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SEBA Class 9 Science: Atoms and Molecules

INTEXT 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 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 = 5.3 g + 6 g = 11.3 g
Mass of products = 2.2 g + 0.9 g + 8.2 g = 11.3 g
Since the total mass of reactants and products is the same, the observations agree with the law of conservation of mass.

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: Hydrogen and oxygen combine in the ratio 1 : 8 by mass.
For 1 g of hydrogen, oxygen required = 8 g.
For 3 g of hydrogen, oxygen required = 3 × 8 = 24 g.

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

Answer: The second postulate of Dalton’s atomic theory is the result of the law of conservation of mass. It states that atoms cannot be created or destroyed in a chemical reaction.

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

Answer: The fifth postulate of Dalton’s atomic theory can explain the law of definite proportions. It states that atoms combine in the ratio of small whole numbers to form compounds.

INTEXT QUESTION

1) Define the atomic mass unit.

Answer: It is defined as the mass unit equal to exactly 1/12th of the mass of one atom of C-12 isotope. Earlier, it was abbreviated as amu, but according to the latest recommendations of IUPAC, it is now written as ‘u’ unified mass.

2) Why is it not possible to see an atom with the naked eye?

Answer: It is not possible to see an atom with the naked eye because atoms are extremely tiny in size. They are so small that even millions of atoms stacked together form a layer only as thin as a sheet of paper.

INTEXT QUESTION

2) Write down the names of compounds represented by the following formulae:
(i) Al₂(SO₄)₃

Answer: Aluminium Sulphate.

(ii) CaCl₂

Answer: Calcium chloride

(iii) K₂SO₄

Answer: Potassium Sulphate.

(iv) KNO₃

Answer: Potassium nitrate.

(v) CaCO₃

Answer: Calcium carbonate.

4) How many atoms are present in
(i) H₂S molecule.

Answer: In the H2S molecule, three atoms. (i.e., 2 atoms of H and 1 atom of S are present)

(ii) PO₄³⁻ ion?

Answer: In the PO₄³⁻ ion, five atoms. (i.e., 1 atom of P and 4 atoms of O are present)

INTEXT QUESTION

1) Calculate the molecular masses of H₂, O₂, Cl₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.

(I) Molecular mass of H2 (Hydrogen)

= atomic mass of hydrogen × 2

= 1 × 2 = 2u

(2) Molecular mass of O2 (Oxygen)

= atomic mass of oxygen × 2

= 16 × 2 = 32u

(3) Molecular mass of Cl2 (Chlorine)

= atomic mass of chlorine × 2

= 35.5 × 2 = 71u

(4) Molecular mass of CO2 (carbon dioxide)

= (atomic mass of carbon × 1) + (atomic mass of oxygen × 2)

= 12 × 1 + 16 × 2 = 44u

(5) Molecular mass of CH4 (methane)

= (atomic mass of carbon × 1) + (atomic mass of hydrogen × 4)

= 12 × 1 + 1 × 4 = 16u

(6) Molecular mass of C2H6 (ethane)

= (atomic mass of carbon × 2) + (atomic mass of hydrogen × 6)

= 12 × 2 + 1 × 6 = 30u

(7) Molecular mass of C2H4 (ethane)

= (atomic mass of carbon × 2) + (atomic mass of hydrogen × 4)

= 12 × 2 + 1 × 4 = 28u

(8) Molecular mass of NH3 (ammonia)

= (atomic mass of nitrogen × 1) + (atomic mass of hydrogen × 3)

= 14 × 1 + 1 × 3 = 17u

(9) Molecular mass of CH3OH (Methanol or methyl alcohol)

= (atomic mass of carbon × 1) + (atomic mass of hydrogen × 3) + (atomic mass of oxygen × 1) +

(atomic mass of hydrogen × 1)

= (12 × 1) + (1 × 3) + (16 × 1) + (1 × 1) = 32u

2) Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65 u, Na = 23 u, K = 39 u, C = 12 u, and O = 16 u.

(1) Formula unit mass of ZnO (zinc oxide)

= (atomic mass of zinc × 1) + (atomic mass of oxygen × 1)

= 65 × 1 + 16 × 1 = 81u

(2) Formula unit of mass of Na2O (sodium oxide)

= (atomic mass of sodium × 2) + (atomic mass of oxygen × 1)

= 23 × 2 + 16 × 1 = 62u

(3) Formula unit of K2CO3 (Potassium carbonate)

= (atomic mass of Potassium × 2) + (atomic mass of carbon × 1) + (atomic mass of Oxygen × 3)

= 39 × 2 + 12 × 1 + 16 × 3 = 138u

IN-TEXT QUESTION

EXERCISES QUESTION

2) When 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?

Answer: When 3.0 g of carbon is burnt in 8.0 g of oxygen, 11.0 g of carbon dioxide is produced.
If 3.0 g of carbon is burnt in 50.0 g of oxygen, the carbon will still combine with only 8.0 g of oxygen because the ratio of combination remains fixed. So the mass of carbon dioxide formed will again be 11.0 g.
This happens because carbon and oxygen always combine in a definite proportion by mass. Therefore, the Law of Definite Proportions governs the reaction.

3) What are polyatomic ions? Give examples.

Answer: Polyatomic ions are groups of atoms that stay together and carry a fixed positive or negative charge.
They behave like a single ion during chemical reactions.
Examples: NH₄⁺ (ammonium), OH⁻ (hydroxide), NO₃⁻ (nitrate), CO₃²⁻ (carbonate), and SO₄²⁻ (sulphate).