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Class 9 Science Case Study Questions Chapter 3 Atoms and Molecules

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Case study Questions in Class 9 Science Chapter 3 are very important to solve for your exam. Class 9 Science Chapter 3 Class 9 Science Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving case study-based questions for Class 9 Science Chapter 3 Atoms and Molecules

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In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Atoms and Molecules Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 9 Science Chapter 3 Atoms and Molecules

Case Study/Passage-Based Questions

Case Study 1: The knowledge of the valencies of various radicals helps us to write the formulae of chemical compounds. The total positive charge on positive ions (cations) is equal to the total negative charge on negative ions (anions) in a molecule. Therefore, in writing the formula of a compound, the positive and negative ions are adjusted in such a way that the total number of positive charges of positive ions (cations) becomes equal to the total number of negative charges of negative ions (anions). There is another simple method for writing the formulae of ionic compounds. In this method, the valencies (or positive or negative charges) of the ions can be ‘crossed over’ to give subscripts. The purpose of crossing over of charges is to find the number of ions required to equalise the number of positive and negative charges.

Element X has two valencies 5 and 3 and Y has valency 2. The elements X and Y are most likely to be respectively (a) copper and sulphur (b) sulphur and iron (c) phosphorus and fluorine (d) nitrogen and iron.

Answer: (d) nitrogen and iron.

The formula of the sulphate of an element X is X 2 (SO 4 ) 3 . The formula of nitride of element X will be (a) X 2 N (b) XN 2 (c) XN (d) X 2 N 3

Answer: (c) XN

The formula of a compound is X 3 Y. The valencies of elements X and Y will be respectively (a) 1 and 3 (b) 3 and 1 (c) 2 and 3 (d) 3 and 2

Answer: (a) 1 and 3

Case Study/Passage Based Questions

Case Study 2: A mole of an atom is a collection of atoms whose total mass is the number of grams equal to the atomic mass. Since an equal number of moles of different elements contain an equal number of atoms it becomes convenient to express the amounts of the elements in terms of moles. A mole represents a definite number of particles viz, atoms, molecules, ions or electrons. This definite number is called the Avogadro number or Avogadro constant which is equal to 6.022 × 1023. Hence a mole represents 6.022 × 1023 particles of the substance. One mole of a substance represents one gram-formula of the substance. One mole of a gas at standard temperature and pressure occupies 22.4 litres.

How many grams of sodium must be taken to get 1 mole of the element? (a) 23 g (b) 35.5 g (c) 63.5 g (d) 46 g

Answer: (a) 23 g

What is the mass in grams of a single atom of chlorine? (Atomic mass of chlorine = 35.5) (a) 6.54 × 10 23 g (b) 5.9 × 10 –23 g (c) 0.0025 g (d) 35.5 g

Answer: (b) 5.9 × 10–23 g

How many number of moles are there in 5.75 g of sodium ? (Atomic mass of sodium = 23) (a) 0.25 (b) 0.5 (c) 1 (d) 2.5

Answer: (a) 0.25

What is the mass in grams of 2.42 mol of zinc? (Atomic mass of Zn = 65.41) (a) 200 g (b) 25 g (c) 85 g (d) 158 g

Answer: (d) 158 g

Case Study 3: According to Dalton’s atomic theory, all matter whether an element, a compound, or a mixture is composed of small particles called atoms which can neither be created nor destroyed during a chemical reaction. Dalton’s theory provides a simple explanation for the laws of chemical combination. He used his theory to explain the law of conservation of masses, the law of constant proportions, and the law of multiple proportions, based on various postulates of the theory. Dalton was the first scientist to use the symbols for the elements in a very specific sense. When he used a symbol for an element he also meant a definite quantity of that element, that is one atom of that element.

Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass? (a) Atoms can neither be created nor destroyed. (b) Each element is composed of extremely small particles called atoms. (c) All the atoms of a given element are identical. (d) During chemical combination, atoms of different elements combine in simple ratios.

Answer: (a) Atoms can neither be created nor destroyed.

Which postulate of Dalton’s atomic theory explains law of definite proportions? (a) Atoms of an element do not change during a chemical reaction. (b) An element consists of atoms having fixed mass and the number and kind of atoms in a given compound is fixed. (c) Different elements have different kind of atoms. (d) Atoms are of various kinds

Answer: (b) An element consists of atoms having fixed mass and the number and kind of atoms in a given compound is fixed.

“If 100 g of calcium carbonate (whether in the form of marble or chalk) is decomposed, 56 g of calcium oxide and 44 g of carbon dioxide are formed.” Which law of chemical combination is illustrated by this statement? (a) Law of constant proportions (b) Law of conservation of mass (c) Law of multiple proportions (d) Law of conservation of energy

Answer: (b) Law of conservation of mass

When 5 g calcium is burnt in 2 g oxygen, 7 g of calcium oxide is produced. When 5 g of calcium is burnt in 20 g of oxygen, then also 7 g of calcium oxide is produced. Which law of chemical combination is being followed? (a) Law of conservation of mass (b) Law of multiple proportions (c) Law of constant proportions (d) No law is being followed.

Answer: (c) Law of constant proportions

Case Study 4: Atoms and molecules are the building blocks of matter. An atom is the smallest unit of an element that retains its chemical properties, while a molecule is a group of two or more atoms held together by chemical bonds. Atoms consist of a positively charged nucleus, which contains protons and neutrons, surrounded by negatively charged electrons in energy levels or shells. The number of protons in an atom determines its atomic number and defines its unique identity as an element. The electrons in an atom occupy specific energy levels, and the outermost shell is known as the valence shell. Atoms gain, lose, or share electrons to achieve a stable electron configuration, forming chemical bonds and giving rise to molecules. Understanding the concept of atoms and molecules is crucial for comprehending various chemical reactions and the composition of substances.

What is the smallest unit of an element that retains its chemical properties? a) Proton b) Electron c) Nucleus d) Atom Answer: d) Atom

What is a group of two or more atoms held together by chemical bonds called? a) Element b) Compound c) Molecule d) Nucleus Answer: c) Molecule

What are the positively charged particles present in the nucleus of an atom called? a) Electrons b) Protons c) Neutrons d) Valence electrons Answer: b) Protons

Which part of an atom contains electrons in energy levels or shells? a) Protons b) Neutrons c) Nucleus d) Valence shell Answer: d) Valence shell

What do atoms do to achieve a stable electron configuration? a) Gain, lose, or share electrons b) Absorb protons c) Increase their atomic number d) Create chemical bonds Answer: a) Gain, lose, or share electrons

Hope the information shed above regarding Case Study and Passage Based Questions for Class 9 Science Chapter 3 Atoms and Molecules with Answers Pdf free download has been useful to an extent. If you have any other queries about the CBSE Class 9 Science Atoms and Molecules Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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Class 9 Science Case Study Questions

Table of Contents

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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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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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Case Study and Passage Based Questions for Class 9 Science Chapter 3 Atoms and Molecules

Last modified on: 2 years ago
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Case Study Questions for Class 9 Science Chapter 3 Atoms and Molecules

In CBSE Class 9 Science Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage based as well. In that, a paragraph will be given, and then questions based on it will be asked.

Here, we have provided case based/passage based questions for Class 9 Science Chapter 3 Atoms and Molecules . Students can practice these questions for their exam.

Case Study/Passage Based Questions

Question 1:

Read the following and answer any four questions from (i) to (iv).

A mole of an atom is a collection of atoms whose total mass is the number of grams equal to the atomic mass. Since equal number of moles of different elements contain an equal number of atoms it becomes convenient to express the amounts of the elements in terms of moles. A mole represents a definite number of particles viz, atoms, molecules, ions or electrons. This definite number is called Avogadro number or Avogadro constant which is equal to 6.022 × 10 23 . Hence a mole represents 6.022 × 10 23 particles of the substance. One mole of substance represents one gram-formula of the substance. One mole of a gas at standard temperature and pressure occupies 22.4 litres.

(i) How many grams of sodium must be taken to get 1 mole of the element ? (a) 23 g (b) 35.5 g (c) 63.5 g (d) 46 g

(ii) What is the mass in grams of a single atom of chlorine? (Atomic mass of chlorine = 35.5) (a) 6.54 × 10 23 g (b) 5.9 × 10 –23 g (c) 0.0025 g (d) 35.5 g

(iii) How many number of moles are there in 5.75 g of sodium ? (Atomic mass of sodium = 23) (a) 0.25 (b) 0.5 (c) 1 (d) 2.5

(iv) What is the mass in grams of 2.42 mol of zinc? (Atomic mass of Zn = 65.41) (a) 200 g (b) 25 g (c) 85 g (d) 158 g

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Case Study Questions Class 9 Science Structure of the Atom

Case study questions class 9 science chapter 4 structure of the atom.

CBSE Class 9 Case Study Questions Science Structure of the Atom. 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 Structure of the Atom.

At Case Study Questions there will given a Paragraph. In where some Questions will made on that respective Case Based Study. There will various types of marks will given 1 marks, 2 marks, 3 marks or 4 marks.

CBSE Case Study Questions Class 9 Science – Structure of the Atom

Dalton’s atomic theory suggested that the atom was indivisible and indestructible. But the discovery of two fundamental particles (electrons and protons) inside the atom, led to the failure of this aspect of Dalton’s atomic theory. It was then considered necessary to know how electrons and protons are arranged within an atom. For explaining this, many scientists proposed various atomic models. J.J. Thomson was the first one to propose a model for the structure of an atom.

J.J. Thomson (1856- 1940) was a British physicist, He was awarded the Nobel Prize in Physics for his work on the discovery of electrons. Thomson proposed the model of an atom to be similar to that of a Christmas pudding. The electrons, in a sphere of positive charge. We can also think of a watermelon, the positive charge in the atom is spread all over like the red edible part of the watermelon, while the electrons are studded in the positively charged sphere, like the seeds in the watermelon. Thomson proposed that: An atom consists of a positively charged sphere and the electrons are embedded in it. The negative and positive charges are equal in magnitude. So, the atom as a whole is electrically neutral.

(1) Identify the correct statement

Statement 1 – Dalton’s atomic theory suggested that the atom was indivisible and indestructible.

Statement 2 – Electrons and protons are present inside the atom.

Statement 3 – J.J. Thomson was the first one to propose a model for the structure of an atom.

Statement 4 – Protons are positively charged particle.

(b) Both 3 & 4

(d) All of the above

(2) According to Dalton’s Atomic Theory, matter consists of indivisible _______

(a) Molecules

(d) Mixtures

(3) Who was the first to propose atomic theory?

(a) J.J. Thomson

(b) John Dalton

(d) Neilsbhore

(4) “Atom is indivisible and indestructible” why this aspect of Dalton’s atomic theory leds to the failure?

(5) Explain the J.J. Thomson’s model for the structure of an atom?

(4) Dalton’s atomic theory suggested that the atom was indivisible and indestructible. But the discovery of two fundamental particles (electrons and protons) inside the atom, led to the failure of this aspect of Dalton’s atomic theory.

(5) Thomson was the first one to propose a model for the structure of an atom:

Postulate 1: An atom consists of a positively charged sphere with electrons embedded in it.

Postulate 2: An atom as a whole is electrically neutral because the negative and positive charges are equal in magnitude

Thomson atomic model is compared to watermelon. Where he considered:

Watermelon seeds as negatively charged particles
The red part of the watermelon as positively charged

Rutherford (1871-1937) was known as the ‘Father’ of nuclear physics. He is famous for his work on radioactivity and the discovery of the nucleus of an atom with the gold foil experiment. Ernest Rutherford was interested in knowing how the electrons are arranged within an atom. Rutherford designed an experiment for this. In this experiment, fast moving alpha (α)-particles were made to fall on a thin gold foil. On the basis of his experiment, Rutherford put forward the nuclear model of an atom, which had the following features:

There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.
The electrons revolve around the nucleus in circular paths.
The size of the nucleus is very small as compared to the size of the atom.

Drawbacks of Rutherford’s model of the atom: The revolution of the electron in a circular orbit is not expected to be stable. Any particle in a circular orbit would undergo acceleration. During acceleration, charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. If this were so, the atom should be highly unstable and hence matter would not exist in the form that we know. We know that atoms are quite stable.

(1) Which of the following scientist was known as the ‘Father of nuclear physics?

(2) Positively charged centre in an atom is termed as

(a) Nucleus

(b) Molecule

(d) Protons

(3) Identify the correct statement

Statement 1 – Positively charged centre in an atom called the nucleus.

Statement 2 – The electrons revolve around the nucleus in circular paths.

Statement 3 – Nearly all the mass of an atom resides in the nucleus.

Statement 4 – The size of the nucleus is very small as compared to the size of the atom.

(4) Write the features of Rutherford’s nuclear model of an atom?

(5) Define Nucleus.

(4) Rutherford put forward the nuclear model of an atom, which had the following features:

(5) There is a positively charged centre in an atom called the nucleus. Nearly all the mass of an atom resides in the nucleus.

Protons are present in the nucleus of an atom. It is the number of protons of an atom, which determines its atomic number. It is denoted by ‘Z’. All atoms of an element have the same atomic number, Z. In fact, elements are defined by the number of protons they possess. For hydrogen, Z = 1, because in hydrogen atom have only one proton is present in the nucleus. Therefore, the atomic number is defined as the total number of protons present in the nucleus of an atom.

The mass of an atom is practically due to protons and neutrons alone. These are present in the nucleus of an atom. Hence protons and neutrons are also called nucleons. Therefore, the mass of an atom resides in its nucleus. For example, mass of carbon is 12 u because it has 6 protons and 6 neutrons, 6 u + 6 u = 12 u. Similarly, the mass of aluminium is 27 u (13 protons+14 neutrons). The mass number is defined as the sum of the total number of protons and neutrons present in the nucleus of an atom. It is denoted by ‘A’.

(1) Atomic number is denoted by

(2) The sum of the total number of protons and neutrons present in the nucleus of an atom.

(a) Atomic number

(b) Mass number

(d) None of the above

(3) Mass number is denoted by

(4) Identify the correct statement

Statement 1 – Protons are present in the nucleus of an atom.

Statement 2 – Atomic number is the number of protons of an atom.

Statement 3 – Atomic number is denoted by ‘Z’.

Statement 4 – The mass of an atom is due to protons and neutrons alone.

(5) Why mass of carbon is 12 u give the reason?

(5) Mass of carbon is 12 u because it has 6 protons and 6 neutrons, 6 u + 6 u = 12 u.

A number of atoms of some elements have the same atomic number but different mass numbers. For example, hydrogen atom, it has three atomic species, namely Protium, Deuterium and Tritium. The atomic number of each one is 1, but the mass number is 1, 2 and 3, respectively. On the basis of these examples, isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers. Therefore, we can say that there are three isotopes of hydrogen atom, namely protium, deuterium and tritium.

Many elements consist of a mixture of isotopes. Each isotope of an element is a pure substance. The chemical properties of isotopes are similar but their physical properties are different.

The mass of an atom of any natural element is taken as the average mass of all the naturally occurring atoms of that element. If an element has no isotopes, then the mass of its atom would be the same as the sum of protons and neutrons in it. But if an element occurs in isotopic forms, then we have to know the percentage of each isotopic form and then the average mass is calculated.

Chemical properties of all the isotopes of an element are the same. Some isotopes have special properties which find them useful in various fields. Such as, an isotope of uranium is used as a fuel in nuclear reactors, isotope of cobalt is used in the treatment of cancer, iodine is used in the treatment of goitre.

(1) The atoms of the same element, having the same atomic number but different mass numbers are termed as __________

(a) Isotopes

(b) Protium

(d) Tritium

(2) Which of the following are the isotopes of hydrogen atom.

(a) Protium

(b) Deuterium

Statement 1 – Chemical properties of all the isotopes of an element are the same.

Statement 2 – Physical properties are different.

Statement 3 – Chemical properties of all the isotopes of an element are different.

Statement 4 – Physical properties are same.

(b)Both 3 & 4

(4) Give any two uses of isotopes.

(5) Define isotopes.

(4) Isotopes have special properties which find them useful in various fields. Such as,

An isotope of uranium is used as a fuel in nuclear reactors,
Isotope of cobalt is used in the treatment of cancer,
Isotope of iodine is used in the treatment of goitre.

(5) Isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers.

In order to overcome the objections raised against Rutherford’s model of the atom, Neil’s Bohr put forward the following postulates about the model of an atom:

Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.
While revolving in discrete orbits the electrons do not radiate energy. These orbits or shells are called energy levels. Energy levels in an atom are shown in Fig. A few energy levels in an atom These orbits or shells are represented by the letters K,L,M,N,… or the numbers, n=1,2,3,4,….

(1) The orbits or shells are represented by

(a) Letters

(b) Numbers

(d) Special symbols

(2) These orbits or shells are called

(a) Energy levels

(b) Discrete orbit

(3) Which of the following book is written by Professor Bohr’s

(a) The Theory of Spectra and Atomic Constitution

(b) Atomic Theory

Statement 1 – The orbits or shells are represented by letters only.

Statement 2 – The orbits or shells are represented by numbers only.

Statement 3 – While revolving in discrete orbits the electrons do not radiate energy.

Statement 4 – Certain special orbits known as discrete orbits of electrons.

(a) Both 1 & 2

(5) Write the postulate of Neil’s Bohr model of an atom?

(5) Neil’s Bohr put forward the following postulates about the model of an atom:

While revolving in discrete orbits the electrons do not radiate energy.

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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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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.

Table of Contents

Class 9 Science: Case Study Questions

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

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.

NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings

NCERT Solutions for Class 9 Science (chemistry) Chapter 1 Matter in Our Surroundings are given below. In these solutions, we have answered all the intext and exercise questions provided in NCERT class 9 science textbook. Class 9 NCERT Solutions Science Chapter 1 provided in this article are strictly based on the CBSE syllabus and curriculum. Students can easily download these solutions in PDF format for free from our app.

Class 9 Science Chapter 1 Textbook Questions and Answers

Intext Questions

Question 1: Which of the following are matter?

Chair, air, love, smell, hate, almonds, thought, cold, cold drink, smell of perfume.

Answer: Chair, air, almonds, and cold-drink are matters.

Explanation: Things that occupy space and have some mass are called matter. Since chair, air, almonds and cold-drink occupy some space and have some mass, so these are matter.

Question 2: Give reasons for the following observation:

The smell of hot sizzling food reaches you several metres away, but to get the smell from cold food you have to go close.

Answer: The smell of hot sizzling food reaches severed meters away, as the particles of hot food have more kinetic energy and hence the rate of diffusion is more than the particles of cold food.

Smell of anything comes because of gases emanating from the given thing. The smell reaches to us because of diffusion of gas. The rate of diffusion increases with increase in temperature. This happens because of higher kinetic energy due to higher temperature. That is why smell of hot sizzling food reaches to us from several feet. On the other hand, the kinetic energy of gases emanating from cold food is low because of lower temperature. Due to this, we need to move closer to a cold food to take its smell.

Question 3: A diver is able to cut through water in a swimming pool. Which property of matter does this observation show?

Answer: A diver is able to cut through water in a swimming pool. This shows that the particles of water have intermolecular space and has less force of attraction.

Question 4: What are the characteristics of particles of matter?

Answer: The characteristics of particles of matter are:

Particles of matter have spaces between them.
Particles of matter are continuously moving.
Particles of mater attract each other.

Question 1: The mass per unit volume of a substance is called density (density = mass/volume). Arrange the following in order of increasing density − air, exhaust from chimney, honey, water, chalk, cotton, and iron.

Answer: The given substances in the increasing order of their densities can be represented as:

Air < Exhaust from chimney < Cotton < Water < Honey < Chalk < Iron

Explanation: Air is the mixture of gases. Chimney exhaust is also a mixture of gases; along with some heavier particles, such as ash. This makes the density of chimney exhaust more than air. Cotton is a porous solid and which has lot of air trapped within pores. This makes its volume more than water. Therefore, it is less dense than water.

Question 2: (a) Tabulate the differences in the characteristics of states of matter. (b) Comment upon the following: rigidity, compressibility, fluidity, filling a gas container, shape, kinetic energy and density.

Answer: (a) The differences in the characteristics of states of matter are given in the following table.

NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings image 1

The difference in the characteristics of the three states of matter.

NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings image 2

(b) Rigidity: The greatest force of attraction between particles and close packing of particles make solids rigid. Rigidity is one of the unique properties of solids. Because of rigidity, a solid can resist from getting distorted. Because of rigidity a solid has definite shape and volume. Rigidity is negligible in fluid and gas.

Compressibility: Compressibility is one of the most important characteristics of gas. Because of lot of space between particles, a gas can be compressed to a great extent.

Liquid and solid cannot be compressed because of the least space between their particles.

Fluidity: The ability to flow is called fluidity. The less force of attraction and more space between particles make liquid and gas to flow. That’s why liquid and gas are called fluid.

Filling of a gas container: Liquids do not fill a gas container completely, while gases fill the gas container completely in which it is kept. This is because the particles of gas can move in all the directions.

Shape: Solids have fixed shape. Liquid and gas take the shape of the container in which they are kept. This happens because of less force of attraction and more kinetic energy between particles of liquids and negligible force of attraction and highest kinetic energy between particles of gas.

Kinetic energy: The kinetic energy of particles of solid is the minimum. They only vibrate at their fixed position. The kinetic energy of particles of liquid is more than that of solid. But they can slide above one another. The kinetic energy of particles of gas is the maximum.

Density: The mass per unit volume of a substance is called density. The density of solid is highest, of liquid is less than solid and of gas is minimum.

Question 3: Give reasons: (a) A gas fills completely the vessel in which it is kept. (b) A gas exerts pressure on the walls of the container. (c) A wooden table should be called a solid. (d) We can easily move our hand in air, but to do the same through a solid block of wood, we need a karate expert.

Answer: (a) There is little attraction between particles of gas. Thus, gas particles move freely in all directions. Therefore, gas completely fills the vessel in which it is kept.

(b) Because of negligible force of attraction between particles of gas, the particles of gas have the highest kinetic energy. These properties enable the particles of gas to move in all directions and hit the walls of container from all sides. Because of this a gas exerts pressure on the walls of the container in which it is kept.

(c) A wooden table has a definite shape and volume. It is very rigid and cannot be compressed i.e., it has the characteristics of a solid. Hence, a wooden table should be called a solid.

(d) Particles of air have large spaces between them. On the other hand, wood has little space between its particles. Also, it is rigid. For this reason, we can easily move our hands in air, but to do the same through a solid block of wood, we need a karate expert.

Since, air is gas, so its particles are loosely packed and there is negligible force of attraction between its particles. Because of that we can easily move our hand in air. But wood is a solid, so the force of attraction between its particles is greatest. The particles of wooden block are closely packed. That’s why we cannot move our hand through a solid block of wood. However, a karate expert can exert required pressure to break the great force of attraction of the particles of a solid wooden block.

Question 4: Liquids generally have lower density as compared to solids. But you must have observed that ice floats on water. Find out why.

Answer: During freezing of water, some space between the particles of water is left vacant with some air trapped between them. These empty spaces having air in them makes the density of ice; lower than that of water. That’s why ice floats on water.

Question 1: Convert the following temperatures into the Celsius scale. (a) 300 K (b) 573 K

Answer: (a) 300 K = (300 − 273)°C = 27°C (b) 573 K = (573 − 273)°C = 300°C

Question 2: What is the physical state of water at (a) 250°C (b) 100°C

Answer: (a) Water at 250°C exists in gaseous state.

(b) At 100°C, water can exist in both liquid and gaseous form. At this temperature, after getting the heat equal to the latent heat of vaporization, water starts changing from liquid state to gaseous state.

Question 3: For any substance, why does the temperature remain constant during the change of state?

Answer: During a change of state, the temperature remains constant. This is because all the heat supplied to increase the temperature is utilized (as latent heat) in changing the state by overcoming the forces of attraction between the particles. Therefore, this heat does not contribute in increasing the temperature of the substance.

Question 4: Suggest a method to liquefy atmospheric gases.

Answer: Atmospheric gas is liquefied by increasing pressure and decreasing temperature.

PAGE NO. 10

Question 1: Why does a desert cooler cool better on a hot dry day?

Answer: Desert cooler works on the basis of evaporation. In hot and dry days the moisture level is very low in atmosphere which increases the rate of evaporation. Because of faster evaporation, cooler works well. That’s why desert cooler cool better on a hot dry day.

When a liquid evaporates, the particles of the liquid absorb energy from the surroundings to compensate the loss of energy during evaporation. This makes the surroundings cool.

In a desert cooler, the water inside it is made to evaporate. This leads to absorption of energy from the surroundings, thereby cooling the surroundings. Again, we know that evaporation depends on the amount of water vapour present in air (humidity). If the amount of water vapour present in air is less, then evaporation is more. On a hot dry day, the amount of water vapour present in air is less. Thus, water present inside the desert cooler evaporates more, thereby cooling the surroundings more. That is why a desert cooler cools better on a hot dry day.

Question 2: How does water kept in an earthen pot (matka) become cool during summers?

Answer: Water from porous wall of earthen pot evaporates continuously, which lowers the temperature of water kept in the earthen pot. In summer moisture level is very low in the atmosphere, which increases the rate of evaporation as evaporation is inversely proportional to the moisture level in atmosphere. That is why in summer water kept in earthen pot becomes cool.

Question 3: Why does our palm feel cold when we put some acetone or petrol or perfume on it?

Answer: When we put some acetone or petrol or perfume on our palm, it evaporates. During evaporation, particles of the liquid absorb energy from the surrounding or the surface of the palm to compensate for the loss of energy, making the surroundings cool. Hence, our palm feels cold when we put some acetone or petrol or perfume on it.

Question 4: Why are we able to sip hot tea or milk faster from a saucer than a cup?

Answer: When hot tea or milk is kept in a saucer, the liquid is exposed over a larger surface area as compared to in case of the liquid being kept in a cup. The larger surface area enables the faster cooling. That’s why we are able to sip hot tea or milk faster from a saucer rather than from a cup.

Question 5: What type of clothes should we wear in summers?

Answer: In summer, it is preferred to wear light-coloured cotton clothes because light colour reflects heat and cotton materials have pores that absorb sweat, facilitating their evaporation hence causing a cooling effect in the skin.

Question 1: Convert the following temperatures into the Celsius scale. (a) 293 K (b) 470 K

Answer: Temperature in Celsius scale = Temperature in Kelvin scale – 273

(a) 293K= (293 – 273)°C = 20°C

(b) 470K= (470 – 273)°C = 197°C

Question 2: Convert the following temperatures into the Kelvin scale. (a) 25°C (b) 373°C

Answer: Temperature in Kelvin scale = Temperature in Celsius scale + 273

(a) 25°C = (25+273)K = 298K

(b) 373°C = (373+273)K = 646K

Question 3: Give reasons for the following observations. (a) Naphthalene balls disappear with time without leaving any solid. (b) We can get the smell of perfume while sitting several metres away.

Answer: (a) At room temperature, naphthalene balls undergo sublimation wherein they directly get converted from a solid to a gaseous state without having to undergo the intermediate state, i.e., the liquid state.

(b) Perfumes vaporize very fast and its vapours diffuse into air easily. That is why we can smell perfume sitting several meters away.

Question 4: Arrange the following in increasing order of forces of attraction between the particles – water, sugar, oxygen.

Answer: Oxygen < Water < Sugar.

Explanation: Oxygen is a gas, thus force of attraction is negligible between particles. Water is a liquid, thus force of attraction between particles is more than liquid and less than solid. Sugar is a solid, thus force of attraction between particles is greatest.

Question 5: What is the physical state of water at — (a) 25°C (b) 0°C (c) 100°C?

Answer: (a) At 25°C – water is in liquid state. (b) At 0°C – water is in solid state. (c) At 100°C – water is in transition state, i.e. in liquid and gas both.

Question 6: Give two reasons to justify: (a) water at room temperature is a liquid. (b) an iron almirah is a solid at room temperature.

Answer: (a) At room temperature (25 °C), water is a liquid because it has the following characteristic of liquid:

(i) Water has definite volume, but not definite shape as it takes the shape of the container in which it is kept. (ii) Water flows at room temperature.

(b) An iron almirah is a solid at room temperature because: (i) It has definite shape. (ii) It has definite volume.

Question 7: Why is ice at 273 K more effective in cooling than water at the same temperature?

Answer: At 273K ice requires more latent heat to melt into water, while water at 273K requires less latent heat; to come to the room temperature. So, ice at 273 K is more effective in cooling than water at the same temperature.

Question 8: What produces more severe burns, boiling water or steam?

Answer: Steam produces more severe burns than boiling water. This is because steam has more energy than boiling water, present in it in the form of latent heat of vaporization.

Question 9: Name A, B, C, D, E and F in the following diagram showing change in its state:

NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings image 3

Answer: A: Melting (or) fusion (or) liquefaction B: Evaporation (or) vaporization C: Condensation D: Solidification E: Sublimation F: Sublimation

Class 9 Science NCERT Solutions Chapter 1 Matter in Our Surroundings

CBSE Class 9 Science NCERT Solutions Chapter 1 helps students to clear their doubts and to score good marks in the board exam. All the questions are solved by experts with a detailed explanation that will help students complete their assignments & homework. Having a good grasp over CBSE NCERT Solutions for Class 9 Science will further help the students in their preparation for board exams and other competitive exams such as NTSE, Olympiad, etc.

NCERT Solutions for Class 9 Science Chapter 1 PDF

Below we have listed the topics discussed in NCERT Solutions for Class 9 Science Chapter 1. The list gives you a quick look at the different topics and subtopics of this chapter.

Class 9 Chemistry Chapter 4 - Structure of the Atom Important Questions with Answers

Class 9 chemistry important questions with answers are provided here for Chapter 4 – Structure of the Atom. These important questions are based on CBSE board curriculum and correspond to the most recent Class 9 chemistry syllabus. By practising these Class 9 important questions, students will be able to quickly review all of the ideas covered in the chapter and prepare for the Class 9 Annual examinations.

Download Class 9 Chemistry Chapter 4 – Structure of the Atom Important Questions with Answers PDF by clicking on the button below. Download PDF

Class 9 Chemistry Chapter 4 – Structure of the Atom Important Questions with Answers

Short answer type questions.

Q1: Is it possible for the atom of an element to have one electron, one proton and no neutron. If so, name the element.

Yes, this is true for hydrogen’s common stable isotope (protium), which is denoted by the letter $\begin{array}{l}_{1}^{1}\textrm{H}\end{array} $ .

Q2: Write any two observations which support the fact that atoms are divisible.

The discovery of electrons and protons serves to strengthen the idea that atoms are divisible. During a chemical reaction, electrons are transferred or shared between distinct atoms, leading atoms to rearrange.

Q3: Will 35 Cl and 37 Cl have different valencies? Justify your answer.

The number of protons or electrons is equal to the valency. Since both chlorine 37 and chlorine 35 have the same number of protons and electrons, they have the same valency.

Chlorine 37 and chlorine 35 are identical isotopes. As a result, they have the same atomic number but have different masses. Their mass numbers differ due to the different amount of neutrons in each of them.

Q4: Why did Rutherford select a gold foil in his α–ray scattering experiment?

Rutherford wanted a metal sheet that could be as thin as possible for the scattering experiment. Gold is the most malleable of all the metals known to man. It’s simple to make thin sheets out of it. As a result, for his alpha-ray scattering experiment, Rutherford used gold foil.

Q5: Find out the valency of the atoms represented by the Fig. 4.3 (a) and (b).

Atom (a) has zero valency because it has established a stable configuration with 8 electrons in the valence shell.

The valency of atom (b) is 1 since the valence shell contains 7 electrons. In order to obtain a stable (octet) configuration, atom (b) can take one additional electron.

Q6: One electron is present in the outer most shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outer most shell?

When an electron is removed from the outermost shell, a cation is formed, and the element’s charge is increased to +1.

When a single electron from the outermost shell of an element X is removed, the atom loses its negative charge and becomes a positively charged ion with a charge of +1.

The net charge on the ion will be equal to the amount of charge present on one electron.

Q7: Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17).

Chlorine has an atomic number of 17. As a result, the innermost shell contains 2 electrons, the second shell has 8 electrons, and the outermost shell has 7 electrons.

K shell – 2 electrons

L shell – 8 electrons

M shell – 7 electrons

The electron configuration of chlorine can be written as 1s 2 2s 2 2p 6 3s 2 3p 5 or as [Ne]3s 2 3p 5 .

Q8: In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?

The electronic configuration of a noble gas is 2,8. It possesses 8 electrons in its outermost orbitals. Element X contains 6 electrons in its outermost shell. Two more electrons are required to obtain the noble gas configuration. As a result, in order to achieve a noble gas configuration, the element should take two electrons, resulting in a charge of -2.

Q9: What information do you get from the Fig. 4.4 about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form.

Atomic number, mass number and valency of atoms X, Y and Z.

Q10: In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer.

The provided statement is not correct.

According to this comment p > n > e.

The number of protons, on the other hand, is never greater than the number of neutrons.

Because mass number is equal to double the atomic number or greater than double the atomic number, the number of neutrons might be equal to or greater than the number of protons.

Because the number of electrons equals the number of protons in the neutral atom, the number of neutrons can be more than the number of electrons.

Q11: Calculate the number of neutrons present in the nucleus of an element X which is represented as 31 X 15 .

Mass number (A) = No. of protons (Z) + No. of neutrons

Given, the mass number as 31 and the number of protons is 15.

No. of protons (Z) + No. of neutrons = 31

∴ Number of neutrons = 31 – number of protons

⇒ Number of neutrons = 31 – 15 = 16 .

Q12: Match the names of the Scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B.

(a) Ernest Rutherford – (iii) Concept of the nucleus

(b) J.J.Thomson – (iv) Discovery of electrons

(d) Neils Bohr – (ii) Stationary orbits

(e) James Chadwick – (vi) Neutron

(f) E. Goldstein – (vii) Canal rays

(g) Mosley – (v) Atomic number

Q13: The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements?

They are called isobars.

Isobars have different atomic numbers (or protons), but the same mass number.

Q14: Complete Table 4.1 on the basis of information available in the symbols given below.

(a) 35 17 Cl

Q15: Helium atom has 2 electrons in its valence shell but its valency is not 2, Explain.

Electronic arrangement of a helium atom is 1s 2 which contains two valence electrons, but because it does not form a chemical bond, its duplet is complete. As a result, valency is zero.

Q16: Fill in the blanks in the following statements:

(a) Rutherford’s α-particle scattering experiment led to the discovery of the _______.

(b) Isotopes have same ________ but different________.

(c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be _______ and ________ respectively.

(d) The electronic configuration of silicon is _______ and that of sulphur is ________.

a) Rutherford’s α-particle scattering experiment led to the discovery of the atomic nucleus .

b) Isotopes have same atomic number but different mass number .

c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be 0 and 1 respectively.

d) The electronic configuration of silicon is 2.8.4 and that of sulphur is 2.8.6

Q17: An element X has a mass number 4 and atomic number 2. Write the valency of this element.

Mass number = 4

Atomic number = 2

X is Helium.

It has 0 valency and it would not react with any other atom because it has its outer shell filled.

Long Answer Type Questions

Q1: Why do Helium, Neon and Argon have a zero valency?

An element’s valency is its ability to combine. The valency of noble gases is zero since there are no free electrons in the valence shell and the elements are already in a stable state.

The outermost orbit of helium has two electrons, filling shell 1 and generating a duplet arrangement in the valence shell.
The valence orbit of neon has eight electrons, completing the duplet structure.
The outermost layer of Argon and Neon has 8 electrons, completing the octet arrangement.

Because these elements have the most electrons in their valence shell, they have a stable electron configuration and do not participate in chemical reactions.

Q2: The ratio of the radii of hydrogen atom and its nucleus is ~ 10 5 . Assuming the atom and the nucleus to be spherical,

(i) What will be the ratio of their sizes?

(ii) If atom is represented by planet earth ‘Re’ = 6.4 ×10 6 m, estimate the size of the nucleus.

Assuming the atom and the nucleus to be spherical,

(i) Atomic size is represented in terms of atomic radius

Volume of the sphere = $\begin{array}{l}\frac{4}{3}\pi r^{3}\end{array} $ $\begin{array}{l}V_{H} = \frac{4}{3}\pi r_{H}^{3}\end{array} $ ………………….. (1)

Ratio of volumes

$\begin{array}{l}\frac{V_{H}}{V_{n}} = 10^{15}\end{array} $

(ii) If an atom is represented by planet earth ‘Re’ = 6.4×10 6 m, then the radius of the nucleus would be r n .

$\begin{array}{l}r_{n} = 64m\end{array} $

Q3: Enlist the conclusions drawn by Rutherford from his α-ray scattering experiment.

The α-particle scattering experiment led Rutherford to the following conclusions:–

(i) Because most α-particles travelled through the gold foil with no deflection, most of the space inside the atom was unoccupied.

(ii) The positive charge occupied very little space inside the atom, as just a few particles were deviated from their route.

(iii) Only a small percentage of α-particles were deflected by 180°, implying that the positive charge and mass of the gold atom are concentrated in a compact volume within the atom.

(iv) He also computed that the radius of the nucleus was approximately 105 times smaller than the radius of the atom based on the data.

Q4: In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?

Rutherford proposed a model in which the electrons moved in well-defined orbits around the nucleus. An atom has a positively charged centre (later named the “nucleus”). He also proposed that the nucleus is very small in comparison to the size of the atom, and that the nucleus contains nearly all of an atom’s mass.

Thomson recommended that the atom be modelled after a Christmas pudding. The electrons are scattered about like currants in a positively charged pudding spherical, and the atom’s mass was expected to be evenly distributed.

Q5: What were the drawbacks of Rutherford’s model of an atom?

Rutherford carried out an experiment in which he bombarded a thin sheet of gold with -particles and then analysed the trajectory of the particles after they collided with the gold foil.

Rutherford’s atomic model could not explain how the moving electrons were able to maintain their orbit.
Any charged particle would give forth energy during acceleration, lose energy while revolving, and eventually fall into the nucleus.

Q6: What are the postulates of Bohr’s model of an atom?

The following are the postulates of Bohr’s atom model:

1. Electrons circle the nucleus of an atom in a fixed path known as the orbit or stationary state of the shell.

2. The shells have varying energy levels, represented by the letters K, L, M, and N.

3. The electron does not absorb or release energy as long as it is in an orbit.

4. The electron can only move in orbits where angular momentum is quantized, that is, where the electron’s angular momentum is an integral multiple of h/2𝜋 .

Q7: Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.

The atomic number (Z) of an element is equal to the number of protons in its atom.

Atomic number of sodium (Z) = 11 and Mass number of sodium (A) = 23

Number of protons in the nucleus = 11

Number of neutrons in the nucleus = 23 − 11 = 12

Number of electrons = 11

Thus, electronic configuration of Na-atom = 2, 8, 1(K,L,M)

An electron is lost from the sodium atom(present in the outermost shell), resulting in the formation of the Na+ ion.

Hence, the electronic configuration is 2, 8(K, L). However, the number of protons and neutrons remains the same.

Q8: In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the α-particles were found to deflect at angles > 50°. If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 50°.

% of α-particles deflected more than 50° = 1% of α-particles.

∴ % of α-particles deflected less than 50° = 100 – 1 = 99%

Number of α-particles bombarded = 1 mole = 6.022 × 10 23 particles

∴ Number of particles that deflected at an angle less than 50° = 99/100 × 6.022 × 10 23

= 5.96 × 10 23

CBSE Class 9 Science Chapter 4 MCQs

1. _______discovered the electron.

(a) Chadwick

(b) Rutherford

(d) Goldstein

Answer: (c) Thomson

2. What property of an element determines its chemical behaviour?

(a) Valency of an element

(b) Size of an element

(d) None of the above

Answer: (a) Valency of an element

3. _______ is the radioactive element used in the treatment of cancer.

(a) Iodine-131

(b) Uranium-234

(d) Cobalt-60

Answer: (d) Cobalt-60

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