Atoms And Molecules

Q38. Give the names of the elements present in the following compounds.

(a) Quick lime

(b) Hydrogen bromide

(c) Baking powder

(d) Potassium sulphate.

Ans. (a) Quick lime or Calcium Oxide (CaO) - Calcium and Oxygen

(b) Hydrogen bromide (HBr) - Hydrogen and Bromine

(c) Baking powder (NaHCO₃) – Sodium, Hydrogen, Carbon and Oxygen

(d) Potassium sulphate (K₂SO₄) – Potassium, Sulphur and Oxygen

Q39. Compute the difference in masses of 10³ moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1×10^–31 kg)

Ans. Mg atom and Mg²⁺ ion differ by electrons = 2

10³ moles of Mg atom and Mg²⁺ ion differ by electrons = 2 x 10³

Mass of 1 electron = 9.1×10^–31 kg

Mass of 2 x 10³ electron = 2 x 10³ x 6.022 x 10²³ x 9.1×10^–31

                                    = 1.096 x 10^-3 kg

Q40. You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

Ans. Upon heating, sugar powder is charred and becomes black while salt does not char. When dissolved in water, salt solution will conduct electricity because it is ionic while sugar solution will not conduct electricity because it is covalent.

Q41. Which postulate of Dalton’s atomic theory can explain the law of definite proportions?

Ans. The last postulate of Dalton’s atomic theory can explain the law of definite proportions.

The last postulate of Dalton’s atomic theory is as follows:

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

Q42. State the number of atoms present in each of the following chemical species

(a) CO₃^2–

(b) PO₄^3–

(c) P₂O₅

(d) CO

Ans. (a) CO₃^2– = 1C + 3O = 4

(b) PO₄^3– = 1P + 4O = 5

(c) P₂O₅ = 2P + 5O = 7

(d) CO = 1C + 1O = 2

Q43. Which of the following represents a correct chemical formula? Name it.

(a) CaCl (b) BiPO₄ (c) NaSO₄ (d) NaS

Ans. (b) BiPO₄

Explanation:

(a) The correct formula is CaCl₂ 

(c) The correct formula is Na₂SO₄

(d) The correct formula is Na₂S 

Q44. Write the cations and anions present (if any) in the following compounds

(a) CH₃COONa

(b) NaCl

(c) H₂

(d) NH₄NO₃

Ans. 

Q45. Which postulate of Dalton’s atomic theory is the result of the law of conservation of mass?

Ans. Second postulate of Dalton’s atomic theory is the result of the law of conservation of mass.

Second postulate of Dalton’s atomic theory is as follows:

‘Atoms are indivisible particles, which cannot be created or destroyed in a chemical reaction.” 

Q46. 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?

Ans. (a) Mass of container containing 5 moles of C atoms=5 x 12 = 60 g

Mass of container containing 5 moles of Na atoms=5 x 23 = 115 g

Hence, container of Krish is heavier.

(b) Both containers have same number of atoms since they contain same number of moles.

Q47. Write the chemical formulae of the following.

(a) Magnesium chloride

(b) Calcium oxide

(c) Copper nitrate

(d) Aluminium chloride

(e) Calcium carbonate

Ans. (a) Magnesium chloride – MgCl₂

(b) Calcium oxide - CaO

(c) Copper nitrate – Cu(NO₃)₂

(d) Aluminium chloride – AlCl₃

(e) Calcium carbonate – CaCO₃

Q48. 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

Ans. (a) Cobalt-Co

(b) Carbon – C

(c) Aluminium-Al

(d) Helium – He (correct)

(e) Sodium-Na

Q49. 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

Ans. They are classified as

Monoatomic - (k) He, (l) Ag

Diatomic - (a) F₂, (i) HCl

Polyatomic - (b) NO₂, (c) N₂O, (d) C₂H₆, (e) P₄, (f) H₂O₂, (g) P₄O₁₀, (h) O₃, (j) CH₄

Q50. 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

Ans. (a) 10³ = kilo

(b) 10^–1 = deci

(c) 10^–2 = centi

(d) 10^–6 = micro

(e) 10^–9 = nano

(f) 10^–12 = pico

Q51. Write down the names of compounds represented by the following formulae:

(i) Al2(SO₄)₃

(ii) CaCl₂

(iii) K₂SO₄

(iv) KNO₃

(v) CaCO₃

Ans. (i) Al₂(SO₄)₃ – Aluminium sulphate

(ii) CaCl₂ – Calcium chloride

(iii) K₂SO₄ – Potassium sulphate

(iv) KNO₃ – Potassium nitrate

(v) CaCO₃ – Calcium carbonate

Q52. 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.

Ans. Molar mass of CaCl₂ = 40+2x35.5 = 111g

111g of CaCl₂ represent = 1 mol

222g of CaCl₂ represent = 222/111 x 1 = 2 mol

No. of molecules in 2 mole of CaCl₂ = 2 x 6.022 x 10²³

1 molecule of CaCl₂ form ions = 3

2 x 6.022 x 10²³ molecule of CaCl₂ form ions = 3 x 2 x 6.022 x 10²³

                                                                  = 3.6132 x 10²⁴

Q54. If one mole of carbon atoms weighs 12 gram, what is the mass (in gram) of 1 atom of carbon?

Ans. 1 mole of carbon atoms weighs 12 gram

1 mole of carbon atoms weighs = 6.022 x 10²³

Therefore, 6.022 x 10²³ atoms of carbon = 12 g

1 atom of carbon = 12/ (6.022 x 10²³)

                          = 1.993 x 10^-23

Q55. 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. When 3.0 g of carbon is burnt in 8.00 g oxygen, 11.00 g of carbon dioxide is produced.

When 3.00 g of carbon is burnt in 50.00 g of oxygen, only 8.00 g of oxygen will react to produce 11.00 g of carbon dioxide, rest 42.00 g of oxygen will remain unreacted.

Here, ‘Law of constant proportion’ governs the answer.

Q56. 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?

Ans. No. of steel screws in 1 mole = 6.022 x 10²³

Mass of 1 steel screw = 4.11g

Mass of 1 mole steel screws = 4.11 x 6.022 x 10²³

                                         = 2.475 x 10²¹ kg

Mass of the earth = 5.98 × 10²⁴ kg

Ratio = Mass of the earth/Mass of the steel screws

         = 2.475 x 10²¹/5.98 × 10²⁴

         = 2.4 x 10³