Numericals
Worked solutions for the examples and calculation questions in this chapter.
EXAMPLE 9.1 · VERIFYING CONSERVATION OF MASS
Reactants: 4.0 + 2.92 = 6.92 g
Products: 1.76 + 0.72 + 4.44 = 6.92 g
Mass of reactants = mass of products, so the law is obeyed.
EXAMPLE 9.2 · CARBON DIOXIDE
12 g C gives 44 g CO₂, so 2.4 g C gives (44/12) × 2.4 = 8.8 g
EXAMPLE 9.3 · SODIUM CHLORIDE
Cl needed = (35.5 / 23) × 46 = 71 g
EXAMPLE 9.4 AND 9.5 · MOLECULAR MASS
H₂O = (1 × 2) + 16 = 18 u
CO₂ = 12 + (16 × 2) = 44 u
EXAMPLE 9.6 AND 9.7 · FORMULA UNIT MASS
Na₂O = (23 × 2) + 16 = 62 u
Ca(NO₃)₂ = 40 + (14 + 48) × 2 = 164 u
PAUSE AND PONDER 2 · HYDROGEN AND OXYGEN
20 g + 160 g = 180 g of water
PAUSE AND PONDER 3 · SULFUR AND OXYGEN
40% sulfur, 60% oxygen. So the mass ratio S : O = 40 : 60 = 2 : 3.
Oxygen for 20 g of sulfur = 20 × 3/2 = 30 g
PAUSE AND PONDER 4 · CARBON MONOXIDE
C : O = 3 : 4; 9 g C needs 9 × 4/3 = 12 g O
PAUSE AND PONDER 6 · OXIDES OF COPPER
X: 4 : 1; Y: 8 : 2 = 4 : 1
Both have the same ratio, so the results support the Law of Constant Proportions.
PAUSE AND PONDER 21–24 · MASSES
HNO₃ = 1 + 14 + (16 × 3) = 63 u
CH₄ = 12 + (1 × 4) = 16 u
KCl = 39 + 35.5 = 74.5 u
Mg(OH)₂ = 24 + 2(16 + 1) = 58 u
REVISE, REFLECT, REFINE 8 · FORMULA UNIT MASS
NH₄NO₃ = (14 × 2) + (1 × 4) + (16 × 3) = 80 u
H₃PO₄ = 3 + 31 + 64 = 98 u (P = 31 u)
NaHCO₃ = 23 + 1 + 12 + 48 = 84 u
REVISE, REFLECT, REFINE 11 · SODIUM CARBONATE AND ACETIC ACID
Reactants: 5.3 + 6.0 = 11.3 g
Products: 2.2 + 0.9 + 8.2 = 11.3 g
The two are equal, so the law of conservation of mass is valid.
REVISE, REFLECT, REFINE 12 · AN ION
11 protons, 12 neutrons, 10 electrons.
Z = 11; A = 11 + 12 = 23; charge = +11 − 10 = +1
REVISE, REFLECT, REFINE 15 · PARTICLES IN IONS
²⁷Al: p = 13, e = 13, n = 27 − 13 = 14
⁸⁰Br⁻: p = 35, e = 36, n = 80 − 35 = 45
²⁰¹Hg²⁺: p = 80, e = 78, n = 201 − 80 = 121
NCERT Q&A
Answers to the Pause and Ponder questions and the Revise, Reflect, Refine exercises. Numerical working is in the Numericals tab.
Pause and Ponder
PAUSE AND PONDER 1
A student burns 10 g of ethanol in an open beaker and no residue is left. Is the Law of Conservation of Mass violated?
Answer: No. The ethanol burns to form carbon dioxide and water vapour, which escape into the air. If all the products and the oxygen used were weighed, the total mass would be unchanged.
PAUSE AND PONDER 2
20 g of hydrogen react completely with 160 g of oxygen. How much water is formed?
Answer: 180 g (20 + 160), by the Law of Conservation of Mass.
PAUSE AND PONDER 3
A compound has 40% sulfur and 60% oxygen. How much oxygen is present in a sample with 20 g of sulfur?
Answer: 30 g.
PAUSE AND PONDER 4
C and O in CO are in the mass ratio 3 : 4. How much oxygen combines with 9 g of carbon?
Answer: 12 g.
PAUSE AND PONDER 5
Why does the Law of Definite Proportions hold for compounds and not for mixtures?
Answer: A compound has a fixed composition, because its elements are chemically combined in a fixed ratio by mass. The components of a mixture are simply mixed and can be in any proportion.
PAUSE AND PONDER 6
Students X and Y make an oxide of copper in ratios 4 : 1 and 8 : 2. Does it justify the Law of Constant Proportions?
Answer: Yes. 8 : 2 = 4 : 1, so both prepared the compound with the same fixed ratio of copper to oxygen by mass.
PAUSE AND PONDER 7
Assertion: 2 g of hydrogen combine with 16 g of oxygen to form 18 g of water. Reason: according to Dalton's Atomic Theory, atoms combine in a simple whole number ratio by mass to form compounds.
Answer: (iii) A is true, but R is false. 2 g + 16 g = 18 g agrees with the Law of Conservation of Mass. Dalton's theory says atoms combine in a simple whole number ratio by number, not by mass, so R as stated is false.
PAUSE AND PONDER 8
Draw the structure of the nitrogen molecule (N₂).
Answer: Each nitrogen atom (2, 5) needs three electrons, so the two atoms share three pairs of electrons: N≡N, a triple bond.
PAUSE AND PONDER 9
Explain the formation of the fluorine molecule (F₂).
Answer: Fluorine (2, 7) needs one electron. Two fluorine atoms each share one electron, forming one shared pair: F–F, a single covalent bond.
PAUSE AND PONDER 10
Show the formation of (i) CO₂, (ii) H₂S, (iii) NH₃.
Answer: (i) Carbon (2, 4) shares two electrons with each of two oxygen atoms: O=C=O. (ii) Sulfur (2, 8, 6) shares one electron with each of two hydrogen atoms: H–S–H. (iii) Nitrogen (2, 5) shares one electron with each of three hydrogen atoms: three N–H single bonds.
PAUSE AND PONDER 11
Neon neither transfers nor shares its valence electrons. Explain.
Answer: Neon (2, 8) already has a complete octet in its outer shell, so it is stable and has no tendency to gain, lose or share electrons.
PAUSE AND PONDER 12
What kind of ion will oxygen form?
Answer: The oxide ion, O²⁻ (it gains two electrons).
PAUSE AND PONDER 13
Fill in the blanks for magnesium and chlorine.
Answer: Mg gives two electrons to become Mg²⁺. Chlorine takes one, so one Mg²⁺ ion and two Cl⁻ ions combine to give magnesium chloride (MgCl₂).
PAUSE AND PONDER 14
Show the formation of cations of K and Ca and their chlorides.
Answer: K (2, 8, 8, 1) loses one electron to form K⁺ (2, 8, 8); K⁺ + Cl⁻ gives KCl. Ca (2, 8, 8, 2) loses two electrons to form Ca²⁺ (2, 8, 8); Ca²⁺ + 2Cl⁻ gives CaCl₂.
PAUSE AND PONDER 15
Illustrate the formation of sodium sulfide (Na₂S).
Answer: Each of two sodium atoms (2, 8, 1) loses one electron to form 2Na⁺. A sulfur atom (2, 8, 6) gains these two electrons to form S²⁻ (2, 8, 8). The ions are held by an ionic bond: Na₂S.
PAUSE AND PONDER 16
Name (i) CO₂, (ii) NO₂, (iii) SF₆, (iv) PCl₃.
Answer: (i) Carbon dioxide. (ii) Nitrogen dioxide. (iii) Sulfur hexafluoride. (iv) Phosphorus trichloride.
PAUSE AND PONDER 17
Write the formula for (i) sodium hydrogencarbonate, (ii) sulfur dioxide, (iii) ferric chloride, (iv) cuprous oxide.
Answer: (i) NaHCO₃. (ii) SO₂. (iii) FeCl₃. (iv) Cu₂O.
PAUSE AND PONDER 18
Write the formulae from (i) Fe³⁺ and OH⁻, (ii) K⁺ and CO₃²⁻.
Answer: (i) Fe(OH)₃. (ii) K₂CO₃.
PAUSE AND PONDER 19
What bond is present in a solid that does not conduct in the solid state but conducts when dissolved in water?
Answer: An ionic bond.
PAUSE AND PONDER 20
Metal M with two valence electrons in its M shell reacts with oxygen to form a compound slightly soluble in water. Predict its formula, type of bond and conductivity of its aqueous solution.
Answer: M is magnesium. (i) MgO. (ii) Ionic bond (Mg²⁺ and O²⁻). (iii) The aqueous solution conducts electricity (weakly, because it is only slightly soluble), as ions are free to move.
PAUSE AND PONDER 21
Molecular mass of nitric acid (HNO₃).
Answer: 63 u.
PAUSE AND PONDER 22
Molecular mass of methane (CH₄).
Answer: 16 u.
PAUSE AND PONDER 23
Formula unit mass of potassium chloride (KCl).
Answer: 74.5 u.
PAUSE AND PONDER 24
Formula unit mass of magnesium hydroxide, Mg(OH)₂.
Answer: 58 u.
Revise, Reflect, Refine
QUESTION 1
Element A has one electron in its third shell; element B has six electrons in its second shell. Find the ions, bond and formula.
Answer: A (2, 8, 1) is sodium. B (2, 6) is oxygen. (i) A gives 1 electron. (ii) It forms a cation (Na⁺). (iii) B takes 2 electrons. (iv) It forms an anion (O²⁻). (v) An ionic bond. (vi) Na₂O.
QUESTION 2
An element X has six electrons in its outer shell and forms a diatomic molecule.
Answer: (i) It needs two electrons to complete its octet, so two atoms each share two electrons. (ii) A double covalent bond. (iii) X=X (like O=O). (iv) Y has two electrons in its second shell (2, 2), so it gives up two electrons and X takes two: an ionic compound of formula YX (Y²⁺ and X²⁻).
QUESTION 3
Which combination gives a total positive charge of 6+ and negative charge of 6−?
Answer: (iii). 2Fe³⁺ and 3O²⁻ give 6+ and 6−. The others do not balance: 2Al³⁺ and 3Cl⁻ (6+, 3−); 3Mg²⁺ and 1PO₄³⁻ (6+, 3−); 3Ca²⁺ and 2SO₄²⁻ (6+, 4−).
QUESTION 4
Choose the correct statements and correct the false ones.
Answer: Only (iv) is correct. (i) is false: elements are made of atoms (some of molecules), and compounds are made of molecules or ions. (ii) is false: the molecule of a compound contains two or more atoms of different kinds. (iii) is false: a molecule of nitrogen gas (N₂) has two nitrogen atoms.
QUESTION 5
Write the formulae of (i) aluminium nitrate, (ii) calcium oxide, (iii) ferric oxide.
Answer: (i) Al(NO₃)₃. (ii) CaO. (iii) Fe₂O₃.
QUESTION 6
Write the formulae from the pairs of ions.
Answer: (i) CaBr₂. (ii) Al₂(CO₃)₃. (iii) K₂SO₄. (iv) NH₄Cl.
QUESTION 7
Which figure in Fig. 9.18 correctly represents the Cl⁻ ion?
Answer: (ii). It has 18 electrons arranged as 2, 8, 8. The others show 2, 7, 8; 2, 8, 9 and 2, 8, 7.
QUESTION 8
Find the formula unit mass of (i) NH₄NO₃, (ii) H₃PO₄, (iii) NaHCO₃.
Answer: (i) 80 u. (ii) 98 u (taking P = 31 u). (iii) 84 u.
QUESTION 9
Write the formulae for compounds formed by (i) magnesium and nitrogen, (ii) lithium and nitrogen, (iii) sodium and sulfur, (iv) aluminium and oxygen.
Answer: (i) Mg₃N₂. (ii) Li₃N. (iii) Na₂S. (iv) Al₂O₃.
QUESTION 10
Complete Table 9.3.
Answer: NH₄⁺: NH₄NO₃, (NH₄)₂SO₄, (NH₄)₃PO₄. Li⁺: LiNO₃, Li₂SO₄, Li₃PO₄. Al³⁺: Al(NO₃)₃, Al₂(SO₄)₃, AlPO₄. Cu²⁺: Cu(NO₃)₂, CuSO₄, Cu₃(PO₄)₂.
QUESTION 11
5.3 g of sodium carbonate and 6.0 g of acetic acid produce 2.2 g of CO₂, 0.9 g of water and 8.2 g of sodium acetate. Is the law of conservation of mass valid?
Answer: Yes. Reactants: 5.3 + 6.0 = 11.3 g. Products: 2.2 + 0.9 + 8.2 = 11.3 g.
QUESTION 12
A species has 11 protons, 12 neutrons and 10 electrons.
Answer: (i) Atomic number 11, mass number 23. (ii) A cation, with charge +1, as it has one electron fewer than protons. (iii) 2, 8. (iv) Sodium ion (Na⁺).
QUESTION 13
Elements A (2, 8, 5) and B (2, 8, 7).
Answer: (i) B is more reactive; it needs only one electron to complete its octet. (ii) Covalent: both are non-metals that need electrons, so they share. (iii) PCl₃ (A is phosphorus, B is chlorine).
QUESTION 14
Assertion: copper sulfate conducts electricity in the molten state but not in the solid state. Reason: copper and sulfate ions are fixed in the lattice in the molten state, while in the solid state they can move freely.
Answer: (iii). A is true. R is false: in the solid state the ions are fixed; in the molten state they are free to move.
QUESTION 15
The species ²⁷Al, ⁸⁰Br⁻ and ²⁰¹Hg²⁺ have 13, 35 and 80 protons. Find the electrons and neutrons.
Answer: ²⁷Al: 13 electrons, 14 neutrons. ⁸⁰Br⁻: 36 electrons, 45 neutrons. ²⁰¹Hg²⁺: 78 electrons, 121 neutrons.