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NCERT Class 9 Science Chapter 3 Atoms and Molecules
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Atoms and Molecules
Chapter – 3
GENERAL SCIENCE
TEXTUAL QUESTIONS AND ANSWERS
INTEX QUESTIONS AND ANSWERS
Textbook Page No. 32-33
1. In a reaction, 5.3 g of sodium carbonate reacted with 6 g of ethanoic acid. The products were 2.2 g of carbon dioxide, 0.9 g water and 8.2 g of sodium ethanoate. Show that these observations are in agreement with the law of conservation of mass. Sodium carbonate + ethanoic acid log → sodium ethanoate + carbon dioxide + water
Ans. Total massa of the reactants
= Mass of sodium carbonate+Mass of ethanoic acid
= 5.3g+6g
= 11.3g
Total mass of the products
= Mass of carbon dioxide + Mass of water + Mass of sodium ethanoate
= 2.2g + 0.9g + 8.2g
= 11.2g
Since the total mass of the reactants is equal to the total mass of the products, these observations are in agreement 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 3g of hydrogen gas?
Ans. 1 g of hydrogen reacts with 8 gm of oxygen to form water. Therefore, according to the law of constant proportions, Mass of oxygen gas required to react completely with 3 g of hydrogen gas is 8 × 3g = 24 g
3. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?
Ans. The postulate of Dalton’s atomic theory is the results of the law of conversation of mass are:
(i) All matter is composed of the indivisible particles called atoms. Atoms of an element cannot be created neither can be destroyed.
(ii) Atoms of the same element have similar shape and mass, but the mass and shape can vary for atoms of other elements.
4. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?
Ans. The postulate of Dalton’s atomic theory that can explain the law of definite proportions is- “The relative mumber and kinds of atoms are constant in a given compound.”
Textbook Page No. 35
1. Define the atomic unit.
Ans. The atomic mass unit (amu) is the quantity of mass equal to exactly one-twelfth (photo) the mass of one atom of carbon-12.
One atomic mass unit (1u)
2. Why is it not possible to see an atom with naked eyes?
Ans. An atom is considered to be the smallest unit of a matter. Its size is smaller than anything that we can imagine or compare with. If more than millions of atoms are stacked, they will make a layer barely as thick as a sheet of paper. Hence, it is not possible to see an atom with naked eyes.
Textbook Page No. 39
1. Write down formula of-
(i) Sodium oxide.
Ans. (i) Na₂O.
(ii) Aluminium chloride.
(ii) AICI₃.
(iii) Sodium sulphide.
(iii) Na₂S.
(iv) Magnesium hydroxide.
Ans: Mg(OH)₂.
2. Write down the names of compounds represented by the following formulae:
(i) AI₂(SO⁴)₃.
Ans. (i) Aluminium sulphate.
(ii) CaCI₂.
Ans. Calcium chloride.
(iii) K₂SO₄.
Ans. Potassium sulphate.
(iv) KNO₃.
Ans. Potassium nitrate.
(v) CaCO₃.
Ans. Calcium carbonate.
3. What is meant by the term chemical formula?
Ans. The chemical formula of a compound is a symbolic representation of its composition. It indicates the numbers and kinds of atoms of different elements present in one molecule of the compound. For example, carbon dioxide is a compound, made up of 1 atom of carbon element and two atoms of oxygen element.
Therefore, the formula of carbon dioxide is written as CO₂.
4. How many atoms are present in a
(i) H₂S molecule.
Ans. H₂S molecule: 3 atoms; 2 atoms of hydrogen and 1 atom of sulphur.
(ii) PO₄³⁻ ion?
Ans: P₄³⁻O ion: 5 atoms; 1 atom of phosphorus and 4 atoms of oxygen.
Textbook Page No. 40
1. Calculate the molecular masses of H₂, O₂, CI₂, CO₂, CH₄, C₂H₆, C₂H₄, NH₃, CH₃OH.
Ans. Molecular mass of H₂ = 2 × atomic of H
= 2 × 1
= 2u
Molecular mass of O₂ = 2 × atomic mass of O
= 2 × 16
= 32u
Molecular mass of CI² = 2 × atomic mass of CI
= 2×35.5
= 71u
Molecular mass of CO₂ = (1×atomic mass of C) + (2 × atomic mass of oxygen)
= (1 × 12)̇ + (2 × 16)
= 12 + 32
= 44u
Molecular mass of CH₄ = (1 x atomic mass of C) + (4 x atomic mass of hydrogen)
= (1 × 12) + (4 × 1)
= 12 + 4
= 16u
Molecular mass of C₂H₆ = (2x atomic mass of C) + (6 × atomic mass of H)
= (2 × 12)+(6 × 1)
= 24 + 6
= 30u
Molecular mass of C₂H₄ = (2 × atomic mass of C) + (4 x atomic mass of H)
= (2 × 12) + (4 × 1)
= 24 + 4
= 28u
Molecular mass of NH₃ = (1 × atomic mass of N) + (3 x atomic mass of H)
= (1 × 14)+(3 × 1)
= 14 + 3
= 17 u
Molecular mass of CH₃OH
= (1x atomic mass of C) + (3 x atomic mass of H)+(1x atomic mass of O) + (1x atomic mass of H)
= (1 × 12)+(31)+(1 × 16) + (1 × 1) =
=12 + 3 + 16 + 1
= 32u.
2. Calculate the formula unit masses of ZnO, Na₂O, K₂CO₃, given atomic masses of Zn = 65u, Na = 23 u, K=39 u, C= 12 u and O = 16 u.
Ans. Formula unit mass of ZnO = (1 × atomic mass of Zn) + (1 x atomic mass of O)
= (1 × 65) + (1 × 16)
= 65+16
= 81 u
Formula unit mass of Na₂O = (2 × atomic mass of Na) + (1 × atomic mass of O)
(2 × 23) + (1 × 16)
= 46 + 16
= 62 u
Formula unit mass of K₂CO₃ = (2 × atomic mass of K) + (1 × atomic mass of C) + (3 × atomic mass of O)
= (2 × 39) + (1 × 12) + (3 × 16)
= 78 + 12 + 48
= 138u.
Textbook Page No. 42.
1. If one mole of carbon atoms weighs 12 grams, what is the mass (in grams) of 1 atom of carbon?
Ans. 1 mole of carbon atoms = 6.022 × 10²³ atoms of carbon Thus, 6.022 × 10²³ atoms of carbon weighs 12 g
∴ Mass of 1 atom of carbon
= 1.99 × 10⁻²³
2. Which has more number of atoms, 100 grams of sodium or 100 grams of iron (given, atomic mass of Na= 23 u, Fe = 56 u)?
Ans. Number of mole atoms of sodium
= 4.35
1 mole atom of sodium contains 6.022 × 10²³ atoms
∴ 4.35 mole atoms of sodium contain = 4.35 × 6.022 × 10²³ atoms
= 2.62 × 10²⁴ atoms
Number of mole atoms of iron mass of iron grams / atomic mass of iron
= 100g/56g
= 1.79
1 mole atom of iron contains 6.022 × 10²³ atoms
1.79 mole atoms of iron contain 1.79 × 6.022 × 1023 atoms
= 1.08 × 10²⁴ atoms
Therefore, 100 g of sodium has more atoms than 100 g of iron.
EXERCISES (Textbook Page No. 43-44)
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 compound = 0.24g
Mass of boron in the compound = 0.096g
= 60%
Therefore, the percentage composition of the compound is 40% boron and 60% oxygen.
2. When 3.0 g of carbon is burnt in 8.00 g 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?
Ans. The law of constant proportions will govern the answer. Carbon and oxygen combine in fixed proportion of 3:8 by mass to produce 11 g of CO₂. Therefore, if we burn 3 g of carbon in 50 g of oxygen, the same mass of carbon dioxide, i.c. 11 g is obtained. The excess oxygen, i.e. 50 g – 8g – 42 g, will remain unreacted.
3. What are polyatomic ions? Give examples.
Ans. A polyatomic ion is a group of atoms having a charge and behaving as a single unit.
For example, hydroxide ion, OH⁻ is a polyatomic ion made up of two types of atoms (oxygen and hydrogen) joined together and carrying a negative charge.
Ammonium ion NH₄⁺, made up of two types of atoms (nitrogen and hydrogen) joined together. and carrying a positive charge.
4. Write the chemical formulae of the following.
(a) Magnesium chloride.
Ans. MgCI₂.
(b) Calcium oxide.
Ans. CaO.
(c) Copper nitrate.
Ans. Cu(NO₃)₂.
(d) Aluminium chloride.
Ans. AICI₃.
(e) Calcium carbonate.
Ans: CaCO₃.
5. Give the names of the elements present in the following compounds.
(a) Quick lime.
Ans. Quick lime: Calcium and oxygen.
(b) Hydrogen bromide.
Ans. Hydrogen bromide: Hydrogen and bromine.
(c) Baking powder.
Ans. Baking powder: Sodium, hydrogen, carbon and oxygen.
(d) Potassium sulphate.
Ans.Potassium sulphate: Potassium, sulphur and oxygen.
6. Calculate the molar mass of the following substances.
(a) Ethyne, C₂H₂
Ans. (a) Molecular mass of ethyne, C₂H₂
= (2 × atomic mass of C) + (2 × atomic mass of H)
= (2 × 12) + (2 × 1)
= 24 + 2
= 26 u
Hence, the molar mass of C₂H₂ is 26 g.
(b) Sulphur molecule, S₈.
Ans. Molecular mass of sulphur molecule, S₈
= (8 atomic mass of S)
= 8×32
= 256 U
Hence, the molar mass of S₈ is 256 g.
(c) Phosphorus molecule, P₄ (Atomic mass of phosphorus = 31)
Ans. Molecular mass of phosphorus molecule, P₄
= (4 × atomic mass of P)
= 4 × 31
= 124 u
Hence, the molar mass of P₄ is 124 g.
(d) Hydrochloride acid, HCI.
Ans. Molecular mass of hydrochloric acid, HCI
= (1 × Limic mass of H) + (1 x atomic mass of CI)
= (1 × 1) + (1 × 35-5)
= 1+35.5
= 36.5u
Hence, the molar mass of HCI is 36.5 g.
(e) Nitric acid, HNO₃.
Ans. Molecular mass of nitric acid, HNO₃
= (1× atomic mass of H) + (1 + atomic mass of N) + (3 + atomic mass of O)
= (1×1)+(1×1)+(3×16)
= 1+14+48
= 63u
Hence, the molar mass of HNO₃ is 63g.
7. What is the mass of-
(a) 1 mole of nitrogen atoms?
Ans. Mass of 1 mole of nitrogen atoms = Gram atomic mass of nitrogen = 14g.
(b) 4 moles of aluminium atoms (Atoms mass of aluminium = 27)?
Ans: Atomic mass of aluminium – 27 u [Given] Therefore, gram atomic mass of aluminium = 27g
Mass of 4 moles of aluminium atoms = 4 × mass of 1 mole of aluminium atom
= 4 × gram atomic mass of aluminium
= 4 × 27g
= 108g.
(c) 10 moles of sodium sulphite (Na₂SO₃).
Ans: Molecular mass of sodium sulphite, Na²SO³
= (2x atomic mass of Na)+(1x atomic mass of S)+(3x atomic mass of O)
= (2×23)+(1×32)+(3×16)
= 46+32+48
= 126u
Therefore, gram molecular mass of Na₂SO₃ is 126 g.
Mass of 10 moles of Na₂SO₃ =10 × mass of 1 mole of Na₂SO₃
= 10 gram molecular mass of Na₂SO₃
= 10 × 126 g
= 1260g.
8. Convert into mole.
(a) 12 g of oxygen gas.
Ans. (a) Molecular mass of oxygen gas, O₂ = 2 × atomic mass of oxygen
= 2 × 16
= 32u
Hence, the gram-molecular mass of oxygen gas is 32g.
∴ Number of moles
Thus, 12g of oxygen gas is equal to 0.375 mole.
(b) 20 g of water.
Ans: Molecular mass of water, H₂O = (2 × atomic mass of H) + (1 x atomic mass of O)
= (2 × 1) + (1 × 16)
= 2+16
= 18 u
Hence, the gram molecular mass of water is 18g.
∴ Number of moles
= 1.11
Thus, 20g of water is equal to 1.11 mole.
(c) 22 g of carbon dioxide.
Ans. Molecular mass of CO₂ = (1× atomic mass of C) + (2 × atomic mass of O)
= (1 × 12) + (2 × 16)
= 12+32
= 44 u
Hence, the gram molecular mass of CO₂ is 44g.
∴ Number of moles
= 0.5
Thus, 22g of CO₂ is equal to 0.5 mole.
9. What is the mass of.
(a) 0.2 mole of oxygen atoms?
Ans. Gram atomic mass of oxygen atom = 16g
∴ 1 mole of oxygen has a mass = 16g
∴ 0.2 mole of oxygen atoms has a mass = 16×0.2g=3.2g
Hence, the mass of 0.2 mole of oxygen atoms is 3.2g.
(b) 0.5 mole of water molecules?
Ans: Molecular Mass of water molecules, H₂O
= (2 × atomic mass of H) + (1 × atomic mass of O)
= (2 × 1) + (1 × 16)
= 2 + 16
= 18u
∴ Gram molecular mass of water molecule is 18g.
Thus, 1 mole of water molecules has a mass = 18g
∴ 0.5 mole of water molecules has a mass = 18 × 0.5g = 9g
Hence, the mass of 0•5 mole of water molecules is 9g.
10. Calculate the number of molecules of sulphur (S₈) present in 16g of solid sulphur.
Ans. We have to find the number of molecules present in 16g of solid sulphur.
The molecular mass of the sulphur molecule is 32 × 8 = 256 u.
1 mole of sulphur molecules is equal to 256 grams.
This means 256 g of solid sulphur contains = 6.022 x 1023 molecules.
Then, 16 g of solid sulphur contains = 6.022 x 1023 / 256 × 16 molecules = 3.76375 x 1022 molecules
Therefore, 16 g of S8 contains 3.76 x 1022 molecules.
11. Calculate the number of aluminium ions present in 0.051g of aluminium oxide.The mass of an ion is the same as that of an atom of the same element. Atomic mass of AI=27u).
Ans. Aluminum (Al) has an atomic mass of 27 u.
Oxygen (O) has an atomic mass of 16 u.
The molar mass of Al₂O₃ is calculated as:
Molar mass of Al2O3 = 2×27u+3×16u = 54u+48u = 102u
convert the mass of Al₂O₃ to moles:
Number of moles of Al2O3 = mass of Al₂O₃ / molar mass of Al₂O₃ – 102g/mol0.051g
= 0.0005mol
Determine the number of aluminum ions (Al³⁺) per formula unit of Al₂O₃: Each formula unit of Al₂O₃ contains 2 aluminum ions.
Calculate the number of moles of aluminum ions:
Moles of Al ions = 2×moles of Al2O3 = 2×0.0005mol = 0.001mol
convert moles of aluminum ions to number of ions:
Use Avogadro’s number (6.022×10236.022 \times 10^ {23} 6.022 × 1023 ions/mol):
Number of Al ions = 0.001 mol × 6.022 ×1023 ions/ mol\ text{Number of Al ions} = 0.001 \, \text{mol} \times 6.022 \times 1023 \, \text{ions/mol}Number of Al ions = 0.001mol× 6.022 × 1023 ions/mol)
Number of Al ions = 0.001mol×6.022×1023 ions/mol = 6.022×1020 ions = 6.022 \times 1020 \, \text {ions} = 6.022×1020 ions
So, the number of aluminum ions present in 0.051 g of aluminum oxide is 6.022×10206.022\times 1020 6.022×1020 ions.
