CBSE Class 9 Science Atoms and Molecules Sure Shot Questions Set 03

CBSE Class 9 Science Atoms and Molecules Sure Shot Questions C. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

ASSIGNMENT QUESTIONS SET - 3
CHAPTER - 3
ATOMS AND MOLECULES

1. Avogadro’s number represents the number of atoms in
(a) 12g of C12
(b) 320g of sulphur
(c) 32g of oxygen
(d) 12.7g of iodine

Answer: (a) 12g of C12

1. 1 mole of any substance contains 6.022 × 10²³ particles.
2. 12 g of carbon-12 is exactly 1 mole of carbon atoms, so it contains 6.022 × 10²³ atoms.
3. 320 g of sulphur is 10 moles, 32 g of oxygen (O₂) is 1 mole of molecules (2 moles of atoms) and 12.7 g of iodine is 0.1 mole of atoms, so none of these contains 6.022 × 10²³ atoms.
4. Hence, the correct option is (a).

Teacher's Note:
a) The printed 'C12' means the carbon-12 isotope; it cannot be chlorine because 12 g of Cl₂ is only 0.17 mole.
b) Avogadro's number is the number of atoms in 12 g of carbon-12.

2. The number of moles of carbon dioxide which contain 8 g of oxygen is
(a) 0.5 mol
(b) 0.20 mol
(c) 0.40 mol
(d) 0.25 mol

Answer: (d) 0.25 mol

1. Moles of oxygen atoms in 8 g = 8 ÷ 16 = 0.5 mole.
2. Each molecule of CO₂ contains 2 atoms of oxygen, so 1 mole of CO₂ contains 2 moles of oxygen atoms.
3. Moles of CO₂ = 0.5 ÷ 2 = 0.25 mol.
4. Hence, the correct option is (d).

Teacher's Note:
a) Always convert the mass of the element to moles of its atoms first.
b) Check: 0.25 mol of CO₂ = 11 g, which contains 8 g of oxygen (32/44 of 11).

3. The total no of ions present in 111 g of CaCl₂ is
(a) One mole
(b) Two mole
(c) Three mole
(d) Four moles

Answer: (c) Three mole

1. Molar mass of CaCl₂ = 40 + 2 × 35.5 = 111 g/mol.
2. 111 g of CaCl₂ = 1 mole.
3. Each formula unit gives 1 Ca²⁺ ion and 2 Cl⁻ ions, that is 3 ions.
4. Total ions = 3 moles.
5. Hence, the correct option is (c).

Teacher's Note:
a) One mole of CaCl₂ gives 3 moles of ions.

4. Which of the following weighs the most ?
(a) one g-atom of nitrogen
(b) One mole of water
(c) One mole of sodium
(d) One molecule of H₂SO₄

Answer: (c) One mole of sodium

1. (a) One g-atom of nitrogen = 14 g.
2. (b) One mole of water = 18 g.
3. (c) One mole of sodium = 23 g.
4. (d) One molecule of H₂SO₄ = 98 ÷ (6.022 × 10²³) g ≈ 1.6 × 10⁻²² g.
5. The largest mass is 23 g.
6. Hence, the correct option is (c).

Teacher's Note:
a) A single molecule has a tiny mass compared with a mole of any substance.

5. 5.0 litre of 0.4 M H₂SO₄ Contains-
(a) 2.0 Mole Of H₂SO₄
(b) 0.4 mole H₂SO₄
(c) 5.0 mole H₂SO₄
(d) 2.0 moles H₂O

Answer: (a) 2.0 Mole Of H₂SO₄

1. Moles of solute = molarity × volume in litres.
2. Moles of H₂SO₄ = 0.4 × 5.0 = 2.0 mol.
3. Hence, the correct option is (a).

Teacher's Note:
a) Molarity is the number of moles of solute per litre of solution.

6. Which of the following correctly represents 360 g of water?
(i) 2 moles of H₂0
(ii) 20 moles of water
(iii) 6.022 × 1023 molecules of water
(iv) 1.2044×1025 molecules of water
(a) (i) (b) (i) and (iv)
(c) (ii) and (iii) (d) (ii) and (iv)

Answer: (d) (ii) and (iv)

1. Molar mass of water = 18 g/mol.
2. Moles in 360 g = 360 ÷ 18 = 20 mol, so (ii) is correct and (i) is wrong.
3. Molecules = 20 × 6.022 × 10²³ = 1.2044 × 10²⁵, so (iv) is correct and (iii) is wrong.
4. (ii) and (iv) are correct.
5. Hence, the correct option is (d).

Teacher's Note:
a) The printed '1023' and '1025' stand for 10²³ and 10²⁵.
b) Always convert grams to moles before counting molecules.

7. Which of the following statements is not true about an atom?
(a) Atoms are not able to exist independently
(b) Atoms are the basic units from which molecules and ions are formed
(c) Atoms are always neutral in nature
(d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch

Answer: (a) Atoms are not able to exist independently

1. Statement (b) is true: atoms are the basic units that form molecules and ions.
2. Statement (c) is true: an atom has equal numbers of protons and electrons, so it is neutral.
3. Statement (d) is true: large numbers of atoms aggregate to form visible matter.
4. Statement (a) is not true as a general statement, because atoms of some elements, such as the noble gases, can exist independently.
5. Hence, the correct option is (a).

Teacher's Note:
a) The textbook sentence says atoms of most elements cannot exist independently; as a blanket statement it is false, which is why this option is chosen.

8. The chemical symbol for nitrogen gas is
(a) Ni (b) N₂ (c) N⁺ (d) N

Answer: (b) N₂

1. Nitrogen gas exists as diatomic molecules, so its formula is N₂.
2. Ni is nickel, N⁺ is an ion and N is the symbol of the atom.
3. Hence, the correct option is (b).

Teacher's Note:
a) The symbol of the element is N, but the gas is represented by the molecular formula N₂.

9. The chemical symbol for sodium is
(a) So (b) Sd (c) NA (d) Na

Answer: (d) Na

1. The symbol of sodium comes from its Latin name natrium: Na.
2. The first letter is capital and the second is small.
3. Hence, the correct option is (d).

Teacher's Note:
a) Na stands for natrium; K (kalium) for potassium is another similar example.

10. Which of the following would weigh the highest?
(a) 0.2 mole of sucrose (C12 H22 O11)
(b) 2 moles of CO₂
(c) 2 moles of CaCO₃
(d) 10 moles of H₂O

Answer: (c) 2 moles of CaCO₃

1. (a) 0.2 mole of sucrose (C₁₂H₂₂O₁₁): 0.2 × 342 = 68.4 g.
2. (b) 2 moles of CO₂: 2 × 44 = 88 g.
3. (c) 2 moles of CaCO₃: 2 × 100 = 200 g.
4. (d) 10 moles of H₂O: 10 × 18 = 180 g.
5. The highest mass is 200 g.
6. Hence, the correct option is (c).

Teacher's Note:
a) Mass = moles × molar mass; compare all four values.

11. Which of the following has maximum number of atoms?
(a) 18g of H₂O
(b) 18g of O₂
(c) 18g of CO₂
(d) 18g of CH₄

Answer: (d) 18g of CH₄

1. Number of atoms = moles × atoms per molecule (in units of Avogadro's number).
2. (a) 18 g H₂O = 1 mol, so atoms = 1 × 3 = 3.
3. (b) 18 g O₂ = 0.5625 mol, so atoms = 0.5625 × 2 = 1.125.
4. (c) 18 g CO₂ = 0.409 mol, so atoms = 0.409 × 3 = 1.23.
5. (d) 18 g CH₄ = 1.125 mol, so atoms = 1.125 × 5 = 5.625.
6. CH₄ has the largest value.
7. Hence, the correct option is (d).

Teacher's Note:
a) The lightest molecule with the most atoms gives the largest count for the same mass.

12. Which of the following contains maximum number of molecules?
(a) 1g CO₂
(b) 1g N₂
(c) 1g H₂
(d) 1g CH₄

Answer: (c) 1g H₂

1. For equal masses, the number of molecules is largest for the smallest molar mass.
2. Molar masses: CO₂ = 44, N₂ = 28, H₂ = 2, CH₄ = 16 g/mol.
3. Hence, 1 g of H₂ = 0.5 mol is the largest number of molecules.
4. Hence, the correct option is (c).

Teacher's Note:
a) Number of molecules = (mass ÷ molar mass) × 6.022 × 10²³.

13. Mass of one atom of oxygen is
(a) \( \frac{16}{6.023\times 10^{23}} \) g
(b) \( \frac{32}{6.023\times 10^{23}} \) g
(c) \( \frac{1}{6.023\times 10^{23}} \) g
(d) 8u

Answer: (a) \( \frac{16}{6.023\times 10^{23}} \) g

1. 1 mole of oxygen atoms has a mass of 16 g and contains 6.023 × 10²³ atoms.
2. Mass of one atom = \( \frac{16}{6.023\times 10^{23}} \) g.
3. (d) 8u is wrong; the mass of an oxygen atom is 16 u.
4. Hence, the correct option is (a).

Teacher's Note:
a) The molar mass of atomic oxygen is 16 g, not 32 g (which is O₂).

14. 3.42 g of sucrose are dissolved in 18g of water in a beaker. The number of oxygen atoms in the solution are
(a) 6.68 × 10²³
(b) 6.09 × 10²²
(c) 6.022 × 10²³
(d) 6.022 × 10²¹

Answer: (a) 6.68 × 10²³

1. Molar mass of sucrose C₁₂H₂₂O₁₁ = 342 g/mol, so 3.42 g = 0.01 mol.
2. Oxygen atoms from sucrose = 0.01 × 11 = 0.11 mol.
3. 18 g of water = 1 mol, which gives 1 mol of oxygen atoms.
4. Total oxygen = 1.11 mol = 1.11 × 6.022 × 10²³ ≈ 6.68 × 10²³ atoms.
5. Hence, the correct option is (a).

Teacher's Note:
a) Do not forget the oxygen atoms of the solvent water.

15. A change in the physical state can be brought about
(a) only when energy is given to the system
(b) only when energy is taken out from the system
(c) when energy is either given to, or taken out from the system
(d) without any energy change

Answer: (c) when energy is either given to, or taken out from the system

1. A change of state (melting, boiling, condensing, freezing) involves absorption or release of heat energy.
2. Heating gives energy (solid to liquid to gas) and cooling removes energy (gas to liquid to solid).
3. So energy may be given to or taken out of the system.
4. Hence, the correct option is (c).

Teacher's Note:
a) A change of state is a physical change that always involves an energy transfer.

16. Which of the following represents a correct chemical formula? Name it.
(a) CaCl (b) BiPO₄ (c) NaSO₄ (d) NaS

Answer: (b) BiPO₄

1. Check the charges: Ca²⁺ with Cl⁻ needs CaCl₂, so CaCl is wrong.
2. Bi³⁺ with PO₄³⁻ balances 1 : 1, so BiPO₄ is correct; it is bismuth phosphate.
3. Na⁺ with SO₄²⁻ needs Na₂SO₄, so NaSO₄ is wrong.
4. Na⁺ with S²⁻ needs Na₂S, so NaS is wrong.
5. Hence, the correct option is (b), BiPO₄ (bismuth phosphate).

Teacher's Note:
a) The total positive charge must equal the total negative charge in a formula.

17. Write the molecular formulae for the following compounds
(a) Copper (II) bromide
(b) Aluminium (III) nitrate
(c) Calcium (II) phosphate
(d) Iron (III) sulphide
(e) Mercury (II) chloride
(f) Magnesium (II) acetate

Answer:
1. (a) Cu²⁺ and Br⁻: CuBr₂.
2. (b) Al³⁺ and NO₃⁻: Al(NO₃)₃.
3. (c) Ca²⁺ and PO₄³⁻: Ca₃(PO₄)₂.
4. (d) Fe³⁺ and S²⁻: Fe₂S₃.
5. (e) Hg²⁺ and Cl⁻: HgCl₂.
6. (f) Mg²⁺ and CH₃COO⁻: Mg(CH₃COO)₂.

Teacher's Note:
a) Use the criss-cross method: the charge number of one ion becomes the subscript of the other.
b) Put brackets around a polyatomic ion when its subscript is more than 1.

18. Write the molecular formulae of all the compounds that can be formed by the combination of following ions
Cu²⁺, Na⁺, Fe³⁺, C1^-, SO₄^2-, PO₄^3-

Answer:
1. Combine each cation with each anion by the criss-cross method.
2. With Cl⁻: CuCl₂, NaCl, FeCl₃.
3. With SO₄²⁻: CuSO₄, Na₂SO₄, Fe₂(SO₄)₃.
4. With PO₄³⁻: Cu₃(PO₄)₂, Na₃PO₄, FePO₄.
5. Hence, nine compounds can be formed.

Teacher's Note:
a) 3 cations × 3 anions = 9 compounds.
b) The printed 'C1' is the chloride ion Cl⁻.

19. Write the cations and anions present (if any) in the following compounds
(a) CH₃COONa
(b) NaCl
(c) H₂
(d) NH₄NO₃

Answer:
1. (a) CH₃COONa: cation Na⁺ and anion CH₃COO⁻ (acetate).
2. (b) NaCl: cation Na⁺ and anion Cl⁻.
3. (c) H₂: it is a molecule of an element; there are no ions.
4. (d) NH₄NO₃: cation NH₄⁺ (ammonium) and anion NO₃⁻ (nitrate).

Teacher's Note:
a) Ionic compounds contain ions; elements such as H₂ are neutral molecules.

20. Give the formulae of the compounds formed from the following sets of elements
(a) Calcium and fluorine
(b) Hydrogen and sulphur
(c) Nitrogen and hydrogen
(d) Carbon and chlorine
(e) Sodium and oxygen
(f) Carbon and oxygen

Answer:
1. (a) Ca²⁺ and F⁻: CaF₂.
2. (b) H⁺ and S²⁻: H₂S.
3. (c) N³⁻ and H⁺: NH₃.
4. (d) Carbon (valency 4) and Cl⁻: CCl₄.
5. (e) Na⁺ and O²⁻: Na₂O.
6. (f) Carbon (valency 4) and O²⁻: CO₂ (carbon also forms CO).

Teacher's Note:
a) Write the symbol of the element with the positive charge first.
b) The common compound is given in (f); CO is also a valid compound.

21. Which of the following symbols of elements are incorrect? Give their correct symbols
(a) Cobalt CO
(b) Carbon c
(c) Aluminium AL
(d) Helium He
(e) Sodium So

Answer:
1. (a) Cobalt: CO is incorrect (it is the formula of carbon monoxide); the correct symbol is Co.
2. (b) Carbon: c is incorrect; the correct symbol is C.
3. (c) Aluminium: AL is incorrect; the correct symbol is Al.
4. (d) Helium: He is correct.
5. (e) Sodium: So is incorrect; the correct symbol is Na.

Teacher's Note:
a) The first letter of a symbol is always capital and the second is small.

22. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
(a) Ammonia
(b) Carbon monoxide
(c) Hydrogen chloride
(d) Aluminium fluoride
(e) Magnesium sulphide

Answer:
1. (a) NH₃: N : H = 14 : (3 × 1) = 14 : 3.
2. (b) CO: C : O = 12 : 16 = 3 : 4.
3. (c) HCl: H : Cl = 1 : 35.5 = 2 : 71.
4. (d) AlF₃: Al : F = 27 : (3 × 19) = 27 : 57 = 9 : 19.
5. (e) MgS: Mg : S = 24 : 32 = 3 : 4.

Teacher's Note:
a) Ratio by mass = (atomic mass × number of atoms) of each element; reduce to lowest terms.
b) Atomic masses used: H = 1, C = 12, N = 14, O = 16, F = 19, Mg = 24, Al = 27, S = 32, Cl = 35.5.

23. State the number of atoms present in each of the following chemical species
(a) CO₃^2-
(b) PO₄^3-
(c) P₂O₅
(d) CO

Answer:
1. (a) CO₃²⁻: 1 + 3 = 4 atoms.
2. (b) PO₄³⁻: 1 + 4 = 5 atoms.
3. (c) P₂O₅: 2 + 5 = 7 atoms.
4. (d) CO: 1 + 1 = 2 atoms.

Teacher's Note:
a) Add the subscripts; a missing subscript means 1.

24. Find the ratio by mass of the combining elements in the compound - C₂H₅OH.

Answer:
1. C₂H₅OH contains 2 C, 6 H and 1 O atoms.
2. Mass of C = 2 × 12 = 24; mass of H = 6 × 1 = 6; mass of O = 16.
3. C : H : O = 24 : 6 : 16 = 12 : 3 : 8.
4. Hence, the ratio by mass is C : H : O = 12 : 3 : 8.

Teacher's Note:
a) Count the hydrogen atoms in both C₂H₅ and OH: 5 + 1 = 6.

25. Give the formula of the compound formed by the elements calcium and fluorine.

Answer:
1. Calcium forms Ca²⁺ and fluorine forms F⁻.
2. Criss-cross the charges: Ca₁F₂.
3. Hence, the formula is CaF₂.

Teacher's Note:
a) The compound is calcium fluoride.

26. What is the acid radical present in sodium peroxide?

Answer:
1. The formula of sodium peroxide is Na₂O₂.
2. It contains Na⁺ ions and the peroxide ion O₂²⁻.
3. Hence, the acid radical is the peroxide ion, O₂²⁻.

Teacher's Note:
a) Do not confuse the peroxide ion O₂²⁻ with the oxide ion O²⁻.

27. Carbon and silicon have the same valency. What is the formula of sodium silicate?

Answer:
1. Carbon forms the carbonate ion CO₃²⁻, so silicon forms the silicate ion SiO₃²⁻.
2. Sodium has valency 1: Na⁺ with SiO₃²⁻ gives Na₂SiO₃.
3. Hence, the formula of sodium silicate is Na₂SiO₃.

Teacher's Note:
a) Sodium carbonate Na₂CO₃ is the model compound for this question.

28. What is the ratio by number of atoms in mercurous chloride?

Answer:
1. Mercurous chloride is Hg₂Cl₂ (mercury(I) chloride).
2. Number of Hg atoms : number of Cl atoms = 2 : 2 = 1 : 1.
3. Hence, the ratio by number of atoms is 1 : 1.

Teacher's Note:
a) Mercurous ion is Hg₂²⁺, so the formula is Hg₂Cl₂, not HgCl.

29. Name the element whose Latin name is Stibium.

Answer:
1. Stibium is the Latin name of antimony.
2. Its symbol is Sb.
3. Hence, the element is antimony.

Teacher's Note:
a) Many symbols come from Latin names, such as Na (natrium) and Fe (ferrum).

30. What is the valency of a sulphide ion?

Answer:
1. The sulphide ion is S²⁻; it has gained 2 electrons.
2. Hence, its valency is 2.

Teacher's Note:
a) Valency is the number of electrons gained or lost to form an ion.

31. How many atoms of oxygen are present in 50g of CaCO₃?

Answer:
1. Molar mass of CaCO₃ = 40 + 12 + 48 = 100 g/mol.
2. Moles of CaCO₃ = 50 ÷ 100 = 0.5 mol.
3. Each formula unit has 3 oxygen atoms, so moles of O atoms = 0.5 × 3 = 1.5 mol.
4. Number of O atoms = 1.5 × 6.022 × 10²³ = 9.033 × 10²³.
5. Hence, 9.033 × 10²³ atoms of oxygen are present.

Teacher's Note:
a) Multiply moles by the number of atoms per formula unit before using Avogadro's number.

32. How many molecules are present in 1 ml of water?

Answer:
1. The density of water is 1 g/mL, so 1 mL of water = 1 g.
2. Moles of water = 1 ÷ 18 = 0.0556 mol.
3. Number of molecules = (1 ÷ 18) × 6.022 × 10²³ = 3.35 × 10²².
4. Hence, 1 mL of water has about 3.35 × 10²² molecules.

Teacher's Note:
a) Density 1 g/mL lets you treat the volume in mL as the mass in grams.

33. What is the unit of measurement of atomic radius?

Answer:
1. Atomic radii are extremely small.
2. They are measured in nanometres (nm), where 1 nm = 10⁻⁹ m.
3. Hence, the unit is the nanometre.

Teacher's Note:
a) Another unit sometimes used is the angstrom, 1 Å = 10⁻¹⁰ m.

34. Name the international organization who approves names of elements.

Answer:
1. The names and symbols of the elements are approved by IUPAC.
2. IUPAC stands for the International Union of Pure and Applied Chemistry.
3. Hence, the organisation is IUPAC.

Teacher's Note:
a) IUPAC also decides the names of new elements.

35. How do we know the presence of atoms if they do not exist independently for most of the elements?

Answer:
1. Atoms of most elements exist as molecules (such as H₂, O₂, P₄) or as ions in compounds.
2. These molecules and ions are made up of atoms in fixed numbers, as the chemical formulae and the laws of chemical combination show.
3. Large aggregates of atoms form the matter we see, and modern techniques such as scanning tunnelling microscopes can image individual atoms.
4. Hence, the presence of atoms is known from molecules, ions and compounds built from them.

Teacher's Note:
a) The formula of a compound, such as H₂O, tells us how many atoms of each kind combine.

36. Give an example to show Law of conservation of mass applies to physical change also.

Answer:
1. Take 50 g of water and dissolve 5 g of sugar in it; this is a physical change.
2. The mass of the sugar solution formed is 55 g.
3. Mass before = 50 + 5 = 55 g and mass after = 55 g.
4. Another example is melting of ice: 18 g of ice gives 18 g of water.
5. Hence, mass is conserved in physical changes too.

Teacher's Note:
a) Dissolving and melting are physical changes in which no new substance forms, yet the mass remains the same.

37. Explain with example that law of conservation of mass is valid for chemical reactions.

Answer:
1. Mix sodium sulphate solution with barium chloride solution: Na₂SO₄ + BaCl₂ → BaSO₄ + 2NaCl.
2. Mass of reactants = 142 g + 208 g = 350 g.
3. Mass of products = 233 g + 117 g = 350 g.
4. The mass of the reactants equals the mass of the products.
5. Hence, the law of conservation of mass holds in chemical reactions.

Teacher's Note:
a) The experiment is done in a closed vessel so that no gas escapes.
b) The law says mass can neither be created nor destroyed in a chemical reaction.

38. Is there any exception to law of conservation of mass?

Answer:
1. Yes, in nuclear reactions (fission and fusion) a small amount of mass is converted into a large amount of energy.
2. Einstein's equation E = mc² relates the energy released to the mass lost.
3. In ordinary chemical reactions, the law holds.
4. Hence, nuclear reactions are the exception.

Teacher's Note:
a) In a nuclear reaction it is the sum of mass and energy that is conserved.

39. 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 → sodium ethanoate + carbon dioxide + water

Answer:
1. Mass of reactants = 5.3 g + 6 g = 11.3 g.
2. Mass of products = 8.2 g + 2.2 g + 0.9 g = 11.3 g.
3. Mass of reactants = mass of products.
4. Hence, the observations agree with the law of conservation of mass.

Teacher's Note:
a) Add the masses on each side separately and then compare.

40. If 12 g of carbon is burnt in the presence of 32 g of oxygen, how much carbon dioxide will be formed?

Answer:
1. C + O₂ → CO₂.
2. 12 g of carbon reacts exactly with 32 g of oxygen.
3. Mass of CO₂ = 12 + 32 = 44 g (law of conservation of mass).
4. Hence, 44 g of carbon dioxide will be formed.

Teacher's Note:
a) Check the mole ratio: 1 mol of C (12 g) reacts with 1 mol of O₂ (32 g) to give 1 mol of CO₂ (44 g).

41. 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.

Answer:
1. Percentage of boron = (0.096 ÷ 0.24) × 100 = 40%.
2. Percentage of oxygen = (0.144 ÷ 0.24) × 100 = 60%.
3. Hence, the compound contains 40% boron and 60% oxygen by weight.

Teacher's Note:
a) The two percentages must add up to 100.

42. 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?

Answer:
1. Carbon and oxygen combine in the fixed ratio 3 : 8 by mass to form CO₂.
2. 3.00 g of carbon needs only 8.00 g of oxygen; the remaining 42.00 g of oxygen is left unreacted.
3. Mass of CO₂ formed = 3.00 + 8.00 = 11.00 g.
4. Hence, 11.00 g of carbon dioxide will be formed, and the answer is governed by the law of constant proportions.

Teacher's Note:
a) The amount of product depends on the limiting reactant, which is carbon here.
b) The law of constant proportions: a compound always contains the same elements in the same fixed ratio by mass.

43. Magnesium and oxygen combine in the ratio of 3 : 2 by mass to form magnesium oxide. How much oxygen is required to react completely with 12 g of magnesium?

Answer:
1. Mg : O = 3 : 2 by mass.
2. Mass of oxygen = (2 ÷ 3) × 12 = 8 g.
3. Hence, 8 g of oxygen is required.

Teacher's Note:
a) Check: 12 : 8 = 3 : 2.

44. Why are Dalton's symbol not used in chemistry?

Answer:
1. Dalton's symbols were pictorial signs (circles with marks) for each element.
2. They were difficult to draw and remember, especially for the large number of elements, and could not show compounds easily.
3. Berzelius suggested using one or two letters from the name of the element instead.
4. Hence, Dalton's symbols are replaced by the simple letter symbols approved by IUPAC.

Teacher's Note:
a) Modern symbols are universal and easy to write.

45. What is the fraction of the mass of water due to neutrons?

Answer:
1. Water is H₂O. The common hydrogen atom has 0 neutrons and oxygen-16 has 8 neutrons.
2. Neutrons in one molecule = 0 + 0 + 8 = 8.
3. Each neutron has a mass of about 1 u, so the mass due to neutrons ≈ 8 u.
4. Mass of one molecule of water = 18 u.
5. Fraction = 8 ÷ 18 = \( \frac{4}{9} \).
6. Hence, about \( \frac{4}{9} \) of the mass of water is due to neutrons.

Teacher's Note:
a) The mass of electrons is negligible; protons and neutrons make up almost all of the mass.

46. Does the solubility of a substance change with temperature? Explain with the help of an example.

Answer:
1. Yes, the solubility of a substance changes with temperature.
2. For most solids, solubility increases on heating. For example, 100 g of water dissolves about 36 g of sodium chloride at 20°C, but a much larger amount of potassium nitrate at 80°C than at 20°C.
3. For gases, solubility decreases when the temperature rises; this is why dissolved oxygen in warm water is lower.
4. Hence, solubility depends on temperature.

Teacher's Note:
a) Mention one solid and one gas to cover both trends.

47. Classify each of the following on the basis of their atomicity.
(a) F₂ (b) NO₂ (c) N₂O (d) C₂H₆ (e) P₄ (f) H₂O₂
(g) P₄O₁₀ (h) O₃ (i) HCl (j) CH₄ (k) He (l) Ag

Answer:
1. (a) F₂: diatomic (2 atoms).
2. (b) NO₂: triatomic (3 atoms).
3. (c) N₂O: triatomic (3 atoms).
4. (d) C₂H₆: polyatomic (8 atoms).
5. (e) P₄: tetratomic (4 atoms).
6. (f) H₂O₂: tetratomic (4 atoms).
7. (g) P₄O₁₀: polyatomic (14 atoms).
8. (h) O₃: triatomic (3 atoms).
9. (i) HCl: diatomic (2 atoms).
10. (j) CH₄: polyatomic (5 atoms).
11. (k) He: monoatomic (1 atom).
12. (l) Ag: monoatomic (1 atom).

Teacher's Note:
a) Atomicity is the number of atoms in one molecule of the substance.
b) Metals and noble gases are monoatomic.

48. You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

Answer:
1. Heat a small amount of the powder on a spatula over a flame.
2. Sugar melts, turns brown and chars, giving black carbon and a smell of burning sugar.
3. Salt does not char; it only crackles and remains white.
4. Another test: dissolve each in water; the salt solution conducts electricity but the sugar solution does not.
5. Hence, the powder can be identified without tasting.

Teacher's Note:
a) Never taste unknown chemicals in a laboratory.

49. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24g mol^-1.

Answer:
1. Number of moles = given mass ÷ molar mass.
2. Moles of Mg = 12 ÷ 24 = 0.5 mol.
3. Hence, the ribbon contains 0.5 mole of magnesium.

Teacher's Note:
a) 0.5 mol corresponds to 3.011 × 10²³ atoms.

50. Verify by calculating that
(a) 5 moles of CO₂ and 5 moles of H₂O do not have the same mass.
(b) 240 g of calcium and 240 g magnesium elements have a mole ratio of 3:5.

Answer:
1. (a) Mass of 5 mol CO₂ = 5 × 44 = 220 g.
2. Mass of 5 mol H₂O = 5 × 18 = 90 g.
3. 220 g ≠ 90 g, so the masses are not the same.
4. (b) Moles of calcium = 240 ÷ 40 = 6 mol.
5. Moles of magnesium = 240 ÷ 24 = 10 mol.
6. Ratio Ca : Mg = 6 : 10 = 3 : 5.
7. Hence, both statements are verified.

Teacher's Note:
a) Equal moles do not mean equal masses unless the molar masses are equal.

51. Find the ratio by mass of the combining elements in the following compounds.
(a) CaCO₃ (d) C₂H₅OH
(b) MgCl₂ (e) NH₃
(c) H₂SO₄ (f) Ca(OH)₂

Answer:
1. (a) CaCO₃: Ca : C : O = 40 : 12 : 48 = 10 : 3 : 12.
2. (b) MgCl₂: Mg : Cl = 24 : 71.
3. (c) H₂SO₄: H : S : O = 2 : 32 : 64 = 1 : 16 : 32.
4. (d) C₂H₅OH: C : H : O = 24 : 6 : 16 = 12 : 3 : 8.
5. (e) NH₃: N : H = 14 : 3.
6. (f) Ca(OH)₂: Ca : O : H = 40 : 32 : 2 = 20 : 16 : 1.

Teacher's Note:
a) Multiply the atomic mass by the number of atoms of each element, then simplify.

52. Calcium chloride when dissolved in water dissociates into its ions according to the following equation.
CaCl₂ (aq) →Ca²⁺ (aq) + 2Cl^- (aq)
Calculate the number of ions obtained from CaCl₂ when 222 g of it is dissolved in water.

Answer:
1. Molar mass of CaCl₂ = 111 g/mol, so 222 g = 2 mol.
2. Each formula unit gives 3 ions (1 Ca²⁺ + 2 Cl⁻).
3. Moles of ions = 2 × 3 = 6 mol.
4. Number of ions = 6 × 6.022 × 10²³ = 3.613 × 10²⁴.
5. Hence, 3.613 × 10²⁴ ions are obtained.

Teacher's Note:
a) Convert grams to moles and then multiply by the ions per formula unit.

53. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002g. Compute the mass of an electron.

Answer:
1. A sodium atom becomes Na⁺ by losing 1 electron, so the difference in mass is the mass of the electrons lost.
2. Number of electrons = 100 × 6.022 × 10²³ = 6.022 × 10²⁵.
3. Mass of one electron = 5.48002 ÷ (6.022 × 10²⁵) = 9.1 × 10⁻²⁶ g, using the printed data.
4. Hence, with the printed data, the mass of an electron works out to 9.1 × 10⁻²⁶ g.

Teacher's Note:
a) The printed data are inconsistent with the real mass of an electron, 9.1 × 10⁻²⁸ g; the correct difference for 100 moles would be 0.0548 g, so the source value 5.48002 g is too large.
b) The method (mass difference ÷ number of electrons) is what is tested.

54. Complete the following crossword puzzle (below Figure) by using the name of the chemical elements. Use the data given in below Table

Table 3.2

[Figure: A crossword grid with eight numbered entries. Across: 2 (4 boxes), 3 (4 boxes), 5 (10 boxes), 7 (8 boxes). Down: 1 (6 boxes), 4 (6 boxes), 6 (7 boxes), 8 (4 boxes). The entries cross each other at shared boxes]

Answer:
1. Across: 2. Gold (Rutherford's α-scattering foil).
2. Across: 3. Iron (rust is hydrated iron oxide).
3. Across: 5. Phosphorus (white phosphorus is stored under water).
4. Across: 7. Hydrogen (burns with a pop sound).
5. Down: 1. Silver (tarnishes black as silver sulphide).
6. Down: 4. Copper (brass and bronze are copper alloys).
7. Down: 6. Mercury (liquid metal at room temperature).
8. Down: 8. Lead (symbol Pb, from plumbum).
9. The words share letters at the crossings; for example, the L of Gold is the L of Silver and the P of Phosphorus is the P of Copper.

Teacher's Note:
a) Check the number of boxes in each answer against the lengths in the grid: Gold 4, Iron 4, Phosphorus 10, Hydrogen 8, Silver 6, Copper 6, Mercury 7, Lead 4.
b) Several symbols come from Latin names: Au (aurum), Fe (ferrum), Ag (argentum), Cu (cuprum), Hg (hydrargyrum) and Pb (plumbum).

55. Fill in the missing data in the below Table

Answer:
1. H₂O: 2 mol has 2 × 6.022 × 10²³ = 1.2044 × 10²⁴ molecules and a mass of 36 g (given, 2 × 18).
2. CO₂: 3.011 × 10²³ ÷ 6.022 × 10²³ = 0.5 mol; mass = 0.5 × 44 = 22 g.
3. Na atom: 115 g ÷ 23 = 5 mol; particles = 5 × 6.022 × 10²³ = 3.011 × 10²⁴.
4. MgCl₂: 0.5 mol has 0.5 × 6.022 × 10²³ = 3.011 × 10²³ formula units; molar mass = 24 + 71 = 95 g/mol, so mass = 0.5 × 95 = 47.5 g.
5. The completed table is given below.

Completed table

Teacher's Note:
a) Moles = particles ÷ 6.022 × 10²³ = mass ÷ molar mass; fill each column using the one value given.
b) The 'particles' of MgCl₂ are formula units, since it is an ionic compound.

56. The visible universe is estimated to contain 10²² stars. How many moles of stars are present in the visible universe?

Answer:
1. Number of moles = number of particles ÷ 6.022 × 10²³.
2. Moles of stars = 10²² ÷ (6.022 × 10²³) = 0.0166 mol.
3. Hence, the visible universe contains about 1.66 × 10⁻² mole of stars.

Teacher's Note:
a) One mole is a huge number; even the stars of the universe are less than a fiftieth of a mole.

57. What is the SI prefix for each of the following multiples and submultiples of a unit?
(a) 10³ (b) 10^-1 (c) 10^-2 (d) 10^-6 (e) 10^-9 (f) 10^-12

Answer:
1. (a) 10³: kilo (k).
2. (b) 10⁻¹: deci (d).
3. (c) 10⁻²: centi (c).
4. (d) 10⁻⁶: micro (μ).
5. (e) 10⁻⁹: nano (n).
6. (f) 10⁻¹²: pico (p).

Teacher's Note:
a) Learn the prefixes in order of power: kilo, hecto, deca, deci, centi, milli, micro, nano, pico.

58. (a) In this crossword puzzle (Fig 3.2), names of 11 elements are hidden. Symbols of these are given below. Complete the puzzle.
1. Cl 7. He
2. H 8. F
3. Ar 9. Kr
4. O 10. Rn
5. Xe 11. Ne
6. N
(b) Identify the total number of inert gases, their names and symbols from this cross word puzzle.

[Figure: A crossword grid for Fig 3.2 with eleven numbered entries. Across: 2 (8 boxes), 3 (5 boxes), 4 (6 boxes), 6 (8 boxes), 9 (7 boxes), 10 (5 boxes), 11 (4 boxes). Down: 1 (8 boxes), 5 (5 boxes), 7 (6 boxes), 8 (8 boxes). The boxes of the inert gases are shaded and a legend marked Inert gases is given below the grid]

Answer:
1. (a) Across: 2. Hydrogen, 3. Argon, 4. Oxygen, 6. Nitrogen, 9. Krypton, 10. Radon, 11. Neon.
2. Down: 1. Chlorine, 5. Xenon, 7. Helium, 8. Fluorine.
3. The words share letters at the crossings; for example, the X of Oxygen is the X of Xenon and the last N of Xenon is the N of Radon.
4. (b) The shaded boxes are the inert (noble) gases. Their total number is 6.
5. The six inert gases are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe) and Radon (Rn).
6. Chlorine, hydrogen, oxygen, nitrogen and fluorine are not inert gases.

Teacher's Note:
a) Inert gases are the elements of Group 18 of the periodic table; their outermost shell is complete.
b) The grid was read from the printed figure; the number of boxes in each word is the check for every answer.

59. Express each of the following in kilograms
(a) 5.84×10^-3 mg
(b) 58.34 g
(c) 0.584g
(d) 5.873×10^-21g

Answer:
1. 1 mg = 10⁻⁶ kg and 1 g = 10⁻³ kg.
2. (a) 5.84 × 10⁻³ mg = 5.84 × 10⁻³ × 10⁻⁶ kg = 5.84 × 10⁻⁹ kg.
3. (b) 58.34 g = 58.34 × 10⁻³ kg = 5.834 × 10⁻² kg.
4. (c) 0.584 g = 0.584 × 10⁻³ kg = 5.84 × 10⁻⁴ kg.
5. (d) 5.873 × 10⁻²¹ g = 5.873 × 10⁻²¹ × 10⁻³ kg = 5.873 × 10⁻²⁴ kg.

Teacher's Note:
a) Multiply by the conversion factor and add the powers of ten.

60. Compute the difference in masses of 103 moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1×10^-31 kg)

Answer:
1. The printed '103 moles' stands for 10³ moles = 1000 moles.
2. A Mg atom forms Mg²⁺ by losing 2 electrons.
3. Number of electrons lost = 1000 × 6.022 × 10²³ × 2 = 1.2044 × 10²⁷.
4. Difference in mass = 1.2044 × 10²⁷ × 9.1 × 10⁻³¹ kg = 1.096 × 10⁻³ kg.
5. Hence, the atoms are heavier by about 1.1 × 10⁻³ kg (1.096 g).

Teacher's Note:
a) The ion is lighter than the atom by the mass of the lost electrons.
b) The charge of Mg²⁺ is 2+, so two electrons are lost per atom.

61. Which has more number of atoms? 100g of N₂ or 100 g of NH₃

Answer:
1. Moles of N₂ = 100 ÷ 28 = 3.57 mol; atoms = 3.57 × 2 = 7.14 mol of atoms.
2. Moles of NH₃ = 100 ÷ 17 = 5.88 mol; atoms = 5.88 × 4 = 23.5 mol of atoms.
3. 23.5 mol of atoms is more than 7.14 mol of atoms.
4. Hence, 100 g of NH₃ has more atoms.

Teacher's Note:
a) Count the atoms per molecule (2 for N₂ and 4 for NH₃) after finding the moles.

62. Compute the number of ions present in 5.85 g of sodium chloride.

Answer:
1. Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol.
2. Moles of NaCl = 5.85 ÷ 58.5 = 0.1 mol.
3. Each formula unit gives 2 ions (Na⁺ and Cl⁻), so moles of ions = 0.2 mol.
4. Number of ions = 0.2 × 6.022 × 10²³ = 1.2044 × 10²³.
5. Hence, 1.2044 × 10²³ ions are present.

Teacher's Note:
a) NaCl has two ions per formula unit, unlike CaCl₂, which has three.

63. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

Answer:
1. Mass of gold in 1 g of the sample = 0.9 g.
2. Molar mass of gold = 197 g/mol.
3. Moles of gold = 0.9 ÷ 197 = 4.568 × 10⁻³ mol.
4. Number of atoms = 4.568 × 10⁻³ × 6.022 × 10²³ = 2.75 × 10²¹.
5. Hence, 2.75 × 10²¹ atoms of gold are present.

Teacher's Note:
a) The copper in the sample does not count.

64. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS? Molar mass of Hg and S are 200.6 g mol^-1 and 32 g mol^-1 respectively.

Answer:
1. Molar mass of HgS = 200.6 + 32 = 232.6 g/mol.
2. 232.6 g of HgS contains 200.6 g of Hg.
3. Mass of Hg in 225 g = (200.6 ÷ 232.6) × 225 = 194.05 g.
4. Hence, about 194 g of mercury is present.

Teacher's Note:
a) The mass fraction of mercury in HgS is 200.6 ÷ 232.6 ≈ 0.862.

65. The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 × 10²⁴kg). Which one of the two is heavier and by how many times?

Answer:
1. Mass of one mole of screws = 4.11 g × 6.022 × 10²³ = 2.475 × 10²⁴ g.
2. In kilograms: 2.475 × 10²⁴ g = 2.475 × 10²¹ kg.
3. Mass of the Earth = 5.98 × 10²⁴ kg.
4. Ratio = 5.98 × 10²⁴ ÷ 2.475 × 10²¹ ≈ 2416.
5. Hence, the Earth is heavier, by about 2.4 × 10³ times.

Teacher's Note:
a) Convert grams to kilograms before comparing.
b) A mole of screws is still much lighter than the Earth.

66. A sample of vitamic C is known to contain 2.58 ×10²⁴ oxygen atoms. How many moles of oxygen atoms are present in the sample?

Answer:
1. Moles = number of atoms ÷ 6.022 × 10²³.
2. Moles of oxygen atoms = 2.58 × 10²⁴ ÷ 6.022 × 10²³ = 4.28 mol.
3. Hence, 4.28 moles of oxygen atoms are present.

Teacher's Note:
a) The printed 'vitamic C' is vitamin C.

67. Raunak took 5 moles of carbon atoms in a container and Krish also took 5 moles of sodium atoms in another container of same weight. (a) Whose container is heavier? (b) Whose container has more number of atoms?

Answer:
1. Mass of 5 mol of carbon = 5 × 12 = 60 g.
2. Mass of 5 mol of sodium = 5 × 23 = 115 g.
3. (a) The empty containers have the same weight, so Krish's container (115 g of sodium) is heavier.
4. (b) Both have 5 moles, that is 5 × 6.022 × 10²³ atoms, so the number of atoms is the same.
5. Hence, (a) Krish's container is heavier and (b) both have the same number of atoms.

Teacher's Note:
a) Equal moles means equal numbers of atoms, but not equal masses.

68. What are ionic and molecular compounds? Give examples.

Answer:
1. Ionic compounds are made of positive and negative ions held by electrostatic attraction. Example: NaCl, CaCl₂, MgO.
2. Molecular compounds are made of molecules in which atoms are joined by sharing electrons. Example: H₂O, CO₂, NH₃, CH₄.
3. Hence, ionic compounds contain ions and molecular compounds contain molecules.

Teacher's Note:
a) Ionic compounds are usually solids with high melting points; molecular compounds are often gases or liquids.

69. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions. (Mass of an electron is 9.1×10^-28 g). Which one is heavier?

Answer:
1. An aluminium atom forms Al³⁺ by losing 3 electrons.
2. Number of electrons lost per mole = 3 × 6.022 × 10²³ = 1.8066 × 10²⁴.
3. Difference in mass = 1.8066 × 10²⁴ × 9.1 × 10⁻²⁸ g = 1.64 × 10⁻³ g.
4. Hence, the atoms are heavier than the ions by about 1.64 × 10⁻³ g.

Teacher's Note:
a) The mass of an Al³⁺ ion is slightly less than that of an Al atom.

70. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

Answer:
1. Mass of silver = m g and mass of gold = 0.01 m g.
2. Atomic masses: Ag = 108 and Au = 197.
3. Number of Au atoms : Ag atoms = (0.01 m ÷ 197) : (m ÷ 108).
4. Ratio = (0.01 × 108) ÷ 197 = 1.08 ÷ 197 ≈ 5.48 × 10⁻³ (about 1 : 182).
5. Hence, the ratio of gold atoms to silver atoms is about 5.48 × 10⁻³.

Teacher's Note:
a) The ratio does not depend on m, because m cancels.

71. A sample of ethane (C₂H₆) gas has the same mass as 1.5 ×10²⁰ molecules of methane (CH₄). How many C₂H₆ molecules does the sample of gas contain?

Answer:
1. Molar masses: CH₄ = 16 g/mol and C₂H₆ = 30 g/mol.
2. Mass of 1.5 × 10²⁰ molecules of CH₄ = 1.5 × 10²⁰ × 16 ÷ (6.022 × 10²³) g.
3. For the same mass, number of C₂H₆ molecules = 1.5 × 10²⁰ × (16 ÷ 30).
4. = 8 × 10¹⁹ molecules.
5. Hence, the sample contains 8 × 10¹⁹ molecules of ethane.

Teacher's Note:
a) For equal masses, the number of molecules is inversely proportional to the molar mass.

72. Fill in the blanks
(a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged. This is called __________.
(b) A group of atoms carrying a fixed charge on them is called __________.
(c) The formula unit mass of Ca₃ (PO₄)₂ is __________.
(d) Formula of sodium carbonate is ____________and that of ammonium sulphate is ------------------.

Answer:
1. (a) This is called the law of conservation of mass.
2. (b) A group of atoms carrying a fixed charge is called a polyatomic ion.
3. (c) Formula unit mass of Ca₃(PO₄)₂ = 3 × 40 + 2 × (31 + 4 × 16) = 120 + 190 = 310 u.
4. (d) The formula of sodium carbonate is Na₂CO₃ and that of ammonium sulphate is (NH₄)₂SO₄.

Teacher's Note:
a) Learn the common polyatomic ions with their charges.

73. Write the formulae for the following and calculate the molecular mass for each one of them.
(a) Caustic potash
(b) Baking powder
(c) Lime stone
(d) Caustic soda
(e) Ethanol
(f) Common salt

Answer:
1. (a) Caustic potash: KOH = 39 + 16 + 1 = 56 u.
2. (b) Baking powder: NaHCO₃ = 23 + 1 + 12 + 48 = 84 u.
3. (c) Lime stone: CaCO₃ = 40 + 12 + 48 = 100 u.
4. (d) Caustic soda: NaOH = 23 + 16 + 1 = 40 u.
5. (e) Ethanol: C₂H₅OH = 24 + 6 + 16 = 46 u.
6. (f) Common salt: NaCl = 23 + 35.5 = 58.5 u.

Teacher's Note:
a) Baking powder is a mixture, but its main component is sodium hydrogencarbonate, NaHCO₃.

74. In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C₆H₁₂O₆. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed assuming the density of water to be 1 g cm^-3.

Answer:
1. 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂.
2. Molar mass of glucose = 6 × 12 + 12 × 1 + 6 × 16 = 180 g/mol.
3. Moles of glucose in 18 g = 18 ÷ 180 = 0.1 mol.
4. 1 mole of glucose needs 6 moles of water, so water needed = 0.6 mol.
5. Mass of water = 0.6 × 18 = 10.8 g.
6. Volume = mass ÷ density = 10.8 ÷ 1 = 10.8 cm³.
7. Hence, 10.8 g of water, that is 10.8 cm³, is required.

Teacher's Note:
a) Use the balanced equation to link moles of glucose with moles of water.

(b) 1g N2

(c) 1g H2

(d) 1g CH4

(a) CO32–

(b) PO43–

(c) P2O5