chapter 1catalogimages.johnwiley.com.au/attachment/07314... · 22s1 19k: 1s 22s22p63s23p64s1 6c: 1s...

© John Wiley & Sons Australia, Ltd 2007

Chapter 1

Page 4 1. Hot air is made up of gas particles that are moving very fast in random directions.

When they collide with the walls of the balloon they exert pressure and the balloon inflates.

2. liquid

3. The gaseous state is compressible as the particles are very far apart.

4. At lower temperatures particles move more slowly.

Page 6 5. (a) gas (b) liquid (c) solid (d) gas (e) gas (f) solid

6. (a) gas (b) liquid

7. (a) melting

(b) sublimation

(c) evaporation

(d) condensation

Page 9 8. See the lower figure on page 7.

9. (a) 1 (b) 10 (c) 47 (d) 79

10. (a) 13 (b) 26 (c) aluminium

11. (a) B (b) Mg (c) Ar (d) Ca

0

12. (a) 22 (b) 18 (c) 4018 Ar

13. number of protons = 35, number of neutrons = 44

14. 21 H

Page 11 15.

Element Number of protons Number of electrons Number of neutrons 126 C 6 6 6 5626 Fe 26 26 30 4018 Ar 18 18 22

1

Studyon Chemistry 1 ANSWERS Chapter 1


Element Number of protons Number of electrons Number of neutrons 23592 U 92 92 143

23892 U 92 92 146

199 F 9 9 10

(a) 23892 U and 235

92 U are isotopes.

(b) 23892 U has 3 more neutrons than 235

92 U .

16. (a) 14 157 7N, N (b) 7 protons (c) 7 and 8 respectively (d) isotopes

17. The atomic number is always the same for a particular element, while the mass number can differ due to the existence of isotopes.

Page 14 18. from top to bottom, symbols: L, M, N; maximum number of electrons: 8, 18, 32

19. He 2; Li 2, 1; B 2, 3; C 2, 4; N 2, 5; O 2, 6; F 2, 7; Ne 2, 8; Mg 2, 8, 2; Al 2, 8, 3; Si 2, 8, 4; P 2, 8, 5; S 2, 8, 6; Ar 2, 8, 8; Ca 2, 8, 8, 2

Page 15 20. (a) The Bohr model was able to explain the emission spectrum of hydrogen.

(b) The spectra of atoms more complex than hydrogen could not be explained satisfactorily.

22. (a) 2, 5 (b) calcium

Page 17 23. Elements cannot be broken down into simpler substances. Compounds can be

decomposed by chemical reactions. Mixtures can be easily separated by physical means.

24. (a) mixture (b) element (c) mixture (d) compound (e) mixture (f) mixture (g) compound (h) mixture (i) compound (j) element (k) compound (l) mixture

Page 18

25. (a) nitrogen, hydrogen, chlorine

(b) potassium, manganese, oxygen

(c) carbon, hydrogen, oxygen

(d) calcium, iodine

(e) carbon, hydrogen, oxygen

26. barium, Ba; beryllium, Be; bromine, Br

27. (a) A (b) C (c) D (d) B, D

2



Page 21 28. (a) For groups 13–18, group number – 10 = number of valence electrons.

(b) period number = number of occupied electron shells

(c) Helium is inert (stable) with a full outer shell. It does not form compounds, and so is similar in behaviour to the other elements in this group, which all have full outer shells.

(d) Hydrogen can be placed in group 1 as it has one valence electron, similar to the metals in group 1. It could also be placed in group 17 as it is a non-metal that may gain an electron, like elements in this group. Since it is difficult to categorise, hydrogen is often positioned slightly away from any particular group in the periodic table.

Page 22 29. (a) Li (b) F

30. Atomic size increases down a group as more electrons are added to successive main shells; thus, the caesium atom is larger than the sodium atom as it is found lower in the group.

31. Metallic character decreases across a period, therefore sodium is more metallic than silicon.

32. Non-metallic character increases across a period, therefore carbon has more non-metallic character than aluminium.

33. (a) beryllium, Be (b) gallium, Ga (c) argon, Ar

34. (a) period 4, group 17 (b) bromine, Br

Multiple choice questions 1. D 2. B 3. C 4. D 5. A 6. D 7. C 8. B 9. C 10. D 11. D 12. B 13. B

Review questions 1. (a) Particles in solids vibrate in fixed positions, while particles in liquids and gases

move around.

3. (a) evaporation (b) freezing (c) sublimation (d) freezing (e) freezing

4. (a) liquid (b) solid (c) gas

5. (a) liquid (b) solid (c) liquid (d) solid (e) solid and gas (f) liquid (g) gas (h) liquid and gas (i) solid and liquid

6. (a) liquid (b) solid

8. Protons and neutrons have almost identical mass while the mass of the electron is negligible. Protons are positively charged. Electrons have negative charge and neutrons have no charge. Protons and neutrons are found in the nucleus while electrons are found in the space around the nucleus.

9. An element is a pure substance that cannot be broken down into simpler substances. The atoms of an element have the same atomic number. A compound is a pure

3



substance in which different elements are chemically combined in fixed proportions by mass.

10. (a) (i) 11 (ii) 23 (iii) 11 (iv) 12 (v) 11 (vi) sodium

(b) (i) 9 (ii) 19 (iii) 9 (iv) 10 (v) 9 (vi) fluorine

(c) (i) 14 (ii) 28 (iii) 14 (iv) 14 v) 14 (vi) silicon

(d) (i) 26 (ii) 56 (iii) 26 (iv) 30 (v) 26 (vi) iron

(e) (i) 79 (ii) 197 (iii) 79 (iv) 118 (v) 79 (vi) gold

(f) (i) 92 (ii) 235 (iii) 92 (iv) 143 (v) 92 (vi) uranium

11.

Name of atom Atomic number Mass number Protons Neutrons Electrons

argon 18 36 18 18 18 sulfur 16 34 16 18 16 argon 18 38 18 20 18 phosphorus 15 31 15 16 15 lead 82 208 82 126 82 potassium 19 39 19 20 19 sulfur 16 36 16 20 16

12. isotopes

13. (a) 6 (b) 12, 13, 14 (c) 6, 7, 8 (d) 6

14. (a) A, C, D

(b) A = 4020 Ca

B = 3717 Cl

C = 4220 Ca

D = 4620 Ca

15. chlorine 37 3517 17A D

magnesium 26 2412 12B E G

cobalt 59 6027 27C F

16. (a) absorption of energy

(b) A photon equal to the difference in energy of the two levels is emitted.

17. One, as each compound contains the same metallic cation Ca2+.

18. When sodium atoms are excited, the electrons jump to a higher energy level. When they return to their normal energy levels they emit the characteristic yellow light.

19. The ground state refers to the most stable state where all electrons are occupying their normal energy levels or shells. When an atom is excited, the electrons can absorb energy and jump to higher energy levels (the excited states) depending on the amount of energy absorbed.

20. An atom becomes an ion when it loses or gains electrons.

21. 60

4



22. (a) oxygen (b) aluminium (c) chlorine (d) argon (e) potassium

23. (a) 2, 7 (b) 2, 8, 8, 1 (c) 2, 6 (d) 2, 8, 2 (e) 2, 8, 8, 2

24 (a) carbon, hydrogen, oxygen

(b) hydrogen, phosphorus, oxygen

(c) sodium, carbon, oxygen

(d) potassium, sulfur, oxygen

26. two hydrogen atoms, one sulfur atom, four oxygen atoms

27. (a) They have similar chemical properties and the same number of electrons in their outer shells.

(b) Their electrons are filling the same electron shell (energy level).

28. (a) decreases (b) increases (c) increases

29. (a) (i) 2 (ii) 15

(b) (i) 2 (ii) 18

(c) (i) 3 (ii) 16

(d) (i) 2 (ii) 2

(e) (i) 4 (ii) 1

5


Chapter 2

Page 39 10. 70.77

11. (b) 24.3

13. (b) 52, 53

(c) 127.6, 126.9

(d) 52, 76, 52; 53, 74, 53

17. (a) 12, 11, 11

(b) 14, 14, 14

(c) 8, 8, 8

(d) 0, 1, 0

(e) 12, 12, 10

18. (b) 14

(c) 28, 29, 30

(d) 28.11

Page 44 21. 3Li: 1s22s1

19K: 1s22s22p63s23p64s1

6C: 1s22s22p2

17Cl: 1s22s22p63s23p5

18Ar: 1s22s22p63s23p6

7N: 1s22s22p3

22. (a) 1s22s22p63s23p3

(b) 32P has 17 neutrons while 31P has only 16.

23. (a) group 16 (b) period 3 (c) sulfur, S

24. (a) lithium — group 1, period 2

(b) neon — group 18, period 2

(c) sodium — group 1, period 3

(d) aluminium — group 13, period 3

(e) argon — group 18, period 3

(f) calcium — group 2, period 4

6



Page 45 (top) 25. (a) Z = 20: 1s22s22p63s23p64s2

Z = 12: 1s22s22p63s2

Z = 4: 1s22s2

Z = 9: 1s22s22p5

Same group: Z = 20, 12, and 4 (group 2)

(b) Z = 5: 1s22s22p1

Z = 6: 1s22s22p2

Z = 8: 1s22s22p4

Z = 16: 1s22s22p63s23p4

Same group: Z = 8 and 16 (group 16)

26. (a) He 1s2; Ne 1s22s22p6; Ar 1s22s22p63s23p6

(b) Al 1s22s22p63s23p1

(c) N 1s22s22p3

(d) Cl 1s22s22p63s23p5

Page 45 (bottom) 29. (a) ground

(b) excited

(c) ground

30. (c) 1s22s22p63s2

31. (a) manganese

(b) calcium

Multiple choice questions

1. C 2. A 3. D 4. B 5. B 6. B 7. D 8. A 9. D 10. C 11. A 12. A 13. B 14. B 15. C 16. C

Review questions 5. 3 RIM, 1 RAM

6. (a) 51.998

7. 99% of 12C, 1% of 13C

10. (a) 6

(b) 2

(c) 10

(d) 18

7



11. (a) 1s22s22p6 3s1

(b) 1s22s22p3

(c) 1s22s22p63s23p6

(d) 1s22s22p63s23p6 3d6 4s2

(e) 1s22s22p63s23p6 3d10 4s1

(f) 1s22s22p63s23p6

(g) 1s22s22p63s23p6

8


Chapter 3

Page 53 1. A mole is a number such that 1 mole of any substance X consists of Avogadro’s

number, 6.02 × 1023 particles of X.

2. (a) N(Cl) = n(Cl) × NA N(Cl) = 2.3 × 6.02 × 1023 N(Cl) = 1.4 × 1024 atoms

(b) N(Cl) = 2 × n(Cl2) × NA N(Cl) = 2 × 15.8 × 6.02 × 1023 N(Cl) = 1.90 × 1025 atoms

(c) N(Cl–) = n(NaCl) × NA N(C–l) = 3.5 × 6.02 × 1023 N(C–l) = 2.1 × 1024 ions

(d) N(Cl–) = 2 × n(MgCl2) × NA N(C–l) = 2 × 0.5 × 6.02 × 1023 N(C–l) = 6 × 1023 ions

3. (a) n(CH3COOH) = 06.602.16

n(CHn(CH3COOH) = 0.270 mole

(b) N(CH3COOH) = n(CH3COOH) × NA n(CH3COOH) = 0.270 × 6.02 × 1023 n(CH3COOH) = 1.62 × 1023 molecules

(c) n(O) = 2 × n(CH3COOH) n(O) = 2 × 1.62 × 1023 n(O) = 0.539 mole

(d) N(O) = 0.539 × 6.02 × 1023 N(O) = 3.25 × 1023 atoms

Page 54 (top)

4. (a) n(CH4) = AN

N

n(CHn(CH4) = 23

24

1002.6102.5××

n(CHn(CH4) = 8.6 mole

(b) n(C) = n(CH4) n(C) = 8.6 mole

(c) n(H) = 4 × n(CH4) n(H) = 4 × 8.6 n(H) = 35 mole

9



Page 54 (bottom)

5. Correct formula to use: n = Mm

(a) n(H2O) = 02.18

46

= 2.6 mole

(b) n(CO2) = 01.444.2

= 0.055 mole

(c) n(Cl2) = 90.70

67

= 0.94 mole

(d) n(NaCl) = 44.580.2

= 0.034 mole

(e) n(CuSO4.5H2O) = 71.249

128

= 0.513 mole

(f) n(Fe2O3) = 38000 g159.7

= 2.4 × 102 mole

Page 55 6. Correct formula to use: m = n × M

(a) m(CO) = 0.41 × 28.01 = 11 g

(b) m(SO2) = 12.0 × 64.06 = 767 g

(c) m(C12H22O11) = 3.84 × 342.34 = 1.31 × 103 g = 1.31 kg

(d) m(Fe) = 58.2 × 55.85 = 3.25 × 103 g = 3.25 kg

(e) m(AgCl) = 0.0051 × 143.32 = 0.73 g

(f) m(Mg3(PO4)2) = 2.53 × 262.87 = 665 g

7. (a) m(H2) = 2.5 × 2.02 = 5.1 g

10



(b) m(Zn) = 0.2 × 65.38 = 13 g

(c) m(CaCl2) = 11.56 g

therefore (b), 0.2 mole of zinc has the greatest mass.

Page 56

8. Convert molecules to mole using n = AN

N , then convert mole to mass using

m = n × M. Combining the two formulae: m = AN

N × M

(a) m(C6H12O6) = 23

24

1002.61025.5

×× × 180.18

= 1.57 × 103 g = 1.57 kg

(b) m(NO2) = 23

21

1002.61083.1

×× × 46.01

= 0.140 g

(c) m(CO2) = 23

14

1002.61056.3

×× × 44.01

= 2.60 × 10–8 g

(d) m(CS2) = 23

28

1002.61013.4

×× × 76.13

= 5.22 × 106 g

(e) m(N2O4) = 23

24

1002.61062.3

×× × 92.02

= 553 g = 5.53 × 102 g

9. (a) m(Mg) = 200 g

(b) m(S) = n × M = 5 × 32.06 = 160.3 g

(c) m(He) = AN

N × M

= 23

24

1002.6102.1×× × 4.00

= 8.0 g

11



(d) m(C3H7O2N) = AN

N × M

= 23

22

1002.6105.3×× × 89.11

= 5.2 g

therefore (a) 200 g Mg has the greatest mass.

10.

Substance Amount

(mass in g)

Molar mass (M)

Number of atoms

in the molecule

Number of moles (n) of substance

Number of molecules

(N) Total number

of atoms

1 water, H2O 3.2 18.02 3 3.2

18.02 = 0.18

0.18×6.02×1023 = 1.1×1023

1.1×1023×3 = 3.2×1023

2 methane, CH4 2.7 16.05 5 2.7

16.05 = 0.17

0.17×6.02×1023 = 1.0×1023

1.0×1023×5 = 5.1×1023

3 ammonia, NH3 0.056 17.04 4 0.056

17.04

= 0.0033

0.0033×6.02×1023 = 2.0×1021

2.0×1021×4 = 7.9×1021

4 acetic acid, CH3COOH

27.3×60.06 = 1.64×103

60.06 8 27.3 27.3×6.02×1023 = 1.64×1025

1.64×1025×8 = 1.31×1026

5 benzene, C6H6 0.56×78.12 = 44

78.12 12 0.56 0.56×6.02×1023 = 3.4×1023

3.4×1023×12 = 4.0×1024

6 octane, C8H18 2.34×114.26 = 267

114.26 26 2.34 2.34×6.02×1023 = 1.41×1024

1.41×1024×26 = 3.66×1025

7 ethanol, CH3CH2OH

10×46.08 = 4.6×102

46.08 9 24

23

6.0 10

6.02 10

×

×

= 10

6.0×1024 6.0×1024 ×9 = 5.4×1025

8 ozone, O3 0.00211 × 48.0 = 0.101

48.0 3 21

23

1.27 10

6.02 10

×

×

= 0.00211

1.27×1021 1.27×1021 ×3 = 3.81×1021

9 sulfuric acid, H2SO4

71×98.08 = 7.0×103

98.08 7 23

25

1002.6103.4××

= 71

263.0 10

7

×

= 4.3× 1025

3.0×1026

10 carbon dioxide, CO2

42×44.01 = 1.8×103

44.01 3 23

25

1002.6105.2××

= 42

257.5 10

3

×

= 2.5 2510×

7.5×1025

12



Page 57 11. (a) M(C3H8) = 3 × 12.01 + 8 × 1.01

M(C3H8) = 44.11 g mol–1

∴ % C = 11.44

01.123× × 1

100

% C = 81.68%

∴ % H = 11.4401.18× ×

1100

% H = 18.32% Alternatively, % H = 100 – 81.68 Alternatively, % H = 18.32%

(b) M(NaHSO4) = 22.99 + 1.01 + 32.06 + 4 ×16.00 M(NaHSO4) = 120.06 g mol–1

∴ % Na = 06.12099.22 ×

1100

% Na = 19.15%

∴ % H = 06.120

01.1 × 1

100

N a = 0.84%

∴ % S = 06.12006.32 ×

1100

% a = 26.70%

∴ % O = 06.12000.164× ×

1100

% Na = 53.31% Alternatively, % O = 100 – (19.15 + 0.84 + 26.70) Alternatively, % H = 53.31%

(c) M(Ca(CH3COO)2) = 40.08 + 4 ×12.01 + 6 × 1.01 + 4 × 16.00 M(Ca(CH3COO)2) = 158.18 g mol–1

∴ % Ca = 18.15808.40 ×

1100

% Na = 25.34%

∴ % C = 18.15801.124× ×

1100

% Na = 30.37%

∴ % O = 18.15800.164× ×

1100

% Na = 40.46%

∴ % H = 18.15801.16× ×

1100

% Na = 3.83% Alternatively, % H = 100 – (25.34 + 30.37 + 40.46) Alternatively, % H = 3.83%

13



(d) M(HCN) = 1.01 + 12.01 + 14.01 M(HCN) = 27.03 g mol–1

∴ % H = 03.27

01.1 × 1

100

% Na = 3.74%

∴ % C = 03.2701.12 ×

1100

% Na = 44.43%

∴ % N = 03.2701.14 ×

1100

% Na = 51.83% Alternatively, % N = 100 – (3.74 + 44.43) Alternatively, % H = 51.83%

Page 59

12. (a) M(NiSO4.6H2O) = 58.71 + 32.06 + 4 × 16.00 + 6 × 18.02 M(NiSO4 6H2O) = 262.89 g mol–1

∴% H2O = 89.26202.186× ×

1100

% H2O = 41.13%

(b) M(Na2CO3.10H2O) = 2 × 22.99 + 12.01 + 3 × 16.00 + 10 × 18.02 M(NiSO4 6H2O) = 286.19 g mol–1

∴% H2O = 19.286

02.1810× × 1

100

% H2O = 62.97%

(c) M(MgCl2.6H2O) = 24.31 + 2 × 35.45 + 6 × 18.02 M(NiSO4 6H2O) = 203.33 g mol–1

∴% H2O = 33.20302.186× ×

1100

% H2O = 53.17%

Page 60 13. symbols C O H

masses 57.7 37.5 4.8

moles 57.71

12.01

37.51

06.00

4.811.01

= 4.804 = 2.344 = 4.752

divide by smallest 344.2804.4

344.2344.2

344.2752.4

= 2.05 = 1.00 = 2.02 ratio 2 1 2 therefore the empirical formula of aspirin is C2OH2.

14



14. symbols C O H masses 65.4 29.1 100 – (65.4 + 29.1) = 5.5

moles 01.124.65 29.1

10

6.00 5.51

1.01

= 5.445 = 1.819 = 5.446


819.1819.1

819.1446.5

= 2.99 = 1.00 = 2.99 ratio 3 1 3 therefore the empirical formula of hydroquinone is C3OH3.

Page 61 15. m(H2O) = 1.124 – 0.889

m(H2O) = 0.235 g Mole ratio of calcium sulfate to water

= 14.136

889.0 : 02.18

235.0

= 0.006 53 : 0.0130

= 0.006 530.006 53

: 0.01300.006 53

= 1 : 2 therefore the empirical formula of the hydrated calcium sulfate is CaSO4.2H2O.

16. m(H2O) = 0.942 – 0.461 m(H2O) = 0.481 g Mole ratio of magnesium sulfate to water

= 37.120

461.0 : 02.18

481.0

= 0.00383 : 0.0267

= 0.003830.003 83

: 0.02670.00383

= 1 : 7 therefore the empirical formula of the hydrated magnesium sulfate is MgSO4.7H2O.

Page 62 17. The empirical formula mass = 2 × 12.01 + 6 × 1.01 + 14.01

The empirical formula mass = 44.09 relative molecular massempirical formula mass

= 09.44

88

ratio = 2 ∴molecular formula of putrescine = 2 × empirical formula molecular formula of putrescine = 2 × C2H6N molecular formula of putrescine = C4H12N2

18. acetylene 26; benzene 78; formaldehyde 30; acetic acid 60; glucose 180 Molecular formulae are whole number multiples of the corresponding empirical formulae.

15



19. The empirical formula mass = 5 × 12.01 + 7 × 1.01 + 14.01 The empirical formula mass = 81.13 relative molecular massempirical formula mass

= 13.81

162

ratio = 2 ∴molecular formula of nicotine = 2 × empirical formula molecular formula of nicotine = 2 × C5H7N molecular formula of nicotine = C10H14N2

Page 63 20. symbols C N O H

masses 49.5 28.9 16.5 5.1

moles 49.51

12.01

28.91

14.01

16.51

16.00

5.111.01

= 4.12 = 2.06 = 1.03 = 5.05


03.106.2

03.103.1

03.105.5

= 4.00 = 2.00 = 1.00 = 4.90 ratio 4 2 1 5 therefore the empirical formula of caffeine is C4N2OH5. The empirical formula mass = 4 × 12.01 + 5 × 1.01 + 2 × 14.01 + 16.00 The empirical formula mass = 97.11 relative molecular massempirical formula mass

= 11.972.194

ratio = 2 ∴molecular formula of caffeine = 2 × empirical formula molecular formula of caffeine = 2 × C4 N2OH5 molecular formula of caffeine = C8N4O2H10

21. symbols C O H masses 58.8 31.4 9.8 moles 58.8

11

2.01 31.4

11

6.00 9.81

1.01

= 4.90 = 1.96 = 9.70


96.196.1

96.170.9

= 2.5 = 1.0 4.9 ratio 5 2 10 therefore the empirical formula of methyl butanoate is C5O2 H10. The empirical formula mass = 5 × 12.01 + 10 × 1.01 + 2 × 16.00 The empirical formula mass = 102.15 relative molecular massempirical formula mass

= 15.102

102

16



ratio = 1 ∴molecular formula of methyl butanoate = 1 × empirical formula molecular formula of methyl butanoate = 1 × C5H10O2 molecular formula of methyl butanoate = C5H10O2

Multiple choice questions 1. B

2. n(Na2CO3.10H2O) = Mm

= 19.286

143

= 0.500 mole

n(O) = 13 × n(Na2CO3.10H2O) n(O) = 13 × 0.500 n(O) = 6.50 mole Answer: D

3. The smallest number of mole of molecules will contain the smallest number of

molecules. As n = Mm the substance with the greatest molar mass will contain the

smallest number of mole.

A. M(N2) = 28.02 g mol–1 B. M(O2) = 32.00 g mol–1 C. M(NO) = 30.01 g mol–1 D. M(NO2) = 46.01 g mol–1 Answer: D

4. A. n(O) = n(H2O)

= 02.18

300

= 16.6 mole

B. n(O) = 9 × n(CuSO4.5H2O) n(O) = 9 × 3.2 n(O) = 28.8 mole

C. n(O) = 6 ×n(C57H110O6) n(O) = 6 × n(C57H110O6)

n(O) = 6 × 3000 g57 12.01 110 1.01 6 16.00× + × + ×

n(O) = 20.19 mole

D. n(O) = 2 × n(NO2)

n(O) = 2 × 23

24

1002.61035.7

××

n(O) = 24.4 mole

17



The largest number of oxygen atoms is contained in the largest number of mole therefore Answer: B

5. % C in acrylic acid, C3H4O2 = 00.16201.1401.123

01.123×+×+×

× × 1

100

= 50.0%

% C in methyl acrylite, C4H6O2 = 00.16201.1601.124

01.124×+×+×

× × 1

100

= 55.8%

% C in acrylonitrile, C3H3N = 01.1401.1301.123

01.123+×+×

× × 1

100

= 67.9% Answer: A

6. B

7. symbols C O masses 27.3 72.7

moles 01.123.27

00.167.72

= 2.273 = 4.544


273.2544.4

= 1 = 1.99 ratio 1 2 therefore the empirical formula is CO2. Answer: A

8. symbols C H masses 83.7 16.3

moles 01.127.83

116.31.01

= 6.97 = 16.1

divide by smallest 97.697.6 716.1

6.97

= 1 = 2.3 ratio 3 7 therefore the empirical formula is C3H7. The empirical formula mass = 3 × 12.01 + 7 × 1.01 The empirical formula mass = 43.1 relative molecular massempirical formula mass

= 1.43

86

ratio = 2

18



∴molecular formula = 2 × empirical formula molecular formula = 2 × C3H7 molecular formula = C6H14 Answer: B

Review questions 1. In 2 moles of (NH4)3PO4 there are:

(a) 3 × 2 = 6 mole nitrogen atoms

(b) 12 × 2 = 24 mole hydrogen atoms

(c) 4 × 2 = 8 mole oxygen atoms

(d) 1 × 2 = 2 mole phosphorus atoms

(e) 6 × 6 × 1023 = 3.6 × 1024 nitrogen atoms

(f) n(PO43–) × M(PO4

3–) = 1 × 2 × (32.97 + 4 × 16.00) n(PO4

3–) M(PO43–) = 193.94 g of phosphate ions

n(PO43–) M(PO4

3–) = 190 g

(g) n (N) × M(N) = 6 × 14.01 n(N) M(N) = 84.06 g of nitrogen atoms n(N) M(N) = 84 g

2.

Name Formula Molar mass (M)

(g mol–1) Mole (n) Mass (m)

(g)

sodium hydroxide NaOH 40.0 3.41

40.0 = 0.0853 3.41

carbon tetrachloride CCl4 153 1.40 1.40 × 153 = 214

sodium carbonate Na2CO3 106 1.00 1.00 ×106 = 106

potassium chloride KCl 74.55 0.25 0.25 × 74.55 = 18.6 = 19

ammonium phosphate (NH4)3PO4 149.12

8.46

149.12 = 0.0567 8.46

3. n = Mm then N = n × NA × total number of atoms in the formula.

(a) n(C2H2) = 04.266.14

n(C2H2) = 0.561 mole N(atoms) = 0.561 × 6.02 × 1023 × 4 n(C2H2) = 1.35 × 1024

(b) n(C3H8) = 11.44

48.0

n(C2H2) = 0.011 mole

19



N(atoms) = 0.011 × 6.02 × 1023 × 11 N(atoms) = 7.2 × 1022

(c) n(C2H5OH) = 08.46

485

n(C2H5OH) = 10.5 mole N(atoms) = 10.5 × 6.02 × 1023 × 9 N(atoms) = 5.70 × 1025

(d) n(CO2) = 01.446.8

n(CO2) = 0.20 mole N(atoms) = 0.20 × 6.02 × 1023 × 3 N(atoms) = 3.5 × 1023

(e) n(I2) = 8.253

67

n(I2) = 0.26 mole N(atoms) = 0.26 × 6.02 × 1023 × 2 N(atoms) = 3.2 × 1023

4. Convert mole to mass using m = n × M. (a) m(Fe) = 160 × 55.85

m(Fe) = 8936 g m(Fe) = 8.94 × 103 g m(Fe) = 8.94 kg

(b) m(SiO2) = 0.075 × 60.09 m(SiO2) = 4.5 g

(c) m(NO2) = 4.23 × 46.01 m(SiO2) = 195 g

5. Sodium fluoride = NaF m(NaF) = 0.013 g

(a) n = Mm

n(NaF) = 99.41

013.0

n(NaF) = 3.1 × 10–4 mole

(b) n(F–) = n(NaF) n(F–) = 3.1 × 10–4 mole N(F–) = n(F–) × NA n(F–) = 3.1 × 10–4 × 6.02 × 1023 n(F–) = 1.9 × 1020 ions

6. m(C6H8O6) = 60 mg m(C6H8O6) = 0.060 g

20



(a) n = Mm

n(C6H8O6) = 14.176

060.0

n(C6H8O6) = 3.4 × 10–4 mole

(b) N(C6H8O6) = n(C6H8O6) × NA N(C6H8O6) = 3.4 × 10–4 × 6.02 × 1023 N(C6H8O6) = 2.1 × 1020 molecules

(c) minimum daily requirement = 60 mg minimum daily requirement = 0.060 g

Therefore the m(spinach) = 5102.1060.0

−× × 10 g

therefore the m(spinach) = 50 000 g therefore the m(spinach) = 50 kg

7. m(C6H8O6) = 500.0 mg m(C6H8O6) = 0.5000 g

(a) n = Mm

n(C6H8O6) = 14.176

5000.0

n(C6H8O6) = 2.839 × 10–3 mole

(b) N(C6H8O6) = n(C6H8O6) × NA N(C6H8O6) = 2.839 × 10–3 × 6.02 × 1023 N(C6H8O6) = 1.709 × 1021 molecules

8. m(C27H46O) = 250 g

(a) n = Mm

n(C27H46O) = )00.16)01.146()01.1227((

250+×+×

n(C27H46O) = 0.646 mole

(b) n(C) = 27 × n(C27H46O) n(C) = 27 × 0.646 n(C) =17.5 mole

(c) m(C) = n(C) × M(C) = 17.5 × 12.01 = 210 g

(d) m(O) = n(O) × M(O) from (a) n(O) = n(C27H46O) from (a) n(O) = 0.646 mole therefore m(O) = 0.646 ×16.00 therefore m(O) = 10.3 g

9. (a) The mass of a 1.00 carat diamond = 0.200 g As diamond is pure carbon the mass of carbon = 0.200 g

21



using the formula n = Mm

n(C) = 01.12

200.0

n(C) = 0.0167 mole therefore N(C) = n(C) × NA therefore N(C) = 0.0167 × 6.02 × 1023 therefore N(C) = 1.00 × 1022 atoms

(b) The mass of a 3.15 carat diamond = 3.15 × 0.200 g the mass of a 3.15 carat diamond = 0.630 g As diamond is pure carbon the mass of carbon = 0.630 g

using the formula n = Mm

n(C) = 01.12

630.0

n(C) = 0.0525 mole therefore N(C) = n(C) × NA therefore N(C) = 0.0525 × 6.02 × 1023 therefore N(C) = 3.16 × 1022 atoms

10. (a) M(CH3COOH) = 4 ×1.01 + 2 × 12.01 + 2 × 16.00 M(CH3COOH) = 60.06 g mol–1

∴% H = 06.6001.14× ×

1100

= 6.73%

∴% C = 06.60

01.122× × 1

100

= 39.99%

∴% O = 06.60

00.162× × 1

100

= 53.28% Alternatively, % O = 100 – (6.73 + 39.99) Alternatively, % O = 53.28%

(b) M(MgCl2.6H2O ) = 24.31 + 2 × 35.45 +12 ×1.01 + 6 ×16.00 = 203.33 g mol–1

∴% Mg = 33.20331.24 ×

1100

= 11.95%

∴% Cl = 33.20345.352× ×

1100

= 34.87%

∴% H = 33.20301.112× ×

1100

= 5.96%

∴% O = 33.20300.166× ×

1100

= 47.21%

22



Alternatively, % O = 100 – (11.95 + 34.87 + 5.96) Alternatively, % O = 47.22%

(c) M(Fe2(SO4)3) = 2 × 55.85 + 3 × 32.06 + 12 ×16.00 M(Fe2(SO4)3) = 399.88 g mol–1

∴% Fe = 88.39985.552× ×

1100

= 27.93%

∴% S = 88.39906.323× ×

1100

= 24.05%

∴% O = 88.399

00.1612× × 1

100

= 48.01% Alternatively, % N = 100 – (27.93 + 24.05) Alternatively, % O = 48.02%

11. (a) M(CdS) = 112.40 + 32.01 = 144.46 g mol–1

∴% Cd = 46.14440.112 ×

1100

= 77.81%

(b) M(CdSe) = 112.40 + 78.96 = 191.36 g mol–1

∴% Cd = 36.19140.112 ×

1100

= 58.74%

(c) M(CdTe) = 112.40 + 127.60 = 240.0 g mol–1

∴% Cd = 0.240

40.112 × 1

100

= 46.83%

12. (a) M(NO) = 14.01 + 16.00 = 30.01 g mol–1

∴% N = 01.3001.14 ×

1100

= 46.68%

∴% O = 01.3000.16 ×

1100

= 53.32%

(b) M(NO2) = 14.01 + 2 ×16.00 = 46.01 g mol–1

∴% N = 01.4601.14 ×

1100

= 30.45%

23



∴% O = 01.46

00.162× × 1

100

= 69.55%

(c) M(N2O4) = 2 ×14.01 + 4 × 16.00 = 92.02 g mol–1

∴% N = 02.92

01.142× × 1

100

= 30.45%

∴% O = 02.92

00.164× × 1

100

= 69.55%

(d) M(N2O) = 2 × 14.01 + 16.00 = 44.02 g mol–1

∴% N = 02.44

01.142× × 1

100

= 63.65%

∴% O = 02.4400.16 ×

1100

= 36.35%

13. (a) M(C21H27NO) = 21 × 12.01 + 27 × 1.01 + 14.01 + 16.00 M(C21H27NO) = 309.49 g mol–1

∴% C = 49.309

01.1221× × 1

100

= 81.49%

∴% H = 49.30901.127× ×

1100

= 8.81%

∴% N = 49.30901.14 ×

1100

= 4.53%

∴% O = 49.309

00.16 × 1

100

= 5.17%

(b) C = 81.6%; H = 8.7%; N = 4.5%; O = 5.2%

14. (a) M(NH4NO3) = 2 × 14.01 + 3 × 16.00 + 4 × 1.01 = 80.06 g mol–1

∴ % N = 06.80

01.142× × 1

100

= 35.00% M(CON2H4) = 2 × 14.01 + 16.00 + 4 × 1.01 + 12.01 = 60.07 g mol–1

∴ % N = 07.60

01.142× × 1

100

= 46.65%

24



(b) Not all compounds with a high nitrogen content would be suitable for use as fertilisers because they may be insoluble, toxic or unable to be taken up by plant roots.

15. symbols C N O H masses 49.48 28.85 16.48 5.19

moles 01.1248.49

01.1485.28

00.1648.16

01.119.5

= 4.12 = 2.06 = 1.03 5.14


03.106.2

03.103.1

03.114.5

= 4.00 = 2.00 = 1.00 = 4.99 ratio 4 2 1 5 therefore the empirical formula of caffeine is C4N2OH5.

16. m(Cu) = m(crucible + contents after heating) – m(crucible) m(Cu) = 27.114 – 27.002 m(Cu) = 0.112 g m(O) = (crucible + contents before heating) – m(crucible + contents after heating) m(O) = 27.128 – 27.114 m(O) = 0.014 g symbols Cu O masses 0.112 0.014

moles 55.63

112.0 0.01416.001

= 0.001 76 = 8.75×10–4

divide by smallest 4

0.001768.75 10−×

4

4

1075.81075.8

−

−

××

= 2.01 = 1.00 ratio 2 1 therefore the empirical formula of the oxide of copper is Cu2O.

17. n(C) = AN

N

n(C) = 23

22

1002.61080.1××

n(C) = 0.0299 mole

n(H) = Mm

= 01.1

0704.0

= 0.0697 mole n(N) = 0.0100 mole To determine the mass of sulfur and the mass of oxygen, convert the number of mole of each of the above elements to a mass and then subtract from the total mass of cysteine:

25



m(C) = 0.0299 × 12.01

= 0.359 g m(H) = 0.0704 g m(N) = 0.0100 × 14.01

= 0.140 g m(S) + m(O) = 1.210 – (0.359 + 0.0704 + 0.140)

= 0.641 g As there are equal masses of sulfur and oxygen present:

∴m(S) = m(O) = 2641.0

= 0.320 g Convert the masses of sulfur and oxygen to moles:

n(S) = 06.32

320.0

= 0.00998 mole

n(O) = 00.16

320.0

= 0.0200 mole symbols C H N O S moles 0.0299 0.0697 0.0100 0.0200 0.00998

divide by smallest 0.02990

9.00998

0.06970

1.00998

0.01000

1.00998

0.02000

1.00998

00998.000998.0

= 3.00 = 6.98 = 1.00 = 2.00 = 1.00 ratio 3 7 1 2 1 therefore the empirical formula of cysteine is C3H7NO2S.

18. symbols C O H masses 37.5 49.9 12.6

moles 37.51

12.01

49.91

16.00

112.61.01

= 3.12 = 3.12 = 12.5


12.312.3

12.35.12

= 1.00 = 1.00 = 4.00 ratio 1 1 4 therefore the empirical formula of methanol is COH4.

19. m(O) = 100 – (22.8 + 21.5) = 55.7 g

26



symbols Na B O masses 22.8 21.5 55.7

moles 22.82

12.99

21.51

10.81

55.71

16.00

= 0.992 = 1.99 = 3.48


01

.992 3.48

01

.992

= 1.00 = 2.00 = 3.51 ratio 2 4 7 therefore the empirical formula of borax is Na2B4O7.

20. m(O) = 100 – (17.04 + 47.41) = 35.55 g

symbols Na S O masses 17.04 47.41 35.55

moles 99.2204.17

06.3241.47

00.1655.35

= 0.7412 = 1.479 = 2.222


01

.7412 2.222

01

.7412

= 1.000 = 1.995 = 2.998 ratio 1 2 3 therefore the empirical formula is NaS2O3. The empirical formula mass = 22.99 + 2 × 32.06 + 3 ×16.00

= 135.11 relative molecular massempirical formula mass

= 11.135

270

ratio = 2 ∴molecular formula = 2 × empirical formula

= 2 × NaS2O3 = Na2S4O6

21. symbols C H O masses 40.00 6.71 53.29

moles 01.1200.40

01.171.6

00.1629.53

= 3.331 = 6.644 = 3.331


331.3644.6

331.3331.3

= 1.000 = 1.995 = 1.000 ratio 1 2 1 therefore the empirical formula of lactic acid is CH2O.

27



The empirical formula mass = 12.01 + 2 × 1.01 + 16.00 = 30.03

relative molecular massempirical formula mass

= 03.30

90

ratio = 3 ∴molecular formula of lactic acid = 3 × empirical formula

= 3 × CH2O = C3H6O3

22. m(H2O) = m(hydrated zinc sulfate) – m(anhydrous zinc sulfate) = 8.2 – 4.6 = 3.6 g

Mole ratio of zinc sulfate (ZnSO4) to water

= 44.1616.4 :

02.186.3

= 0.028 : 0.20

= 028.0028.0 :

028.020.0

= 1 : 7.14 = 1 : 7 therefore the empirical formula of the hydrated zinc sulfate is ZnSO4.7H2O. ∴ × = 7

23. m(Br) = 100 – (12.8 + 2.13) = 85.07 g

symbols C H Br masses 12.8 2.13 85.07

moles 12.81

12.01

01.113.2

90.7907.85

= 1.07 = 2.11 = 1.06


06.111.2

06.106.1

= 1.01 = 1.99 = 1.00 ratio 1 2 1 therefore the empirical formula is CH2Br. The empirical formula mass = 12.01 + 2 × 1.01 + 79.90


= 93.93

188

ratio = 2 ∴molecular formula = 2 × empirical formula

= 2 × CH2Br = C2H4Br2

28



24. m(H) = 0.282 – (16.00 × 10–3 + 0.228) = 0.038 g

symbols C H O masses 0.228 0.038 0.016

moles 0.22812.011

0.0381.011

0.01611

6.001

= 0.0190 = 0.0376 = 0.00100

divide by smallest 0.01900

1.00100

0.03760

1.00100

00100.000100.0

= 19 = 37.6 = 1.0 ratio 19 38 1 therefore the empirical formula of disparlure is C19H38O. The empirical formula mass = 19 × 12.01 + 38 × 1.01 + 16.00


= 57.282

282

ratio = 1 ∴molecular formula of disparlure = 1 × empirical formula

= 1 × C19H38O = C19H38O

Exam questions (page 67)

Extended response questions

1. (a) Mr(C8H18) = 8 × 12.01 + 18 × 1.01 = 114.26

(b) m(C8H18) = n × M = 3.20 × 114.26 = 366 g

(c) n(C8H18) = Mm

= 26.1145.2

= 0.022 mole

therefore N(C8H18) = n × NA = 0.022 × 6.02 × 1023 = 1.3 × 1022 molecules

(d) n(C8H18) = Mm

= 26.1140.5

= 0.044 mole

29



therefore N(C8H18) = n × NA = 0.044 × 6.02 × 1023 = 2.6 × 1022 molecules

N(atoms) = 2.6 × 1022 × 26 = 6.8 × 1023 individual atoms

(e) M(C8H18) = 8 ×12.01 + 18 × 1.01 M(C8H18) = 114.26 g mol–1

∴ % H = 26.11401.118× ×

1100

= 15.91%

(f) % C = 100 – 15.91 = 84.09%

∴m(C) = 100

09.84 × 150

= 126 g Alternative method:

n(C8H18) = Mm

= 26.114

150

= 1.31 mole n(C) = 8 × n(C8H18) = 8 × 1.31 = 10.5 mole ∴ m(C) = n × M = 10.5 × 12.01 = 126 g

2. symbols C H O masses 54.5 9.1 36.4

moles 54.51

12.01

9.111.01

36.41

16.00

= 4.54 = 9.00 = 2.28


28.200.9

28.228.2

= 1.99 = 3.95 = 1.00 ratio 2 4 1 therefore the empirical formula of butyric acid is C2H4O. The empirical formula mass = 2 × 12.01 + 4 × 1.01 + 16.00


= 06.44

88

ratio = 2

30



∴molecular formula of butyric acid = 2 × empirical formula = 2 × C2H4O = C4H8O2

31


Chapter 4

Page 72 1. Ca and Al each form ions in order to attain a full outer shell and hence stability.

2. (a) calcium Ca → Ca2+ + 2e–

2, 8, 8, 2 2, 8, 8

aluminium Al → Al3+ + 3e–

2, 8, 3 2, 8

(b) nitrogen N + 3e– → N3–

2, 5 2, 8

fluorine F + e– → F–

2, 7 2, 8

3. (a) barium ion, Ba2+ (b) potassium ion, K+

(c) phosphide ion, P3– (d) chloride ion, Cl–

(e) sulfide ion, S2–

4. Period Group 1 Group 2 Group 13 Group 15 Group 16 Group 17

2 +1 +2 +3 –3 –2 –1

3 +1 +2 +3 –3 –2 –1

Page 75

5. (a) Ca2+ + O2– → CaO

(b) Be2+ + 2Cl– → BeCl2

(c) Li+ + F– → LiF

(d) 2Al3+ + 3S2– → Al2S3

(e) 3Na+ + N3– → Na3N

(f) Mg2+ + S2– → MgS

6. (a) K+ 2, 8, 8; Ca2+ 2, 8, 8; Al3+ 2, 8

(b) F– 2, 8; O2– 2, 8; N3– 2, 8

7. Neon

32



Page 77 8. (a) The ions Mg2+ + 2Cl– make up MgCl2.

The ions K+ + Cl– make up KCl.

(b) MgCl2 has the ratio 1:2 as two chloride ions are necessary to balance the charge on one Mg2+ ion. KCl has the ratio of 1:1 as one K+ ion balances one Cl– ion.

(c) The lattice is maintained by the strong electrostatic attraction between the positive and negative ions. This is called the ionic bond.

Page 78 9. (a) Calcium chloride has a high melting point as it is an ionic compound with strong

electrostatic attractive forces holding the ions together in a three-dimensional lattice.

(b) It will shatter when pressure is applied as ions of like charge repel when aligned with one another.

(c) It will conduct electricity only in the liquid state where the ions are free to move. In the solid state, the ions are held strongly in place by the strong ionic bond and so will not conduct electricity.

10. (a) (i) NaCl has one Na+ ion to each Cl– ion; therefore the ratio is 1:1.

(ii) MgO has one Mg2+ ion to each O2– ion; therefore the ratio is 1:1.

(b) MgO has a higher melting point as the electrostatic attraction is greater between its ions due to their higher charges.

Page 79 12. (a) X2Y (b) ionic (c) hard, brittle, high melting point

Page 81 13. (a) AlCl3 (b) BaO (c) Na2S (d) Mg3P2

14. Ions K+ Ca2+ Al3+

F– KF CaF2 AlF3

O2– K2O CaO Al2O3

N3– K3N Ca3N2 AlN

15. potassium fluoride, calcium fluoride, aluminium fluoride, potassium oxide, calcium oxide, aluminium oxide, potassium nitride, calcium nitride, aluminium nitride

16. (a) potassium chloride (b) silver sulfide (c) sodium nitride

Page 82 17. (a) iron(II) sulfide (b) iron(III) sulfide

(c) copper(II) chloride (d) tin(II) oxide (e) copper(I) oxide (f) lead(II) bromide

33



18. (a) SnF4 (b) PbS (c) HgO (d) FeN (e) Cu2S (f) SnO

Page 85 19.

Valency of cation Valency of anion Empirical formula

Ag+

K+

Mg2+

Al3+

Fe3+

Ba2+

NH4+

Cl–

S2–

O2–

Br–

CO32–

PO43–

SO42–

AgCl

K2S

MgO

AlBr3

Fe2(CO3)3

Ba3(PO4)2

(NH4)2SO4

20. (a) Na2SO3 (b) Ca(NO2)2 (c) Cu(HCO3)2

21. (a) aluminium carbonate, Al2(CO3)3

(b) sodium nitrate, NaNO3

(c) mercury(II) phosphate, Hg3(PO4)2

(d) lead(II) sulfate, PbSO4

Page 86 22. (a) Na2CO3.10H2O

(b) magnesium sulfate heptahydrate

(c) MgCl2.6H2O

(d) barium chloride dihydrate

Multiple choice questions 1. A 2. C 3. A 4. C 5. C 6. D 7. A 8. B 9. C 10. B 11. A 12. C 13. C

Review questions

1. Name of atom Symbol for atom Electron configuration of atom

lithium Li 2, 1

beryllium Be 2, 2

nitrogen N 2, 5

oxygen O 2, 6

fluorine F 2, 7

sodium Na 2, 8, 1

34



Name of atom Symbol for atom Electron configuration of atom

magnesium Mg 2, 8, 2

aluminium Al 2, 8, 3

phosphorus P 2, 8, 5

sulfur S 2, 8, 6

chlorine Cl 2, 8, 7

potassium K 2, 8, 8, 1

calcium Ca 2, 8, 8, 2

Name of ion Symbol for ion Electron configuration of ion

lithium Li+ 2

beryllium Be2+ 2

nitride N3– 2, 8

oxide O2– 2, 8

fluoride F– 2, 8

sodium Na+ 2, 8

magnesium Mg2+ 2, 8

aluminium Al3+ 2, 8

phosphide P3– 2, 8, 8

sulfide S2– 2, 8, 8

chloride Cl– 2, 8, 8

potassium K+ 2, 8, 8

calcium Ca2+ 2, 8, 8

2. (a) 2, 8, 8, 2 (b) 2, 7

(c) Two fluorine atoms each accept one electron from a calcium atom, resulting in two fluoride ions, 2F–, and one calcium ion, Ca2+.

(e) the ionic bond — the strong electrostatic attraction between the oppositely charged ions

(f) 1:2

3. Metal atoms lose electrons to attain a full outer shell and become stable ions or cations. The resultant charge of the ion is positive as there are more protons than electrons.

4. The ratio of the different ions in potassium oxide is 2:1. This is because two K+ ions are necessary to balance the O2– charge.

5. (a) K+ + F–→ KF

(b) 2Al3+ + 3O2–→ Al2O3

(c) Be2+ + 2Cl–→ BeCl2

35



6. Ionic salts conduct electricity in the molten and aqueous states as the ions are mobile. In the solid state they are rigidly held by strong ionic bonds and therefore cannot conduct electricity.

8. Ionic substances are usually solids and have high melting points due to the strong attractive electrostatic forces holding the ions together in a lattice.

9. (a) XY (b) ionic (c) X2+ + Y2–→ XY

10. (a) MgSO4 (b) ZnO (c) Fe(OH)2 (d) AgCl (e) AlN

(f) CaCO3 (g) Ca(HCO3)2 (h) PbI2 (i) KHSO4

(j) (NH4)2CO3 (k) Ag2SO4 (l) SnCl2 (m) K2SO4

(n) NaF (o) Ba(NO3)2 (p) Fe(OH)3 (q) Na2S (r) Al2O3

(s) CaH2 (t) CuSO4 (u) NH4OH (v) Cr2O3

(w) Ca(NO3)2 (x) LiCl (y) KCN (z) Na2HPO4

11. (a) iron(III) oxide (b) aluminium sulfate

(c) calcium chloride (d) magnesium nitrate

(e) barium sulfate (f) zinc chloride

(g) aluminium carbonate (h) sodium sulfate

(i) silver nitrate (j) sodium hydroxide

(k) potassium nitrate (l) lead(II) chloride

12. (a) sodium carbonate, Na2CO3

(b) sodium hydrogen carbonate, NaHCO3

(c) calcium carbonate, CaCO3

13. (a) MgSO4.7H2O (b) Na2CO3.10H2O (c) ZnCl2.6H2O (d) BaCl2.2H2O

14. (a) barium chloride trihydrate

(b) lithium chloride tetrahydrate

(c) cobalt(II) chloride pentahydrate


Multiple choice questions 1. B 2. A

Extended response questions 1. (a) Ca2+ (b) P3– (c) Al3+

2. Oxygen atoms have the electron configuration of 2, 6. For two atoms of oxygen to bond ionically, each would need to gain two electrons to attain a full outer shell. This is not possible.

36


Chapter 5

Page 94 1. (a) 2, 1 (b) Li+, 2

(c) Solid lithium contains lithium cations in a ‘sea’ of delocalised valence electrons. Electrostatic forces of attraction hold the lattice together.

2. b, d, e

3. The structure of a metal is stable as metallic atoms, when losing their valence electrons to form cations, attain full outer shells. Each aluminium atom gives up three electrons when forming the lattice producing the cation Al3+ with the stable electron configuration of 2, 8.

Page 95 4. (a) Iron is hard since its structure is very dense and the electrostatic attraction

between the cations and delocalised electrons is very strong.

(b) When iron is hammered into a sheet (malleable) or drawn into a wire (ductile), layers of atoms move past one another without disrupting the force between the cations and the negative ‘sea’ of electrons. This is due to the non-directional nature of the metallic bond.

(c) Under the force of an electric field, the ‘sea’ of electrons is mobile whether the metal is in solid or molten form.

Page 96 5. (a) The metallic lattice consists of magnesium cations, Mg2+, in a ‘sea’ of delocalised

electrons. The electrostatic attraction between the cations and the electrons bonds them together.

(b) The mobile electrons within the lattice are able to reflect light.

Page 98 6. (a) iron (b) aluminium (c) copper (d) lead

7. Aluminium is cheaper and lighter than copper.

Page 100 10. It does not explain the differences in magnetism, density and strength of metals.

11. A metal that has large grains will have fewer dislocations and will bend more easily than a metal with small grains which has many dislocations.

37



12. (a) Tempering disrupts the metallic lattice reducing brittleness while retaining hardness.

(b) quenching and annealing

13. The work-hardened copper becomes more brittle; annealing it restores ductility.

14. The metal clip has been work hardened — the crystal grains are now smaller and the metal is more brittle, making it more difficult to bend it back to its original shape.

Page 102 15. (a) Solder can be made by heating 30% tin and 70% lead together.

(b) substitutional

(c) Solder has a lower melting point than either tin or lead and therefore can be used to join metal parts together.

16. (a) brass (b) aluminium alloy (c) steel (d) steel

Multiple choice questions 1. C 2. C 3. D 4. A 5. B 6. A 7. B 8. A 9. B

Review questions 1. B

2. a, e

6. Silver — bonds to the mercury and helps the amalgam set. Tin — helps the amalgam stay free from tarnish. Mercury — can be poured into the tooth easily. Copper and zinc — gives the amalgam strength. Gold — will not tarnish or react with food or drink.

7. A work-hardened metal is difficult to bend as it becomes brittle due to smaller crystal grains and increased dislocations.

8. Annealing is the process of heating a metal and then cooling it slowly. This produces larger crystals and a softer metal. Tempering means to warm a quenched metal and allow it to cool slowly. This reduces brittleness but retains hardness.

9. Quenching is the process where the red hot metal is cooled quickly by plunging into cold water. This produces smaller crystals and hence a harder but more brittle metal.

10. (a) A metal is a pure substance (an element) while an alloy is a mixture of metals or of metal and non-metal (such as carbon) heated to molten state and then cooled.

11. (a) element (b) alloy (c) element (d) alloy (e) alloy (f) element (g) alloy (h) compound (i) compound (j) element (k) alloy (l) alloy (m) compound

12. The properties of an alloy are different from the properties of the constituent metals.

13. Alloys often have properties that make them better suited for a particular use than the original metal(s). Steel, for example, is stronger and more resistant to rusting than iron.

38




Extended response questions 1. (a) C (b) A (c) B

2. (b) steel, resistant to corrosion

39


Chapter 6

Page 113

1.

Group Element

Valence

electrons

Lewis

diagram

Lone

pairs

Bonding

electrons

16 S 6

2 2

14 Si 4

0 4

15 P 5

1 3

17 Cl 7

3 1

17 Br 7

3 1

16 Se 6

2 2

18 Ar 8

4 0

Page 115

2. (a) This is a triple covalent bond.

(b) 2

(c) N N

3.

4. iodine

Page 116

5. (a)

(b)

6.

40



7. (a)

(b)

(c)

(d)

(e)

(f)

(g)

Page 118

8.

Compound

Electron diagram

of molecule

Number of lone

pairs around

central atom

Number of

bonding pairs

around central

atom

Shape diagram and

name of shape

NI3

1 3

pyramidal around

N atom

CF4

0 4

tetrahedral around

C atom

OF2

2 2

V-shaped around

O atom

41



Compound

Electron diagram

of molecule

Number of lone

pairs around

central atom

Number of

bonding pairs

around central

atom

Shape diagram and

name of shape

HOF

2 2

V-shaped around

O atom

C2F2

0 (for each atom) 2 (for each atom)

linear around each

C atom

C2Cl4

0 (for each atom) 3 (for each atom) planar around each

C atom

Page 120

9. nitrogen trihydride, dihydrogen dioxide, trioxygen

10. (a) nitrogen trichloride

(b) sulfur hexafluoride

(c) oxygen dichloride

(d) silicon tetrahydride

(e) phosphorus trihydride

11. (a) CO (b) SiCl4 (c) P2O5 (d) NO (e) SO3

12. from top to bottom: N2O; hydrogen chloride; tetraphosphorus decoxide; HF

Page 122

13. (a) H F, polar (b) O H, polar (c) C H polar

(d) N H, polar (e) C—C, non-polar

14. (a) polar covalent (b) ionic (c) non-polar covalent

15. N—N, Cl—N, F—N, Li—F

Page 123

16. (a) H I polar

(b)

non-polar

42



(c) S═C═S non-polar

(d)

(e)

(f)

(g)

(h)

(i)

(j)

(k) N N non-polar

Page 127

17. (a) dispersion forces

(b) dispersion forces

(c) dispersion and dipole–dipole forces

(d) dispersion and dipole–dipole forces

(e) dispersion forces and hydrogen bonding

18. Although dispersion and dipole–dipole intermolecular forces exist in all three

substances, as the number of electrons in each molecule increases (from 18 in HCl, to

36 in HBr, and 54 in HI) so does the strength of the dispersion forces. This is reflected

in the increasing boiling points.

Page 128

19. Cl2 has the higher boiling point as dispersion forces increase with increasing number of

electrons.

43



20. Helium atoms have only two electrons while argon atoms have 18. Thus the dispersion

forces acting on argon atoms are stronger and its boiling point is higher.

21. HBr is a polar molecule and experiences dipole–dipole forces as well as dispersion

forces, while Kr experiences only dispersion forces.

22. HF has hydrogen bonding, which is stronger than other dipole–dipole forces.

23. Glucose dissolves in water as both glucose and water are polar substances (like

dissolves like). It does not conduct electricity in solid or aqueous form as it has no free

ions or electrons.

24. They are all gases as only weak dispersion forces act between the molecules.

25. I2 is a non-polar substance and therefore will dissolve in non-polar carbon tetrachloride

more readily than in polar water.

26. (a) Candle wax is relatively soft and can be scratched as it is a non-polar covalent

molecular compound with weak dispersion forces between its molecules.

(b) It is not soluble in water as it is non-polar.

(c) It is a non-conductor as it has no free-moving charged particles (electrons or

ions).

Page 129

27. Cl2 has only dispersion forces between its molecules while BrF has dipole–dipole

forces as well as dispersion forces.

Page 132

28. (a) Graphite can be used for printing as it has a covalent layer lattice structure. There

are only weak dispersion forces between the layers, allowing them to slide over

each other.

(b) Diamond has a covalent network structure with strong covalent bonding in three

dimensions. This makes it very hard and hence its use in drilling.

29. There are no free electrons in the covalent network structure of diamond. Graphite,

however, has delocalised electrons that can move across its layers under an electric

field.

Page 135

34. (a) covalent molecular (b) metallic

(c) covalent molecular (d) ionic

35. (a) O (b) M (c) L (d) N

Page 137

36. (a) metallic (b) ionic

(c) dispersion forces between atoms of argon

(d) covalent (e) covalent (f) metallic (g) covalent

(h) covalent (i) covalent (j) ionic

44



37. (a) covalent molecular (b) covalent network

(c) ionic network lattice (d) metallic lattice

(e) metallic lattice (f) covalent molecular

38. (a) covalent within the molecule; dispersion forces between the molecules

(b) covalent within the molecule; dispersion forces and hydrogen bonding between

molecules

(c) covalent within the molecule; dispersion forces between molecules

(d) covalent within the molecule, dispersion forces and hydrogen bonding between

molecules

39. (a) O═C═O

(b)

(c)

(d)

40. H—F

41. (a) ionic bonding

(b) Pressure forces like charges to align, causing repulsion and shattering of the

crystal.

(c) Solubility in water — water molecules are able to move between the ions and free

them by disrupting the rigid ionic lattice.

42. (a) Gold is used in jewellery as it is shiny, durable, highly ductile and malleable.

These properties are due to its metallic lattice (cations in a ‘sea’ of delocalised

electrons). The non-directional nature of the metallic bond allows distortion

without disrupting the lattice, while the lustre is due to light being reflected from

the delocalised electrons.

(b) Graphite has a covalent layer lattice with weak dispersion forces between the

layers — this accounts for its use as a lubricant as it is slippery due to the layers

sliding over one another.


1. D 2. A 3. A 4. A 5. D 6. B 7. C 8. D 9. C 10. D 11. D 12. B 13. C

14. A 15. C 16. C 17. C 18. (i) B (ii) C 19. D 20. B 21. D 22. D 23. C

45



Review questions

1. (a) They already have complete outer shells of electrons.

2. (a) A fluorine molecule is composed of two fluorine atoms covalently bonded

together. Each fluorine atom shares one electron pair so that both atoms attain a

full outer shell of eight electrons, complying with the octet rule.

(b) A hydrogen molecule is composed of two hydrogen atoms covalently bonded

together and sharing two electrons between them. The octet rule is not used as the

first shell only is involved, needing just two electrons to be full and therefore

stable.

3. (a) group 14 (b) group 18 (c) group 17 (d) group 15

4. (a)

(b)

(c)

(d)

(e)

(f)

7. (a) O = O linear

(b)

(c)

(d)

(e)

46



(f) H—C≡N linear

(g)

(h)

(i) H—C≡C—H linear (around each C atom)

(j)

(k) N N linear

8. (a) tetrahedral (b) pyramidal (c) linear (d) V-shaped

(e) pyramidal (f) planar around both carbon atoms

9. (a) non-polar (b) polar (c) polar (d) polar (e) ionic

10. (b) Each atom has the same attraction for the bonding pair of electrons.

(c) Polarity depends on the shape of the molecule as well as bond dipoles. If the bond

dipoles cancel each other out due to the shape of the molecule, it will be a non-

polar molecule.

11. In question 7: b, d, e, f, g and j are polar

In question 8: b, c, d and e are polar

12. (a) polar (b) non-polar (c) non-polar (d) polar (e) polar (f) non-polar

13. Ammonia, NH3, has a molecular dipole due to its pyramidal shape. Carbon dioxide,

CO2, is non-polar because its shape is linear and the bond dipoles cancel each other out.

14. (a) Iodine has the higher number of electrons and hence stronger dispersion forces.

(b) Fluorine is a gas at room temperature due to its weak dispersion forces. Iodine is

a solid at room temperature as it has stronger dispersion forces.

15. CH3F has dispersion and dipole–dipole forces acting between its molecules; however,

dispersion forces and the stronger hydrogen bonding between molecules of CH3OH

cause it to have a higher boiling point.

16. (a) OCS has the stronger intermolecular forces (dispersion and dipole–dipole forces)

while non-polar CO2 has dispersion forces only.

(b) HF has both dispersion forces and hydrogen bonding, which is stronger than other

dipole–dipole forces, while HBr has dispersion and dipole–dipole forces.

17. Ar < F2 < HCl

18. a, e, f, g, h

47



19. Although both methanol and ethanol have hydrogen bonding and dispersion forces,

ethanol has the higher boiling point as it has more electrons and therefore stronger

dispersion forces.

20. HCl is a covalent molecular compound. It has no free-moving charged particles

(electrons or ions) to allow it to conduct electricity in the liquid state. However, when

dissolved in water it ionises to produce free mobile ions which can conduct electricity.

21.

Substance

Melting

point

Electrical conductivity

Solid Molten Dissolved in water

Mg(NO3)2 high no yes yes

Cu high yes yes insoluble

CH3OH low no no no

Ne low no no no

Cl2 low no no no

22. from top to bottom: nitrogen monoxide; nitrogen dioxide; SO2; CO; O3; phosphorus;

NH3; methane (carbon tetrahydride); H2S

23. (a) nitrogen tribromide (b) dinitrogen tetroxide

(c) dinitrogen monoxide (d) dinitrogen pentoxide

(e) dinitrogen trioxide (f) phosphorus trichloride

(g) phosphorus pentachloride (h) diphosphorus pentoxide

(i) sulfur hexafluoride (j) sulfur dichloride

24. (a) Diamond has a covalent network lattice structure and graphite has a covalent

layer lattice structure.

(b) Both have strong covalent bonding between the carbon atoms in the lattice;

however, graphite has strong covalent bonding in only two dimensions with weak

dispersion forces between layers while diamond has strong covalent bonding in

all three dimensions.

(c) Both consist of countless atoms held together by covalent bonds.

(d) Graphite is used as a lubricator, in pencil ‘lead’ and electrodes; diamond is used

in jewellery, drilling and as an abrasive (cutting glass).

25. (a) lubricant — layers can slide over one another

(b) There are only weak dispersion forces between the layers.

26. The purpose of the clay is to bind the graphite powder. The ratio of clay to graphite

determines the hardness of the pencil. The more clay, the harder the pencil; the less

clay, the softer the pencil.

27. The property of hardness due to the strong covalent bonding in three dimensions.

28. When enough pressure is applied to break the strong covalent bonds, the lattice is

distorted and the diamond will shatter.

48



30. They all have a full outer shell.

31. Na, 2, 8, 1, and K, 2, 8, 8, 1, both have the same number of valence electrons (1), so

their chemical properties are similar.

32. oxide O2–

, fluoride F– , aluminium Al

3+

33. (a) ionic (b) XY2

34. (a) iii (b) v (c) i (d) iv (e) ii

35. b and d

36. (a) ionic bonding

(b) covalent within molecule

(c) ionic bonding

(d) covalent within molecule, dispersion forces and hydrogen bonding between

molecules in the solid state

(e) covalent within molecules

(f) covalent within molecules; dispersion forces between molecules in the solid state

(g) covalent bonds in three-dimensional network lattice

(h) metallic bonding

(i) weak dispersion forces between atoms

(j) covalent bonds within the molecule

(k) covalent bonds within molecules, dispersion forces and hydrogen bonding

between the molecules in the liquid phase

(l) ionic bonding

(m) covalent bonding within the molecule

(n) ionic bonding

(o) ionic bonding

(p) metallic bonding

(q) metallic bonding

37. (a) covalent molecular

(b) metallic

(c) ionic

(d) covalent molecular

(e) covalent molecular

(f) ionic

(g) covalent molecular

(h) covalent molecular

(i) ionic

(j) covalent molecular

49



(k) metallic

(l) covalent molecular

(m) covalent molecular

38. Discrete molecules are often volatile (evaporate) at room temperature due to weak

intermolecular forces. Metallic, ionic and covalent network substances have strong

bonds and remain solids at room temperature, hence do not have a smell.

39. (a) (i) covalent molecular — benzoic acid, paraffin wax and sugar; ionic —

magnesium sulfate and sodium fluoride; metallic — aluminium

(ii) aluminium

40. Ionic solids do not have free mobile electrons, whereas metals have free delocalised

electrons which can move under an electric field.

41. (a) (i) ionic (ii) covalent molecular (iii) metallic (iv) covalent network

(b) CH3OH, NaF, Fe, SiO2

42. KI, as ionic bonds are stronger and thus harder to break than the dispersion forces and

hydrogen bonding between ammonia molecules.

43. (a) KF, MgBr2, CaO (b) P4O10, CCl4, C2H4, SiO2

44. The metallic bonding is non directional allowing distortion without disrupting the

lattice of cations in the ‘sea’ of electrons. Calcium chloride is brittle because its rigid

ionic lattice will shatter if pressure aligns the ions of like charge.

45. Diamond has a covalent network structure with strong covalent bonds in three

dimensions. Tungsten has a metallic lattice structure consisting of close-packed cations

in a ‘sea’ of electrons. The bonding is strong in both substances, and extreme heat is

needed to break the bonds.

46. (a) covalent, covalent molecular lattice

(b) ionic, ionic lattice

47. (a) G (b) H (c) E, F

48. A — covalent molecular

B — metallic

C — ionic

D — covalent network

Exam questions (pages 145–6)

Multiple choice

1. C 2. C 3. B 4. A

Extended response

1. Silicon dioxide has a covalent network structure. Its strong three-dimensional bonding

results in a rigid solid at room temperature. Carbon dioxide is a discrete covalent

50



molecular compound. It has weak dispersion forces between its molecules and hence is

a gas at room temperature.

2. (i) (a) dispersion forces (b) molecules

(ii) (a) covalent (b) atoms

(iii) (a) ionic (b) ions

(iv) (a) covalent (b) atoms

(v) (a) metallic (b) ions

(vi) (a) metallic (b) ions

(vii) (a) dispersion forces and hydrogen bonding

(b) molecules

4. (a) HF (b) Na (c) KF (d) He (e) Si (f) H2S

51


Chapter 7

Page 161 11. (a) period 2, group 16

(b) period 2, group 1

(c) period 4, group 8

(d) period 4, group 18

(e) period 4, group 3

(f) period 4, group 12

12. (a) O (b) Li (c) Fe (d) Kr (e) Sc (f) Zn

Page 162 13. (a) 1s22s22p63s23p5 period 3, group 17

(b) 1s22s22p63s23p63d6 4s2 period 4, group 8

(c) 1s22s22p63s23p63d104s24p5 period 4, group 17

Page 167 16. (a) Ca (b) Be (c) Cl

Multiple choice questions 1. C 2. D 3. C 4. D 5. B 6. C 7. A 8. D 9. D 10. C 11. D

52


Chapter 8

Page 175 (top) 1. C16H34

2. C25H52

Page 175 (bottom)

4. (a) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

(b) C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g)

5.

6. 2C6H14(g) + 19O2(g) → 12CO2(g) + 14H2O(g)

Page 177

10. (a)

(b)

Page 178

12. (a)

(b)

(c)

53



(d)

13. carbon dioxide and water

14. 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

15. C21H42

16. C12H24

17. the presence of the double bond

18. any two of combustion, addition, self-addition

Page 179

19. (a) C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)

(b) C6H12(g) + 9O2(g) → 6CO2(g) + 6H2O(g)

20.

21.

22. Dispersion forces increase with molecular size.

Page 181 (top)

24. (a)

(b)

(c)

54



25. (a) 2,2-dimethylhexane (b) octane

26. (a)

(b)

Page 181 (bottom) 27. (1) (a) CH3CHCHCH2CH2CH3 (b) CH2CCH3CH3

(c) CH3CCCH2CH3

(2) (a) (CH3)3C(CH2)3CH3

(b) CH3(CH2)6CH3

(3) (a) CH3CHCHCHCH3CH3

(b) CH3CHCH3CH2CHCH3CCH2CH3

28. CH3CH(CH3)CH(CH3)CH2CH2CH3

Page 183

29. (a)

(b)

(c)

(d)

55



(e)

30. Yes, each successive member differs by a CH2 group.

Page 185 31. ethyl ethanoate

32.

33. The sulfuric acid acts as a catalyst. It reacts with the water formed.

34. COOH

35.

36. Refer to table 7.5 on page 158.

37. (a)

(b)

Page 188

40. Lighter fractions are gases, which burn more readily than liquids.

41. The viscosity increases with increasing molecular size because of the stronger intermolecular dispersion forces and the increased possibility of entanglement of the long molecules.

42. The molecules with fewest carbon atoms in them are the smaller molecules. They are lighter and have fewer dispersion forces acting between them. They are gaseous.

43. The bubble caps act to condense the vapour before the next temperature rise.

Page 190 45. Catalysts increase the rate of reaction.

46. less heat required, quicker reaction time

56



Multiple choice questions 1. D 2. C 3. C 4. D 5. D 6. C 7. D 8. B 9. C 10. A, D 11. C 12. B 13. C

14. A 15. D 16. D 17. D 18. A

Review questions 1. (a) chemical compounds that contain only carbon and hydrogen

(b) propane, C3H8; propene, C3H6; propyne, C3H4

2. (b) 2C3H6(g) + 9O2(g) 6CO2(g) + 6H2O(g) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

7. (a) C22H46 (b) C17H34 (c) C13H24

8. (a) 2-methylbutane (b) 3-methylpentane

(c) 3-ethylhexane (d) 3,3-dimethylhexane

(e) 2,4-dimethylhexane (f) 2,2,4,4-tetramethylpentane

9. (a)

(b)

(c)

(d)

(e)

57



(f)

10. (a)

(b)

(c)

(d)

(e)

(f)

(g)

58



(h)

11. (a) 2,4-dimethyl-1-heptene

(b) pent-1-yne

(c) oct-3-ene

(d) 3,4-dimethylpent-1-ene

(e) 2,3-dimethylbutane

(f) 4-ethyl-3-methylheptane

12. The longest chain has seven carbon atoms. The compound should have been named 3,3-dimethylheptane.

13. Successive members of the alcohol series differ by a CH2 group.

15. (a)

(b)

17. (a)

(b) esterification

(c) (i) pungent odour (iii) fruity odour

59



18. (a)

(b)

(c)


Multiple choice questions 1. C 2. B

Extended response questions 1. Boiling points reflect the strength of the intermolecular forces operating. The

intermolecular forces operating are dispersion forces, which increase with molecule size.

2. (a) petrol (b) petrol (c) paraffin (d) further refine by heating

60


Chapter 9

Page 200

1. (a)

(b) polytetrafluoroethylene, PTFE

(c) Teflon, nonstick coating on frying pans and saucepans

2. (a)

(b) polyvinylacetate (c) glue

Page 208

7. (a) H2N(CH2)6NH2 + ClOC(CH2)8COCl→ —HN(CH2)6NH—OC(CH2)8CO—HN(CH2)6NH —OC(CH2)8CO— + nHCl

(b)

Multiple choice questions 1. B 2. C 3. B 4. C 5. D 6. D 7. A 8. C 9. B 10. A 11. B 12. C

61



Review questions

3. (a)

4. See figures (a) and (b) at the top of page 206.

9. (a) and (b) the number of carbon atoms in the two monomers

(c) weak dispersion forces and hydrogen bonding

(d) hydrogen bonding

(e) nylon 6:10; being the larger molecule, the weak dispersion forces would have greater effect than those operating in nylon 6.6.

10. Nylon is a larger molecule with side groups that enhance the polarity of the chains, increasing the effect of dispersion forces acting on it.

12. noxious fumes, melting

13. (a)

(b) addition

(c) dispersion

14. (i) (a)

(b) condensation polymerisation

(ii) (a)

(b) condensation polymerisation



1. (a)

62



(b)

(c)

2. (a) (i)

(ii)

(b) but-2-ene

(c)

63


Chapter 10

Page 219 1. energy needed to create a new surface by breaking bonds within a substance

2. (a) Magnesium oxide, being an ionic compound, has very strong ionic bonds holding anions and cations in its lattice and thus has a high surface energy. Mercury atoms are less strongly held together by metallic bonds within the liquid.

(b) Mercury atoms are held together by metallic bonds whereas helium atoms are attracted by weaker dispersion forces.

(c) Diamond is made up of carbon atoms that are held together by strong covalent bonds in three dimensions and thus have a high surface energy because these bonds are difficult to break. Copper atoms are held together in a lattice by weaker metallic bonding, so copper has a lower surface energy than diamond.

(d) Iron atoms are held together in a strong metallic lattice whereas Teflon is an organic compound consisting of molecules held together by weaker intermolecular forces. Iron thus has a higher surface energy than Teflon.

(e) Water molecules are held together by relatively strong hydrogen bonding whereas octane molecules are held together by relatively weak dispersion forces. Thus, water has a higher surface energy than octane.

Page 220 3. Higher surface energies correspond to greater difficulty in breaking bonds to create new

surfaces and hence higher boiling and melting points. Generally, substances with the highest surface energies are very hard solids whereas substances with the lowest surface energies are gases.

4. (a) Particles are closer together in solids than in liquids so that the interparticle forces are therefore stronger.

(b) Paraffin wax molecules are larger than those of octane and have more electrons, thereby increasing the strength of dispersion forces between molecules.

(c) Atoms in mercury are held together by strong metallic bonds whereas molecules of Teflon are held together by weaker intermolecular dispersion forces.

(d) Water molecules are held together by hydrogen bonding and dispersion forces whereas heptane molecules are held together only by dispersion forces.

(e) Magnesium oxide and iron both have strong bonds (ionic and metallic bonds respectively) holding particles together in their lattice structures.

5. Gold has a lower surface energy than iron, indicating that the bonding between gold atoms in the metallic lattice is weaker than the metallic bonding holding iron atoms in a metallic lattice. It is therefore easier to separate gold atoms than iron atoms so that gold is more malleable than iron.

64



Page 222 7. Water molecules, being polar, have a greater attraction to each other than to air (since

air consists of a number of non-polar substances) and thus will form a droplet to achieve lowest possible surface area-to-volume ratio. Water molecules have a higher surface energy than the major components of air, namely N2, O2, CO2 and He.

Page 224 8. spread: clean glass, washed car body, washed skin drop: greasy plate, Teflon pan,

waxed car body, leaf, flower petal, unwashed skin

9. (a) no (b) yes

Page 226 10. Ethanol has a lower surface energy than glass and will therefore wet it. Since there is a

smaller downward force acting on the ethanol in the thinner tube, the ethanol will rise higher in this tube than in the wider tube.

11. Particles in detergent solution have equal attraction to the bulk of the liquid and glass.

Page 227 12. (a) Since carbon dioxide molecules are non-polar molecules and grapes have a

hydrophobic surface, they are able to attract.

(b) Hydrophobic reactions: attachment of carbon dioxide bubbles to grapes; carbon dioxide bubbles released into the atmosphere (no hydrophilic reactions occur).

Page 229 14. (b) Dipalmitoyllecithin has a polar head and a non-polar tail and lowers surface

tension of water on the alveoli surfaces.

Page 231

16. An ester is a molecule that is formed when an organic acid reacts with an alcohol.

17.

Page 237 18. See table 10.2 on page 236.

19. (a) See figures (a) and (b) on page 235.

65



(b) Property Milk Butter feel cool greasy conductivity high low solubility in water high low addition of food dye coloured evenly not coloured evenly

20. conductivity; placement on skin; addition of food colouring to a sample; solubility in water

21. (a) insoluble

(b) soluble

22. (a) Marine bird and animal feathers and fur become coated with oil, limiting movement; oxygen is prevented from dissolving into sea water and oil may be toxic to marine life.

(b) See figure (a) on page 235.

(c) Surfactant may itself be toxic to marine life; since the surfactant is used to disperse the oil, surfactants may be found a long way from where they were added at the site of the oil spill.

Page 238 23. 12.5 times

24. 47 times

Page 244 25. Nanoscale particles do not reflect and react with light in the same way as microscopic

particles.

26. Carbon nanotubes are covalently bonded, are non-polar and have a low surface energy, therefore polar water droplets will not adhere to a carbon nanotube’s surface.

27. A material with low surface energy and nanoscale roughness

28. The water droplets make low contact angles with the nanoscale-rough surface of the leaf and spread rapidly from peak to peak instead of bonding with the surface. Dirt particles on the surface are collected by the droplets and carried away when the droplets roll off the leaf.

29 Nanoparticles of titanium oxide are placed on the glass to form an ultrahydrophilic coating.

Page 246

30. Many small dispersion forces form between the surface and the fine hairs on the gecko’s feet.

31. Removal of all the tape at once may damage the surface as there are many bonding forces at work.

66



Multiple choice questions 1. C 2. C 3. D 4. B 5. A 6. A 7. C 8. A 9. A 10. C 11. C 12. B 13. B

Review questions 1. (a) cup resting on a table

(b) water in a glass

(c) air surrounding a bottle

(d) oil floating on water, e.g. ‘French dressing’

(e) boiling water

2. Gases would mix into each other with no ‘boundary’, or interface, between them.

3. Surface energy is the amount of energy needed to create a new surface (e.g. the amount of energy needed to split a log of wood).

4. silicon dioxide (covalent network lattice) > nickel (metallic) > phosphorus trichloride (dipole–dipole forces) > decane (liquid, long molecule, dispersion forces) > liquid oxygen (small molecule, dispersion)

5. Surface energy is the amount of energy needed to create a new surface in any substance (solid, liquid or gas); surface tension is the surface energy of a liquid.

6. Liquid particles in the bulk of a liquid are surrounded by liquid, thus only liquid/liquid interactions are possible. Particles at the surface of a liquid may react with other liquid particles or particles in the air and may consequently experience liquid/liquid as well as liquid/air interactions.

7.

Water displays hydrogen bonding between molecules, whereby the oxygen atom from

one water molecule attracts a hydrogen atom from another water molecule.

8. (a) The secretion of oil enables insects to stay above the water since the oil will not mix with, or dissolve, in the water, i.e. oil will not ‘wet’ the water so the insect remains on the surface.

(b) standing on fewer legs; coating feet with a surfactant before walking on water

10.

11. Water molecules are able to ‘wet’ the glass surface since glass has a higher surface

energy than water and will thus form a concave meniscus. Mercury atoms have a greater attraction to the bulk of the liquid than the glass (since mercury has a higher surface energy than glass) and thus form a convex meniscus (see the figures on page 224 and the top of page 225).

67



12. Place equal amounts of each of the four solutions into separate beakers. Put a capillary tube in each beaker, clamping it to a retort stand so that it is upright. Measure the height each liquid rises to determine relative surface tension. Higher rise corresponds to greater surface tension.

13. hydrophobic: substances that do not mix well with water (e.g. oil, butter, oxygen gas)

hydrophilic: substances that dissolve in water (e.g. sodium chloride; copper(II) chloride)

14. (a) hydrophobic CH3(CH2)12; hydrophilic COO–

(b) hydrophobic CH3C(CH3)2; hydrophilic NH3+

(c) hydrophobic CH3(CH2)4; hydrophilic Cl

(d) hydrophobic CH3(CH2)17; hydrophilic COOH

(e) hydrophobic CH3(CH2)16; hydrophilic N+

15. Water ‘spreads’ further after the addition of a surfactant, thus enabling it to ‘wet’ the surface better.

16. A surfactant is a ‘surface active agent’ that changes the properties of surfaces from being hydrophilic to hydrophobic, or vice versa. It is able to do this due to its characteristic hydrophilic, water-loving ‘head’ and hydrophobic, water-hating ‘tail’.

17. (a) The high surface tension of water behaves like a ‘skin’, preventing the needle from sinking.

(b) needle sinks, since surface tension of water has been lowered

18. (a) test tube without detergent: oil and water separate test tube with detergent: oil and water remain mixed

(b) surfactant/emulsifier

(c) (i) add surfactant to mix oil spill into water

(ii) mixture itself may be a hazard to marine life

19. (a) long-chain carboxylic acid

(b) molecule formed when an organic acid reacts with an alcohol

(c) ester formed by the reaction of three fatty acids and one glycerol molecule

(d) sodium salts of fatty acids

(e) reaction between an alcohol and an organic acid

(f) hydrolysis of a triglyceride using sodium hydroxide to produce glycerol and three soap molecules

20.

21. (a)

68



(b)

22. (a) hydrophobic CH3(CH2)10; hydrophilic COO–

(b)

23. Bile is an emulsifier that acts to react with digested fat molecules such that the fat

molecules, after reaction with bile, become water soluble and can dissolve in the bloodstream.

24. (a)

(b)

25. ammonia

26. (a) The different alkalis used during saponification affect the ‘softness’ of a soap.

(b) Availability and cost of ingredients affect the cost of a soap.

27. Petrochemicals form the raw material for many other products needed in society and are a limited resource.

28. (a) Polar substances dissolve in polar solvents and non-polar substances dissolve in non-polar solvents.

69



(b) Water is a polar substance and so will dissolve only polar substances including ionic compounds and polar covalent molecules. Non-polar molecules do not dissolve in water.

29. Similarities: both are used as cleaning agents; both have surfactant characteristics i.e. hydrophilic polar ‘head’ and hydrophobic non-polar ‘tail’.

Differences: soaps are manufactured from plant or animal oils or fats whereas detergents are largely manufactured from cheaper products obtained from the refining of petroleum; manufacture of soap is via saponification whereas in detergent manufacture, surfactant ‘head’ and ‘tail’ components are joined.

31. (a) Homogenisation prevents separation of aqueous and fat layers.

(b) Cream is the ‘dispersed phase’ and the lower aqueous phase is the ‘dispersion medium’ since the cream is mixed into the more abundant aqueous phase.

(c) Electrical conductivity: if the milk sample conducts electricity it is an oil-in-water (O/W) emulsion. Addition of water-soluble dye will spread and colour the emulsion.

32. (a) Emulsion prevents water loss from the skin since the ‘oil’ component of the moisturiser remains on the skin, preventing loss of existing skin moisture.

(b) (i) Most of the emulsion is oily, with small amounts of water which can evaporate.

(ii) O/W emulsions wash off, or are removed as you perspire.

(c) (i) W/O emulsion is preferable when there is a need to remove oil-based make-up (e.g. waterproof mascara).

(ii) O/W emulsion is preferable when there is a need to remove water-based make-up (e.g. water-based foundation).

33. (a) 50 μm = 50 × 10–3 mm 50 μm = 5 × 10–2 mm

(b) 6

950 10 m40 10 m

−

−

××

= 1.25 × 103

= 1250

34. Self-cleaning coating is not appropriate for a surface on which a layer of some other material might be placed later, e.g. a surface that might be painted.

35. Nanocrystals of waxy substances (which can be deposited from solution) can be laid down to produce hydrophobic regions, while nanoscale silica stripes produce hydrophilic regions. The glass surface can be masked using lithographic methods similar to those used in silicon chip manufacture to coat alternate stripes with the different materials.

36. (a) 65 cm2

37. 108

70




Extended response questions 1. C8H18; C2H5OH; CHCl3; H2O

2. (a) prevents separation of water and oil phases of ice-cream

(b) hydrophilic CH2OH; hydrophobic (CH2)16CH3

(c) See figure (a) on page 235.

71


Chapter 11

Page 258

1. (a) 4Al(s) + 3O2(g) → 2Al2O3(s)

(b) 3 2

4

2KClO (s) 3O (g) 2KCl(s) MnO

Δ→ +

2. (a) When solutions of lead(II) nitrate and sulfuric acid are mixed, solid lead(II) sulfate and aqueous nitric acid are produced.

(b) When ammonia gas and oxygen gas react in the presence of a platinum catalyst, nitrogen monoxide gas and water vapour are produced.

Page 260

3. (a) N2(g) + 3H2(g) → 2NH3(g)

(b) SiO2(s) + 2C(s) → Si(s) + 2CO(g)

(c) 4FeO(s) + O2(g) → 2Fe2O3(s)

(d) 16Cr(s) + 3S8(s) → 8Cr2S3(s)

(e) 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

4. SiCl4(l) + 2Mg(s) → Si(s) + 2MgCl2(s)

5. (a) Hydrogen forms H2 (rather than H4) molecules. CH4 + O2 → CO2 + 2H2

(b) Iodochloride has the formula ICl. I2 + Cl2 → 2ICl

(c) Potassium oxide has the formula K2O. 4K + O2 → 2K2O

(d) Sodium chloride has the formula NaCl; bromine has the formula Br2. Cl2 + 2NaBr → 2NaCl + Br2

Page 263 6. (a) insoluble

(b) soluble

(c) moderately soluble in hot water

(d) soluble

7. (a) all

(b) none

72



(c) nitrate and nitrite

8. (a) 2KOH(aq) + Ca(NO3)2(aq) → Ca(OH)2(s) + 2KNO3(aq)

(b) Na2S(aq) + Pb(CH3COO)2(aq) → 2NaCH3COO(aq) + PbS(s)

(c) 2(NH4)3PO4(aq) + 3CaCl2(aq) → 6NH4Cl(aq) + Ca3(PO4)2(s)

Page 264

9. (a) LiOH(aq) + HCl(aq) → LiCl(aq) + H2O(l)

(b) NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l)

(c) KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)

10. (a) HNO3(aq) + LiOH(aq) → LiNO3(aq) + H2O(l)

(b) H2SO4(aq) + Na2CO3(aq) → H2O(l) + CO2(g) + Na2SO4(aq)

(c) H2SO4(aq) + CuO(s) → CuSO4(s) + H2O(l)

(d) 2HCl(aq) + Mg(s) → MgCl2(aq) + H2(g)

Page 265

11. (a) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)

(b) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g)

(c) 2C2H6(g) + 5O2(g) → 4CO(g) + 6H2O(g)

12. (a) very limited oxygen supply

(b) 2C4H10(g) + 5O2(g) → 8C(s) + 10H2O(g)

(c) 2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)

Page 266

13. (a) 2H O

NaOH(s) Na (aq) OH (aq)+ −→ +

(b) 2H O

33 4 4Na PO (s) 3Na (aq) PO (aq)−+→ +

14. (a) Ba2+(aq) + SO42–(aq) → BaSO4(s)

(b) Ca2+(aq) + CO32–(aq) → CaCO3(s)

(c) (i) Ca2+(aq) + 2OH–(aq) → Ca(OH)2(s)

(ii) Pb2+(aq) + S2–(aq) → PbS(s)

(iii) 3Ca2+(aq) + 2PO43–(aq) → Ca3(PO4)2(s)

15. (a) no reaction

(b) Mg(s) + ZnCl2(aq) → Zn(s) + MgCl2(aq)

(c) Fe(s) + 2AgNO3(aq) → Fe(NO3)2(aq) + 2Ag(s)

73



Page 267

16. (a) HNO3(l) + H2O(l) → H3O+(aq) + NO3–(aq)

(b) H2SO4(l) + H2O(l) → HSO4–(aq) + H3O+(aq)

(c) NH3(g) + H2O(l) → NH4+(aq) + OH–(aq)

17. (a) CO32–(aq) + 2H+(aq) → H2O(l) + CO2(g)

(b) HCO3–(aq) + H+(aq) → H2O(l) + CO2(g)

(c) SO42–(aq) + 2H+(aq) + CaCO3(s) → H2O(l) + CO2(g) + CaSO4(s)

Multiple choice questions 1. A 2. D 3. C 4. C 5. C 6. D 7. B 8. A or C

Review questions 1. (a) Something appears to have changed visually; energy loss or gain is involved.

(b) Reactants can be reobtained in a physical change; new substances are produced in chemical changes.

2. colour change (e.g. rusting); temperature change (e.g. rotting compost); light emittted (e.g. firefly); odour produced (e.g. unleaded petrol combustion); gas produced (e.g. fizzy tablet dissolved in water)

3. A chemical equation is a shorthand representation of a chemical reaction that must be balanced to show what has happened to each reactant component (matter is not created or destroyed).

4. The law of conservation of mass applies to chemical equations since the total numbers and types of atom in the reactants must equal the total numbers and types of atom in the products.

5. (a) states of reactants and products; identification of reactants and products

(b) how much of reactants and products were involved in the reaction (i.e. reaction quantities or reaction conditions, such as temperature, catalyst)

6. (a) N2(g) + 3H2(g) → 2NH3(g)

(b) 6HCl(aq) + 2Al(s) → 2AlCl3(aq) + 3H2(g)

(c) 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)

(d) 2NO(g) + O2(g) → 2NO2(g)

(e) 12Na(s) + P4(s) → 4Na3P(s)

7. (a) 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)

(b) HCl(g) + NH3(g) → NH4Cl(aq)

(c) 2NO(g) + O2(g) → 2NO2(g)

(d) 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

74



8. (a) combination: 2Cu + O2 → 2CuO

(b) decomposition: 2HgO → 2Hg + O2

(c) double replacement: 2AsCl3 + 3H2S → As2S3 + 6HCl

(d) single replacement: Fe2O3 + 3H2 → 2Fe + 3H2O

(e) decomposition: 2NaCl → 2Na + Cl2

(f) single replacement: 2Al + 3H2SO4 → 3H2 + Al2(SO4)3

(g) combustion: 2C8H18 + 25O2 → 16CO2 + 18H2O

9. (a) 2HNO3(aq) + Zn(s) → Zn(NO3)2(aq) + H2(g)

(b) 2HCl(aq) + MgO(s) → H2O(l) + MgCl2(aq)

(c) 2HCl(aq) + Ca(OH)2(s) → 2H2O(l) + CaCl2(s)

(d) H2SO4(aq) + Na2CO3(aq) →Na2SO4(aq) + H2O(l) + CO2(g)

(e) 2Ca(s) + O2(g) → 2CaO(s)

10. (a) Zn(s) + H2SO4(aq) → ZnSO4(s) + H2(g)

(b) 2Ba(s) + O2(g) → 2BaO(s)

(c) 2HCl(aq) + Mg(OH)2(s) → MgCl2(aq) + 2H2O(l)

(d) CaO(s) + H2O(l) → Ca(OH)2(s)

11. Cl2(g) + 2KI(aq) → I2(g) + 2KCl(aq)

12. (a) Potassium metal reacts with water to produce a solution of potassium hydroxide, with the evolution of hydrogen gas.

(b) Carbon can react with iron(II) oxide, when heated, to produce metallic iron and carbon monoxide gas.

13. (a) 3Ca(OH)2(s) + 2H3PO4(aq) → Ca3(PO4)2(s) + 6H2O(l)

(b) 3H2SO4(aq) + 2Al(OH)3(s) → Al2(SO4)3(s) + 6H2O(l)

(c) Fe(OH)3(s) + 3HCl(aq) → FeCl3(aq) + 3H2O(l)

(d) BaCl2(aq) + (NH4)2CO3(aq) → BaCO3(s) + 2NH4Cl(aq)

(e) 2AgNO3(aq) + H2S(g) → Ag2S(s) + 2HNO3(aq)

(f) K2CrO4(aq) + Pb(NO3)2(aq) → 2KNO3(aq) + PbCrO4(s)

14. (a) C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(g) 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(g) C4H8(g) + 6O2(g) → 4CO2(g) + 4H2O(g)

(b) C2H4(g) + 2O2(g) → 2CO(g) + 2H2O(g) C3H6(g) + 3O2(g) → 3CO(g) + 3H2O(g) C4H8(g) + 4O2(g) → 4CO(g) + 4H2O(g)

15. (a) pentane: C5H12(g) + 8O2(g) → 5CO2(g) + 6H2O(g)

75



2-pentene: 2C5H10(g) + 15O2(g) → 10CO2(g) + 10H2O(g)

2-pentyne: C5H8(g) + 7O2(g) → 5CO2(g) + 4H2O(g)

(b) pentane: 2C5H12(g) + 11O2(g) → 10CO(g) + 12H2O(g)

2-pentene: C5H10(g) + 5O2(g) → 5CO(g) + 5H2O(g)

2-pentyne: 2C5H8(g) + 9O2(g) → 10CO(g) + 8H2O(g)

16. (a) C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

(b) 2C8H18(l) + 17O2(g) → 16CO(g) + 18H2O(g)

17. (a) 2C3H8O3(l) + 7O2(g) → 6CO2(g) + 8H2O(g)

(b) C12H22O11(l) + 12O2(g) → 12CO2(g) + 11H2O(g)

(c) CH3COOH(aq) + 2O2(g) → 2CO2(g) + 2H2O(g)

18. soluble: a, b, f, g, h

19. (a), (b) yes (c), (d) no (e), (f) yes (g) no (h) yes (i), (j) no

20. (a) Ag+(aq) + Cl–(aq) → AgCl(s)

(b) Ba2+(aq) + SO42–(aq) → BaSO4(s)

(e) 3Ca2+(aq) + 2PO43–(aq) → Ca3(PO4)2(s)

(f) Pb2+(aq) + S2–(aq) → PbS(s)

(h) Hg2+(aq) + 2Br–(aq) → HgBr2(s)

21. (a) NaOH(aq) + HNO3(aq) → NaNO3(aq) + H2O(l) OH–(aq) + H+(aq) → H2O(l)

(b) K2SO4(aq) + Ca(NO3)2(aq) → 2KNO3(aq) + CaSO4(s) SO42–(aq) + Ca2+(aq) →

CaSO4(s)

(c) Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(aq) + PbI2(s) Pb2+(aq) + 2I–(aq) → PbI2(s)

(d) HCl(aq) + NaHCO3(aq) → NaCl(aq) + CO2(g) + H2O(l) H+(aq) + HCO3–(aq) →

CO2(g) + H2O(l)

(e) 3MgCl2(aq) + 2Na3PO4(aq) → 6NaCl(aq) + Mg3(PO4)2(s) 3Mg2+(aq) + 2PO43–(aq) →

Mg3(PO4)2(s)

(f) K2S(aq) + ZnCl2(aq) → ZnS(s) + 2KCl(aq) S2–(aq) + Zn2+(aq) → ZnS(s)

(g) (NH4)2CO3(aq) + 2HNO3(aq) → 2NH4NO3(aq) + H2O(l) + CO2(g)

CO32–(aq) + 2H+(aq) → H2O(l) + CO2(g)

(h) no reaction

(i) 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) 2Li(s) + 2H2O(l) → 2Li+(aq) + 2OH–(aq) + H2(g)

22. (a) Ba2+ (b) Pb2+ (c) Ba2+ (d) Ba2+

76



23. (a) Chlorides, bromides and iodides will precipitate silver but not zinc.

(b) Sulfates precipitate barium but not magnesium.

(c) Sulfates precipitate lead but not silver.

(d) Barium precipitates sulfates but not chlorides.

24. H2C2O4(aq) + CaCl2(aq) → 2HCl(aq) + CaC2O4(s) A reaction will occur according to the ionic equation: C2O4

2–(aq) + Ca2+(aq) → CaC2O4(s)


Extended response questions 1. (a) BaSO4

(b) Ba(OH)2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaOH(aq)

2. (a) The one with barium chloride

(b) BaCl2(aq) + (NH4)2CO3(aq) → BaCO3(s) + 2NH4Cl(aq)

(c) Ba2+(aq) + CO32–(aq) → BaCO3(s)

(d) Spectator ions are present in the reaction but do not take part in the formation of the product. Cl–(aq) and NH4

+

(e) NH4Cl(s) — as the water is evaporated, the solution will form solid crystals of ammonium chloride.

77


Chapter 12

Multiple choice questions 1. D 2. D 3. B 4. A 5. D 6. B 7. B 8. D 9. A 10. B 11. A

Review questions 2. (a) quantitative (b) qualitative (c) quantitative

9. H2O(l) + CO2(g) → H2CO3(aq)

10. N2(g) + O2(g) → 2NO(g)

2NO(g) + O2(g) → 2NO2(g)

3NO2(g) + H2O(g) → 2HNO3(aq) + NO(g)

12. (a) CO(g) (b) SO2(g)


Multiple choice question 1. B

Extended response question 1. (a) HNO3, H2CO3, H2SO4

78


Chapter 13

Page 295 2. Heat is taken from the body for water to change from a liquid to a gaseous state.

Page 298 3. (a) yes (b) yes (c) no (d) no (e) yes (f) no

4. because of the polarity of the water molecule

5. 12 g

Page 299 6. From the graph on page 299 the solubility curve for potassium chloride (KCl) is used to

determine the solubility of KCl at 80 °C:

Solubility = 48 g/100 g water therefore the mass of KCl that will dissolve in 50 g of water at 80 °C

= 248

= 24 g

7. sodium sulfate

Page 302 9. (a) Convert 25.0 mL to a volume in litres.

V = 1000

0.25

V = 0.0250 L

C = mass of solute in gramsvolume of solution in litres

c = 0250.0425.0

c = 17.0 g L–1 c = 17.0 × 1000 mg L–1 c = 1.70 × 104 mg L–1

(b) Convert 55.0 mL to a volume in litres.

V = 1000

0.55

V = 0.0550 L Convert 26.4 mg to a mass in grams.

79



m = 1000

4.26

m = 0.0264 g


c = 0550.00264.0

c = 0.480 g L–1 c = 0.480 × 1000 mg L–1 c = 480 mg L–1

Page 303 10. Convert 475 mL to a volume in litres.

V = 1000475

V = 0.475 L

Manipulating the concentration equation to find mass: mass of solute (g) = concentration (g L–1) × volume of solution (L) therefore the mass of sodium chloride = 0.200 × 0.475 therefore the mass of sodium chloride = 0.0950 g

11. Convert mg to g. mass of solute = 200 mg mass of solute = 0.200 g Manipulating the concentration equation to find volume:

V = 1

mass of solute (g)concentration (g L )−

V =0.50

200.0

V = 0.004 00 L V = 4.00 mL

Page 304 (top)

12. Use the equation for concentration in ppm ( gμ g–1)

C = mass of solute in microgramsmass of solution in grams

mass of solute = 7.50 mg × 103 mass of solute = 7.50 × 103 gμ volume of solution = 2.00 L volume of solution = 2000 mL as the density of water = 1 g mL–1 then the mass of water = 2000 g

80



therefore C = 2000

1050.7 3×

therefore c = 3.75 gμ g–1 therefore c = 3.75 ppm

13. volume of pond water = 4.00 L volume of pond water = 4000 mL as the density of water = 1 g mL–1 then the mass of pond water = 4000 g

mass of dissolved oxygen = concentration (μ g g–1) × mass of pond water (g) mass of dissolved oxygen = 88.00 × 4000 mass of dissolved oxygen = 35 2000 gμ mass of dissolved oxygen = 0.352 g

14. concentration of dissolved oxygen = 5.60 ppm concentration of dissolved oxygen = 5.60 gμ g–1

mass water tested = 1

mass dissolved oxygen ( g)concentration ( g g )

μμ −

mass water tested = 60.540.1

mass water tested = 0.250 g

Page 304 (bottom)

15. % w/w = mass of solute in gmass of solution in g

× 100%

(a) concentration of NaCl = 0.1504.38 × 100

concentration of NaCl = 25.6% w/w

(b) concentration of benzalkonium chloride = 0.50

25.0 × 100

concentration of benzalkonium chloride = 0.50% w/w

(c) mass of sugar = 144 mg mass of sugar = 0.144 g

concentration of sugar = 200144.0 × 100

concentration of sugar = 0.0720% w/w

Page 305 (top)

16. mass of sodium hydrogen carbonate = % w/w × mass of solution (g) × 100

1

mass of sodium hydrogen carbonate = 0.5% × 250 g × 100

1

mass of sodium hydrogen carbonate = 1.25 g mass of sodium hydrogen carbonate = 1 g

81



17. mass of solution = mass of solute in g%w/w

× 100%

mass of cleaner solution = 0.65.15 × 100

mass of cleaner solution = 258 g mass of cleaner solution = 2.6 × 102 g


× 100%

concentration of NaCl = 50

5.24 × 100

concentration of NaCl = 49% w/w

Page 305 (bottom)

19. % w/v = mass of solute in gvolume of solution in mL

× 100%

(a) mass potassium chloride = 200 mg mass potassium chloride = 0.200 g

concentration of KCl = 0.50

200.0 × 100

concentration of KCl = 0.400% w/v

(b) mass of sodium bromide = 50 kg

mass of sodium bromide = 5.0 × 104 g volume of solution = 250 L volume of solution = 2.50 × 105 mL

concentration of sodium bromide = 5

4

1050.2100.5×× × 100

concentration of sodium bromide = 20% w/v

(c) concentration = 500025.0 × 100

concentration = 0.0050% w/v

20. mass of sodium hypochlorite = % w/v × volume of solution (mL) × 100

1

mass of sodium hypochlorite = 1.5% × 375 mL × 100

1

mass of sodium hypochlorite = 5.6 g

21. mass of plant food = 100 mg mass of plant food = 0.100 g

volume of solution = mass of solute in g%w/v

× 100%

82



volume of solution = 5.0

100.0 × 100

volume of solution = 20 mL volume of solution = 2 × 101 mL

22. mass of sodium hydroxide = 10 mg

mass of sodium hydroxide = 0.010 g

volume of solution = mass of solute in g%w/v

× 100%

volume of solution = 10010.0 × 100

volume of solution = 0.10 mL

Page 306 23. volume of solution = 0.455 L

volume of solution = 455 mL

% v/v = volume of solute in mLvolume of solution in mL

× 100%

% v/v = 455

0.10 × 100%

% v/v = 2.20% v/v

24. (a) volume of vinegar = 0.750 L

volume of vinegar = 750 mL

volume of acetic acid = % v/v × volume of solution (mL) × 100

1

volume of acetic acid = 5.0% × 750 mL × 100

1

volume of acetic acid = 38 mL

(b) volume of vinegar = volume of acetic acid in mL%v/v

× 100%

volume of vinegar = 0.5

25 × 100%

volume of vinegar = 500 mL volume of vinegar = 5.0 × 102 mL

25. volume of ethanol = % v/v × volume of solution (mL) × 100

1

volume of ethanol = 11.5% × 750 mL × 100

1

volume of ethanol = 86.3 mL

83



Multiple choice questions 1. C 2. D 3. A 4. B 5. C 6. A 7. B 8. B

9. 35.0 g of potassium nitrate in 50.0 g water = 70.0 g potassium nitrate in 100 g water from the graph: 70.0 g of potassium nitrate will dissolve in 100 g water at 45 °C

Answer: C

10. B

11. 80 g ammonium chloride in 200 g water = 40 g ammonium chloride in 100 g water from the graph this represents an unsaturated solution at 100 °C.

The solution containing 40 g ammonium chloride in 100 g water is saturated at 40 °C, therefore this is the temperature at which crystals will start to appear.

Answer: B

12. Manipulating the concentration equation to find mass:

mass of solute (g) = concentration (g L–1) × volume of solution (L) therefore the mass of sodium chloride = 28.5 × 2.0 therefore the mass of sodium chloride = 57 g Answer: B

13. volume of water (L) = 1

mass of solute in mgconcentration mgL−

volume of water(L) = 250150

volume of water(L) = 0.60 L Answer: A

14. volume of water = 1 L

as the density of water = 1 g mL–1 then the mass of water = 1000 g mass of cadmium = concentration ( gμ g–1) × mass of water (g) mass of cadmium = 0.01 × 1000 mass of cadmium = 10 gμ mass of cadmium = 0.010 mg Answer: D

15. volume of solution = 1 L


% v/v = volume of solute in mLvolume of solution in mL

× 100%

% v/v = 1000

50 × 100%

% v/v = 5% w/v Answer: C

16. A

84



Review questions

8. (a) C = mass of solute in gramsvolume of solution in litres

c = 500

2.0

c = 4 × 10–4 g L–1

(b) Use the equation for concentration in ppm ( gμ g–1)

C = mass of solute in microgramsmass of solution in grams

mass of solute = 0.2 g mass of solute = 2 × 105 gμ volume of solution = 500 L volume of solution = 500 000 mL as the density of water = 1 g mL–1 then the mass of water as the density of water = 500 000 g

therefore C = 52 10

500000×

therefore c = 0.4 gμ g–1 therefore c = 0.4 ppm

9. (a) (i) 48 °C (ii) 58 °C (iii) 24 °C

(b) 65 °C

(c) (i) 36 g/100 g (ii) 99 g/100 g (iii) 55 g/100 g


× 100%

(a) concentration of KNO3 = 455 × 100

concentration of KNO3 = 11% w/w concentration of KNO3 = 1 × 101% w/w

(b) mass of ammonium nitrate = 250 mg mass of ammonium nitrate = 0.250 g

concentration of benzalkonium chloride = 5.24

250.0 × 100

concentration of benzalkonium chloride = 1.02% w/w

12. mass of solution = mass of solute in g%w/w

× 100%

mass of KCl solution = 5.84.3 × 100

mass of KCl solution = 40 g

85



13. mass of ammonium nitrate = % w/w × mass of solution (g) × 100

1

mass of ammonium nitrate = 7.5% × 125 g × 100

1

mass of ammonium nitrate = 9.4 g

14. % v/v = volume of solute in mLvolume of solution in mL

× 100%

% v/v = 120

5.3 × 100%

% v/v = 2.9% v/v

15. (a) volume of solution = 4.0 L


volume of ethanol = % v/v × volume of solution (mL) × 100

1

volume of ethanol = 10% × 4000 mL × 100

1

volume of ethanol = 400 mL

(b) volume of ethanol = % v/v × volume of solution (mL) × 100

1

volume of ethanol = 40% × 350 mL × 100

1

volume of ethanol = 140 mL

16. volume of saline = 2.0 L

volume of saline = 2000 mL

mass of salt = % w/v × volume of solution (mL) × 100

1

mass of salt = 2.5% × 2000 mL × 100

1

mass of salt = 50 g


Extended response questions 1. (a) V, W (b) T, X, Z (c) U, Y

86



2. (i) (a) volume of solution = 100 mL volume of solution = 0.100 L


c = 100.03

c = 30 g L–1 c = 3 × 101 g L–1

(b) C = 3 × 104 mg L–1

(ii) (a) volume of solution = 800 mL

volume of solution = 0.800 L


c = 800.0004.0

c = 0.005 g L–1

(b) C = 5 mg L–1

(iii) (a) volume of solution = 200 mL volume of solution = 0.200 L

mass of solute = 500 mg mass of solute = 0.500 g


= 200.0500.0

= 2.50 g L–1

(b) C = 2.50 × 103 mg L–1

87


Chapter 14

Page 319

1. 2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g) Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g) 2HCl(aq) + CaCO3(s) → CaCl2(aq) + CO2(g) + H2O(l) 2H+(aq) + CaCO3(s) → Ca2+(aq) + CO2(g) + H2O(l)

2. (a) H2SO4(aq) + Mg(s) → MgSO4(aq) + H2(g) Mg(s) + 2H+(aq) → Mg2+((aq) + H2(g)

(b) 3H2SO4(aq) + 2Al(s) → Al2(SO4)3(aq) + 3H2(g) 2Al(s) + 6H+(aq) → 2Al3+((aq) + 3H2(g)

(c) 2HNO3(aq) + CuCO3(s) → Cu(NO3)2(aq) + CO2(g) + H2O(l) 2H+(aq) + CuCO3(s) → Cu2+(aq) + CO2(g) + H2O(l)

(d) H2SO4(aq) + 2KHCO3(s) → K2SO4(aq) + 2CO2(g) + 2H2O(l) 2H+(aq) + 2KHCO3(s) → 2K+(aq) + 2CO2(g) + 2H2O(l)

(e) 2HNO3(aq) + Na2SO3(aq) → 2NaNO3(aq) + SO2(g) + H2O(l) 2H+(aq) + SO3

2–(aq) → SO2(g) + H2O(l)

(f) 2H3PO4(aq) + 3PbS(s) → Pb3(PO4)2(aq) + 3H2S(g) 2H+(aq) + PbS(s) → Pb2+(aq) + H2S(g)

(g) 2HCl(aq) + CuO(s) → CuCl2(aq) + H2O(l) 2H+(aq) + CuO(s) → Cu2+(aq) + H2O(l)

(h) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + H2O(l) H+(aq) + OH–(aq) → H2O(l)

Page 321 3. a, b, d, e, g

Page 323

4. For example H2SO4(aq) + 2NH4OH(aq) → (NH4)2SO4(aq) + 2H2O(l)

Page 324 5. (a) HSO4

– (b) HS– (c) S2– (d) OH– (e) NH3

6. (a) H2O (b) H2CO3 (c) HS– (d) H3O+ (e) HCN

7. c

88



8. (a) H3PO4(aq) + H2O(l) → H3O+(aq) + H2PO4–(aq)

22 4 2 3 4H PO (aq) H O(l) H O (aq) HPO (aq)− + −+ +

2 34 2 3 4HPO (aq) H O(l) H O (aq) PO (aq)− + −+ +

(b) H3PO4(aq)/H2PO4–(aq); H3O+(aq)/H2O(aq)

H2PO4–(aq)/HPO4

2–(aq); H3O+(aq)/H2O(aq) HPO4

2–(aq)/PO43–(aq); H3O+(aq)/H2O(aq)

9. (a) HS–(aq) + H2O(l) → H3O+(aq) + S2–(aq) HS–(aq) + H2O(l) → OH–(aq) + H2S(aq)

(b) HSO4–(aq) + H2O(l) → H3O+(aq) + SO4

2–(aq) HSO4

–(aq) + H2O(l) → OH–(aq) + H2SO4(aq)

Page 327

10. HNO3(aq) + H2O(l) → H3O+(aq) + NO3–(aq)

Strong conductor of electricity 2

2 3HS (aq) H O(l) H O (aq) S (aq)− + −+ +

Weak conductor of electricity

11. 3 2 2 2 3 2HN(CH ) H O(l) H N(CH ) (aq) OH (aq)+ −+ +

12. (a) HNO3/NO3–; HCl/Cl–

Multiple choice questions 1. C 2. D 3. D 4. B 5. D 6. C 7. C 8. B 9. B 10. B 11. A

Review questions

5. (a) 2H3PO4(aq) + 3PbS(s) → Pb3(PO4)2(aq) + 3H2S(g) 2H+(aq) + PbS(s) → Pb2+(aq) + H2S(g)

(b) 2HCl(aq) + CuO(s) → CuCl2(aq) + H2O(l) 2H+(aq) + CuO(s) → Cu2+(aq) + H2O(l)

(c) H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + H2O(l) H+(aq) + OH–(aq) → H2O(l)

(d) 6HCl(aq) + 2Al(s) → 2AlCl3(aq) + 3H2(g) 2Al(s) + 6H+(aq) → 2Al3+(aq) + 3H2(g)

(e) Ca(s) + H2CO3(aq) → CaCO3(s) + H2(g) Ca(s) + 2H+(aq) + CO3

2–(aq) → CaCO3(s) + H2(g)

(f) 2HNO3(aq) + Li2CO3(aq) → 2LiNO3(aq) + CO2(g) + H2O(l) 2H+(aq) + CO3

2–(aq) → CO2(g) + H2O(l)

11. NH3(g) + HCl(g) → NH4Cl(g)

12. (b) (i) HCl (ii) HCO3– (iii) H2SO4 (iv) H3O+ (v) H2O

(c) (i) F– (ii) CO32– (iii) SO4

2– (iv) OH– (v) HS–

89



13. (a) HS–/S2–; NH4+/NH3

(b) NH4+/NH3; CH3COOH/CH3COO–

(c) HSO4–/SO4

2–; H3O+/H2O

(d) H2PO4–/HPO4

2–; H2O/OH–

(e) HNO3/NO3–; H2O/OH–

19. (b) (i) acidic (ii) neutral (iii) basic (iv) acidic



1. (a) can only donate one proton

(b) CH3COOH(aq) + H2O(l) → CH3COO–(s) + H3O+(aq)

(c) CH3COOH(aq) + NaOH(aq) → CH3COO–(aq) + H2O(l) + Na+(aq)

(d) The equation in (b) does not react completely.

2 (a) can donate or receive a proton

(b) HCO3–(aq) + H2O(l) → H2CO3(aq) + OH–(aq)

HCO3–(aq) + H2O(l) → CO3

2–(aq) + H3O+(aq)

(c) The first reaction occurs to a greater extent than the second.

90


Chapter 15

Page 338

1. (a) n(Ag2S) = n(Ag) × 42

n(Ag2S) = 1 × 42

n(Ag2S) = 0.5 mol

(b) n(H2S) = n(Ag) × 42

n(H2S) = 1 × 42

n(H2S) = 0.5 mol

(c) n(Ag2S) = n(H2S) n(Ag2S) = 3.5 mol

2. (a) n(CH4) = n(O2) × 21

n(CH4) = 1 × 21

n(CH4) = 0.5 mol

(b) n(O2) = n(CH4) × 12

n(O2) = 0.1 × 12

n(O2) = 0.2 mol

(c) n(CO2) = n(CH4) n(CO2) = 0.1 mol

(d) n(H2O) = n(CH4) × 12

n(H2O) = 0.1 × 12

n(H2O) = 0.2 mol

(e) n(CO2) = n(O2) × 21

n(CO2) = 0.1 × 21

n(CO2) = 0.05 mol

91



(f) n(O2) = n(CH4) × 12

n(O2) = 0.25 × 12

n(O2) = 0.50 mol

(g) n(H2O) = n(O2) n(H2O) = 8 mol

Page 339

3. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (g)

n(CH4) = Mm

n(CH4) = 04.168.2

n(CH4) = 0.18 mol From the equation:

n(H2O) = n(CH4) × 12

n(H2O) = 0.18 × 12

n(H2O) = 0.35 mol m(H2O) = m × M m(H2O) = 0.35 × 18.02 m(H2O) = 6.3 g

4. n(Al) = Mm

n(Al) = 98.26

100

n(Al) = 3.70 mol From the equation:

n(Al2O3) = n(Al) × 31

n(Al2O3) = 3.70 × 31

n(Al2O3) = 1.24 mol ∴ m(Al2O3) = m × M m(Al2O3) = 1.24 × 101.96 m(Al2O3) = 126 g

92



From the equation:

n(AlCl3) = n(Al) × 31

n(AlCl3) = 3.70 × 31

n(AlCl3) = 1.24 mol ∴ m(AlCl3) = m × M ∴ m(AlCl3) = 1.24 × 133.33 ∴ m(AlCl3) = 165 g From the equation: n(NO) = n(Al) n(NO) = 3.70 mol ∴ m(NO) = m × M n m(NO) = 3.70 × 30 n m(NO) = 111 g From the equation:

n(H2O) = n(Al) × 36

n(H2O) = 3.70 × 36

n(H2O) = 7.41 mol ∴ m(H2O) = m × M ∴ m(H2O) = 7.41 × 18.02 ∴ m(H2O) = 134 g

Page 340

5. n(Ca3P2) = Mm

n(Ca3P2) = 18.182

25

n(Ca3P2) = 0.14 mol From the equation:

n(PH3) = n(Ca3P2) × 12

n(PH3) = 0.14 × 12

n(PH3) = 0.27 mol ∴ m(PH3) = m × M ∴ m(PH3) = 0.27 × 34.00 ∴ m(PH3) = 9.3 g

93



6. (a) n(O2) = Mm

n(O2) = 00.325.6

n(O2) = 0.20 mol From the equation:

n(H2O) = n(O2) × 54

n(H2O) = 0.20 × 54

n(H2O) = 0.16 mol ∴ m(H2O) = m × M ∴ m(H2O) = 0.16 × 18.02 ∴ m(H2O) = 2.9 g

(b) n(C3H8) =Mm

n(C3H8) = 11.447.1

n(C3H8) = 0.039 mol From the equation:

n(O2) = n(C3H8) × 15

n(O2) = 0.039 × 15

n(O2) = 0.19 mol ∴ m(O2) = m × M ∴ m(O2) = 0.19 × 32.00 ∴ m(O2) = 6.2 g

(c) From the equation:

n(CO2) = n(C3H8) × 13

n(CO2) = 0.50 × 13

n(CO2) = 1.5 mol ∴ m(CO2) = m × M ∴ m(CO2) = 1.5 × 44.01 ∴ m(CO2) = 66 g

(d) n(CO2) = 01.44

92.5

n(CO2) = 0.135 mol

94



From the equation:

n(C3H8) = n(CO2) × 31

n(C3H8) = 0.135 × 31

n(C3H8) = 0.0448 mol ∴ m(C3H8) = m × M ∴ m(C3H8) = 0.0448 × 44.11 ∴ m(C3H8) = 1.98 g

(e) n(C3H8) = Mm

n(C3H8) = 11.44

5000

n(C3H8) = 113.4 n(C3H8) = 1.1 × 102 mol From the equation:

n(CO2) = n(C3H8) × 13

n(CO2) = 113.4 × 13

n(CO2) = 340.0 n(CO2) = 3.4 × 102 mol ∴ m(CO2) = m × M ∴ m(CO2) = 340.0 × 44.01 ∴ m(CO2) = 14 966 g ∴ m(CO2) = 15 kg

7. (a) 16.6 kg

(b) B2: moles of haematite; C2: moles of iron; D2: mass of iron

(c) 19.6 kg

(d) 666 kg

Page 342

8. Calculate the amount, in mole, of both reactants:

n(K2MnF6) = 14.247

32.5

n(K2MnF6) = 0.0215 mol

n(SbF5) = 75.216

75.8

n(SbF5) = 0.0404 mol 2K2MnF6(aq) + 4SbF5(s) → 4KSbF6(aq) + 2MnF3(s) + F2(g)

95



Divide the amount of each reactant, in mole, by its coefficient in the chemical equation:

K2MnF6 : 20215.0 = 0.0108 mol

SbF5: 40404.0 = 0.0101 mol

The two reactants are now in a 1:1 ratio, therefore the limiting reactant is the one with the lowest number of mole, in this case = SbF5. Use the n(SbF5) to calculate the mass of F2 produced.

n(F2) = n(SbF5) × 41

n(F2) = 0.0404 × 41

n(F2) = 0.0101 mol m(F2) = 0.0101 × 38.00 m(F2) = 0.384 g

9. n(Hg) = 59.200

15.2

n(Hg) = 0.0107 mol

n(Br2) = 8.159

56.1

n(Br2) = 0.009 76 mol Hg(l) + Br2(l) → HgBr2(s) The two reactants are already in a 1:1 ratio, therefore the limiting reactant is the one with the lowest number of mole, in this case = Br2. Use the n(Br2) to calculate the mass of HgBr2 produced. (a) From the equation:

n(HgBr2) = n(Br2) n(HgBr2) = 0.009 76 mol ∴ m(HgBr2) = 0.0098 × 360.39 ∴ m(HgBr2) = 3.52 g

(b) n(Hg) used = n(Br2)

n(Hg) used = 0.009 76 mol n(Hg) in excess = 0.0107 – 0.009 76 n(Hg) in excess = 0.000 956 mol ∴ m(Hg) in excess = 0.000 956 × 200.59 ∴ m(Hg) in excess = 0.192 g

96



Page 344

10. Step 1: Calculate number of mole using the formula n = Mm .

Step 2: Calculate concentration using the formula c = Vn where the volume must be in

litres.

(a) n(CuSO4.5H2O) = 71.249

200

n(CuSO4.5H2O) = 0.801 mol

∴c = 1801.0

∴ = 0.8 mol L–1

(b) n(CuSO4.5H2O) = 71.249

250

n(CuSO4.5H2O) = 1.00 mol

∴ c = 0.2

00.1

∴ c = 0.50 mol L–1

(c) n(CuSO4.5H2O) = 71.249

750

n(CuSO4.5H2O) = 3.00 mol volume = 700 mL volume = 0.700 L

∴ c = 700.000.3

∴ c = 4.29 mol L–1

(d) n(CuSO4.5H2O) = 71.249

50

n(CuSO4.5H2O) = 0.200 mol volume = 125 mL volume = 0.125 L

c = 125.0200.0

c = 1.60 mol L–1

Page 345

11. (a) n(NaOH) = Mm

n(NaOH) = 00.40

20

n(NaOH) = 0.50 mol

97



volume = 250 mL volume = 0.250 L

∴ c = Vn

∴ c = 250.050.0

∴ c = 2.0 mol L–1

(b) NaOH (s) → Na+(aq) + OH–

(aq) n(Na+) = n(OH–) n(Na+) = n(NaOH) ∴ c(Na+) = c(OH–) ∴ c(Na+) = c(NaOH) ∴ c(Na+) = 2.0 mol L–1

12. Mg(NO3)2 (s) → Mg2+(aq) + 2NO3

– (aq)

From the equation:

n(Mg(NO3)2 = n(NO3–) ×

21

∴ c(Mg(NO3)2 = c(NO3–) ×

21

∴ c(Mg(NO3)2 = 0.030 × 21

∴ c(Mg(NO3)2 = 0.015 mol L–1

13. (NH4)3PO4 (s) → 3NH4+

(aq) + PO43–

(aq)

From the equation:

n((NH4)3PO4)) = n(NH4+) ×

31

∴ c( (NH4)3PO4)) = c(NH4+) ×

31

= 1.25 × 31

= 0.417 mol L–1

Page 346

14. V1 = 1

22

cVc ×

V2 = 250 mL V2 = 0.250 L

98



∴V1 = 0.250 2.1512×

∴V1 = 0.048 L ∴V1 = 45 mL

15. c2 = 2

11

VVc ×

V2 = 2.0 L + 500 mL V2 = 2.5 L

∴c2 = 2.0 2.02.5×

∴c2 = 1.6 mol L–1

Page 349

16. Zn(s) + H2SO4 (aq) ───> ZnSO4 (aq) + H2 (aq) V(H2SO4) = 400 mL V(H2SO4) = 0.400 L n(H2SO4) = c × V n(H2SO4) = 0.250 × 0.400 n(H2SO4) = 0.100 mol

From the equation: n(ZnSO4) = n(H2SO4) n(ZnSO4) = 0.100 n(ZnSO4) = 0.100 mol ∴ m(ZnSO4) = m × M ∴ m(ZnSO4) = 0.100 × 161.44 ∴ m(ZnSO4) = 16.1 g

17. (a) V(Na2S) = 235 mL

V(Na2S) = 0.235 L n(Na2S) = c × V n(Na2S) = 0.178 × 0.235 n(Na2S) = 0.0418 mol

From the equation: n(CdS) = n(Na2S) n(CdS) = 0.0418 mol ∴ m(CdS) = m × M ∴ m(CdS) = 0.0418 × 144.46 ∴ m(CdS) = 6.04 g

(b) Ensures all Na2S reacts.

99



18. (a) All HCl has reacted.

(b) m(Mg) used = 2.56 – 0.350 m(Mg) used = 2.21 g

n(Mg) used = Mm

n(Mg) used = 31.2421.2

n(Mg) used = 0.0909 mol Mg(s) + 2HCl(aq) → MgCl2 (aq) + H2 (g)

From the equation:

n(HCl) used = n(Mg) × 12

n(HCl) used = 0.0909 × 12

n(HCl) used = 0.182 mol V(HCl) = 200 mL V(HCl) = 0.200 L

∴ c(HCl) = 200.0182.0

∴ c(HCl) = 0.909 mol L–1

(c) incomplete drying of magnesium strip; inaccuracies in weighing magnesium and measuring acid volume

(d) Wear gloves, laboratory coat and safety goggles when handling acid; treat acid spills with copious quantities of water.

Page 350

19. (a) Na2CO3 (aq) + 2HCl (aq) → 2NaCl (aq) + H2O (l) + CO2 (g)

(b) V(Na2CO3) = 25.0 mL V(Na2CO3) = 0.025 L n(Na2CO3) = c × V n(Na2CO3) = 0.055 × 0.025 n(Na2CO3) = 0.0014 mole

From the equation:

n(HCl) = n(Na2CO3) × 12

n(HCl) = 0.0014 × 12

n(HCl) = 0.0028 mole V(HCl) = 15.5 mL V(HCl) = 0.0155 L

100



∴ c(HCl) = Vn

∴ c(HCl) = 0155.00028.0

∴ c(HCl) = 0.18 mol L–1

20. HCl(aq) + KOH(aq) → KCl(aq) + H2O(l) V(HCl) = 20.0 mL V(HCl) = 0.0200 L n(HCl) = c × V n(HCl) = 0.300 × 0.0200 n(HCl) = 0.006 00 mole

From the equation:

n(KOH) = n(HCl) × 11

n(KOH) = 0.006 00 × 11

n(KOH) = 0.006 00 mole c(KOH) = 0.200 mol L–1

∴ V(HCl) = cn

∴ V(HCl) = 200.0

00600.0

∴ V(HCl) = 0.0300 L ∴ V(HCl) = 30.0 mL

21. H2SO4(aq) + 2KOH(aq) → K2SO4(aq) + 2H2O(l)

V(H2SO4) = 20.0 mL V(H2SO4) = 0.0200 L n(H2SO4) = c × V n(H2SO4) = 0.300 × 0.0200 n(H2SO4) = 0.006 00 mole

From the equation:

n(KOH) = n(H2SO4) × 12

n(KOH) = 0.006 00 × 12

n(KOH) = 0.0120 mole c(KOH) = 0.200 mol L–1

∴ V(KOH) = cn

101



∴ V(KOH) = 200.00120.0

∴ V(KOH) = 0.0600 L ∴ V(KOH) = 60.0 mL

22. H3PO4(aq) + 3KOH(aq) → K3PO4(aq) + 3H2O(l)

V(H3PO4) = 20.0 mL V(H3PO4) = 0.0200 L n(H3PO4) = c × V n(H3PO4) = 0.300 × 0.0200 n(H3PO4) = 0.006 00 mole From the equation:

n(KOH) = n(H3PO4) × 13

n(KOH) = 0.006 00 × 13

n(KOH) = 0.0180 mole c(KOH) = 0.200 mol L–1

∴ V(KOH) = cn

∴ V(KOH) = 200.00180.0

∴ V(KOH) = 0.0900 L ∴ V(KOH) = 90.0 mL

Page 351

23. V(AgNO3) = 22.40 mL V(AgNO3) = 0.022 40 L n(Ag+) = n(AgNO3) n(Ag+) = c × V n(Ag+) = 0.500 × 0.022 40 n(Ag+) = 0.0112 mole

From the equation:

n(Cl–) = n(Ag+) × 11

n(Cl–) = 0.0112 × 11

n(Cl–) = 0.0112 mole V(seawater) = V(Cl–) V(seawater) = 20.00 mL V(seawater) = 0.020 00 L

102



∴ c(NaCl) = c(Cl–)

∴ c(NaCl) = Vn

∴ c(NaCl) = 0.01120.02000

∴ c(NaCl) = 0.560 mol L–1

Page 352

24. (b) n(Na2C2O4) = Mm

n(Na2C2O4) = 0.134

183.3 = 0.023 75 mol

V(Na2C2O4) = 250 mL V(Na2C2O4) = 0.250 L

∴ c(Na2C2O4) = Vn

= 0.023750.250

= 0.0950 mol L–1

25. (a) (ii) n(Na2CO3) = Mm

= 99.105

461.1

= 0.013 78 mol

c(Na2CO3) = Vn

= 0.013 780.250

= 0.0551 mol L–1

(b) (i) Na2CO3 (aq) + 2HNO3 (aq) → 2NaNO3 (aq) + CO2 (g) + H2O(l)

(ii) V(Na2CO3) = 20 mL = 0.020 L

n(Na2CO3) = c × V = 0.0551 × 0.020 = 0.0011 mole

From the equation:

103



n(HNO3) = n(Na2CO3) × 12

= 0.0011 × 12

= 0.0022 mole

V(HNO3) = 22.17 mL = 0.022 17 L

∴ c(HNO3) = Vn

= 0.00220.02217

= 0.099 mol L–1

Page 356 28. (a) pH = –log 10–5

= –(–5) = 5

(b) [H3O+] = 0.1 mol L–1 = 10–1 mol L–1

∴ pH = –log 10–1 = –(–1) = 1

29. [H3O+] = [HClO4] = 0.001 mol L–1 = 10–3 mol L–1

∴ pH = –log 10–3 = –(–3) = 3

30. (a) [OH–] = 10–4 mol L–1 As [H3O+] × [OH–] = 10–14 (mol L–1)2

then [H3O+] = 1410

[OH ]

−

−

[H3O+] = 4

14

1010

−

−

= 10–10 mol L–1

104



∴ pH = –log 10–10 = –(–10) = 10

(b) [OH–] = 0.1 = 10–1 mol L–1

As [H3O+] × [OH–] = 10–14 (mol L–1)2

then [H3O+] = 1410

[OH ]

−

−

[H3O+] = 1

14

1010

−

−

= 10–13 mol L–1

∴ pH = –log 10–13 = –(–13) = 13

31. If pH = 2 then [H3O+] = 10–2 mol L–1

c2 = 2

11

VVc ×

V2 = 90 mL + 10 mL = 100 mL

∴c2 = 100

1010 2 ×−

= 10–3 mol L–1

∴ pH = –log 10–3 = –(–3) = 3

Page 357

32. (a) n(Al) = 0.27

34.2

= 0.0867 mol

n(Cu) = 0.0867 × 23

= 0.130 mol

m(Cu)max = 0.130 × 63.5 = 8.26 g

105



% yield = 3.89 1008.26×

= 47.1%

(b) n(Fe2O3) = 8.1595.63

= 0.397 mol

n(Fe) = 0.397 × 12

= 0.795 mol

m(Fe)max = 0.795 × 55.9 = 44.4 g

% yield = 35.6 10044.4×

= 80.2%

Multiple choice questions 1. C 2. D

3. n(CO2) = Mm

= 01.44

44

= 1.0 mole n(C3H8) : n(O2) : n(CO2) : n(H2O) = 1 : 5 : 3 : 4

Therefore if n(CO2) =1.0 mole, the above ratio will be 31 :

35 : 1:

34 .

Answer: A

4. m(FeS2) = 60 tonnes = 6.0 × 107 g

n(FeS2) = 97.119100.6 7×

= 5.0 ×105 mole From the equation:

n(Fe2O3) = 42 × n(FeS2)

= 42 × 5.0 × 105

= 2.50 × 105 mole ∴m(Fe2O3) = n × M(Fe2O3)

= 2.50 × 105 × 159.70

106



= 3.99 × 107 g = 39.9 tonnes

Answer: B

5. m(Ca3(PO4)2) = 1000 kg = 1.000 × 106 g

n(Ca3(PO4)2) = Mm

= 18.31010000.1 6×

= 3224 mole From the equation:

n(P4) = 21 × n(Ca3(PO4)2)

= 21 × 3224

= 1612 mole ∴m(P4) = n × M(P4)

= 1612 × 123.88 = 199 691 g = 200 kg

Answer: C

6. n(KOH) = c × V = 0.25 × 0.200 = 0.050 mole

∴m(KOH) = n × M(KOH) = 0.050 × 56.11 = 2.8 g

Answer : B

7. HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

n(NaOH) = n(HCl) = 0.001 mole

10 mL of 0.1 mole L–1 NaOH = 0.001 mole Answer: A

8. c(KNO3) = c(NO3–)

= 0.600 mole L–1 ∴c(K+) from this solution = 0.600 mole L–1

107



c(K+) from the K2SO4 solution = 1.40 – 0.600 = 0.800 mole L–1

∴c(SO42–) =

21× c(K+)

= 0.400 mole L–1 Answer: A

9. n(NaOH) = Mm

= 00.4000.2

= 0.0500 mole

A: c(NaOH) = Vn

= 00.1

0500.0

= 0.0500 mole L–1

B: c(NaOH) = 100500.0

= 0.005 00 mole L–1

C: c(naOH) = 0.0050010

= 0.000 500 mole L–1 Answer: D

10. D 11. B 12. D

13. c(OH–) = c(NaOH) = 0.01 = 10–2 mole L–1

c(H3O+) = 2

14

1010

−

−

= 10–12 mole L–1 pH = –log 10–12

= –(–12) = 12

Answer: C

14. 75% of theoretical yield = 16.0 g

10075 × theoretical yield = 16.0

108



∴theoretical yield = 16.0 × 75

100

theoretical yield = 21.3 g Answer : B

Review questions

1. (a) n(H2O2) = 1.5 × 7 = 10.5 mol

(b) n(HNO3) = 1.5 × 2 = 3.0 mol

n(H2O) = 1.5 × 8 = 12 mol

2. (a) N2 (g) + 3Cl2 (g) → 2NCl3 (l)

n(N2) n(Cl2) n(NCl3) (i) 2.0 6.0 4.0 (ii) 4.0 12.0 8.0 (iii) 3.2 9.6 6.4

(b) C2H8 (g) + 4O2 (g) → 2CO2 (g) + 4H2O (l)

n(C2H8) n(O2) n(CO2) n(H2O) (i) 0.0750 0.300 0.150 0.300 (ii) 2.00 8.00 4.00 8.00 (iii) 6.00 24.0 12.0 24.0

3. (a) n(N2) = 1.52 × 21

= 0.760 mol

(b) n(O2) = 1.52 × 21

= 0.760 mol

∴ m(O2) = 0.760 × 32.00 = 24.3 g

4. (a) 2C (s) + O2 (g) → 2CO (g)

n(C) = Mm

= 01.125.3

= 0.29 mol From the equation:

n(O2) = 21 × n(C)

109



∴ n(O2) = 21 × 0.29

= 0.15 mole

(b) C (s) + O2 (g) → CO2 (g)

From (a): n(C) = 0.29 mol

From the equation:

n(CO2) = n(C) = 0.29 mol

∴ m(CO2) = 0.29 × 44.01 = 12.8 g = 13 g

5. n(HCl) = 0.540 mol From the equation:

n(Fe(OH)3) = 31 × n(HCl)

= 0.540 × 31

= 0.180 mol m(Fe(OH)3) = n × M(Fe(OH)3) ∴ m(Fe(OH)3) = 0.180 × 106.8

= 19.2 g

6. Mg(OH)2(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)

n(HCl) = Mm

= 46.36

242.0

= 0.006 64 mol From the equation:

n(Mg(OH)2) = 21× n(HCl)

= 0.006 63 × 21

= 0.003 32 mol m(Mg(OH)2) = n × M(Mg(OH)2) ∴ m(Mg(OH)2) = 0.003 32 × 58.33

= 0.194 g

110



7. n(C6H12O6) = Mm

= 18.180

90.8

= 0.0494 mole (a) From the equation:

n(O2) = 16 × n(C6H12O6)

= 16 × 0.0494

= 0.296 mole

m(O2) = n × M(O2)

∴ m(O2) = 0.296 × 32.00

= 9.48 g

(b) From the equation:

n(CO2) = 16 × n(C6H12O6)

= 16 × 0.0494

= 0.296 mole

m(CO2) = n × M(CO2)

∴ m(CO2) = 0.296 × 44.01

= 13.0 g

8. (a) 2Na(s) + Cl2(g) → 2NaCl(s)

(b) n(Na) = Mm

= 99.220.10

= 0.435 mole

From the equation:

n(NaCl) = 22× n(Na)

= 0.435 mole

∴ m(NaCl) = n × M(NaCl)

= 0.435 × 58.44

= 25.4 g

9. m(aspirin, C9H8O4) = 155 kg = 155 000 g

111



n(C9H8O4) = Mm

= 155000180.17

= 860 mole From the equation:

n(C7H6O3) = 11× n(C9H8O4)

= 860 mole ∴ m(C7H6O3) = n × M(C7H6O3)

= 860 × 138.13

= 118 833 g

= 119 kg

10. n(Ag) = Mm

= 87.107

025.0

= 2.3 × 10–4 mole From the equation:

n(Ag2S) = 42× n(Ag)

= 1.2 × 10–4 mole

∴ m(Ag2S) = n × M(Ag2S)

= 1.2 × 10–4 × 247.80

= 0.029 g

11. Equation for the reaction:

2H2(l) + O2(l) → 2H2O(g)

n(O2) = Mm

= 00.32

2030

= 63.4 mole

From the equation:

n(H2) = 12× n(O2)

= 127 mole

∴ m(H2) = n × M(H2)

= 127 × 2.02

= 256 g

= 0.256 kg

112



12. m(H2S2O7) = 5.00 kg = 5000 g

n(H2S2O7) = Mm

= 14.178

5000

= 28.1 mole

From the equation:

n(H2SO4) = 12× n(H2S2O7)

= 56.1 mole

∴ m(H2SO4) = n × M(H2SO4)

= 56.1 × 98.08

= 5506 g

= 5.51 kg

13. (a) n(WO3) = Mm

= 85.231

200

= 0.863 mole

From the equation:

n(W) = 11× n(WO3)

= 0.863 mole

∴ m(W) = n × M(W)

= 0.863 × 183.85

= 159 g

(b) n(H2) = 13× n(WO3)

= 0.863 × 3 = 2.59 mole

∴ m(H2) = n × M(H2)

= 2.59 × 2.02

= 5.23 g

14. (a) 2NH3(g) + H2SO4(aq) → (NH4)2SO4(s)

(b) n(NH3) = Mm

= 04.17

700

= 41.1 mole

113



From the equation:

n((NH4)2SO4) = 21× n(NH3)

= 21 × 41.1

= 20.5 mole

∴ m((NH4)2SO4) = n × M((NH4)2SO4)

= 20.5 × 132.16

= 2715 g

= 2.72 kg

15. (a) Cu(s) + 2AgNO3(aq) → 2Ag(s) + Cu(NO3)2(aq)

(b) m(Cu) reacting = 4.36 – 2.21 = 2.15 g

n(Cu) = Mm

= 55.6315.2

= 0.0338 mole

From the equation:

n(Ag) = 12 × n(Cu)

= 12 × 0.0338

= 0.0677 mole

∴ m(Ag) = n × M(Ag)

= 0.0677 × 107.87

= 7.30 g

16. (a) Actual yield CaO = 13.1 g

Calculate the theoretical yield of CaO:

m(CaCO3) = 24.8 g

n(CaCO3) = Mm

= 09.1008.24

= 0.248 mole

From the equation:

n(CaO) = 11 × n(CaCO3)

= 0.248 mole

114



∴ m(CaO) = n × M(CaO)

= 0.248 × 56.08

= 13.9 g

(b) % yield CaO = actual yieldtheoretical yield

× 100

= 9.131.13 × 100

= 94.3%

17. m(O2) = 4.0 × 109 tonnes = 4.0 × 1015 g

n(O2) = Mm

= 00.32100.4 15×

= 1.3 × 1014 mole (a) From the equation:

n(CO2) = 11 × n(O2)

= 1.3 × 1014 mole

m(CO2) = n × M(CO2)

∴ m(O2) = 1.3 × 1014 × 44.01

= 5.5 × 1015 g

= 5.5 × 109 tonnes


n(C6H12O6) = 61 × n(O2)

= 61 × 1.3 × 1014

= 2.1 × 1013 mole

m(C6H12O6) = n × M(C6H12O6)

∴ m(C6H12O6) = 2.1 × 1013 × 180.18

= 3.8 × 1015 g

= 3.8 × 109 tonnes

18. Pb2+(aq) + SO42–(aq) → PbSO4(s)

m(PbSO4) = 0.0806 g

115



n(PbSO4) = Mm

= 26.303

0806.0

= 2.66 × 10–4 mole From the equation:

n(Pb2+) = 11 n(PbSO4)

= 2.66 × 10–4 mole

m(Pb2+) = n × M(Pb2+)

m(Pb2+) = 2.66 × 10–4 × 207.2

= 0.0551 g

∴% Pb in the paint pigment = 2(Pb )

(pigment)m

m

+

× 100

= 50.1

0551.0 × 100

= 3.67%

19. (a) 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)

(b) n(Na) = 99.222.9

= 0.40 mole

From the equation:

n(H2) = 21 × n(Na)

= 21 × 0.40

= 0.20 mole

∴ m(H2) = 0.20 × 2.02 = 0.40 g

(c) n(Na) = 99.226.4

= 0.20 mole

From the equation:

n(H2O) = 11 × n(Na)

= 0.20 mole

∴ m(H2O) = 0.20 × 18.02 = 3.6 g

116



(d) m(Na) = 1.0 kg = 1000 g

n(Na) = 99.22

1000

= 44 mole

As the density of H2O = 1 g mL–1 the mass of 300 mL of H2O = 300 g

n(H2O) = 02.18

300

= 16 mole

From the equation:

n(Na) : n(H2O) = 1 : 1

∴H2O will be used up first.

20. (a) Calculate the amount, in mole, of both reactants:

n(Na) = 99.225.12

= 0.544 mol

n(Cl2) = 9.705.25

= 0.360 mol 2Na(s) + Cl2(g) → 2NaCl(s)

Divide the amount of each reactant, in mole, by its coefficient in the chemical equation:

Na : 2544.0 = 0.272 mol

Cl2: 1360.0 = 0.360 mol

The two reactants are now in a 1:1 ratio, therefore the limiting reactant is the one with the lowest number of mole, in this case Na. Use the n(Na) to calculate the mass of NaCl produced.

(b) n(NaCl) = n(Na) × 22

= 0.544 mol

∴ m(NaCl) = n × M(NaCl)

= 0.544 × 58.44 = 31.8 g

117




n(SO2) = 06.64

32

= 0.50 mol

n(H2O) = 02.18

18

= 1.0 mol

SO2(g) + H2O(l) → H2SO3(aq)

From the equation, the two reactants are in a 1:1 ratio, therefore the limiting reactant is the one with the lowest number of mole, in this case SO2. Use the n(SO2) to calculate the mass of H2SO3 produced.

n(H2SO3) = n(SO2) × 11

= 0.50 mol

∴ m(H2SO3) = n × M(H2SO3)

= 0.50 × 82.08 = 41 g

(b) n(H2O) used = n(SO2) = 0.50 mole

n(H2O) in excess = 1.0 – 0.50 = 0.50 mole

∴m(H2O) in excess = n × M(H2O) = 0.50 × 18.02

= 9.0 g

22. (a) n(N2H4) = 06.3229.2

= 0.0714 mol

n(O2) = 00.3214.3

= 0.0981 mol

N2H4(l) + O2(g) → N2(g) + 2H2O(l)

The two reactants are already in a 1:1 ratio, therefore the limiting reactant is the one with the lowest number of mole, in this case N2H4. Use the n(N2H4) to calculate the mass of H2O produced in (c).

(b) n(O2) used = n(N2H4) = 0.0714 mol

n(O2) in excess = 0.0981 – 0.0714 = 0.0267 mol

118



∴ m(O2) in excess = 0.0267 × 32.00 = 0.854 g

(c) From the equation:

n(H2O) = 12 × n(N2H4)

= 12 × 0.0714

= 0.143 mole

∴ m(H2O) = n × M(H2O) = 0.143 × 18.02 = 2.57 g


n(Mg) = 31.24

20

= 0.82 mol

n(Cl2) = 9.70

20

= 0.28 mol

Mg(s) + Cl2(g) → MgCl2(s)

From the equation, the two reactants are in a 1:1 ratio, therefore the limiting reactant is the one with the lowest number of mole, in this case Cl2. Use the n(Cl2) to calculate the mass of MgCl2 produced.

(b) n(MgCl2) = n(Cl2) × 11

= 0.28 mol

∴ m(MgCl2) = n × M(MgCl2)

= 0.28 × 95.21 = 27 g

24. Using the formula: m = c × V × M where V = volume in litres (a) V = 300 mL

= 0.300 L

∴ m(NaOH) = 1.5 × 0.300 × 40.00

= 18 g

(b) V = 250 mL = 0.250 L

∴ m(H2SO4) = 2.0 × 0.250 × 98.08

= 49 g

119



(c) V = 17.5 L

∴ m(Na2CO3) = 1.5 × 17.5 × 105.99 = 2782 g = 2.8 kg

(d) V = 200 mL = 0.200 L

∴ m(CuSO4.5H2O) = 2.5 × 0.200 × 249.71 = 125 g = 1.3 × 102 g

(e) V = 120 mL = 0.120 L

∴ m(Mg(NO3)2) = 1.7 × 0.120 × 148.33 = 30 g

25. Convert mass of solute to mole using n = Mm .

Use the calculated value of n to determine c using c = Vn where V = volume in L.

(a) n(NaBr) = 89.102

17

= 0.17 mole

V = 250 mL = 0.250 L

∴ c = 250.017.0

= 0.66 mole L–1

(b) n(MgSO4) = 37.120

200

= 1.66 mole

V = 500 mL = 0.500 L

∴ c = 500.066.1

= 3.32 mole L–1

(c) n(Mg(NO3)2) = 33.148

1500

= 10.11 mole

V = 1.5 L

120



∴ c = 5.111.10

= 6.7 mole L–1

(d) n(Na3PO4) = 94.163

1000

= 6.100 mole

V = 100 mL = 0.100 L

∴ c = 100.0100.6

= 61.0 mole L–1

(e) n(K2CO3) = 21.138

500

= 3.62 mole

V = 2.00 L

∴ c = 00.262.3

= 1.81 mole L–1

(f) n(AlCl3) = 33.1335.12

= 0.0938 mole

V = 40.0 mL = 0.0400 L

∴ c = 0400.00938.0

= 2.34 mole L–1

26. Convert mass of solute to mole using n = Mm .


(a) n(NaOH) = 00.400.4

= 0.10 mole

V = 200 mL = 0.200 L

∴ c = 200.010.0

= 0.50 mole L–1

121



(b) n(Na2CO3) = 99.1056.12

= 0.119 mole

V = 350 mL = 0.350 L

∴ c = 350.0119.0

= 0.340 mole L–1

(c) n(MgCO3) = 32.8435.5

= 0.0634 mole

V = 500 mL = 0.500 L

∴ c = 500.00634.0

= 0.127 mole L–1

27. Using the formula: n = c × V where V = volume in litres (a) V = 30 mL

= 0.030 L

∴ n(AgNO3) = 0.10 × 0.030 = 0.0030 mole

(b) V = 300 mL = 0.300 L

∴ n(Mg(NO3)2) = 1.5 × 0.300 = 0.45 mole

(c) V = 230 mL = 0.230 L

∴ n(KCl) = 0.40 × 0.230 = 0.092 mole

(d) V = 2.5 L

∴ n(KNO3) = 0.2 × 2.5 = 0.5 mole

28. Using the formula: m = c × V × M where V = volume in litres (a) V = 250 mL

= 0.250 L

122



∴ m(NaBr) = 1.5 × 0.250 × 102.89 = 39 g

(b) V = 500 mL = 0.500 L

∴ m(MgSO4) = 1.75 × 0.500 × 120.37 = 105 g

(c) V = 1.50 L

∴ m(Mg(NO3)2) = 1.50 × 0.575 × 148.33 = 128 g

(d) V = 100 mL = 0.100 L

∴ m(Na3PO4) = 0.850 × 0.100 × 163.94 = 13.9 g

(e) V = 2.00 L

∴ m(K2CO3) = 0.005 × 2.00 × 138.21 = 1.38 g = 1 g

(f) V = 40.0 mL = 0.0400 L

∴ m(AlCl3) = 2.30 × 0.0400 × 133.33 = 12.3 g

29. Using the formula: m = c × V × M where V = volume in litres

V = 400.0 mL = 0.400 L

∴ m(CH3COONa) = 0.500 × 0.400 × 82.04 = 16.4 g

30. Convert mass of solute to mole using n = Mm


(a) n(H2SO4) = 08.985.58

= 0.596 mole

V = 2.00 L

123



∴ c = 00.2

596.0

= 0.298 mole L–1

(b) n(HCl) = 46.367.2

= 0.074 mole

V = 500 mL = 0.500 L

∴ c = 500.0074.0

= 0.15 mole L–1

(c) n(KNO3) = 11.101

04.4

= 0.0400 mole

V = 150.0 mL = 0.1500 L

∴ c = 1500.00400.0

= 0.266 mole L–1

(d) n(NaCl) = 44.58

234

= 4.00 mole

V = 6.00 L

∴ c = 00.600.4

= 0.667 mole L–1

31. (a) n(NH4NO3) = Mm

= 06.80

150

= 1.87 mole

∴ c(NH4NO3) = Vn

= 5.1

87.1

= 1.25 mole L–1 = 1.3 mole L–1

124



(b) NH4NO3(aq) → NH4+ (aq) + NO3

–(aq)

c(NH4+) = c(NH4NO3)

= 1.3 mole L–1

c(NO3–) = c(NH4NO3)

= 1.3 mole L–1

32. (a) n(AlCl3) = Mm

= 33.133

200

= 1.50 mole

V = 700 mL = 0.700 L

∴ c(AlCl3) = Vn

= 700.050.1

= 2.14 mole L–1

(b) AlCl3(aq) → Al3+ (aq) + 3Cl–(aq)

c(Al3+) = c(AlCl3) = 2.14 mole L–1

c(Cl–) = 3 × c(AlCl3) = 3 × 2.14 = 6.43 mole L–1

33. Ca(OH)2(aq) → Ca2+(aq) + 2OH–(aq) c(OH–) = 0.050 mole L–1

∴ c(Ca(OH)2) = 21 × c(OH–)

= 21× 0.050

= 0.025 mole L–1

34. Fe2(SO4)3(aq) → 2Fe3+(aq) + 3SO42–(aq)

c(SO42–) = 0.25 mole L–1

∴ c(Fe2(SO4)3) = 31 × c(SO4

2–)

= 31× 0.25

= 0.083 mole L–1

125



35. Dilution formula: c1V1 = c2V2 V(stock solution) = V1

therefore use the formula V1 = 1

22

cVc

(a) V2 = 500 mL = 0.500 L

∴ V1 = 0.500 0.75018×

= 0.021 L = 21 mL

(b) V2 = 200 mL = 0.200 L

∴ V1 = 0.200 2.515×

= 0.033 L = 33 mL

(c) V2 = 350 mL = 0.350 L

∴ V1 = 0.350 0.15018×

= 0.0029 L = 2.9 mL

36. Use the formula: c2 = 2

11

VVc ×

(a) V2 = 100 mL + 10 mL = 110 mL = 0.110 L

∴c2 = 0.100 0.10.110

×

= 0.09 mol L–1

(b) V2 = 16 mL + 4.0 mL = 20 mL = 0.020 L

∴c2 = 020.0

20.0016.0 ×

= 0.16 mol L–1

(c) V2 = 750 mL + 150 mL = 900 mL = 0.900 L

126



∴c2 = 0.750 1.500.900

×

= 1.25 mol L–1

37. Use the formula V2 = 2

11

cVc

then V(H2O) added = V2 – V1

(a) V1 = 100 mL = 0.100 L

V2 = 0.100 155.0×

= 0.3 L = 300 mL

∴V(H2O) = 300 – 100 = 200 mL

(b) V1 = 130 mL = 0.130 L

V2 = 0.130 3.501.00×

= 0.455 L = 455 mL

∴V(H2O) = 455 – 130 = 325 mL

(c) V1 = 170 mL = 0.170 L

V2 = 0.170 2.600.250×

= 1.768 L = 1768 mL

∴V(H2O) = 1768 – 170 = 1598 mL = 1.60 L

38. Dilution formula: c1V1 = c2V2 V(stock solution) = V1

therefore use the formula V1 = 1

22

cVc

V2 = 500 mL = 0.500 L

127



V1 = 0.500 0.10.5×

= 0.1 L

∴ V(0.5 mol L–1 H2SO4) required = 100 mL

39. m(NaCl) = 2.35 g

n(NaCl) = Mm

= 44.5835.2

= 0.0402 mole

V(NaCl) = cn

= 00.2

0402.0

= 0.0201 L = 20.1 mL

40. (a) m(MgCl2) = 3.50 g

n(MgCl2) = Mm

= 21.95

50.3

= 0.0368 mole V = 200 mL

= 0.200 L

c(NaCl) = Vn

= 200.00368.0

= 0.184 mole L–1

(b) Use the formula: c2 = 2

11

VVc ×

V2 = 200 mL + 150 mL = 350 mL = 0.350 L

∴c2 = 0.184 0.2000.350×

= 0.105 mol L–1

128



41. n(Ca(OH)2) = Mm

= 10.745.2

= 0.034 mole From the equation:

n(HCl) = 12 × n(Ca(OH)2)

= 12 × 0.034

= 0.067 mole

∴ V(HCl) = cn

= 3.1

067.0

= 0.052 L = 52 mL

42. V(AgNO3) = 250 mL = 0.250 L

n(AgNO3) = c × V = 0.100 × 0.250 = 0.0250 mole

From the equation:

n(Cu) = 21 × n(AgNO3)

= 21 × 0.0250

= 0.0125 mole ∴m(Cu) = n × M(Cu)

= 0.0125 × 63.55 = 0.794 g

43. V(NaBr) = 120 mL = 0.120 L

n(NaBr) = c × V = 0.300 × 0.120 = 0.0360 mole

From the equation:

129



n(Cl2) = 21 × n(NaBr)

= 21 × 0.0360

= 0.0180 mole

44. (a) n(As(OH)3) = Mm

= 95.125

15.3

= 0.0250 mole


n(NaOH) = 13 × n(As(OH)3)

= 13 × 0.0250

= 0.0750 mole

(c) V(AsCl3) = 25.0 mL = 0.0250 L

From the equation:

n(AsCl3) = n(As(OH)3) = 0.0250 mole

∴ c(AsCl3) = Vn

= 0250.00250.0

= 1.00 mole L–1

(d) From (b): n(NaOH) = 0.0750 mole

V(NaOH) = cn

= 200.00750.0

= 0.375 L = 375 mL

45. Ag+(aq) + Cl–(aq) → AgCl(s)

n(AgCl) = Mm

= 32.143

26.1

= 0.008 79 mole

130



From the equation: n(Cl–) = n(AgCl)

= 0.008 79 mole V(sample) = 500 mL

= 0.500 L

∴ c(Cl–) = Vn

= 0.008790.500

= 0.0176 mole L–1

46. n(KBr) = Mm

= 0.1190.32

= 0.269 mole V(KBr) = 400 mL

= 0.400 L

∴c(KBr) = Vn

= 400.0269.0

= 0.672 mole L–1 Use the dilution formula to calculate V2:

V2 = 2

1

cVc ×

= 100.0

400.0672.0 ×

= 2.69 L ∴V(H2O) added = 2.69 – 0.400

= 2.29 L

47. V(KOH) = 400 mL = 0.400 L

n(KOH) = c × V = 0.250 × 0.400 = 0.100 mole

From the equation:

n(Cr(OH)3) = 31 × n(KOH)

= 31 × 0.100

131



= 0.0333 mole ∴m(Cr(OH)3) = n × M(Cr(OH)3)

= 0.0333 × 103.03 = 3.43 g

48. n(Al(OH)3) = Mm

= 01.78

287


n(Ba(OH)2) = 23 × n(Al(OH)3)

= 23 × 3.68

= 5.52 mole

∴ V(Ba(OH)2) = cn

= 55.152.5

= 3.56 L

49. (a) NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

(b) V(NaCl) = 120 mL = 0.120 L

n(NaCl) = c × V = 1.5 × 0.120 = 0.18 mole

From the equation:

n(AgNO3) = 11 × n(NaCl)

= 0.18 mole

∴V(AgNO3) = cn

= 0.218.0

= 0.090 L = 90 mL

50. H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) V(H2SO4) = 50 mL

= 0.050 L

132



n(H2SO4) = c × V = 1.2 × 0.050 = 0.060 mole

From the equation:

n(NaOH) = 12 × n(H2SO4)

= 2 × 0.060 = 0.12 mole

V(NaOH) = 30 mL = 0.030 L

∴c(NaOH) = Vn

= 030.012.0

= 4.0 mole L–1

51. V(MnO4–) = 15.0 mL

= 0.0150 L n(MnO4

–) = c × V = 0.0100 × 0.0150 = 1.5 × 10–4 mole

From the equation:

n(Fe2+) = 15 × n(MnO4

–)

= 5 × 1.5 ×10–4 = 7.5 ×10–4 mole

V(Fe2+) = 10.0 mL = 0.0100 L

∴c(Fe2+) = Vn

= 0100.0

105.7 4−×

= 0.0750 mole L–1

52. V(Ca(ClO3)2) = 560 mL = 0.560 L

n(Ca(ClO3)2) = c × V = 3.25 × 0.560 = 1.82 mole

From the equation:

133



n(Na3PO4) = 32 × n(Ca(ClO3)2)

= 32 × 1.82

= 1.21 mole

∴V(Na3PO4) = cn

= 30.221.1

= 0.528 L = 528 mL

53. V(H2S) = 280 mL = 0.280 L

n(H2S) = c × V = 0.200 × 0.280 = 0.0560 mole

From the equation:

n(HNO3) = 32 × n(H2S)

= 0.0373 mole

∴V(HNO3) = cn

= 250.0

0560.0

= 0.149 L = 149 mL

54. (The question in text should read ‘the concentration of ethanoic acid in vinegar … Find the concentration of ethanoic acid in the vinegar.’) V(NaOH) = 43.20 mL

= 0.043 20 L n(NaOH) = c × V

= 0.350 × 0.043 20 = 0.0151 mole

From the equation:

n(CH3COOH) = 11 × n(NaOH)

= 0.0151 mole V(CH3COOH) = 20.00 mL

= 0.020 00 L

134



∴c(CH3COOH) = Vn

= 0.01510.02000

= 0.756 mole L–1

55. H2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2H2O(l) V(NaOH) = 24.00 mL

= 0.02400 L n(NaOH) = c × V

= 0.620 × 0.024 00 = 0.0149 mole

From the equation:

n(H2SO4) = 21 × n(NaOH)

= 21 × 0.0149

= 0.007 44 mole

∴V(H2SO4) = cn

= 0.007 440.460

= 0.0162 L = 16.2 mL

56. Calculate the concentration of the Na2CO3 solution first:

n(Na2CO3) = Mm

= 99.105

365.1

= 0.012 88 mole V(Na2CO3) solution = 250.0 mL

= 0.2500 L

c(Na2CO3) = Vn

= 0.012880.2500

= 0.051 51 mole L–1 Na2CO3(aq) + 2HNO3(aq) → 2NaNO3(aq) + H2O(l) + CO2(g) V(Na2CO3) aliquots = 20.00 mL

= 0.020 00 L

135



n(Na2CO3) = c × V = 0.051 51 × 0.020 00 = 0.001 030 mole

From the equation:

n(HNO3) = 12 × n(Na2CO3)

= 12 × 0.001 030

= 0.002 060 mole V(HNO3) = 21.95 mL

= 0.021 95 L

∴ c(HNO3) = Vn

= 0.0020600.02195

= 0.09388 mole L–1

57. (a) A monoprotic acid is an acid that can donate one proton only.

(b) HX (aq) + H2O(l) → H3O+(aq) + X–(aq)

(c) If the acid HX is a strong acid then the concentration of H3O+ should be equal to the acid concentration, 0.10 mole L–1. As the concentration of H3O+ is less than this (0.010 mole L–1) then the acid has only partially ionised, therefore HX is an example of a weak acid.

(d) Conductivity is directly proportional to the concentration of ions in the solution. As HX is a weak acid and only partially ionises, the concentration of ions (H3O+ and X–) will be small, therefore HX will conduct but not very well.

58. (a) [H3O+] = 10–3 mol L–1 ∴ pH = –log 10–3

= –(–3) = 3

(b) 9

59. (a) [H3O+] = 10–4 mol L–1

∴ pH = –log 10–4 = –(–4) = 4

(b) [H3O+] = 10–6 mol L–1

∴ pH = –log 10–6 = –(–6) = 6

136



(c) [H3O+] = 10–9 mol L–1

∴ pH = –log 10–9 = –(–9) = 9

(d) [H3O+] = 0.001 mol L–1 = 10–3 mol L–1

∴ pH = –log 10–3 = –(–3) = 3

(e) [H3O+] = 1 mol L–1 = 100 mol L–1

∴ pH = –log 100 = 0

60. pH = –log [H3O+]

Find [H3O+] using the formula [H3O+] = 1410

[OH ]

−

−

(a) [H3O+] = 14

14

1010

−

−

= 1 = 100 mole L–1

∴pH = –log 100 = –(0) = 0

(b) [H3O+] = 10

14

1010

−

−

= 10–4 mole L–1

∴pH = –log 10–4 = –(–4) = 4

(c) [H3O+] = 1

14

1010

−

−

= 10–13 mole L–1

∴pH = –log 10–13 = –(–13) = 13

(d) [H3O+] = 01.0

10 14−

= 10–12 mole L–1

137



∴pH = –log 10–12 = –(–12) = 12

(e) [H3O+] = 1

10 14−

= 10–14 mole L–1

∴pH = –log 10–14 = –(–14) = 14

61. [H3O+] = 10–pH mole L–1 (a) pH = 1

∴[H3O+] = 10–1 mole L–1

(b) pH = 2 ∴[H3O+] = 10–2 mole L–1

(c) pH = 3 ∴[H3O+] = 10–3 mole L–1

(d) pH = 10 ∴[H3O+] = 10–10 mole L–1

62. [H3O+] = 10–pH mole L–1

Then [OH–] = 14

3

10[H O ]

−

+ mole L–1

(a) pH = 3 [H3O+] = 10–3 mole L–1

∴ [OH–] = 3

14

1010

−

−

= 10–11 mole L–1

(b) pH = 12 [H3O+] = 10–12 mole L–1

∴ [OH–] = 12

14

1010

−

−

= 10–2 mole L–1

(c) pH = 7 [H3O+] = 10–7 mole L–1

∴ [OH–] = 7

14

1010

−

−

= 10–7 mole L–1

138



(d) pH = 1 [H3O+] = 10–1 mole L–1

∴ [OH–] = 1

14

1010

−

−

= 10–13 mole L–1

63. (a) [H3O+] = 100 mol L–1 = 102 mol L–1

∴ pH = –log 102 = –(2) = –2

(b) 2

(c) [H3O+] = 10–2 mol L–1

∴ pH = –log 10–2 = –(–2) = 2

(d) [H3O+] = 0.001 mol L–1 = 10–3 mol L–1

∴ pH = –log 10–3 = –(–3) = 3

(e) 11

64. (a) n(HCl) = Mm

= 46.36

073.0

= 0.0020 mole

c(HCl) = Vn

= 00.2

0020.0

= 0.0010 mole L–1 [H3O+] = c(HCl)

= 0.0010 = 10–3 mole L–1

∴ pH = –log 10–3 = –(–3) = 3

139



(b) n(NaOH) = Mm

= 00.40

80.0

= 0.020 mole

c(NaOH) = Vn

= 00.2

020.0

= 0.010 mole L–1

[OH–] = c(NaOH) = 0.010 = 10–2 mole L–1

[H3O+] = 2

14

1010

−

−

= 10–12 mole L–1

∴ pH = –log 10–12 = –(–12) = 12

65. (a) basic

(b) pH = 7.4

∴ [H3O+] = 10–7.4 = 4.0 × 10–8 mole L–1

66. pH = 3 ∴ c1 = 10–3 mole L–1 pH = 5 ∴ c2 = 10–5 mole L–1 V1 = 30 mL

= 0.030 L

V2 = 2

11

cVc ×

= 5

3

10030.010

−

− ×

= 3.0 L ∴V(H2O) = 3.0 – 0.030

= 2.97 L = 2970 mL

67. If pH = 2 then [H3O+] = 10–2 mol L–1

c2 = 2

11

VVc ×

140



V2 = 450 mL + 50 mL = 500 mL

∴c2 = 500

5010 2 ×−

= 10–3 mol L–1 ∴ pH = –log 10–3

= –(–3) = 3

68. 5% of reactants remain unused

69. Actual yield CaO = 12.6 g Calculate the theoretical yield of CaO: m(CaCO3) = 25.9 g

n(CaCO3) = Mm

= 09.1009.25


n(CaO) = 11 × n(CaCO3)

= 0.259 mole ∴ m(CaO) = n × M(CaO)

= 0.259 × 56.08 = 14.5 g

% yield CaO = actual yieldtheoretical yield

× 100

= 5.146.12 × 100

= 86.8%

70. Actual yield Fe = 41.8 g Calculate the theoretical yield of Fe: m(Fe2O3) = 67.3 g

n(Fe2O3) = Mm

= 70.1593.67


141



n(Fe) = 12 × n(Fe2O3)

= 0.843 mole ∴ m(Fe) = n × M(Fe)

= 0.843 × 55.85 = 47.1 g

% yield Fe = actual yieldtheoretical yield

× 100

= 1.478.41 × 100

= 88.8%

71. SiO2(s) + 2C(s) → SiC(s) + 2CO(g) Actual yield SiC = 32.5 g Calculate the theoretical yield of SiC: m(SiO2) = 54.8 g

n(SiO2) = Mm

= 09.608.54


n(SiC) = 11 × n(SiO2)

= 0.912 mole ∴ m(SiC) = n × M(SiC)

= 0.912 × 40.10 = 36.6 g

% yield SiC = actual yieldtheoretical yield

× 100

= 6.365.32 × 100

= 88.9%

72. Actual yield Cu = 2.58 g Calculate the theoretical yield of Cu: m(Al) = 1.56 g

n(Al) = Mm

142



= 98.26

56.1


n(Cu) = 23 × n(Al)

= 0.0867 mole ∴ m(Cu) = n × M(Cu)

= 0.0867 × 63.55 = 5.51 g

% yield Cu = actual yieldtheoretical yield

× 100

= 51.558.2 × 100

= 46.8%



1. m(aspirin, C9H8O4) = 10.0 g

n(C9H8O4) = Mm

= 17.1800.10


n(C7H6O3) = 11 × n(C9H8O4)

= 0.0555 mole ∴ m(C7H6O3) = n × M(C7H6O3)

= 0.0555 × 138.13 = 7.67 g

2. (a) V(Na2CO3) solution = 500.00 mL = 0.500 00 L

n(Na2CO3) = c × V = 0.0500 × 0.500 00 = 0.0250 mole

143



∴ m(Na2CO3) = n × M = 0.0250 × 105.99 = 2.65 g

(b) c(Na2CO3) = 0.0500 mole L–1

V(Na2CO3) used in the reaction = 25.00 mL = 0.025 00 L

Na2CO3(aq) + 2HCl(aq) → 2NaCl(aq) + H2O(l) + CO2(g)

n(Na2CO3) = c × V = 0.0500 × 0.025 00 = 0.00125 mole

From the equation:

n(HCl) = 12 × n(Na2CO3)

= 12 × 0.001 25

= 0.002 50 mole

V(HCl) = 23.50 mL = 0.023 50 L

∴ c(HCl) = Vn

= 0.002500.02350

= 0.106 mole L–1

144


Chapter 16

Page 369 1. (a) oxidant: S; reductant: Mg

(b) oxidant: Ag+; reductant: Fe

(c) reductant: Br–; oxidant: Cl2

(d) The reaction is not redox as there is no loss or gain of electrons.

Page 371 2. (a) H +1; Br –1

(b) Na +1; O –2 (c) C –4; H +1

(d) Na +1; Cl +5; O –2 (e) Al +3; O –2

(f) H +1; P +5; O –2

3. (a) +6 (b) +4 (c) +6 (d) –2

Page 372

4. clockwise from green solution: +3, +6, +3, +2, +6

5. (a) N –3; H +1 (b) Mn +7; O –2 (c) H +1; S –2

(d) V +4; O –2 (e) I +5; O –2 (f) P +5; O –2

Page 373

6. a, c, g, h

7. (a) Fe is oxidised, Cl2 is reduced.

(c) N is oxidised, O2 is reduced.

145



(g) C is oxidised, O2 is reduced.

(h) H2 is oxidised, C is reduced.

Page 374

9. (a) oxidation: Pb(s) → Pb2+(aq) + 2e– reduction: Ag+(aq) + e– → Ag(s)

(b) Ag+ accepts electrons.

Page 375 (c) Pb is oxidised to Pb2+.

(d) Pb is the reductant.

(e) Electrons are taken from Pb.

10. (a) H+ accepts electrons. (d) Electrons are taken from Al.

(b) H+ is reduced to H2. (e) Al is oxidised to Al3+.

(c) Al is the reductant. (f) H+ is the oxidant.

Page 378

11. (a) MnO4–(aq) + 8H+(aq) + 5e–→ Mn2+(aq) + 4H2O(l)

(b) +7 and +2

(c) 5

(d) The change in oxidation number is equal to the number of electrons used.

12. (a) I2(s) + H2S(g) → 2I–(aq) + S(s) + 2H+(aq)

(b) 2NO3–(aq) + 3H2S(g) +2H+(aq) → 2NO(g) + 3S(s) + 4H2O(l)

(c) Cr2O72–(aq)+ 3H2S(g) + 8H+(aq) → 2Cr3+(aq) + 3S(s) + 7H2O(l)

(d) 5SO32–(aq) +2MnO4

–(aq) + 6H+(aq) → 5SO42–(aq) + 2Mn2+(aq) + 3H2O(l)

(e) 3Cu(s) + 2NO3–(aq) +8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

Page 380

13. (a) Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) (reductant Mg, oxidant Cu2+)

(b) no reaction

(c) 2Al(s) + 3Pb2+(aq) → 2Al3+(aq) + 3Pb(s) (reductant Al, oxidant Pb2+)

(d) no reaction

14. (a) The aluminium metal may have had an oxide coating.

(b) Use steel wool or emery paper to strip off the coating to expose the metal. Repeat the experiment.

146



Page 383

15. (a) Pb is oxidised, Cu2+ is reduced.

(b) see diagram above

(c) from anode to cathode

(d) Pb(s) → Pb2+(aq) + 2e– Cu2+(aq) + 2e–→ Cu(s)

(e) Pb(s) + Cu2+(aq) → Pb2+(aq) + Cu(s)

16. (a)

(b) oxidation: Zn(s) → Zn2+(aq) + 2e– reduction: Ag+(aq) + e– → Ag(s)

(c) Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

17.

oxidation (at anode): Mg(s) → Mg2+(aq) + 2e–

reduction (at cathode): Zn2+(aq) + 2e–→ Zn(s)

overall reaction: Mg(s) + Zn2+(aq) → Mg2+(aq) + Zn(s)

Zn2+ is the oxidant, Mg is the reductant.

147



Page 384 18. Ag+(aq), Cu2+(aq), Pb2+(aq), Zn2+(aq), Al3+(aq), Mg2+(aq), Na+(aq)

19. (a) The species will be reduced.

(b) cathode

20. (a) Pb2+(aq)/Pb

(b) Mg2+(aq)/Mg

(c) Pb2+(aq)/Pb

(d) Zn2+(aq)/Zn

Page 386

22. (a) Fe(s) + Pb2+(aq) → Fe2+(aq) + Pb(s)

(b) no reaction

(c) no reaction

Page 387

(d) Zn(s) + Sn2+(aq) → Zn2+(aq) + Sn(s)

(e) 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s)

(f) no reaction

23. (a) Al(s) → Al3+(aq) + 3e– Cu2+(aq) + 2e–→ Cu(s) 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)

(b) Sn2+(aq) + 2e–→ Sn(s) Fe(s) → Fe2+(aq) + 2e– Sn2+(aq) + Fe(s) → Fe2+(aq) + Sn(s)

(c) Pb2+(aq) + 2e–→ Pb(s) Zn(s) → Zn2+(aq) + 2e– Pb2+(aq) + Zn(s) → Pb(s) + Zn2+(aq)


1. D 2. A 3. B 4. D 5. D 6. A 7. D 8. B 9. B 10. B 11. C 12. D 13. B

14. B 15. D 16. A 17. D 18. A 19. C 20. B 21. B 22. C 23. B 24. A 25. B

26. B 27. D 28. A

Review questions 3. (a) oxidant: Ce4+; reductant: Sn2+

(b) oxidant: Pb2+; reductant: Cd

(c) oxidant: Cl2; reductant: Al

(d) oxidant: H+; reductant: Mg

(e) oxidant: H2O; reductant: Na

148



4. (a) Ca +2; O –2 (b) C –3; H +1 (c) Cr +7; O –2 (d) H +1; S +6; O –2

5. (a) 0 (b) +2 (c) +7 (d) +4

6. (a) +5 (b) +6 (c) +5 (d) +6 (e) +6 (f) +5

7. (a) reduction (b) oxidation (c) oxidation (d) reduction

8. (a) Zn is oxidised, HCl is reduced.

(b) NO is oxidised, O2 is reduced.

(c) Mg is oxidised, H2SO4 is reduced.

(d) Al is oxidised, Cl2 is reduced.

9. (a) Na is the reductant, S is the oxidant.

(b) K is the reductant, Cl2 is the oxidant.

(c) Al is the reductant, O2 is the oxidant.

(d) HBr is the reductant, Cl2 is the oxidant.

(e) C is the reductant, O2 is the oxidant.

(f) Zn is the reductant, MnO2 is the oxidant.

10. c, d, f, h, i

11. (a) oxidation (b) reduction (c) reduction

12. (a) Ag → Ag+ + e–; oxidation; O2 + 4e–→ 2O2–; reduction

(b) Fe→ Fe2+ + 2e–; oxidation S + 2e–→ S2–; reduction

149



(c) Pb→ Pb2+ + 2e–; oxidation Cl2 + 2e–→ 2 Cl–; reduction

(d) Hg→ Hg2+ + 2e–; oxidation O2 + 4e–→ 2O2–; reduction

13. (a) HNO3(aq) + H+(aq) + e–→ NO2(g) + H2O(l)

(b) Cl2(g) +6H2O(l) → 2ClO3–(aq) + 12H+(aq) + 10e–

(c) 2NH3(g) → N2(g) + 6H+(aq) + 6e–

(d) MnO4–(aq) + 4H+(aq) + 3e–→ MnO2(s) + 2H2O(l)

14. (a) 2Br–(aq) + SO42–(aq) + 4H+(aq) → Br2(l) + SO2(g) + 2H2O(l)

(b) 2Al(s) + 3Cl2(g) → 2AlCl3(s)

(c) I2(s) + H2S(g) → 2I–(aq) + S(s) + 2H+(aq)

(d) 3Cu(s) + 2NO3–(aq) + 8H+(aq) → 3Cu2+(aq) + 2NO(g) + 4H2O(l)

(e) Cu(s) + 2NO3–(aq) + 4H+(aq) → Cu2+(aq) + 2NO2(g) + 2H2O(l)

(f) 3CuO(s) + 2NH3(g) → 3Cu(s) + 3H2O(l) + N2(g)

(g) PbS(s) + 4H2O2(l) → PbSO4(s) + 4H2O(l)

(h) 2Cr2O72–(aq) + 16H+(aq) + 3CH3CH2OH(g) → 4Cr3+(aq) + 11H2O(l)

+ 3CH3COOH(aq)

15. Gold and silver are less reactive than zinc and iron.

16. Displacement occurs when metallic ions are converted to the metal by another metal. For example zinc displaces copper ions in the reaction: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

17. The better the reducing strength of a metal the more reactive the metal.

18. Groups 1 and 2 of the periodic table contain the most reactive metals.

19. Zn(s) + Pb2+(aq) → Zn2+(aq) + Pb(s)

20. (a) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

(b) Fe(s) + Sn2+(aq) → Fe2+(aq) + Sn(s)

(c) Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s)

(d) 2Al(s) + 3Zn2+(aq) → 2Al3+(aq) + 3Zn(s) In each case the reductant has been placed first in the equation.

21. Tin will react with copper ions and will eventually wear away.

22. (a) AgNO3(aq), CuSO4(aq), Pb(NO3)2(aq)

(b) Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s) Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s) Fe(s) + Pb2+(aq) → Fe2+(aq) + Pb(s)

23. Na > X > Cu

24. (a) Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s) Cd(s) + Sn2+(aq) → Cd2+(aq) + Sn(s)

150



(b) Cd > Sn > Cu

(c) only (iv) will occur

25. (a) A+, B2+, C+, D3+

(b) D is second in the series.

(c) B > D > A > C

(d) B2+

26. (a) E > A > F > C > D > B

(b) Mg, Zn, Fe, Sn, Pb, Cu

29. (a) Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)

(c) anode reaction: Zn(s) → Zn2+(aq) + 2e– cathode reaction: Ag+(aq) + e–→ Ag(s)

30. (a) Zn(s) → Zn2+(aq) + 2e–

(b) Fe2+(aq) + 2e–→ Fe(s)

(c) Zn(s) + Fe2+(aq) → Zn2+(aq) + Fe(s)

31. (a) Electrons flow from zinc to copper.

(b) Zinc is the anode.

(c) The copper container is the cathode.

(d) Zn(s) → Zn2+(aq) + 2e–

(e) Cu2+(aq) + 2e–→ Cu(s)

(f) Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

(g) porous pot

32. oxidation reaction at anode: Al(s) → Al3+(aq) + 3e– reduction reaction at cathode: Cu2+(aq) + 2e–→ Cu(s) overall reaction: 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s) Electrons flow from Al to Cu. NO3

– ions flow to the anode, K+ ions flow to the cathode.

(b) The reaction would stop.

(c) Colour fades as blue copper ions are used up.

33. (a) Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s)

(b) no reaction

(c) no reaction

(d) Mg(s) + Sn2+(aq) → Mg2+(aq) + Sn(s)

(e) no reaction

(f) no reaction

34. (a) anode reaction: Pb(s) → Pb2+(aq) + 2e– cathode reaction: Ag+(aq) + e–→ Ag(s) overall reaction: Pb(s) + 2Ag+(aq) → Pb2+(aq) + 2Ag(s) Electrons flow from anode, Pb, to cathode, Ag. K+ ions flow to the Ag half-cell and NO3

– ions flow to the Pb half-cell.

151



(b) anode reaction: Sn(s) → Sn2+(aq) + 2e– cathode reaction: Cu2+(aq) + 2e–→ 2Cu(s) overall reaction: Sn(s) + Cu2+(aq) → Sn2+(aq) + Cu(s) Electrons flow from anode, Sn, to cathode, Cu. K+ ions flow to the Cu half-cell and NO3

– ions flow to the Sn half-cell.

(c) anode reaction: Ni(s) → Ni2+(aq) + 2e– cathode reaction: Pb2+(aq) + 2e–→ Pb(s) overall reaction: Ni(s) + Pb2+(aq) → Ni2+(aq) + Pb(s) Electrons flow from anode, Ni, to cathode, Pb. K+ ions flow to the Pb half-cell and NO3

– ions flow to the Ni half-cell.

Exam questions (pages 401–402)

Multiple choice questions 1. C 2. C


1. (a) Br2(l) + 2e–→ 2Br–(aq); reduction

(b) Fe3+(aq) + e–→ Fe2+(aq); reduction

(c) O2(g) + 2H+(aq) + 2e–→ H2O2(l); reduction

(d) IO3–(aq) + 6H+(aq) + 6e–→ I–(aq) + 3H2O(l); reduction

(e) MnO4–(aq) + 8H+(aq) + 5e–→ Mn2+(aq) + 4H2O(l); reduction

(f) 2Cr3+(aq) + 7H2O(l) → Cr2O72–(aq) + 14H+(aq) + 6e–; oxidation

2. (a) anode reaction: Mg(s) → Mg2+(aq) + 2e– cathode reaction: Ag+(aq) + e–→ Ag(s)

(b) Mg(s) + 2Ag+(aq) → Mg2+(aq) + 2Ag(s)

(c) Electrons flow from Mg to Ag.

(e) Mg electrode loses mass.

(f) Ag electrode gains mass.

(g) It completes the circuit and allows anions to flow to the anode and cations to flow to the cathode.

152


Chapter 17

Page 408 3. nitrogen gas N2; nitrate ion NO3

–; nitrogen monoxide NO

Page 419 14. The range of boiling points reflects the range in molecular size. Greater dispersion

forces act between molecules of larger size.

Multiple choice questions 1. C 2. D 3. C 4. A, C 5. B 6. B 7. A 8. C 9. D 10. C

Review questions 2. oxygen


Multiple choice questions 1. C 2. C

Extended response question 1. (a) carbon dioxide

(b) helium

(c) oxygen

(d) nitrogen

(e) water

153


Chapter 18

Page 427

5. (a) 780 mm Hg = 760780 × 1 atm

= 1.03 atm

(b) 4.0 atm = 0.10.4 × 101 325 Pa

= 405 300 Pa = 4.1 × 105 Pa

(c) 1000 mm Hg = 760

1000 × 101 325 Pa

= 133 322 Pa = 1.333 × 105 Pa

(d) 1250 mm Hg = 760

1250 × 101.325 kPa

= 166.7 kPa

(e) 200 °C = 200 + 273 = 473 K

(f) 500 K = 500 – 273 = 227 °C

(g) 3.0 m3 = 3.0 × 1000 = 3.0 × 103 L

(h) 250 L = 250 × 1000 = 2.50 × 105 mL

(i) 1600 mL = 1600 × 10–3 = 1.600 L

(j) 3 × 106 mL = 3 × 106 × 10–3 = 3 × 103 L

(k) 5 × 103 mL = 5 × 103 × 10–6 = 5 × 10–3 m3

(l) 600 mL = 600 × 10–6 = 6.00 × 10–4 m3

Page 430 8. P1 = 78.0 kPa

V1 = ? P2 = 275 kPa V2 = 35.0 L

154



V1= 2 2

1

P VP×

= 0.78

0.35275×

= 123 L

∴V(air) required = 123 L

9. P1 = 560 kPa

V1 = ? P2 = 105 kPa V2 = 100 mL

V1= 2 2

1

P VP×

= 560

100105×

= 18.8 mL

∴V of barrel occupied by the gas = 18.8 mL

10. P1 = ?

V1 = 223 mL P2 = 1.00 atm V2 = 669 mL

P1= 2 2

1

P VV×

= 223

66900.1 ×

= 3.00 atm

∴P(gas) = 3.00 atm

11. P1 = ?

V1 = 1.89 L P2 = 745 mm Hg V2 = 1.32 L

P1 = 2 2

1

P VV×

= 89.1

32.1745×

= 520 mm Hg

∴pressure at that altitude = 520 mm Hg

155



Page 433 13. V1 = 0.156 L

T1 = 12 °C = 12 + 273 = 285 K

V2 = ?

T2 = 24 °C = 24 + 273 = 297 K

V2 = 1

1

TV × T2

= 285156.0 × 297

= 0.16 L

∴ New volume of H2 gas = 0.16 L

14. V1 = 325 mL

T1 = 17 °C = 17 + 273 = 290 K

V2 = 392 mL

T2 = ?

T2 = 1

1

VT × V2

= 325290

× 392

= 350 K

∴ Final temperature of the gas = 350 – 273 = 77 °C

15. V1 = 20.0 L

T1 = 30 °C = 30 + 273 = 303 K

V2 = 10.0 L

T2 = ?

156



T2 = 1

1

VT × V2

= 0.20

303× 10.0

= 152 K

∴ Final temperature of the gas = 152 – 273 = –121 °C

Page 434 (bottom) 18. P1 = 1.20 × 104 kPa

V1 = 10.0 L

T1 = 21 + 273 = 294 K

P2 = 101 kPa V2 = ?

T2 = 43 + 273 = 316 K

V2 = 1 1

1

P VT× × 2

2

TP

so V2 = 294

0.101020.1 4 ×× × 101316

= 1.28 × 103 L

∴ Volume of air provided = 1.28 × 103 L

Page 435

19. P1 = 145 kPa V1 = 4500 L

T1 = 20 + 273 = 293 K

P2 = ? V2 = 60.0 L

T2 = 12 + 273 = 285 K

P2 = 1 1

1

P VT× ×

2

2

VT

so P2 = 293

4500145× × 0.60

285

= 1.06 × 104 kPa

∴ Pressure required = 1.06 × 104 kPa

157



20. P1 = 1.05 atm V1 = 1.55 L

T1 = 25.0 + 273 = 298 K

P2 = 1.02 atm V2 = 1.38 L T2 = ?

T2 = 1

1 1

TP V×

× 2 2

1P V×

so T2 = 55.105.1

298×

× 1

38.102.1 ×

= 258 K = 258 – 273 = –15 °C

∴ Temperature in the freezer = –15 °C

Page 437

21. total pressure = 1.36 atm = 1.36 × 760 mm Hg = 1033.6 mm Hg

∴ partial pressure of the second gas = 1033.6 – 750 = 283.6 mm Hg = 284 mm Hg

22. P1 = 758 – 17.5

= 740.5 mm Hg

V1 = 98 mL

T1 = 20 + 273 = 293 K

P2 = 760 mm Hg V2 = ?

T2 = 0 + 273 = 273 K

V2 = 1 1

1

P VT× × 2

2

TP

so V2 = 293

985.740 × × 760273

= 89 mL

Page 438 23. The total pressure is equal to the pressure exerted by nitrogen when allowed to fill both

flask A and B.

158



P1 = 60 kPa V1 = 100 mL P2 = ?

V2 = 100 + 450 = 550 mL

P2 = 1 1

2

P VV×

so P2 = 550

10060×

= 11 kPa

∴ total pressure = 11 kPa

24. Calculate the partial pressure exerted by each gas.

For nitrogen: P = 1000 mm Hg

For neon: P1 = 500 mm Hg

V1 = 5.0 L P2 = ? V2 = 3.0 L

P2 = 2

11

VVp ×

so P2 = 0.3

0.5500×

= 833 mm Hg

The total pressure = P(nitrogen) + P(neon) = 1000 + 833 = 1833 mm Hg = 1.8 × 103 mm Hg

Page 441

25. (a) n(O2) at STP = m

VV

= 4.22

15

= 0.67 mole

n(O2) at SLC = m

VV

= 5.24

15

= 0.61 mole

159



(b) n(Cl2) at STP = m

VV

= 4.22

25

= 1.1 mole

n(Cl2) at SLC = m

VV

= 5.24

25

= 1.0 mole

26. (a) V(H2) at STP = n(H2) × Vm

= 1.3 × 22.4 = 29 L

V(H2) at SLC = n(H2) × Vm = 1.3 × 24.5 = 32 L

(b) n(CH4) = Mm

= 3.616.05

= 0.22 mole

V(CH4) at STP = n(CH4) × Vm = 0.22 × 22.4 = 5.0 L

V(CH4) at SLC = n(CH4) × Vm = 0.22 × 24.5 = 5.5 L

(c) n(Ar) = Mm

= 0.3539.95

= 0.0088 mole

V(Ar) at STP = n(Ar) × Vm = 0.0088 × 22.4 = 0.20 L

V(Ar) at SLC = n(Ar) × Vm = 0.0088 × 24.5 = 0.21 L

160



27. (a) n(Ne) at SLC = m

VV

= 5.245.16

= 0.673 mole

∴ m(Ne) = n(Ne) × M(Ne) = 0.673 × 20.18 = 13.6 g

(b) V(SO2) = 1050 mL = 1.050 L

n(SO2) at SLC = m

VV

= 5.24

050.1

= 0.042 86 mole

∴ m(SO2) = n(SO2) × M(SO2) = 0.042 86 × 64.06 = 2.745 g

28. n(CO) at STP = m

VV

= 4.22

850

= 37.9 mole

∴ m(CO) = n(CO) × M(CO) = 37.9 × 28.01 = 1063 g = 1.06 kg

29. n(gas) at STP = m

VV

= 4.22

941.0

= 0.0420 mole

∴ M(gas) = (gas)(gas)

mn

= 0420.020.3

= 76.2 g mole–1

161



Page 442 30. General gas equation : PV = nRT

∴use V = Rn Tp

for all parts in question 30.

(a) n = 3.5 mole P = 100 kPa

T = 50 °C = 50 + 273 = 323 K

V(O2) = 100

32331.85.3 ××

= 94 L

(b) n = 12.8 mole

P = 10 atm = 10 × 101.325 = 1013.25 kPa

T = 60 °C = 60 + 273 = 333 K

V(CH4) = 25.1013

33331.88.12 ××

= 35 L

(c) n(Ar) = Mm

= 95.395.6

= 0.16 mole

P = 50 kPa

T = 100 °C = 100 + 273 = 373 K

V(Ar) = 50

37331.816.0 ××

= 10 L

(d) n(CO2) = Mm

= 01.44

56.0

= 0.013 mole

P = 50 atm = 50 × 101.325 = 5066.25 kPa

162



T = 20 °C = 20 + 273 = 293 K

V(CO2) = 25.5066

29331.8013.0 ××

= 0.0062 L

(e) n(He) = Mm

= 00.4103.1 3−×

= 3.25 × 10–4 mole

P = 60 kPa

T = –50 °C = –50 + 273 = 323 K

V(He) = 43.25 10 8.31 32360

−× × ×

= 0.015 L

(f) n(Ne) = AN

N

= 23

21

1002.6105.1××

= 0.0025 mole

P = 40 kPa

T = 200 °C = 200 + 273 = 473 K

V(Ne) = 40

47331.80025.0 ××

= 0.25 L

(The question in the text should read ‘1.5 × 1021 ATOMS of Ne’, not molecules.)

31. n(N2) = Mm

= 02.280.42

= 1.50 mole

P = 200 kPa

T = 77 °C = 77 + 273 = 350 K

163



V(N2) = Rn TP

= 200

35031.850.1 ××

= 21.8 L

Page 443 (top) 32. n = 0.200 mole

V = 5.00 L P = 120 kPa T = ?

T(air) = R

PVn

= 31.8200.0

00.5120××

= 361 K = 361 – 273 °C = 88 °C

33. n = 55 mole

V = 10 L P = ? kPa

T = 7 °C = 7 + 273 K = 280 K

P(H2) = Rn TV

= 10

28031.855 ××

= 1.3 × 104 kPa

Page 443 (bottom) 34. Convert mass of H2O to mole:

n(H2O) = Mm

= 02.1800.5

= 0.277 mole

From the equation:

164



n(C2H2) = 21 × n(H2O)

= 21 × 0.277

= 0.139 mole

Convert mole C2H2 to volume using PV = nRT:

T = 25 °C = 25 + 273 = 298 K

P = 745 mm Hg

= 760745 × 101.325

= 99.3 kPa

R = 8.31 J K–1 mol–1

n = 0.139 mole

V(C2H2) = Rn TP

= 3.99

29831.8139.0 ××

= 3.46 L

∴ volume of acetylene released from 5.00 g of H2O = 3.46 L

Page 444 (top) 35. Convert mass of glucose to mole:

n(C6H12O6) = Mm

= 18.1805.1

= 0.0083 mole

From the equation:

n(CO2) = 16 × n(C6H12O6)

= 16 × 0.0083

= 0.050 mole

Convert mole CO2 to volume using PV = nRT:

T = 28 °C = 28 + 273 = 301 K

165



P = 1.3 atm = 1.3 × 101.325 = 131.7 kPa

R = 8.31 J K–1 mol–1

n = 0.050 mole

V(CO2) = Rn TP

= 7.131

30131.8050.0 ××

= 0.95 L

∴ volume of carbon dioxide released from 1.5 g of glucose = 0.95 L

Page 444 (bottom) 36. Convert volume of O2 to mole using PV = nRT:

V = 14.5 L

P = 1.00 atm = 1.00 × 101.325 = 101 kPa

T = 25.0 °C = 25.0 + 273 = 298 K

n(O2) = RPV

T

= 29831.8

5.14101××

= 0.591 mole

From the equation:

n(KNO2) = 12 × n(O2)

= 12 × 0.591

= 1.18 mole

m(KNO2) = n × M(KNO2) = 1.18 × 85.11 = 101 g

∴mass KNO2 formed when 14.5 L O2 is formed = 101 g

Page 445 (top) 37. Convert volume of H2 to mole using PV = nRT:

V = 5.0 L P = 1.2 kPa

166



T = 26.0 °C = 26.0 + 273 = 299 K

n(H2) = RPV

T

= 29931.8

0.52.1××

= 0.0024 mole

From the equation:

n(Mg) = 11 × n(H2)

= 11 × 0.0024

= 0.0024 mole

m(Mg) = n × M(Mg) = 0.0024 × 24.31 = 0.059 g

∴mass Mg formed when 5.0 L H2 is formed = 0.059 g

Page 445 (bottom)

38. CaCO3(s) → CaO(s) + CO2(g) convert mass CaCO3 to mole:

n(CaCO3) = Mm

= 09.100

152

= 1.52 mole

From the equation:

n(CO2) = 11 × n(CaCO3)

= 1.52 mole

V(CO2) = n × Vm = 1.52 × 22.4 = 34.0 L

∴ Volume of carbon dioxide gas produced when 152 g CaCO3 decomposes = 34.0 L

39. 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(g)

convert mass C8H18 to mole:

167



n(C8H18) = Mm

= 26.114

125

= 1.09 mole

From the equation:

n(O2) = 225 × n(C8H18)

= 225 × 1.09

= 13.7 mole

V(O2) = n × Vm = 13.7 × 22.4 = 306 L

∴ Volume of oxygen gas needed to burn 125 g C8H18 = 306 L

Page 447 40. Since temperature and pressure are the same throughout, the volume ratio is equal to

the mole ratio.

(a) V(O2) = 12 × V(CH4)

= 12 × 25

= 50 mL

(b) V(CO2) = 11 × V(CH4)

= 11 × 25

= 25 mL

(c) V(H2O) = 12 × V(CH4)

= 12 × 25

= 50 mL

41. Since temperature and pressure are the same throughout, the volume ratio is equal to

the mole ratio.

(a) V(O2) = 12 × V(N2)

= 12 × 20

= 40 mL

168



(b) Initial volume of reactants = V(N2) + V(O2) = 20 mL + 40 mL = 60 mL

(c) Final volume of products = V(NO2)

= 12 × V(N2)

= 12 × 20

= 40 mL

(d) 60 mL reactants 40 mL product ∴ an overall decrease of 20 mL in the volume of gas on completion of the reaction.

Multiple choice questions 1. B 2. C 3. A

4. P1 = 100 kPa V1 = 6 L P2 = 150 kPa V2 = ?

V2 = 2

11

pVp ×

= 150

6100×

= 4 L

Answer: B

5. D

6. P is constant V1 = 2 L T1 = 200 K V2 = ? T2 = 400 K

V2 = 1

21

TTV ×

= 200

4002×

= 4 L

Answer: D

7. P1 = 700 mm Hg V1 = 200 mL

169



T1 = 27 + 273 = 300 K

P2 = 800 mm Hg V2 = 210 mL T2 = ?

T2 = 11

1

VpT×

× 1

22 Vp ×

so T2 = 200700

300×

× 1

210800×

= 360 K = 360 – 273 = 87 °C

Answer: B

8. C

9. P(O2) = P(atmosphere) – P(water vapour) = 765 – 21 = 744 mm Hg

Answer: B

10. Calculate the partial pressure exerted by each gas. For nitrogen: P = 2.0 atm

For hydrogen: P1 = 6.0 atm V1 = 1.0 L P2 = ? V2 = 2.0 L

P2 = 2

11

VVp ×

so P2 = 0.2

0.10.6 ×

= 3.0 atm

The total pressure = P(nitrogen) + P(hydrogen) = 2.0 + 3.0 = 5.0 atm

Answer: C

11. partial pressure of hydrogen = )300100(

300+

× 101.3

= 76 kPa

Answer: B

170



Review questions

7. (a) 1.34 atm = 0.134.1 × 101.325 Pa

= 136 kPa

1.34 atm = 0.134.1 × 760 mm Hg

= 1.02 × 103 mm Hg

(b) 365 mm Hg = 760365 × 101 325 Pa

= 48 663 Pa = 4.87 × 104 Pa

365 mm Hg = 760365 × 1 atm

= 0.480 atm

(c) 102 576 Pa = 102576101325

× 1 atm

= 1.012 35 atm

102 576 Pa = 102576101325

× 760 mm Hg

= 769.383 mm Hg

8. (a) 100 °C = 100 + 273 = 373 K

(b) –20 °C = –20 + 273 = 253 K

(c) –200 °C = –200 + 273 = 73 K

(d) 345 °C = 345 + 273 = 618 K

9. (a) 300 K = 300 – 273 = 27 °C

(b) 427 K = 427 – 273 = 154 °C

(c) 173 K = 173 – 273 = –100 °C

(d) 392 K = 392 – 273 = 119 °C

(e) 73 K = 73 – 273 = –200 °C

171



10. (a) 125 L = 125 × 10–3 = 0.125 m3

(b) 125 L = 125 × 1000 = 1.25 × 105 mL

(c) 300 mL = 300 × 10–3 = 0.300 L

(d) 2.6 mL = 2.6 × 10–3 L

(e) 2 m3 = 2 × 1000 = 2 × 103 L

(f) 3 × 103 mL = 3 × 103 × 10–3 = 3 L

14. P1 = 360 kPa V1 = ? P2 = 90.0 kPa V2 = 13.0 L

V1 = 1

22

pVp ×

= 360

0.130.90 ×

= 3.25 L

∴V(Neon) = 123 L

15. P1 = 150 kPa V1 = ? P2 = 5.0 ×104 kPa V2 = 1.00 L

V1 = 1

22

pVp ×

= 150

00.1100.5 4 ××

= 3.3 × 102 L

∴Volume of the balloon = 3.3 × 102 L

16. V1 = 700 mL

T1 = 27 °C + 273 = 300 K

V2 = 14.0 mL T2 = ?

172



(a) T2 = 1

1

VT × V2

= 700300

× 14.0

= 6.0 K

∴ Temperature needed = 6.0 – 273 = –267 °C

(b) V1 = 700 mL

T1 = 27 °C + 273 = 300 K

V2 = 420.0 mL T2 = ?

T2 = 1

1

VT × V2

= 700300

× 420.0

= 180 K

∴ Temperature needed = 180 – 273 = –93 °C

17. P1 = 101.3 kPa V1 = 10 m3

T1 = 21 + 273 = 294 K

P2 = 2 × 101.3 = 202.6 kPa

V2 = ?

T2 = 47 + 273 = 320 K

V2 = 1

11

TVp ×

× 2

2

pT

so V2 = 294

103.101 × × 6.202

320

= 5.4 m3

∴ Volume of system = 5.4 m3

18. P1 = 2.0 × 105 Pa V1 = 15 L

T1 = 15 + 273 = 288 K

173



P2 = ? V2 = 10 L

T2 = 40 + 273 = 313 K

P2 = 1

11

TVp × ×

2

2

VT

so P2 = 288

15100.2 5 ×× × 10313

= 3.3 × 105 Pa

∴ Pressure required = 3.3 × 105 Pa

19. P1 = 4500 kPa V1 = 15 L

T1 = 15 + 273 = 288 K

P2 = 100 kPa V2 = ?

T2 = 25 + 273 = 298 K

V2 = 1

11

TVp ×

× 2

2

pT

so V2 = 288

154500× × 100298

= 698 L = 7.0 × 102 L

∴ Volume of oxygen gas = 7.0 × 102 L

20. total pressure = (760200 × 101.325) + (1.52 × 101.325) + 102.4

= 283 kPa

21. P1 = 749 – (760 ×325.1018.1 )

= 749 – 14 = 735 mm Hg

V1 = 325 mL

T1 = 15 + 273 = 288 K

174



P2 = 101 150 Pa

= 101150101325

× 760

= 758.7 mm Hg

V2 = ?

T2 = 27 + 273 = 300 K

V2 = 1 1

1

P VT× × 2

2

TP

so V2 = 288

325735× × 7.758

300

= 328 mL

22. P1 = 790 – 24 = 766 mm Hg

V1 =300 mL

T1 = 25 + 273 = 298 K

P2 = 760 mm Hg V2 = ?

T2 = 0 + 273 = 273 K

V2 = 1 1

1

P VT× × 2

2

TP

so V2 = 298

300766× × 760273

= 277 mL

23. 2C3H8(g) + 7O2(g) → 6CO(g) + 8H2O(g) As temperature and pressure are the same throughout, the volume ratio equals the mole ratio.

(a) V(CO) = 26 × V(C3H8)

= 26 × 5.35

= 16.05 L = 16.1 L

(b) V(O2) = 27 × V(C3H8)

= 27 × 5.35

= 18.7 L

175



24. The total pressure is equal to the pressure exerted by neon when allowed to fill both flasks attached by valve A.

P1 = 1 atm V1 = 2 L P2 = ?

V2 = 2 + 1 = 3 L

P2 = 2

11

VVp ×

so P2 = 3

21×

= 0.67 atm

∴ total pressure = 0.7 atm

(b) The total pressure is equal to the pressure exerted by neon when allowed to fill all three flasks.

P1 = 1 atm V1 = 2 L P2 = ?

V2 = 2 + 1 + 0.5 = 3.5 L

P2 = 1 1

2

P VV×

so P2 = 5.321×

= 0.57 atm


25. The total pressure is equal to the pressure exerted by oxygen when allowed to fill both

flasks.

P1 = 1.0 atm V1 = 5.0 L P2 = ?

V2 = 15 + 5.0 = 20 L

P2 = 1 1

2

P VV×

so P2 = 20

0.50.1 ×

= 0.25 atm


176



26. (a) n(O2) at STP = m

VV

= 4.22

5.1

= 0.067 mole

n(O2) at SLC = m

VV

= 5.24

5.1

= 0.061 mole

(b) n(H2) at STP = m

VV

= 4.22

56.2

= 0.114 mole

n(H2) at SLC = m

VV

= 5.24

56.2

= 0.104 mole

(c) V(N2) = 250 mL = 0.250 L

n(N2) at STP = m

VV

= 4.22

250.0

= 0.0112 mole

n(N2) at SLC = m

VV

= 5.24

250.0

= 0.0102 mole

27. (a) V(H2) at STP = n(H2) × Vm

= 1.53 × 22.4 = 34.3 L

V(H2) at SLC = n(H2) × Vm = 1.53 × 24.5 = 37.5 L

177



(b) n(CH4) = Mm

= 05.166.13

= 0.847 mole

V(CH4) at STP = n(CH4) × Vm = 0.847 × 22.4 = 19.0 L

V(CH4) at SLC = n(CH4) × Vm = 0.847 × 24.5 = 20.8 L

(c) n(N2) = AN

N

= 23

30

1002.6105.2××

= 4.2 × 106 mole

V(N2) at STP = n(N2) × Vm = 4.2 × 106 × 22.4 = 9.3 × 107 L

V(N2) at SLC = n(N2) × Vm = 4.2 × 106 × 24.5 = 1.0 × 108 L

28. (a) V(O2) = 150 mL

= 0.150 L

n(O2) at STP = m

VV

= 4.22

150.0

= 0.006 70 mole

∴ m(O2) = n(O2) × M(O2) = 0.006 70 × 32.00 = 0.214 g

(b) n(CO2) at STP = m

VV

= 4.225.4

= 0.20 mole

∴ m(CO2) = n(CO2) × M(CO2) = 0.20 × 44.01 = 8.8 g

178



29. (a) n(N2) at STP = m

VV

= 4.220.26

= 1.16 mole

(b) N(N2) = n(N2) × NA = 1.16 × 6.02 × 1023 = 6.99 × 1023 molecules

30. P1 = 101.325 kPa

V1 = 11.2 L T1 = 273 K

P2 = 2.5 atm = 2.5 × 101.325 = 253.313 kPa

V2 = ?

T2 = 127 °C + 273 = 400 K

V2 = 1 1

1

P VT× × 2

2

TP

so V2 = 273

2.11325.101 × × 313.253

400

= 6.6 L

∴ Volume of gas = 6.6 L

31. V = 2.60 L

T = 20.0 °C = 20.0 + 273 = 293 K

P = 102 kPa

m = 506.31 – 500.00 = 6.31 g

n(butane) = RpV

T

= 29331.860.2102

××

= 0.109 mole

179



∴ M(butane) = nm

= 109.031.6

= 57.9 g mole–1

32. m(CO2) = 4.4 kg

= 4400 g

∴ n(CO2) = Mm

= 440044.01

= 1.0 × 102 mole

At STP: V(CO2) = n × 22.4 = 1.0 × 102 × 22.4 = 2.2 × 103 L

33. P = 150 kPa

V = 4.155 × 10–3 m3 = 4.155 L

T = ?

n = 0.25 mole

T = R

PVn

= 31.825.0

155.4150××

= 300 K = 300 – 273 = 27 °C

34. P = 125 mm Hg

= 760125 × 101.325 kPa

= 16.7 kPa

V = 16.64 L

T = –73 °C = –73 + 273 = 200 K

n = ? mole

180



n(gas) = RPV

T

= 20031.8

64.167.16××

= 0.17 mole

m(gas) = n × M = 0.17 × 30.0 = 5.0 g

∴ mass of gas present in container = 5.0 g

35. n = 3.0 mole V = 20 L P = ? kPa

T = 27 °C = 27 + 273 K = 300 K

P(NH3) = Rn TV

= 20

30031.80.3 ××

= 373.95 kPa = 3.7 × 102 kPa

36. P = 101.3 kPa

V = 1.0 L

T = 127 °C = 127 + 273 = 400 K

n = ? mole

n(gas) = RPV

T

= 40031.8

0.13.101××

= 0.030 mole

37. (a) P = 9.0 × 105 Pa

P = 9.0 × 102 kPa

V = 120 mL = 0.120 L

T = 27 °C = 27 + 273 = 300 K

n = ? mole

181



n(gas) = RPV

T

= 30031.8

120.0100.9 2

×××

= 0.043 mole

∴ N = n × NA = 0.043 × 6.02 × 1023 = 2.6 × 1022

particles

(b) n = 0.043 mole

V = 200 mL = 0.200 L

P = 6.0 × 105 Pa = 6.0 × 102 kPa

T = ?

T = R

PVn

= 31.8043.0200.0100.6 2

×××

= 333 K = 333 – 273 °C = 60 °C

38. P = 770 mm Hg

= 760770 × 101.325

= 102 kPa

V = 200 mL = 0.200 L

T = 17.0 °C = 17.0 + 273 = 290 K

n = ? mole

n(gas) = RPV

T

= 29031.8200.0102

××

= 0.008 52 mole

m(gas) = 85.084 – 84.845 = 0.239 g

182



∴M(gas) = nm

= 0.2390.008 52

= 28.1 g mole–1

39. P1 = 1.5 atm

V1 = 1.5 L

T1 = 25 °C = 25 + 273 = 298 K

P2 = ? V2 = 1.5 L (glass bottle of fixed volume)

T2 = 100 °C = 100 + 273 = 373 K

P2 = 1 2

1

P TT×

= 298

3735.1 ×

= 1.88 atm

∴ pressure increases by 1.88 – 1.5 atm = 0.38 atm

40. P = 1.0 × 104 Pa

= 1.0 × 101 kPa

V(O2) = 20% of 400 mL of air

= 10020 × 400

= 80 mL = 0.080 L

T = 25 °C = 25 + 273 = 298 K

n = ? mole

n(O2) = RPV

T

= 29831.8

080.0100.1 1

×××

= 3.2 × 10–4 mole

∴ N(O2) = n × NA = 3.2 × 10–4 × 6.02 × 1023 = 1.9 × 1020

molecules

183



41. (a) Convert mass of CaCO3 to mole:

n(CaCO3) = Mm

= 09.1000.35

= 0.350 mole

From the equation:

n(CO2) = 11 × n(CaCO3)

= 11 × 0.350

= 0.350 mole


T = 10.0 °C = 10.0 + 273 = 283 K

P = 100 kPa R = 8.31 J K–1 mol–1 n = 0.216 mole

V(CO2) = Rn TP

= 100

28331.8350.0 ××

= 8.22 L

∴ volume of carbon dioxide released when 35.0 g of CaCO3 is produced = 8.22 L

(b) Convert mass of CaCO3 to mole:

n(CaCO3) = Mm

= 09.1002.15

= 0.152 mole

From the equation:

n(CO2) = 11 × n(CaCO3)

= 11 × 0.152

= 0.152 mole

Convert mole CO2 to volume using V = n × Vm:

V(CO2) = 0.152 × 22.4 = 3.40 L

184



∴ volume of carbon dioxide released when 15.2 g of CaCO3 is produced = 3.40 L

42. Convert volume of O2 to mole using n = m

VV

:

n(O2) = 4.22

80.5

= 0.259 mole

From the equation:

n(Mg) = 12 × n(O2)

= 12 × 0.259

= 0.518 mole

m(Mg) = n × M(Mg) = 0.518 × 24.31 = 12.6 g

∴mass Mg combining with 5.80 L O2 = 12.6 g

43. (a) Convert mass of CO(NH2)2 to mole:

n(CO(NH2)2) = Mm

= 07.60

600

= 9.99 mole

From the equation:

n(NH3) = 12 × n(CO(NH2)2)

= 12 × 9.99

= 20.0 mole

Convert mole NH3 to volume using PV = nRT:

T = 37 °C = 37 + 273 = 310 K


185



V(NH3) = Rn TP

= 105

31031.80.20 ××

= 490 L

∴ volume of ammonia gas required to produce 600 g of urea = 490 L

(b) from (a) n(CO(NH2)2) = Mm

= 07.60600

= 9.99 mole

From the equation:

n(CO2) = 11

× n(CO(NH2)2)

= 11

× 9.99

= 9.99 mole


T = 37 °C = 37 + 273 = 310 K


V(CO2) = Rn TP

= 9.99 8.31 310105

× ×

= 245 L ∴ volume of carbon dioxide gas required to produce 600 g of urea = 245 L

Alternatively: Since temperature and pressure are the same the volume ratio is equal to the mole ratio. ∴ 2 mole NH3 : 1 mole CO2 = 490 L : 245 L


the mole ratio.

(a) V(O2) = 12 × V(CH4)

= 12 × 12.5

= 25.0 L

186



(b) V(air) = 25.0 × 20

100

= 125 L 45. 2C3H8(g) + 7O2(g) → 6CO(g) + 8H2O(g)

Since temperature and pressure are the same throughout, the volume ratio is equal to the mole ratio.

V(CO) = 26 × V(C3H8)

= 26 × 4.25

= 12.75 L = 12.8 L


the mole ratio.

The oxygen gas is in excess, therefore hydrogen gas is the limiting reagent.

V(H2O) = 22 × V(H2)

= 22 × 4.0

= 4.0 L

V(O2) reacting = 21 × V(H2)

= 21 × 4.0

= 2.0 L

∴ V(O2) unreacted = 10 – 2.0 = 8.0 L


the mole ratio.

(a) N2(g) + 3H2(g) → 2NH3(g)

(b) V(N2) = 31 × V(H2)

= 31 × 45

= 15 m3

(c) V(NH3) = 32 × V(H2)

= 32 × 45

= 30 m3

187





Extended response questions 1. Convert mass nitroglycerine to mole:

m(C3N3H5O9) = 1.00 kg = 1000 g

n(C3N3H5O9) = Mm

= 11.2271000

= 4.40 mole

n(gaseous products) = n (N2) + n(CO2) + n(H2O) + n(O2) From the equation:

n(gaseous products) = 294

× n(C3N3H5O9)

= 294

× 4.40

= 31.9 mole Convert mole of nitroglycerine to volume using PV = nRT: P = 300 kPa R = 8.31 T = 250 °C

= 250 + 273 = 523 K

n = 31.9 mole

V(C3N3H5O9) = Rn TP

= 31.9 8.31 523300

× ×

= 462 L ∴ total volume of gaseous products released when 1.00 kg of nitroglyerine explodes = 462 L


the mole ratio.

V(NO2) = 22 × V(NO)

= 22 × 10

= 10 mL

188


Chapter 19

Page 458 1. Green chemistry is the design of chemical processes and products in order to eliminate

the use and generation of hazardous substances.

3. The total RAM of the products is 75% of the total RAM of the reactants. This figure does not take equilibrium considerations into account.

Page 462 4. (a) Advantages: high diffusion rate, better flavours and aromas, no organic solvents,

easy recovery of product. Disadvantages: cost of producing scCO2.

5. (a) Three forms of the substance exist at the one time: solid, liquid and gas.

(b) evaporation of scCO2

6. (a) A solid is produced.

(b) The CO2 could change from a gas to a liquid.

(c) Critical point is achieved but at a higher pressure.

Page 463 7. (a) 1.96 g

(b) 15 mL

(c) Attraction between molecules causes non-ideal gas properties.

Multiple choice questions 1. B 2. B

Review questions 3. (a) Biohazard: a biological material that poses a threat to human health; do not

contact.

(b) Flammable: able to combust readily; keep away from ignition source.

(c) Corrosive: deterioration of intrinsic properties of a material; do not contact, store safely.

7. Ionic liquids contain cations and anions that are unable to form a solid lattice at ambient temperatures. They can dissolve a large range of polar substances but are non-flammable, have good thermal stability and are easy to manufacture.

8. (a) Advantages: non-toxic, biodegradable, non-explosive, efficient. Disadvantages: energy consumption and release of CO2 in manufacture, produces more NOx.

189



(b) Advantages: biodegradable. Disadvantages: water-soluble, expensive compared to oil-based plastics.



190

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