NCERT Solutions Class 9 Science Chapter 3

NCERT Solutions Class 9 Science Chapter 3: Atoms and Molecules

Science is an exciting subject and equally challenging. It plays a vital role in students' lives, and Class 9 is essential for their academic growth. It prepares them for the upcoming classes as well as opens many opportunities. Students need to have a strategic approach towards learning Science to understand  the concepts and new information added in the chapter.  

Class 9 Science Chapter 3: Atoms and Molecules  educates students about the formation of atoms and the laws of chemical combinations. Some exciting topics include molecules of elements, the mole concept, calculating simple compounds and molecular mass, etc.

Students can visit the Extramarks website and access NCERT Solutions for Class 9 Science Chapter 3. The subject matter experts appointed by the Extramarks prepare the Solutions. It allows students to prepare for their  examinations with theoretical notes, diagrams ,  examples and previous years' question papers. Students can understand the topics once they follow the systematic approach and regular practice to master the topic . 

Extramarks is one of the most trusted online learning platforms among students as they offer the best NCERT Solutions and learning materials. The motive is to help students develop the essential skills for learning and don’t need to look for answers elsewhere. That’s the reason why Extramarks Solutions is trusted and used by children and their numbers have been growing steadily. s.

Students can also visit the Extramarks website for the latest  syllabus update and related news. Besides, students can also refer to other primary and secondary class Solutions, including NCERT Solutions Class 10, NCERT Solutions Class 11 and NCERT Solutions Class 12. 

Key Topics Covered In NCERT Solutions Class 9 Science Chapter 3:

NCERT Solutions Class 9 Science Chapter 3 introduces new concepts of atoms and molecules. Students will learn more about protons, neutrons, and electrons. They will explore the formation of the compound and chemical formulas. 

Some of the key topics featured in NCERT Solutions Class 9 Science Chapter 3 are as follows: 

3.1 Law of Chemical Combination:

Atoms and molecules are essential for the formation of everything in our universe. From tiny sand particles to substantial black holes, everything consists of  atoms and  molecules. An atom is a unit of matter, makes  up everything we see around us. It is small and measures less than 0.1 to 0.5 nanometers. 

Chemicals Reactions

  • Two or more molecules interact in a chemical reaction to produce new compounds called reactants. 
  • A chemical change must occur in a chemical reaction, generally observed with physical changes like precipitation, heat production, and colour change.

Law of conservation of mass

  • The law of conservation of mass states that matter cannot be created or destroyed by chemical reactions. Although, it can be changed and preserved.
  • The mass of reactants will always be equal to the mass of products.

Law of constant proportions:

  • The law of definite proportions states that a pure chemical compound contains all the elements in the same proportion.

For Example, water taken from a river or an ocean will have the same amount of oxygen and hydrogen.

Students can get NCERT Solutions Class 9 Science Chapter 3, Revision Notes, Important Questions, etc., to understand the chemical reactions and laws of conservation of mass and constant proportion.

Atoms

Atoms are responsible for the formation of tiny particles. The following are the fundamentals of atoms elaborated in NCERT Solutions Class 9 Science Chapter 3: 

  • An atom is the fundamental structure of an element and cannot be altered by chemical means.
  • The three-part atomic symbol is Z^A X.
  • The symbol X is the symbol for an element.
  • The atomic number A is equal to the number of protons.
  • The mass number Z equals the number of protons or neutrons contained in an element.

3.2 Dalton's Atomic Theory:

The critical elements of Dalton's Atomic Theory explained in NCERT Solutions Class 9 Science Chapter 3 are as follows: 

  • All the properties of an element's atoms are identical, even their mass. This is also true if all the atoms in an element have identical mass and chemical properties, and atoms from different elements have different masses and chemical properties.
  • The atoms can form compounds from different elements combined in fixed proportions.
  • Atoms can't be created or destroyed. Rearranging existing atoms in chemical reactions results in the formation of new products (compounds).
  • A compound's relative number and types of atoms are constant.

3.3 Atomic Mass:

Atomic mass and Atomic Mass Unit:

The characteristics of atomic mass in NCERT Solutions Class 9 Science Chapter 3 are: 

  • The total mass of all electrons, neutrons and protons within an atom or group of atoms is the atomic mass.
  • This is often expressed in the form of an international agreement using the unified atomic weight unit (AMU).
  • It is 1/12th of the mass in the ground state of a carbon-12-atom.

Molecular mass:

  • The sum of all elements in a molecule is called the molecular mass.
  • The atomic mass of an element is multiplied by the number of atoms in a molecule, and then the mass of all elements in the molecule is calculated.

3.4 Mole Concept: 

  • A substance's number of entities, such as atoms, molecules, and ions, is defined as a mole. 
  • In other words, a mole is a combination of atoms, molecules and ions. 
  • A mole is any substance that has a dimension of 6.022x10 23 moles.
  • The mole concept is one way to express the reactants and products in a reaction.
  • Avogadro's number has a value of approximately 6.022x10. Avogadro number refers to the number of particles found in a substance's mole (or Mol). These particles could be electrons, molecules, or atoms.

3.5 Molar Mass: 

A substance is a thing that has mass and occupies space. The molar mass/molecular mass is the sum of all the atoms that make up a mole of a molecule. Grams/mole is the unit of molar weight.

3.6 Atomic Valency: 

Molecules and Atomicity

A molecule is the smallest unit in a compound with chemical properties. An element's atomicity is the number of elements in a single molecule. 

For Example, hydrogen, oxygen, chlorine and iodine all have two atoms within their molecules. Further, the atomicity of hydrogen and nitrogen, oxygen, chlorine, oxygen, chlorine, iodine and bromine are two.

Structure of atom

  • The three components of an Atom are electron, proton, and neutron.
  • The nucleus is the centre of an atom. The entire mass of an atom is contained in the nucleus.
  • The shells and orbitals are used to arrange electrons in an atom.

Valency

These electrons are found in the outermost orbits of an atom and are called valence electrons. The atom's valency is determined by its ability to gain, lose or share valence electrons to complete its octet.

To learn more about the valency properties, students can visit the Extramarks website and access NCERT Solutions Class 9 Science Chapter 3. The academics team at Extramarks have elaborated valency properties with pictorial representation. 

3.7 Writing Chemical Formulae: 

Compound:

  • A compound is formed in a chemical reaction in which two or more elements are combined in a fixed mass ratio.
  • A compound is a substance that contains two or more types of elements in a fixed proportion of its atoms.

Ions

  • An ion can be defined as an atom, molecule, or other substance that has lost or gained one of its valence electrons. This gives it a net negative or positive charge.
  • An anion is a negatively charged particle, while a cation is a positively charged particle.

Ionic compounds-chemical formula

  • Ionic compounds are chemical compounds that contain ions held together by special bonds called ionic bonds.
  • An Ionic compound has equal amounts of both positive and negative charges.

Example: Calcium chloride is an example of an ionic bond. It is formed by the oppositely charged calcium ions and chloride ions. 

Students can refer to the Extramarks NCERT Solutions Class 9 Science Chapter 3 and other study resources where the subject matter experts have explained the formation of ionic compounds in detail. 

NCERT Solutions Class 9 Science Chapter 3: Exercise &  Solutions

Extramarks NCERT Solutions assist students in their examination preparation. Students can access it   by registering on the website. NCERT Solutions Class 9 Science Chapter 3 offers theoretical explanations and step-by-step answers to the questions. Referring to the Class 9 Science Chapter 3 Solution will help students revise the complete syllabus and score high marks in their Examinations. 

Chapter 3- Atoms and Molecules has numerous questions, including multiple-choice questions, match the following, and true or false questions. Students will also find long answer type and short answer type questions. 

Students will benefit from miscellaneous questions at the end of the NCERT Solutions Class 9 Science Chapter 3 Question Answer. It helps in clearing the concepts and improving students' learning experience. 

Students can click on the links below to access specific questions and their solutions that are covered in NCERT Solutions Class 9 Science Chapter 3: 

  • Chapter 3: Exercise 3.1 Solutions: 4 Questions
  • Chapter 3: Exercise 3.2 Solutions: 4 Questions 
  • Chapter 3: Exercise 3.3 Solutions: 4 Questions 
  • Chapter 3: Exercise 3.4 Solutions: 4 Questions

Students can also explore NCERT Solutions for other  classes on the Extramarks website:

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NCERT Exemplar Class 9 Science

Science has some complex concepts. Therefore, students may find it difficult to understand them. To  grasp these concepts, it is better to solve diverse questions. It helps to build a strong foundation. . Also, it is beneficial for the students to solve questions of  different levels of complexity. 

NCERT exemplar helps students understand the core concepts quickly and easily. It is enriched with high-order thinking questions, conceptual questions, multiple-choice questions, and other different types of Important questions. Questions include fill in the blanks, match the following, and true or false questions. In addition, the exemplar has objective questions, very short-answer types, short answer types and long-type questions for Class 9 Chapter 3.

It is highly recommended for the students who wish to appear for competitive examinations. The NCERT Solutions Class 9 Science Chapter 3 has questions from the exemplar. Here, students can explore various topic-wise questions and their Solutions as well.

Key Features of NCERT Solutions Class 9 Science Chapter 3

Extramarks has experienced faculty and subject matter experts who write appropriate and accurate Solutions for every topic while adhering to the CBSE guidelines. Students who face difficulty solving the questions can refer to the answer to clear any doubts.

The characteristic features of the Extramarks NCERT Solutions Class 9 Science Chapter 3 are:

  • After going through the Class 9 Solutions: atoms, and molecules, students will get a thorough understanding of the topics and develop an interest in learning and mastering the topic with ease.
  • Each Solution is explained in an easy-to-follow language, thus enabling students to understand the topics quickly so that they can study from them independently without any assistance from teachers or parents.
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Q.1 A 0.24 g sample of compound of oxygen and boron was found by analysis to contain 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition of the compound by weight.

Ans-

Mass of the given compound = 0.24 g

Mass of boron = 0.096 g

Mass of oxygen = 0.144g

So,

Percentage of boron by weight = Mass of boron Mass of the compound x 100 = 0.096g 0.24g x 100 = 40 % MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@C702@

Percentage of oxygen by weight = Mass of oxygen Mass of the compound x 100 = 0.144g 0.24g x 100 = 60 % MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@CCCA@

Q.2 When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00g of carbon dioxide is produced. What mass of carbon dioxide will be formed when 3.00 g of carbon is burnt in 50.00 g of oxygen? Which law of chemical combination will govern your answer?

Ans-

Given that,

C + O 2 CO 2 3g 8g 11g MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaiaaboeacaqGGaGaae4kaiaabccacaqGpbWaaSbaaSqaaiaaikdaaeqaaOGaaeiiaiabgkziUkaabccacaqGdbGaae4tamaaBaaaleaacaaIYaaabeaakiaabccaaeaacaqGZaGaae4zaiaabccacaqGGaGaaeioaiaabEgacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabgdacaqGXaGaae4zaaaaaa@4FB8@

The given data shows that in this reaction, 3.0 g of carbon and 8.0 g of oxygen react completely to produce 11.0 g of carbon dioxide.

In the second experiment, excess of oxygen (50 g) is present. When 3.00 g of carbon is burnt in 50.00 g of oxygen, it will use just 8.00 g of oxygen from the total of 50 g of oxygen to produce 11 g of carbon dioxide and the rest of 42 g (50 g – 8 g) of oxygen will be left behind without any change.

The above answer is governed by the following two laws of chemical combination:

1. Law of conservation of mass which states that mass can neither be created nor destroyed in a chemical reaction i.e., total mass of reactants remains same as the total mass of products.
Here, in terms of mass 3g of carbon and 50 g of oxygen combine to give 11 g of carbon dioxide and 42 g of unused oxygen keeping the total mass of combining chemicals by weight constant before and after the reaction.
Total mass of reactants = 3 + 50= 53 g

Total mass of products = 11 + 42 (un-used oxygen) = 53g
Total mass of reactants = total mass of products

Hence, the law of conservation of mass is proved.

2. Law of constant proportions states that in a chemical compound the elements are always present in definite proportions by mass. Here, irrespective of the excess quantity of oxygen available for burning of same weight of carbon, the amount of carbon dioxide will remain same. This means that (50 – 8) = 42 g of oxygen will remain unreacted.

Q.3 What are polyatomic ions? Give examples.

Ans-

Polyatomic ions are the group of atoms carrying a net charge on them. This charge may be positive or negative.

Examples:

Ammonium ion (NH4+), Nitrate ion (NO3), Carbonate ion (CO32–), Sulphate ion (SO42–)

Q.4 Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

Ans-

(e) Calcium carbonate

(a) Magnesium chloride: MgCl2

(b) Calcium oxide: CaO

(c) Copper nitrate: Cu(NO3)2

(d) Aluminium chloride: AlCl3

(e) Calcium carbonate: CaCO3

Q.5 Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Ans-

Compound

Elements present

(a) Quick lime (CaO)

Carbon, Oxygen

(b) Hydrogen bromide (HBr)

Hydrogen, Bromine

(c) Baking powder (NaHCO3)

Sodium, Hydrogen, Carbon, Oxygen

(d) Potassium sulphate (K2SO4)

Potassium, Sulphur, Oxygen

Q.6 Calculate the molar mass of the following substances.

(a) Ethyne, C2H2

(b) Sulphur molecule, S8

(c) Phosphorus molecule, P4 (Atomic mass of phosphorus = 31)

(d) Hydrochloric acid, HCl

(e) Nitric acid, HNO3

Ans-

(a) Gram atomic mass of carbon = 12 g

Gram atomic mass of hydrogen = 1g

Molar mass of ethyne (C2H2) = 2 x 12 + 2 x 1

= 26 g/mol

(b) Gram atomic mass of sulphur = 32 g

Molar mass of sulphur molecule (S8) = 8 x 32

= 256 g/mol

(c) Gram atomic mass of phosphorus = 31 g

Molar mass of phosphorus molecule (P4) = 4 x 31

= 124 g/mol

(d) Gram atomic mass of hydrogen = 1g

Gram atomic mass of chlorine = 35.5 g

Molar mass of hydrochloric acid (HCl) = 1 + 35.5

= 36.5 g/mol

(e) Gram atomic mass of hydrogen = 1g

Gram atomic mass of nitrogen = 14 g

Gram atomic mass of oxygen = 16 g

Molar mass of nitric acid (HNO3) = 1 + 14 + 3 x 16

= 63 g/mol

Q.7 What is the mass of:

(a) 1 mole of nitrogen atoms?

(b) 4 moles of aluminium atoms (Atomic mass of aluminium = 27)?

(c) 10 moles of sodium sulphite (Na2SO3) ?

Ans-

a. Mass of 1 mole of nitrogen atoms = Atomic mass of nitrogen expressed in grams = 14 g
b. Mass of 4 moles of aluminium atoms = 4 x Atomic mass of aluminium expressed in grams = 4 x 27 = 108 g
c.Molar mass of sodium sulphite (Na2SO3) = (2 x 23 + 32 + 3 x 16) g/mol

= (46 + 32 + 48) g/mol = 126 g/mol Mass of 10 moles of sodium sulphite = 10 mol x Molar mass of sodium sulphite = 10 mol x 126 g/mol = 1260 g MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@729D@

Q.8 Convert into mole:

(a) 12 g of oxygen gas

(b) 20 g of water

(c) 22 g of carbon dioxide

Ans-

(a)

Given mass of oxygen gas = 12 g Molar mass of oxygen gas (O2) = 2 x 16 g/mol = 32 g/mol The number of moles (n) = Given mass Molar mass = 12 g 32 g/mol = 0.375 molMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@FDC2@

Hence, 12 g of oxygen gas = 0.375 mol of oxygen gas

(b)

Given mass of water = 20 g Molar mass of water (H2O) = (2 x 1 + 1 x 16) g/mol = 18 g/mol The number of moles (n) = Given mass Molar mass = 20 g 18 g/mol = 1.11 mol

Hence, 20 g of water = 1.11 mol of water

(c) MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaaKqzagGaae4raiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeyBaiaabggacaqGZbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqG3bGaaeyyaiaabshacaqGLbGaaeOCaiaabccacaqG9aGaaeiiaiaabkdacaqGWaGaaeiiaiaabEgaaOqaaKqzagGaaeytaiaab+gacaqGSbGaaeyyaiaabkhacaqGGaGaaeyBaiaabggacaqGZbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqG3bGaaeyyaiaabshacaqGLbGaaeOCaiaabccacaqGOaGaaeisaiaabkdacaqGpbGaaeykaiaabccacaqG9aGaaeiiaiaabIcacaqGYaGaaeiiaiaabIhacaqGGaGaaeymaiaabccacaqGRaGaaeiiaiaabgdacaqGGaGaaeiEaiaabccacaqGXaGaaeOnaiaabMcacaqGGaGaae4zaiaab+cacaqGTbGaae4BaiaabYgaaOqaaKqzagGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqG9aGaaeiiaiaabgdacaqG4aGaaeiiaiaabEgacaqGVaGaaeyBaiaab+gacaqGSbaakeaajugGbiaabsfacaqGObGaaeyzaiaabccacaqGUbGaaeyDaiaab2gacaqGIbGaaeyzaiaabkhacaqGGaGaae4BaiaabAgacaqGGaGaaeyBaiaab+gacaqGSbGaaeyzaiaabohacaqGGaGaaeikaiaab6gacaqGPaGaaeiiaiaab2dacaqGGaWaaSaaaOqaaKqzagGaae4raiaabMgacaqG2bGaaeyzaiaab6gacaqGGaGaaeyBaiaabggacaqGZbGaae4CaaGcbaqcLbyacaqGnbGaae4BaiaabYgacaqGHbGaaeOCaiaabccacaqGTbGaaeyyaiaabohacaqGZbaaaaGcbaqcLbyacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaab2dacaqGGaWaaSaaaOqaaKqzagGaaeOmaiaabcdacaqGGaGaae4zaaGcbaqcLbyacaqGXaGaaeioaiaabccacaqGNbGaae4laiaab2gacaqGVbGaaeiBaaaacaqGGaGaaeypaiaabccacaqGXaGaaeOlaiaabgdacaqGXaGaaeiiaiaab2gacaqGVbGaaeiBaaaaaa@EFED@

Given mass of carbon dioxide = 22 g Molar mass of carbon dioxide (CO2) = (12 + 2 x 16) g/mol = 44 g/mol The number of moles (n) = Given mass Molar mass = 22 g 44 g/mol = 0.5 mol

Hence, 20 g of carbon dioxide = 0.5 mol of carbon dioxide

Q.9 What is the mass of:

(a) 0.2 mole of oxygen atoms?

(b) 0.5 mole of water molecules?

Ans-

( a ) Mass of 0.2 mol atoms of oxygen = 0.2 mol x Molar mass of oxygen atom = 0.2 mol x 16 g/mol = 3.2 g ( b ) Molar mass of water ( H 2 O ) = (2 x 1 + 16) g/mol = 18 g/mol Mass of 0.5 mol molecules of water = 0.5 mol x Molar mass of water = 0.5 mol x 18 g/mol = 9 g MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaciGacaGaaeqabaWaaqaafaaakqaabeqaamaabmaabaGaaeyyaaGaayjkaiaawMcaaaqaaiaab2eacaqGHbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeimaiaab6cacaqGYaGaaeiiaiaab2gacaqGVbGaaeiBaiaabccacaqGHbGaaeiDaiaab+gacaqGTbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGVbGaaeiEaiaabMhacaqGNbGaaeyzaiaab6gacaqGGaGaaeypaiaabccacaqGWaGaaeOlaiaabkdacaqGGaGaaeyBaiaab+gacaqGSbGaaeiiaiaabIhacaqGGaGaaeytaiaab+gacaqGSbGaaeyyaiaabkhacaqGGaGaaeyBaiaabggacaqGZbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGVbGaaeiEaiaabMhacaqGNbGaaeyzaiaab6gacaqGGaGaaeyyaiaabshacaqGVbGaaeyBaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaab2dacaqGGaGaaeimaiaab6cacaqGYaGaaeiiaiaab2gacaqGVbGaaeiBaiaabccacaqG4bGaaeiiaiaabgdacaqG2aGaaeiiaiaabEgacaqGVaGaaeyBaiaab+gacaqGSbaabaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeypaiaabccacaqGZaGaaeOlaiaabkdacaqGGaGaae4zaaqaamaabmaabaGaaeOyaaGaayjkaiaawMcaaaqaaiaab2eacaqGVbGaaeiBaiaabggacaqGYbGaaeiiaiaab2gacaqGHbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaae4DaiaabggacaqG0bGaaeyzaiaabkhacaqGGaWaaeWaaeaacaqGibWaaSbaaSqaaiaaikdaaeqaaOGaae4taaGaayjkaiaawMcaaiaabccacaqG9aGaaeiiaiaabIcacaqGYaGaaeiiaiaabIhacaqGGaGaaeymaiaabccacaqGRaGaaeiiaiaabgdacaqG2aGaaeykaiaabccacaqGNbGaae4laiaab2gacaqGVbGaaeiBaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeypaiaabccacaqGXaGaaeioaiaabccacaqGNbGaae4laiaab2gacaqGVbGaaeiBaaqaaiaab2eacaqGHbGaae4CaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaaeimaiaab6cacaqG1aGaaeiiaiaab2gacaqGVbGaaeiBaiaabccacaqGTbGaae4BaiaabYgacaqGLbGaae4yaiaabwhacaqGSbGaaeyzaiaabohacaqGGaGaae4BaiaabAgacaqGGaGaae4DaiaabggacaqG0bGaaeyzaiaabkhacaqGGaGaaeypaiaabccacaqGWaGaaeOlaiaabwdacaqGGaGaaeyBaiaab+gacaqGSbGaaeiiaiaabIhacaqGGaGaaeytaiaab+gacaqGSbGaaeyyaiaabkhacaqGGaGaaeyBaiaabggacaqGZbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqG3bGaaeyyaiaabshacaqGLbGaaeOCaaqaaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeypaiaabccacaqGWaGaaeOlaiaabwdacaqGGaGaaeyBaiaab+gacaqGSbGaaeiiaiaabIhacaqGGaGaaeymaiaabIdacaqGGaGaae4zaiaab+cacaqGTbGaae4BaiaabYgaaeaacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaab2dacaqGGaGaaeyoaiaabccacaqGNbGaaeiiaiaabccacaqGGaGaaeiiaiaabccaaaaa@9D04@

Q.10 Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur.

Ans-

Mass of solid sulphur = 16 g

Molar mass of sulphur molecule (S8) = 8 × 32g/mol = 256 g/mol

Avogadro’s number = 6.022 × 1023

Number of moles of S 8 molecules = Given mass of sulphur Molar mass of S 8 = 16 g 256 g/mol = 16 256 mol No . of S 8 molecules in the sample = No . of moles of S 8 x Avogadro’s number = 16 256 x 6 .022 x 10 23 = 3 .76 x 10 22 molecules MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFr0lbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@504E@

Q.11 Calculate the number of aluminium ions present in 0.051 g of aluminium oxide.

Ans-

(Hint: The mass of an ion is the same as that of an atom of the same element. Atomic mass of Al = 27 u)

Mass of aluminium oxide = 0.051 g Molar mass of aluminium oxide ( Al 2 O 3 )=( 2×27+3×16 ) g/mol = 102 g/mol Number of moles of aluminium oxide molecules = Given mass Molar mass = 0.051g 102 g/mol =5.0× 10 4 molMathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaqGnbGaaeyyaiaabohacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGSbGaaeyDaiaab2gacaqGPbGaaeOBaiaabMgacaqG1bGaaeyBaiaabccacaqGVbGaaeiEaiaabMgacaqGKbGaaeyzaiaabccacaqG9aGaaeiiaiaabcdacaqGUaGaaeimaiaabwdacaqGXaGaaeiiaiaabEgaaeaacaqGnbGaae4BaiaabYgacaqGHbGaaeOCaiaabccacaqGTbGaaeyyaiaabohacaqGZbGaaeiiaiaab+gacaqGMbGaaeiiaiaabggacaqGSbGaaeyDaiaab2gacaqGPbGaaeOBaiaabMgacaqG1bGaaeyBaiaabccacaqGVbGaaeiEaiaabMgacaqGKbGaaeyzaiaabccadaqadaqaaiaabgeacaqGSbWaaSbaaSqaaiaaikdaaeqaaOGaae4tamaaBaaaleaacaaIZaaabeaaaOGaayjkaiaawMcaaiabg2da9maabmaabaGaaGOmaiabgEna0kaaikdacaaI3aGaey4kaSIaaG4maiabgEna0kaaigdacaaI2aaacaGLOaGaayzkaaGaaeiiaiaabEgacaqGVaGaaeyBaiaab+gacaqGSbaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caqG9aGaaeiiaiaabgdacaqGWaGaaeOmaiaabccacaqGNbGaae4laiaab2gacaqGVbGaaeiBaaqaaiaab6eacaqG1bGaaeyBaiaabkgacaqGLbGaaeOCaiaabccacaqGVbGaaeOzaiaabccacaqGTbGaae4BaiaabYgacaqGLbGaae4CaiaabccacaqGVbGaaeOzaiaabccacaqGHbGaaeiBaiaabwhacaqGTbGaaeyAaiaab6gacaqGPbGaaeyDaiaab2gacaqGGaGaae4BaiaabIhacaqGPbGaaeizaiaabwgacaqGGaGaaeyBaiaab+gacaqGSbGaaeyzaiaabogacaqG1bGaaeiBaiaabwgacaqGZbGaaeiiaiaab2dadaWcaaqaaiaabEeacaqGPbGaaeODaiaabwgacaqGUbGaaeiiaiaab2gacaqGHbGaae4CaiaabohaaeaacaqGnbGaae4BaiaabYgacaqGHbGaaeOCaiaabccacaqGTbGaaeyyaiaabohacaqGZbaaaaqaaiaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7cqGH9aqpdaWcaaqaaiaaicdacaGGUaGaaGimaiaaiwdacaaIXaGaaGjbVlaabEgaaeaacaaIXaGaaGimaiaaikdacaqGGaGaae4zaiaab+cacaqGTbGaae4BaiaabYgaaaaabaGaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlaaysW7caaMe8UaaGjbVlabg2da9iaaiwdacaGGUaGaaGimaiabgEna0kaaigdacaaIWaWaaWbaaSqabeaacqGHsislcaaI0aaaaOGaaeiiaiaab2gacaqGVbGaaeiBaaaaaa@E1F9@

From the stoichiometry,

Al2O3 ≡ 2 Al3+
∴ 1 mol of Al2O3 contains = 2 mol of Al3+ ions
∴ 5.0 × 10–4 mol Al2O3 contains = 2 × 5.0 × 10–4 mol of Al3+ ions = 10–3 mol of Al3+ ions

So,
Number of Al3+ ions present in 10–3 mol
= 10–3 mol × 6.022 × 1023 ions mol–1
= 6.022 × 1020 ions

Hence, 0.051 g of aluminium oxide (Al2O3) contains 6.022 × 1020 Al3+ ions.

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