Exploration | Class 9 Science | Chapter 9 - Atomic Foundations of Matter | Q&A
CHAPTER 9

Atomic Foundations of Matter

Questions & Solutions | Exploration | Class 9 Science

Activity 9.1 Page 183

Activity 9.1
Place a clean and dry 100 mL beaker on a digital weighing balance. Set the balance reading to zero by pressing the tare or reset button. Pour about 50 mL of water into the beaker. Add a spatula full of common salt to the water contained in the beaker. Record the reading on the weighing balance (Fig. 9.1a). Swirl until the added salt dissolves and record your observations (Fig. 9.1b). What do you observe? You may notice that the mass of the solution is equal to the sum of the masses of water and salt taken. This shows that there is practically no change in the mass during the formation of a solution, which is a physical change. This is true for all physical changes. You can repeat the above activity by weighing a piece of paper before and after tearing it into pieces, and observe whether its mass changes or not.
Solution
Mass remains conserved during physical changes. The mass of the solution equals the sum of masses of solute and solvent.

Activity 9.2 Pages 183-185

Activity 9.2
Let us investigate a chemical change. Experimental set-up 1: Place a clean, dry 100 mL conical flask and a medium-sized balloon on a weighing balance. Set the balance reading to zero. Pour about 20 mL of vinegar or lemon juice into the conical flask. Take about 2 g of baking soda and put it into the balloon. Keep the balloon filled with baking soda on the weighing balance next to the conical flask. Record the initial reading (Fig. 9.2a). Carefully transfer the baking soda from the balloon into the conical flask containing vinegar (Fig. 9.2b). Place the conical flask and balloon back on the weighing balance, and record the final reading (Fig. 9.2c). What do you observe? Are the initial and the final readings same? A brisk effervescence is observed. The final reading does not match the initial reading. What can be the reason for this? Repeat the above experiment in a slightly modified way as explained below. Experimental set-up 2: Place a clean, dry 100 mL conical flask and a medium-sized balloon on a weighing balance. Set the balance reading to zero. Pour about 20 mL of vinegar or lemon juice into the conical flask. Place about 2 g of baking soda in the balloon. Fix the balloon to the mouth of the conical flask using a thread, without allowing the baking soda to mix with the vinegar. Weigh the conical flask containing vinegar and the balloon containing baking soda, and record the reading (Fig. 9.3a). Lift the other end of the balloon upwards, allowing the baking soda to fall into the vinegar (Fig. 9.3b). What do you observe? As in experimental set-up 1, a brisk effervescence occurs, which inflates the balloon during the reaction. Record the final reading (Fig. 9.3c). Are the initial and the final readings same in this case? In this experiment, the final reading matches the initial reading. You have noticed that the total mass of vinegar and baking soda before the chemical reaction is equal to the total mass of carbon dioxide and other substances formed after the reaction. In experimental set-up 1, the mass difference occurs because the gas produced by the chemical reaction escapes, resulting in a difference between the initial and the final readings.
Solution
Mass is conserved in a chemical reaction when the system is closed (no gas escapes). In open setup, gas (CO₂) escapes, causing apparent mass loss.

Activity 9.3 Page 185

Activity 9.3
Let us verify the law - Group activity. Place two clean and dry 100 mL conical flasks on a weighing balance, and mark them A and B. Set the balance reading to zero by pressing the tare or reset button. Pour about 10 mL of 1% m/v sodium sulfate solution into the Conical Flask marked A. In the Conical Flask B, pour about 10 mL of 1% m/v barium chloride solution. Leave both Conical Flasks A and B on the weighing balance undisturbed, and record the total mass of both the solutions (Fig. 9.4a). Transfer the solution from Conical Flask B to Conical Flask A and mix the two solutions carefully. What do you observe? Now, place both the Conical Flasks A and B on the weighing balance again as shown in Fig. 9.4b, and note the reading. Do you observe any change in the reading after mixing the solutions?
Solution
A white precipitate of barium sulfate is formed. The mass remains the same before and after the reaction, verifying the Law of Conservation of Mass.

Think as a Scientist Page 186

Think as a Scientist
You are given a chemical reaction in which zinc reacts with dilute hydrochloric acid to form zinc chloride and hydrogen gas. Zinc + Hydrochloric acid (dilute) → Zinc chloride + Hydrogen. Design and perform an experiment to test the hypothesis that mass is conserved during the chemical reaction. You may use a set-up different from the one shown in Activity 9.2.
Solution
Use a closed container (like a conical flask with a balloon to collect hydrogen gas). Weigh the apparatus before and after the reaction. If mass remains constant, the law is verified.

Meet a Scientist

Page 185 - Antoine Lavoisier
Antoine Lavoisier is known as the Father of Modern Chemistry. He proposed the Law of Conservation of Mass. This law applies to every chemical reaction. Lavoisier continued to study this and proposed that "...in every operation an equal quantity of matter exists both before and after the operation".
Solution
Lavoisier established the Law of Conservation of Mass, which states that mass is neither created nor destroyed in a chemical reaction.
Page 187 - Joseph Louis Proust
Joseph Louis Proust was a prominent French chemist known for his careful experimental work. He contributed to the Law of Definite Proportions by showing that chemical compounds always contain elements in fixed ratios by mass. For example, Proust studied the composition of copper carbonate. He showed that copper carbonate always contains copper, carbon and oxygen in the same proportion by mass, no matter how it was prepared or where it was found. His work laid an important foundation that helped shape modern chemistry.
Solution
Proust's Law of Definite Proportions states that a compound always contains the same elements in the same proportion by mass, regardless of its source.
Page 188 - John Dalton
John Dalton was born in England. In 1793, Dalton moved to Manchester to teach mathematics, physics and chemistry at a college. He spent most of his life teaching and researching there. In 1808, he presented his atomic theory, which proved to be a turning point in the study of matter.
Solution
Dalton's Atomic Theory proposed that all matter is made of indivisible atoms, atoms of the same element are identical, and atoms combine in simple whole-number ratios to form compounds.

Bridging Science and Society Page 191

Bridging Science and Society
Atoms can release enormous energy when their nuclei split or combine to form new elements. This is called atomic or nuclear energy, and it plays a vital role in modern life. Beyond electricity generation, it is used in medicine, scientific research and space exploration. In nuclear power plants, the thermal energy from nuclear reactions produces steam that drives turbines and generates electricity, a cleaner alternative to fossil fuels. In India, scientists like Raja Ramanna (often called the Father of the Indian Nuclear Programme) made significant contributions in developing the nation's Nuclear Energy Programme and promoting its peaceful use for development.
Solution
Nuclear energy is released during nuclear fission or fusion. It is used for electricity generation, medical treatments, and scientific research.

Threads of Curiosity

Page 187
In many ancient civilisations, red pigment derived from rocks was widely used in painting and as a colouring agent for various objects. In India, it was known as hingula, and in Latin and English, as cinnabar. Over the centuries, it was discovered in many civilisations that heating cinnabar could yield two elements- mercury and sulfur in mass percentage of around 86.22% and 13.78%, respectively. Interestingly, most civilisations also found that grinding mercury and sulfur together in this ratio could form cinnabar, although the toxic nature of both prevented this process from becoming widespread.
Solution
Cinnabar (HgS) always contains mercury and sulfur in a fixed mass ratio (86.22% : 13.78%), illustrating the Law of Definite Proportions.
Page 193
Ionic compounds usually do not remain as single units. They form three-dimensional (3-D) crystals in which ions are arranged in a repeating pattern. For example, in sodium chloride (NaCl), each sodium ion (Na⁺) is surrounded by six chloride ions (Cl⁻), and each chloride ion is surrounded by six sodium ions (Fig. 9.14a). These oppositely charged ions are arranged in a regular, repeating 3-D pattern known as a crystal structure (Fig. 9.14b). The crystal structure is represented as a crystal lattice, with ions depicted as points or dots (Fig. 9.14c). It helps to visualise the arrangement of ions in the crystal. You will learn more about the crystal structure in higher grades.
Solution
Ionic compounds form crystal lattices due to strong electrostatic forces between oppositely charged ions. The repeating pattern gives crystals their characteristic shape.

Pause and Ponder

Page 187 (Q3)
A compound consists of 40% sulfur and 60% oxygen by mass. In a sample of the same compound containing 20 g of sulfur, what mass of oxygen must be present to satisfy the Law of Constant Proportions?
Solution
S:O = 40:60 = 2:3. For 20 g sulfur, oxygen = (3/2) × 20 = 30 g.
Page 187 (Q4)
Carbon monoxide (CO) contains carbon and oxygen in the mass ratio of 3:4. How much oxygen will combine with 9 g of carbon to form carbon monoxide?
Solution
C:O = 3:4. For 9 g carbon, oxygen = (4/3) × 9 = 12 g.
Page 187 (Q5)
The Law of Definite Proportions holds true for compounds but not for mixtures. Give reason.
Solution
Compounds have fixed composition by mass, while mixtures can have variable proportions of their components.
Page 187 (Q6)
Students X and Y, both prepared an oxide of copper by combining copper and oxygen in the ratios of 4:1 and 8:2, respectively. Do their results justify the Law of Constant Proportions? Explain.
Solution
4:1 simplifies to 4:1. 8:2 simplifies to 4:1. Both are the same ratio, so the law is justified.
Page 188 (Q7)
Assertion (A): 2 g of hydrogen combines with 16 g of oxygen to form 18 g of water. Reason (R): According to Dalton's Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds. Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
Solution
Option (i) is correct. The mass ratio 2:16 = 1:8 is a simple whole number ratio (1:8).
Page 190 (Q8)
Nitrogen has five valence electrons. Draw the structure of the nitrogen molecule (N₂).
Solution
N≡N (triple bond) — each nitrogen shares three electrons with the other.
Page 190 (Q9)
The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F₂).
Solution
Each fluorine atom has 7 valence electrons. They share one electron each to form a single covalent bond (F–F).
Page 191 (Q10)
Show the formation of the following molecules: (i) Carbon dioxide (CO₂) (ii) Hydrogen sulfide (H₂S) (iii) Ammonia (NH₃)
Solution
(i) O=C=O (double bonds). (ii) H–S–H (single bonds). (iii) NH₃ with single bonds (three N–H bonds).
Page 191 (Q11)
Neon (atomic number 10) neither transfers nor shares its valence electrons. Explain.
Solution
Neon has a complete octet (2,8), making it chemically inert and stable.
Page 194 (Q12)
What kind of ion will oxygen (O) form?
Solution
Oxygen will gain 2 electrons to form O²⁻ (oxide ion).
Page 194 (Q13)
Fill in the blanks. Among magnesium and chlorine, magnesium atom can give two electrons to become Mg²⁺. However, chlorine can take only one electron to become ______. Now, ______ ion of magnesium and ______ ions of chlorine combine to give magnesium chloride.
Solution
Cl⁻; one; two.
Page 194 (Q14)
Show the formation of cations of potassium (K) and calcium (Ca) atoms, and the formation of their corresponding chlorides using diagrams.
Solution
K → K⁺ + e⁻ (KCl). Ca → Ca²⁺ + 2e⁻ (CaCl₂).
Page 194 (Q15)
Illustrate how sodium sulfide (Na₂S) is formed.
Solution
Two Na atoms each lose one electron to become Na⁺; one S atom gains two electrons to become S²⁻. Electrostatic attraction forms Na₂S.
Page 197 (Q16)
Name the following: (i) CO₂ (ii) NO₂ (iii) SF₆ (iv) PCl₃
Solution
(i) Carbon dioxide. (ii) Nitrogen dioxide. (iii) Sulfur hexafluoride. (iv) Phosphorus trichloride.
Page 197 (Q17)
Write the formula for the following: (i) Sodium hydrogencarbonate (ii) Sulfur dioxide (iii) Ferric chloride (iv) Cuprous oxide
Solution
(i) NaHCO₃ (ii) SO₂ (iii) FeCl₃ (iv) Cu₂O
Page 197 (Q18)
Write the formulae for the compounds formed from the following pairs of ions: (i) Fe³⁺ and OH⁻ (ii) K⁺ and CO₃²⁻
Solution
(i) Fe(OH)₃ (ii) K₂CO₃
Page 199 (Q19)
What type of chemical bond is present in a solid compound that does not conduct electricity in the solid state but conducts electricity when dissolved in water?
Solution
Ionic bond. In solid state, ions are fixed; in solution, they are free to move and conduct electricity.
Page 199 (Q20)
Metal M, with two electrons in its valence shell (M shell), reacts with oxygen to form a compound that is slightly soluble in water. Predict its: (i) formula (ii) type of bond (iii) electrical conductivity of its aqueous solution.
Solution
Metal is magnesium (Mg). (i) MgO. (ii) Ionic bond. (iii) Conducts electricity (ions in solution).
Page 199 (Q21)
Find the molecular mass of nitric acid (HNO₃). Atomic mass - H = 1 u; N = 14 u; O = 16 u.
Solution
Molecular mass = 1 + 14 + (16 × 3) = 1 + 14 + 48 = 63 u.
Page 199 (Q22)
Find the molecular mass of methane (CH₄). Atomic mass - C = 12 u; H = 1 u.
Solution
Molecular mass = 12 + (1 × 4) = 12 + 4 = 16 u.
Page 200 (Q23)
Find the formula unit mass of potassium chloride (KCl). Atomic mass—K = 39 u; Cl = 35.5 u.
Solution
Formula unit mass = 39 + 35.5 = 74.5 u.
Page 200 (Q24)
Find the formula unit mass of magnesium hydroxide, Mg(OH)₂. Atomic mass—Mg = 24 u; O = 16 u; H = 1 u.
Solution
Formula unit mass = 24 + (16 + 1) × 2 = 24 + 34 = 58 u.

END-OF-CHAPTER EXERCISES Pages 201-203

1.
A particular element (A) has one electron in its third shell. There is another element (B) with six electrons in its second shell. (i) How many electrons does A tend to give or take to become stable? (ii) What kind of ion would it form? (iii) How many electrons does B tend to give or take to become stable? (iv) What kind of ion would it form? (v) If A and B were to combine, what kind of bond would be formed? (vi) What would be the formula for the compound thus formed?
Solution
A = Sodium (Na) - 1 valence electron. B = Oxygen (O) - 6 valence electrons. (i) Gives 1 electron. (ii) Cation (Na⁺). (iii) Takes 2 electrons. (iv) Anion (O²⁻). (v) Ionic bond. (vi) Na₂O.
2.
An element X has six electrons in its outer shell and forms a diatomic molecule. (i) Why would that be so? (ii) What kind of bond would it form? (iii) Draw the structure of the molecule it would form. (iv) A certain other element Y has two electrons in its second shell. Draw the structure of the molecule that X would form with Y.
Solution
X = Oxygen (O). (i) Needs 2 more electrons to complete octet, so shares 2 electrons with another oxygen atom. (ii) Covalent bond (double bond). (iii) O=O. (iv) Y = Beryllium (Be) — BeO forms ionic bond? Actually Be and O form BeO (ionic/covalent).
3.
You want to design a new ionic compound, where the total positive charge is 6+ and the total negative charge is 6-. Which of the following combinations gives the correct number of ions? (i) 2 Al³⁺ and 3 Cl⁻ (ii) 3 Mg²⁺ and 1 PO₄³⁻ (iii) 2 Fe³⁺ and 3 O²⁻ (iv) 3 Ca²⁺ and 2 SO₄²⁻
Solution
Option (iii): 2 × 3+ = 6+, 3 × 2- = 6-. Also option (i): 2 × 3+ = 6+, 3 × 1- = 3- (not 6-). Option (ii): 3 × 2+ = 6+, 1 × 3- = 3- (not 6-). Option (iv): 3 × 2+ = 6+, 2 × 2- = 4- (not 6-). Only (iii) works.
4.
Choose the correct statement(s) and correct the false statement(s). (i) Elements are made up of molecules and compounds are made up of atoms. (ii) The molecule of a compound is always made up of two or more atoms of the same kind. (iii) One molecule of nitrogen gas contains three nitrogen atoms. (iv) Water is made of two hydrogen atoms, covalently bonded with one oxygen atom.
Solution
(i) False — Elements are made of atoms; compounds are made of molecules. (ii) False — Molecule of a compound has atoms of different elements. (iii) False — N₂ has two nitrogen atoms. (iv) True.
5.
Write the chemical formulae for the following compounds. (i) Aluminium nitrate (ii) Calcium oxide (iii) Ferric oxide
Solution
(i) Al(NO₃)₃ (ii) CaO (iii) Fe₂O₃
6.
Write the formulae of the compounds formed from the following pairs of ions. (i) Ca²⁺ and Br⁻ (ii) Al³⁺ and CO₃²⁻ (iii) K⁺ and SO₄²⁻ (iv) NH₄⁺ and Cl⁻
Solution
(i) CaBr₂ (ii) Al₂(CO₃)₃ (iii) K₂SO₄ (iv) NH₄Cl
7.
Which of the following, in Fig. 9.18, correctly represents Cl⁻ ion (Atomic number of chlorine = 17). (i) (ii) (iii) (iv)
Solution
Cl⁻ has 18 electrons (17 protons + 1 extra electron). The correct diagram will show 2,8,8 electronic configuration.
8.
Determine the formula unit mass of the following substances. (i) Ammonium nitrate (NH₄NO₃), used as a nitrogen fertiliser. (ii) Phosphoric acid (H₃PO₄), used to make phosphate fertiliser and detergents. (iii) Sodium hydrogencarbonate (NaHCO₃), used to relieve acidity and helps in digestion.
Solution
(i) NH₄NO₃ = 14 + (4×1) + 14 + (16×3) = 14+4+14+48 = 80 u. (ii) H₃PO₄ = (3×1) + 31 + (16×4) = 3+31+64 = 98 u. (iii) NaHCO₃ = 23 + 1 + 12 + (16×3) = 23+1+12+48 = 84 u.
9.
Write the formulae for the compounds formed by the reaction of: (i) Magnesium and nitrogen (ii) Lithium and nitrogen (iii) Sodium and sulfur (iv) Aluminium and oxygen
Solution
(i) Mg₃N₂ (ii) Li₃N (iii) Na₂S (iv) Al₂O₃
10.
Complete the Table 9.3 by writing the formulae of the compounds formed by the cations on the left and the anions at the top. LiNO₃ is given as an example.
Solution
NH₄NO₃, (NH₄)₂SO₄, (NH₄)₃PO₄; LiNO₃, Li₂SO₄, Li₃PO₄; Al(NO₃)₃, Al₂(SO₄)₃, AlPO₄; Cu(NO₃)₂, CuSO₄, Cu₃(PO₄)₂.
11.
5.3 g of sodium carbonate and 6.0 g of acetic acid react to produce 2.2 g of carbon dioxide, 0.9 g of water, and 8.2 g of sodium acetate. Verify whether the law of conservation of mass is valid.
Solution
Mass of reactants = 5.3 + 6.0 = 11.3 g. Mass of products = 2.2 + 0.9 + 8.2 = 11.3 g. Mass is conserved, so law is valid.
12.
If a species has 11 protons, 12 neutrons and 10 electrons then (i) what is its atomic number and mass number? (ii) is it neutral, a cation or an anion? Explain. (iii) write its electronic configuration. (iv) name the species.
Solution
(i) Atomic number = 11, Mass number = 23. (ii) Cation (more protons than electrons). (iii) Electronic configuration: 2,8. (iv) Sodium ion (Na⁺).
13.
Two elements, A and B, have the following configurations — A: 2,8,5; B: 2,8,7. (i) Which element is more reactive? (ii) Will A and B form ionic or covalent bonds when they combine? Explain using electron transfer or sharing. (iii) Predict the formula of the compound they would form.
Solution
(i) B is more reactive (halogen). (ii) Both are non-metals, so they will form covalent bonds by sharing electrons. (iii) AB₃ (e.g., PCl₃).
14.
Assertion (A): Copper sulfate conducts electricity in the molten state but not in the solid state. Reason (R): Copper and sulfate ions are fixed in the lattice in molten state, while in solid state they can move freely. Choose the correct option: (i) Both A and R are true, and R is the correct explanation of A. (ii) Both A and R are true, but R is not the correct explanation of A. (iii) A is true, but R is false. (iv) A is false, but R is true.
Solution
Option (iii) is correct. A is true, but R is false — ions are fixed in solid state, free to move in molten state (opposite of what R says).
15.
The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons, respectively. How many electrons and neutrons do they have?
Solution
²⁷Al: 13 e⁻, 14 n. ⁸⁰Br⁻: 36 e⁻, 45 n. ²⁰¹Hg²⁺: 78 e⁻, 121 n.

Chapter Summary

  • Law of Conservation of Mass: Mass is neither created nor destroyed in a chemical reaction.
  • Law of Definite Proportions: A compound always contains the same elements in the same proportion by mass.
  • Dalton's Atomic Theory: Atoms are indivisible; atoms of same element are identical; atoms combine in simple whole-number ratios.
  • Molecule: Electrically neutral group of atoms that can exist independently.
  • Covalent bond: Formed by sharing of electrons.
  • Ionic bond: Formed by transfer of electrons (cation + anion).
  • Molecular mass: Sum of atomic masses of atoms in a molecule.
  • Formula unit mass: Sum of atomic masses in the simplest ratio of ions in an ionic compound.

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Chapter 9: Atomic Foundations of Matter — All Questions with Solutions

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