $12$ grams contain $6.023\times {{10}^{23}}\text{ C-atoms}$

Now for the mass of one carbon atom –

$6.023\times {{10}^{23}}\text{ C-atoms=}12g\text{ }of\text{ }C$

$1\text{ C-atom = }\dfrac{12}{6.023\times {{10}^{23}}}g\text{ =1}\text{.993}\times {{10}^{-23}}g$

Thus, the mass of one carbon atom is $\text{1}\text{.993}\times {{10}^{-23}}g$

A $0.24g$ sample of 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.

Mass of sample compound: $0.24g$

Mass of boron in the sample: $0.096g$

Mass of oxygen in the sample: $0.144g$

To find: Percentage composition of boron and oxygen in the compound by weight.

$Percentage\text{ }of\text{ }element\text{ present }in\text{ }a\text{ }compound=\dfrac{mass\text{ }of\text{ }element\text{ }in\text{ }compound}{mass\text{ }of\text{ }compound}\times 100\%$

$Percentage\text{ }of\text{ Boron }in\text{  }compound=\dfrac{mass\text{ }of\text{ Boron }in\text{ }compound}{mass\text{ }of\text{ }compound}\times 100\%$

$=\dfrac{0.096}{0.24}\times 100\%=0.4\times 100\%=40\%$

\[Percentage\text{ }of\text{ Oxygen }in\text{ }compound=\dfrac{mass\text{ }of\text{ Oxygen }in\text{ }compound}{mass\text{ }of\text{ }compound}\times 100\%\]

$=\dfrac{0.144}{0.24}\times 100\%=0.6\times 100\%=60\%$

The percentage of Boron by weight in the compound is $40\%$ and the percentage of Oxygen by weight in the compound is $60\%$.

Important Question of Atoms And Molecules Class 9

Topics covered for class 9 science ch 3 important questions.

The ch 3 science class 9 important questions notes have been prepared by the experts who have been teaching in their respective subjects for quite some time. They know all the ins and outs of the topics and the chapters. Also, with these notes, you will get some insights about the topics which you might not find in your textbook or from your class teacher. Below are some of the topics covered in this chapter, and we have discussed them briefly to give students a revision about the topics they need to learn to answer these questions.

Chemical Reactions 

When a chemical reaction occurs between two or more molecules, then a new compound is formed, and then the two molecules are called reactants whereas the newly formed compound will be called products. 

During the chemical reaction, a chemical change needs to occur which can be seen by a physical change like precipitation, or production of heat, or in some cases change of colour. 

Law of Conservation of Mass

A matter can neither be created nor destroyed in any chemical reaction; it only remains conserved; this is the law of conservation of mass. 

On the other hand, the mass of the given reactants in the chemical reaction will be equal to the mass of the products formed from the chemical reaction. 

Law of Constant Proportions 

A chemical compound that is said to be pure contains the same amount of elements combined in a fixed proportion by mass is said to be the law of definite proportions. A great example of the law of constant proportions is that when we take out water from the lake present in the mountains and take out the water from the ocean both have the same number of oxygen and hydrogen molecules present in them. 

Atoms 

Atoms are said to be the building blocks of the world that we see around us. They are present everywhere, and they are present inside our body. An atom is the most fundamental part of the element, and it cannot be broken by any chemical means. 

Dalton's Atomic Theory 

The matter present everywhere in the universe is made up of tiny particles that cannot be divisible into other smaller particles, and they are called atoms. 

When comparing the properties of the atoms of a given element, they tend to be the same, meaning their mass is also the same. As a result, we can state that an element's atoms have the same mass and chemical properties. On the other hand, atoms of a different element have different mass, showing different chemical properties. 

Compounds are formed when the atoms of different elements combine in fixed ratios. 

Atoms are the particles that can neither be created, nor they can be destroyed. The formation of the new compounds occurs from the rearrangement of the existing atoms in a chemical reaction. 

Lastly, in a given compound, the relative number and kinds of atoms are constant. 

Atomic Mass 

Atomic mass is said to be the total mass of all the neutrons, protons, and electrons present in a given atom or a group of atoms. Atomic mass is also said to be the average mass. 

The mass of the atomic particle can be defined as the atomic mass. 

Molecular Mass 

In the important question of atoms and molecules, class 9 molecular mass of a given element can be presented as the sum of the masses of the elements present in the molecule. 

Concept of Molar Mass

In any given substance, the number of atoms, molecules, ions present is defined as a mole. A mole of any substance is said to be 6.022×10 23 molecules. It is one of the easiest ways to express the number of reactants and products in the reaction. On the other hand, there is an Avogadro's number that approximately has the same value as one mole. It tells us about the number of particles present in one mole. These particles which are represented in mole could be electrons, protons, and neutrons. 

Molar Mass 

Any particle or a substance present in the universe has some mass to it and acquires some space. The molar mass or molecular weight is the sum of the total mass in grams of the atoms present inside the atom that makes up the molecule per mole. The unit of molar mass is grams/mole. 

Concept of Atomicity 

A molecule is the smallest unit of a compound that can represent all the chemical properties of a compound. The atomicity of a given element is measured by the number of atoms present in its one molecule. 

These are some of the important concepts that students need to learn about before they tackle the class 9 science ch 3 important questions that Vedantu has created. 

Important Questions for Class 9 Science Chapter 3 Exam Point of View 

Now you get the idea of the concepts and definitions you need to learn for chapter 3 of the science textbook. Now let's move on to the interesting part, which is the important questions from the exam point of view. Given below, we have ten questions which can help you prepare for your upcoming science exams and clear lots of written concepts in this chapter. 

Q1) Name the scientist who laid the foundation of the chemical sciences and led to chemistry studies. Also, provide the answer to how did he find it?

Q2) Write down the law of conservation of mass by giving the example of a chemical reaction.

Q3) What is the law of constant proportion, and how does it help students perform chemical reactions?

Q4) Which world organization approves all the names of the elements we see in our textbooks and use in our daily lives? Also, write the symbol of the mercury. 

Q5) Write down the symbols of Oxygen and Hydrogen.

Q6) "Atoms of most of the elements tend to not exist on their own" Name the two atoms which can exist as an independent atom. 

Q7) Provide one relevant reason in your answer for why the scientists have chosen 1/16 of the mass of an atom of a naturally found oxygen as the atomic mass unit. 

Q8) Which of the postulates from Dalton's atomic theory results from the law of conservation of mass?

Q9) What are the two drawbacks of Dalton's atomic theory and how were they corrected? 

Q10) How will you be able to differentiate between the molecule of a given element and the molecule of a given compound?

In addition to this, we appreciate students to write down the answer to these questions on their own and don't take help from our solved questions Pdf. But in case there are some questions you might find difficult you can look at its solution without a problem. 

Benefits of Learning Ch 3 Science Class 9 Important Questions

Chapter 3 science class 9 important questions Pdf will help students learn the various complicated topics, including the molar mass and valency of the atoms. 

In addition to this, students get to knows step by step solution of the numerical answers, so they don't lose any extra marks for skipping a step. 

Likewise, the answer to chapter 3 science class 9 important questions are written by well trained and experienced teachers from all around India. They know the core of the subject and have been teaching it for a while. So when you are looking at the solution of the question, you are reading the answer which is written by an expert in the field of science.

CBSE Class 9 Science Atoms and Molecules Extra Questions

Name two scientists who established the law of chemical combination.

What is the unit used to measure the size of an atom?

Why is Avogadro's number also known as  Avogadro’s constant? 

Write 6 postulates of Dalton’s atomic theory?

Find the mass percentage of oxygen present in HNO 3 .

So there you go, these are some of the chapter 3 science class 9 important questions to which students must have the answers. To get to their solution, a student must prepare the chapter thoroughly and know every topic to the core. In addition to this, these questions are quite important from an exam point of view, so it's better to solve them first before you feel prepared for your exams. If you are stuck somewhere and don't know what the answer to the question should be, then open Vedantu's solved question answer Pdf and find out what were you missing from the solution. This chapter's numerical questions need to be practised regularly, so students know which formulas need to use for a given numerical problem. 

Important Related Links for CBSE Class 9 

FAQs on Important Questions for CBSE Class 9 Science Atoms and Molecules 2024-25

1. What is an atom according to CBSE Chapter 3 of Class 9 Science?

Atoms may be defined as the most important element that forms the basic unit of things. It is present everywhere. It is the most important structure and it cannot be broken further. Our body is also made of atoms. Students of Class 9 can study atoms in detail through NCERT book. They should read NCERT Chapter 3 carefully to understand the concept of atoms. They can also refer to the  Important Questions For CBSE Class 9 Science Atoms And Molecules  provided by Vedantu to learn more about atoms. 

2. What are chemical reactions according to CBSE Chapter 3 of Class 9 Science?

A chemical reaction is  defined as  a reaction between two or more substances to form a new substance. A chemical change occurs in the substances that react with each other during a chemical reaction. The change is visible in the form of change in colour, production of heat, or precipitation.

3. What is given in the law of proportions in Chapter 3 of Class 9 Science?

According to the law of proportions given in Chapter 3 of Class 9 Science, a pure chemical compound consists of the same number of elements that are combined in a fixed proportion by mass. For example, if you take out water from your fridge or a lake, it will consist of the same number of hydrogen and oxygen atoms by mass. The number of hydrogen and oxygen atoms that combine to form water remains the same in all forms of water. 

4. What are the important topics covered in Chapter 3 of Class 9 Science?

In  Class 9 Science Chapter 3,  students will study the definitions of atoms, molecules, chemical reactions, chemical formulas, etc. Students will also study different theories related to the discovery of an atom. They will also study how to write a chemical formula for a given compound. Students will also study the law of constant proportion and other laws related to chemical reactions. Mole concept and molecule mass are also important parts of Chapter 3 of Class 9 Science.

5. Are numerical problems given in Chapter 3 of Class 9 Science important for exams?

Yes, numerical problems given in Chapter 3 of Class 9 Science are important for students. They should practice all numerical given in Chapter 3 of Class 9 Science for scoring good marks in exams. They can practice the numerical  questions  from the NCERT book. They can also refer to the  Important Questions For CBSE Class 9 Science Atoms And Molecules provided by Vedantu  to understand the concepts for solving numerical problems.

CBSE Class 9 Science Important Questions

Cbse study materials.

Case Study Questions Class 9 Science The Fundamental Unit of Life

Case study questions class 9 science chapter 5 the fundamental unit of life.

CBSE Class 9 Case Study Questions Science The Fundamental Unit of Life. Important Case Study Questions for Class 9 Exam. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions The Fundamental Unit of Life.

CBSE Case Study Questions Class 9 Science – The Fundamental Unit of Life

All living Organisms are made up of cells and these cells perform all the functions essential for the survival of the Organism eg. Respiration, digestion, excretion etc. In Unicellular organisms, a single cell carries out all these functions and in multicellular organisms different group of cells carry out different functions.

(1) Who discovered the cell?

(d) T. Schwann

(3) Who coined the term ‘Protoplasm’?

(b) Cell organelle

(c) Fluid substance of the cell.

(b) Cell is the basic unit of life

(a) British zoologist

(b) German zoologist

(d) Antony Von Leeuwenhoek

(c) Rudolph Virchow

(a) Proteins

(4) Why cell membrane is known as selectively permeable membrane?

(a) Cell membrane

Statement 2 – Cell wall is mainly composed of cellulose.

(a) Protoplasm

(a) Cisternae, tubules and vesicles.

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