Tag: Chapter 3 – Atoms and Molecules

  • Class 9th Science Chapter-3 Exercises

    Atoms and Molecules

    Exercises – Answers and Solutions


    Question 1

    A 0.24 g sample of a compound of oxygen and boron contains 0.096 g of boron and 0.144 g of oxygen. Calculate the percentage composition by weight.

    Solution:

    Total mass of compound = 0.24 g

    Percentage of boron:

    0.0960.24×100=40%

    Percentage of oxygen:

    0.1440.24×100=60%

    Answer:

    • Boron = 40%

    • Oxygen = 60%


    Question 2

    When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is formed.
    What mass of carbon dioxide will be formed when 3.0 g of carbon is burnt in 50.0 g of oxygen?
    Which law governs this?

    Solution:

    Carbon reacts with oxygen in a fixed ratio.

    Given:

    • 3 g carbon + 8 g oxygen → 11 g CO₂

    Even though oxygen is present in excess (50 g), only 8 g oxygen will react with 3 g carbon.

    So,

    Mass of CO₂ formed=11 g

    Law governing this:

    Law of Constant (Definite) Proportions

    Answer:

    • Mass of CO₂ formed = 11 g

    • Law: Law of Constant Proportions


    Question 3

    What are polyatomic ions? Give examples.

    Answer:

    Polyatomic ions are groups of two or more atoms that carry a net charge and act as a single ion.

    Examples:

    • Ammonium (NH₄⁺)

    • Hydroxide (OH⁻)

    • Carbonate (CO₃²⁻)

    • Sulphate (SO₄²⁻)


    Question 4

    Write the chemical formulae of the following:

    Compound Formula
    (a) Magnesium chloride

    MgCl₂

    (b) Calcium oxide CaO
    (c) Copper nitrate

    Cu(NO₃)₂

    (d) Aluminium chloride AlCl₃
    (e) Calcium carbonate

    CaCO₃


    Question 5

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

    (a) Quick lime (CaO)

    • Calcium

    • Oxygen

    (b) Hydrogen bromide (HBr)

    • Hydrogen

    • Bromine

    (c) Baking powder

    • Sodium

    • Hydrogen

    • Carbon

    • Oxygen

    (d) Potassium sulphate (K₂SO₄)

    • Potassium

    • Sulphur

    • Oxygen


    Question 6

    Question:
    Calculate the molar mass of the following substances.

    (Atomic masses used: H = 1, C = 12, N = 14, O = 16, S = 32, P = 31, Cl = 35.5)


    (a) Ethyne (C₂H₂)

    (2×12)+(2×1)=24+2=26 g mol1


    (b) Sulphur molecule (S₈)

    8×32=256 g mol1


    (c) Phosphorus molecule (P₄)

    4×31=124 g mol1


    (d) Hydrochloric acid (HCl)

    1+35.5=36.5 g mol1


    (e) Nitric acid (HNO₃)

    1+14+(3×16)=1+14+48=63 g mol1

  • Class 9th Science Chapter-3 In-Text Questions

    Atoms and Molecules

    In-Text Question Answers

    Page 27 – Questions – Answers

    Question 1

    In a reaction, 5.3 g of sodium carbonate reacted with 6 g of acetic acid. The products were 2.2 g of carbon dioxide, 0.9 g of water and 8.2 g of sodium acetate. Show that these observations are in agreement with the law of conservation of mass.

    Answer:

    Mass of reactants
    = Mass of sodium carbonate + Mass of acetic acid
    = 5.3 g + 6.0 g
    = 11.3 g

    Mass of products
    = Mass of carbon dioxide + Mass of water + Mass of sodium acetate
    = 2.2 g + 0.9 g + 8.2 g
    = 11.3 g

    Since the total mass of reactants = total mass of products, the reaction obeys the Law of Conservation of Mass.


    Page 28 – Questions

    Question 2

    Hydrogen and oxygen combine in the ratio of 1 : 8 by mass to form water. What mass of oxygen gas would be required to react completely with 3 g of hydrogen gas?

    Answer:

    Given ratio (H : O) = 1 : 8

    If 1 g hydrogen reacts with 8 g oxygen
    Then 3 g hydrogen will react with:

    3×8=24 g

    Required mass of oxygen = 24 g


    Question 3

    Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

    Answer:

    The postulate stating that:

    “Atoms can neither be created nor destroyed in a chemical reaction.”

    This explains the law of conservation of mass.


    Question 4

    Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

    Answer:

    The postulate stating that:

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

    This explains the law of definite (constant) proportions.


    Page 30 – Questions

    Question 1

    Define the atomic mass unit.

    Answer:
    One atomic mass unit (u) is defined as one-twelfth (1/12th) of the mass of one atom of carbon-12.

    Question 2

    Why is it not possible to see an atom with naked eyes?

    Answer:
    An atom is extremely small in size (of the order of 1010), which is much smaller than the resolving power of the human eye. Therefore, atoms cannot be seen with naked eyes.


    Page 34 – Questions

    Question 1

    Write down the formulae of the following:

    (i) Sodium oxideNa₂O
    (ii) Aluminium chlorideAlCl₃
    (iii) Sodium sulphideNa₂S
    (iv) Magnesium hydroxideMg(OH)₂


    Question 2

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

    Formula Name of compound
    (i) Al₂(SO₄)₃

     

    Aluminium sulphate

    (ii) CaCl₂

     

    Calcium chloride

    (iii) K₂SO₄

     

    Potassium sulphate

    (iv) KNO₃

     

    Potassium nitrate

    (v) CaCO₃ Calcium carbonate

    Question 3

    What is meant by the term chemical formula?

    Answer:
    A chemical formula is a symbolic representation of a compound which shows the elements present in it and the number of atoms of each element.


    Question 4

    How many atoms are present in:

    (i) H₂S molecule

    • Hydrogen atoms = 2

    • Sulphur atoms = 1

    Total atoms = 3


    (ii) PO₄³⁻ ion

    • Phosphorus atoms = 1

    • Oxygen atoms = 4

    Total atoms = 5


    Page 35 – Questions

    Question 1

    Calculate the molecular masses of the following:

    (Atomic masses used: H = 1 u, C = 12 u, N = 14 u, O = 16 u, Cl = 35.5 u)


    (i) H₂

    2×1=2 u


    (ii) O₂

    2×16=32 u


    (iii) Cl₂

    2×35.5=71 u


    (iv) CO₂

    12+(2×16)=12+32=44 u


    (v) CH₄

    12+(4×1)=12+4=16 u


    (vi) C₂H₆

    (2×12)+(6×1)=24+6=30 u


    (vii) C₂H₄

    (2×12)+(4×1)=24+4=28 u


    (viii) NH₃

    14+(3×1)=14+3=17 u


    (ix) CH₃OH

    12+(4×1)+16=12+4+16=32 u


    Question 2

    Calculate the formula unit masses of the following:

    (Given: Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, O = 16 u)


    (i) ZnO

    65+16=81 u


    (ii) Na₂O

    (2×23)+16=46+16=62 u


    (iii) K₂CO₃

    (2×39)+12+(3×16)=78+12+48=138 u