Question 1

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

Accepted Answer

Given,C 3 g+O2 8 g⟶CO211 g\underset{	ext{ 3 g}}{	ext{C}} + \underset{	ext{ 8 g}}{	ext{O}_2} \longrightarrow \underset{	ext{11 g}}{	ext{CO}_2} 3 gC​+ 8 gO2​​⟶11 gCO2​​When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is produced.Now,C 3 g+O2 50 g⟶CO2?\underset{	ext{ 3 g}}{	ext{C}} + \underset{	ext{ 50 g}}{	ext{O}_2} \longrightarrow \underset{	ext{?}}{	ext{CO}_2} 3 gC​+ 50 gO2​​⟶?CO2​​Carbon dioxide contains carbon and oxygen in a fixed ratio by mass, which is 3:8. Hence, 3 g of carbon will react with 50 g of oxygen but only 8 g of it will be used in producing 11 g of CO2 and 42 g [i.e., 50 g - 8 g] of oxygen will be left unused.The above answer is governed by the law of constant proportions.

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 3: Atoms and MoleculesWeb Content &mdash; Quick Reference
  
    Chapter 3: Atoms and Molecules &mdash; Quick Reference

Dalton theory
      atomic mass
      molecular mass
      mole concept
      chemical formula
      valency
      Avogadro number

Quick Revision Points
      
        Law of conservation of mass: total mass unchanged in a reaction
Law of constant proportions: elements in fixed mass ratio in a compound
1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
Molecular mass = sum of atomic masses of all atoms in molecule
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Moles = mass / molar mass; Particles = moles × Nₐ
Criss-cross method for writing chemical formulae using valencies
Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻

Exam Tips for Chapter 3
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch03-atoms-and-molecules.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

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Question 2

Question 4Write the chemical formula of the following.(a) Magnesium chloride(b) Calcium oxide(c) Copper nitrate(d) Aluminium chloride(e) Calcium carbonate

Accepted Answer

(a) Magnesium chlorideStep 1 — Write each symbol with its valencyMg2+↗Cl1−\text{Mg}^{2+} \phantom{\nearrow} \text{Cl}^{1-}Mg2+↗Cl1−Step 2 — Interchange the valenciesMg22 ↘↙ Cl1⇒Mg11 ↘↙ Cl2\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{Cl}} \[0.5em]Mg22​ ↘↙ Cl1⇒11Mg​ ↘↙ 2Cl​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Magnesium chloride : MgCl2\bold{MgCl_2}MgCl2​(b) Calcium oxideStep 1 — Write each symbol with its valencyCa2+↗O2−\text{Ca}^{2+} \phantom{\nearrow} \text{O}^{2-}Ca2+↗O2−Step 2 — Interchange the valenciesCa22 ↘↙ O2⇒Ca12 ↘↙ O2\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}} \[0.5em]Ca22 ↘↙ O2⇒12Ca​ ↘↙ 2O​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Calcium oxide : CaO\bold{CaO}CaO(c) Copper nitrateStep 1 — Write each symbol with its valencyCu2+↗NO31−\text{Cu}^{2+} \phantom{\nearrow} \text{NO}_3^{1-}Cu2+↗NO31−​Step 2 — Interchange the valenciesCu22 ↘↙ NO31⇒Cu11 ↘↙ NO32\overset{\phantom{2}{2}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{NO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Cu}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{NO}_3} \[0.5em]Cu22 ↘↙ NO3​1​⇒11Cu​ ↘↙ 2NO3​​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of copper nitrate : Cu(NO3)2\bold{Cu(NO_3)_2}Cu(NO3​)2​(d) Aluminium chlorideStep 1 — Write each symbol with its valencyAl3+↗Cl1−\text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-}Al3+↗Cl1−Step 2 — Interchange the valenciesAl23 ↘↙ Cl1⇒Al11 ↘↙ Cl3\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \[0.5em]Al23 ↘↙ Cl1⇒11Al​ ↘↙ 3Cl​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Aluminium chloride : AlCl3\bold{AlCl_3}AlCl3​(e) Calcium carbonate –Step 1 — Write each symbol with its valencyCa2+↗CO31−\text{Ca}^{2+} \phantom{\nearrow} \text{CO}_3^{1-}Ca2+↗CO31−​Step 2 — Interchange the valenciesCa22 ↘↙ CO31⇒Ca11 ↘↙ CO32\overset{\phantom{2}{2}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{CO}_3} \Rightarrow \underset{\phantom{1}{1}}{\text{Ca}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{CO}_3} \[0.5em]Ca22 ↘↙ CO3​1​⇒11Ca​ ↘↙ 2CO3​​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of copper nitrate : Ca(CO3)2\bold{Ca(CO_3)_2}Ca(CO3​)2​

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 3: Atoms and MoleculesWeb Content &mdash; Quick Reference
  
    Chapter 3: Atoms and Molecules &mdash; Quick Reference

Dalton theory
      atomic mass
      molecular mass
      mole concept
      chemical formula
      valency
      Avogadro number

Quick Revision Points
      
        Law of conservation of mass: total mass unchanged in a reaction
Law of constant proportions: elements in fixed mass ratio in a compound
1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
Molecular mass = sum of atomic masses of all atoms in molecule
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Moles = mass / molar mass; Particles = moles × Nₐ
Criss-cross method for writing chemical formulae using valencies
Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻

Exam Tips for Chapter 3
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch03-atoms-and-molecules.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 3

Question 5Give the names of the elements present in the following compounds.(a) Quick lime(b) Hydrogen bromide(c) Baking powder(d) Potassium sulphate

Accepted Answer

(a) Quick lime (CaO) — Calcium and oxygen(b) Hydrogen bromide (HBr) — Hydrogen and bromine(c) Baking powder (NaHCO3) — Sodium, hydrogen, carbon and oxygen(d) Potassium sulphate (K2SO4) — Potassium, sulphur and oxygen,

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 3: Atoms and MoleculesWeb Content &mdash; Quick Reference
  
    Chapter 3: Atoms and Molecules &mdash; Quick Reference

Dalton theory
      atomic mass
      molecular mass
      mole concept
      chemical formula
      valency
      Avogadro number

Quick Revision Points
      
        Law of conservation of mass: total mass unchanged in a reaction
Law of constant proportions: elements in fixed mass ratio in a compound
1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
Molecular mass = sum of atomic masses of all atoms in molecule
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Moles = mass / molar mass; Particles = moles × Nₐ
Criss-cross method for writing chemical formulae using valencies
Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻

Exam Tips for Chapter 3
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch03-atoms-and-molecules.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 4

Question 6Calculate 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

Accepted Answer

(a) Molar mass of Ethyne C2H2 = (2 x Mass of C) + (2 x Mass of H) = (2 × 12) + (2 × 1) = 24 + 2 = 26 g(b) Molar mass of Sulphur molecule S8 = (8 x Mass of S) = 8 x 32 = 256 g(c) Molar mass of Phosphorus molecule, P4 = 4 x Mass of P = 4 x 31 = 124 g(d) Molar mass of Hydrochloric acid, HCl = Mass of H + Mass of Cl = 1 + 35.5 = 36.5 g(e) Molar mass of Nitric acid, HNO3 = Mass of H + Mass of N + (3 x Mass of O) = 1 + 14 + (3 × 16) = 15 + 48 = 63 g

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 3: Atoms and MoleculesWeb Content &mdash; Quick Reference
  
    Chapter 3: Atoms and Molecules &mdash; Quick Reference

Dalton theory
      atomic mass
      molecular mass
      mole concept
      chemical formula
      valency
      Avogadro number

Quick Revision Points
      
        Law of conservation of mass: total mass unchanged in a reaction
Law of constant proportions: elements in fixed mass ratio in a compound
1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
Molecular mass = sum of atomic masses of all atoms in molecule
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Moles = mass / molar mass; Particles = moles × Nₐ
Criss-cross method for writing chemical formulae using valencies
Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻

Exam Tips for Chapter 3
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch03-atoms-and-molecules.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 5

Question 3Which postulate of Dalton's atomic theory is the result of the law of conservation of mass?

Accepted Answer

One of the postulates of Dalton's atomic theory states that : 'Atoms cannot be created nor be destroyed in a chemical reaction'. This law is the result of law of conservation of mass.

Source: This question is from Atoms and Molecules, Science &mdash; Class 9, CBSE Board.
Key Concepts CoveredThis question tests your understanding of the following concepts from the chapter Atoms and Molecules: Question, Postulate, Dalton&#x27;s, Atomic, Theory, Result. These are fundamental topics in Science that students are expected to master as part of the CBSE Class 9 curriculum.A thorough understanding of these concepts will help you answer similar questions confidently in your CBSE examinations. These topics are frequently tested in both objective and subjective sections of Science papers. We recommend revising the relevant section of your textbook alongside practising these solved examples to build a strong foundation.
How to Approach This QuestionRead the question carefully and identify what is being asked. Break down complex questions into smaller parts. Use the terminology and concepts discussed in this chapter. Structure your answer logically — begin with a definition or key statement, then provide supporting details. Review your answer to ensure it addresses all parts of the question completely.
Key Points to RememberRead the question carefully and identify all parts before answering.
Use the terminology specific to this subject and chapter.
Review the textbook content for this chapter before attempting questions.
Practice writing concise, well-structured answers within time limits.
Related Questions from Atoms and MoleculesQ1: Question 1 Write down the formulae of (i) sodium oxide (ii) aluminium chloride (iii) sodium...
Q2: Question 2 When 3.0 g of carbon is burnt in 8.00 g of oxygen, 11.00 g of carbon dioxide is...
Q4: Question 4 Write the chemical formula of the following. (a) Magnesium chloride (b) Calcium oxide...
Q5: Question 5 Give the names of the elements present in the following compounds. (a) Quick lime (b)...
Q6: Question 6 Calculate the molar mass of the following substances. (a) Ethyne, C 2 H 2 (b) Sulphur...
Practice more questions from Atoms and Molecules &mdash; Science, Class 9 CBSE

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 3: Atoms and MoleculesWeb Content &mdash; Quick Reference
  
    Chapter 3: Atoms and Molecules &mdash; Quick Reference

Dalton theory
      atomic mass
      molecular mass
      mole concept
      chemical formula
      valency
      Avogadro number

Quick Revision Points
      
        Law of conservation of mass: total mass unchanged in a reaction
Law of constant proportions: elements in fixed mass ratio in a compound
1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
Molecular mass = sum of atomic masses of all atoms in molecule
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Moles = mass / molar mass; Particles = moles × Nₐ
Criss-cross method for writing chemical formulae using valencies
Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻

Exam Tips for Chapter 3
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch03-atoms-and-molecules.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Question 6

Question 1Write down the formulae of(i) sodium oxide(ii) aluminium chloride(iii) sodium sulphide(iv) magnesium hydroxide

Accepted Answer

(i)	Sodium oxide:Step 1 — Write each symbol with its valencyNa1+↗O2−\text{Na}^{1+} \phantom{\nearrow} \text{O}^{2-}Na1+↗O2−Step 2 — Interchange the valenciesNa21 ↘↙ O2⇒Na12 ↘↙ O1\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{O}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{O}} \[0.5em]Na21 ↘↙ O2⇒12Na​ ↘↙ 1O​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Sodium oxide : Na2O\bold{Na_2O}Na2​O(ii) Aluminium chlorideStep 1 — Write each symbol with its valencyAl3+↗Cl1−\text{Al}^{3+} \phantom{\nearrow} \text{Cl}^{1-}Al3+↗Cl1−Step 2 — Interchange the valenciesAl23 ↘↙ Cl1⇒Al11 ↘↙ Cl3\overset{\phantom{2}{3}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{Cl}} \Rightarrow \underset{\phantom{1}{1}}{\text{Al}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{Cl}} \[0.5em]Al23 ↘↙ Cl1⇒11Al​ ↘↙ 3Cl​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Aluminium chloride : AlCl3\bold{AlCl_3}AlCl3​(iii) sodium sulphideStep 1 — Write each symbol with its valencyNa1+↗S2−\text{Na}^{1+} \phantom{\nearrow} \text{S}^{2-}Na1+↗S2−Step 2 — Interchange the valenciesNa21 ↘↙ S2⇒Na12 ↘↙ S1\overset{\phantom{2}{1}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}} \Rightarrow \underset{\phantom{1}{2}}{\text{Na}} \space {\searrow}\mathllap{\swarrow} \space \underset{1}{\text{S}} \[0.5em]Na21 ↘↙ S2⇒12Na​ ↘↙ 1S​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Sodium sulphide : Na2S\bold{Na_2S}Na2​S(iv) Magnesium hydroxideStep 1 — Write each symbol with its valencyMg2+↗OH1−\text{Mg}^{2+} \phantom{\nearrow} \text{OH}^{1-}Mg2+↗OH1−Step 2 — Interchange the valenciesMg22 ↘↙ OH1⇒Mg11 ↘↙ OH2\overset{\phantom{2}{2}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{OH}} \Rightarrow \underset{\phantom{1}{1}}{\text{Mg}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{OH}} \[0.5em]Mg22​ ↘↙ OH1⇒11Mg​ ↘↙ 2OH​Step 3 — Write the interchanged number, ignore equal numbers & hence the formulaTherefore, we getFormula of Magnesium hydroxide : Mg(OH)2\bold{Mg(OH)_2}Mg(OH)2​

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BRIGHT TUTORIALS
  BRIGHT TUTORIALSCBSE Class IX | Academic Year 2026-20279403781999Excellence in Education
  Science | Chapter 3: Atoms and MoleculesWeb Content &mdash; Quick Reference
  
    Chapter 3: Atoms and Molecules &mdash; Quick Reference

Dalton theory
      atomic mass
      molecular mass
      mole concept
      chemical formula
      valency
      Avogadro number

Quick Revision Points
      
        Law of conservation of mass: total mass unchanged in a reaction
Law of constant proportions: elements in fixed mass ratio in a compound
1 u = 1/12 mass of C-12 = 1.66 × 10⁻²⁴ g
Molecular mass = sum of atomic masses of all atoms in molecule
1 mole = 6.022 × 10²³ particles (Avogadro's number)
Moles = mass / molar mass; Particles = moles × Nₐ
Criss-cross method for writing chemical formulae using valencies
Polyatomic ions: NH₄⁺, OH⁻, SO₄²⁻, NO₃⁻, CO₃²⁻, PO₄³⁻

Exam Tips for Chapter 3
      
        Read the detailed chapter notes for complete coverage of all NCERT topics.
        Practice all NCERT in-text and back exercise questions &mdash; they are frequently asked in exams.
        Focus on comparison tables, diagrams, and definitions &mdash; these are high-scoring areas.
        For numericals (if applicable), practice at least 20 problems of varying difficulty.
        Refer to the practice question bank (200+ questions) for thorough preparation.

Related Resources
      
        Detailed Notes: ch03-atoms-and-molecules.html
        Practice Questions: 100+ questions with answers in 05-practice-questions/
        Chapter Test: 30-mark test paper in 06-tests/chapter-tests-30marks/
        Formula Sheet: Complete formula reference in 03-teacher-aid/formula-sheet.html

BRIGHT TUTORIALSContact: 9403781999 | CBSE Class IX Coaching | 2026-2027

Atoms and Molecules

Exercises

Intext Questions 1

Intext Questions 3