1. 2 The Mole Concept 2.1 The Mole 2.2 Molar Volume and Avogadro’s Law
2.3 Ideal Gas Equation
2.4 Determination of Molar Mass
2.5 Dalton’s Law of Partial Pressures

2. 2.1
The Mole

3. Mole An undercover agent, a counterspy, a double agent
A burrowing mammal with fossorial forefeet
A small congenital pigmented spot on the skin
A breakwater
P. 3 / 66

4. A mole is the number of atoms in exactly 12.00 g ofpure isotope.
This number, known as the Avogadro’s constant,
can be determined by mass spectrometry.
B is the magnetic field strength.
V is the accelerating potential.
k is a constant of the instrument.

5. At fixed e, k, B and V
m can be determined.

6. Q.5 = 6.022  1023 mol1 Mass of one mole of atoms = 12.00 g mol1
= Mass of an Avogadro’s number of atoms
= Avogadro’s number   g
Avogadro’s number
=  1023 mol1

7. Particles in Chemistry
2.1 The mole (SB p.18)
What is “mole”?
Item
Unit used to count
No. of items per unit
Shoes
pairs 2
Eggs
dozens 12
Paper
reams
500
Particles in Chemistry
moles
6.022  1023
for counting common objects
for counting particles like atoms, ions, molecules

8. 千進制 1 mole ~ 602.2 sextillions ~602,200,000,000,000,000,000,000 mol1
quadrillion
billion
~602,200,000,000,000,000,000,000 mol1
million
trillion
quintillion
千進制
1 mole ~
602.2 sextillions
P. 8 / 66

9. The fastest supercomputer can count 1.7591015 atoms per second.
Calculate the time taken for the superconductor to count mole of carbon-12 atoms.
 years
P. 9 / 66

10. We can count the number of coins by weighing if the mass of one coin is known.
Similarly, we can count the number of 12C by weighing if the mass of one 12C is known.
P. 10 / 66

11. Molar mass is the mass, in grams, of 1 mole of a substance

12. Relative isotopic mass
Q.6
Relative isotopic mass
Relative intensity
12.000
100.00
13.003
1.12
Relative atomic mass

13. Relative isotopic mass
Q.6
Relative isotopic mass
Relative intensity
12.000
100.00
13.003
1.12
Molar mass of carbon
= g mol1

14. Relative isotopic mass
Q.6
Relative isotopic mass
Relative intensity
12.000
100.00
13.003
1.12
Relative isotopic mass is not exactly
equal to mass number of the isotope

15. Q.7 Number of moles of a substance number of particles
mass
molar mass =
Q.7
Number of moles of oxygen atoms =
number of oxygen atoms
6.022  1023 mol1
2 g
16 g mol1

16. Q.7 Number of moles of oxygen atoms = number of oxygen atoms
16. g mol1
Number of oxygen atoms =
Number of atoms
=  1019

17. Molar mass is the same as the relative atomic mass in grams.
The mole (SB p.20)
Molar mass is the same as the relative atomic mass in grams.
Molar mass is the same as the relative molecular mass in grams.
Molar mass is the same as the formula mass in grams.
Example 2-1A
Example 2-1B
Example 2-1C
Check Point 2-1
Example 2-1D
Example 2-1E

18. Molar Volume and Avogadro’s Law
2.2
Molar Volume and Avogadro’s Law

19. What is molar volume of gases?
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
Volume occupied by one mole of molecules of a gas.

20. What is molar volume of gases?
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
Depends on T & P
Two sets of conditions

21. What is molar volume of gases?
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
at 298 K & 1 atm
(Room temp & pressure / R.T.P.)

22. What is molar volume of gases?
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
22.4 dm3
at 273 K & 1 atm
(Standard temp & pressure / S.T.P.)

23. Gas Molar mass/g Molar volume at R.T.P./dm3 Molar volume at S.T.P./dm3
Gas
Molar mass/g
Molar volume at R.T.P./dm3
Molar volume at S.T.P./dm3 O2 32
24.0
22.397 N2 28
22.402 H2 2
24.1
22.433 He 4
22.434
CO2 44
24.3
22.260 17
22.079
NH3
~ 24
~ 22.4
Not constant

24. Avogadro’s Law At fixed T & P, V  n If n = 1, V = molar volume
Molar volume and Avogadro’s law (SB p.24)
Avogadro’s Law
Equal volumes of ALL gases at the same temperature and pressure contain the same number of moles of molecules.
At fixed T & P, V  n
If n = 1, V = molar volume

25. Avogadro’s Law V  n V = Vm  n
Molar volume and Avogadro’s law (SB p.24)
Avogadro’s Law
V  n
V = Vm  n

26. Interconversions involving number of moles
Molar volume and Avogadro’s law (SB p.24)
Interconversions involving number of moles
Example 2-2B
Example 2-2C
Example 2-2A
Check Point 2-2
Example 2-2D

27. 2.3
Ideal Gas Equation

28. 2.3 Ideal gas equation (SB p.27)
Boyle’s law
At fixed n and T,
PV = constant or
n = number of moles of gas molecules

29. Schematic diagrams explaining Boyle’s law
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Boyle’s law

30. 1/P

31. 2.3 Ideal gas equation (SB p.28)
A graph of volume against the reciprocal of pressure for a gas at constant temperature

32. 2.3 Ideal gas equation (SB p.28)
Charles’ law
At fixed n and P,
T is the absolute temperature in Kelvin, K

33. Schematic diagrams explaining Charles’ law
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Charles’ law

34. A graph of volume against temperature for a gas at constant pressure
2.3 Ideal gas equation (SB p.28)
Volume oC
0oC
Temperature / oC
A graph of volume against temperature for a gas at constant pressure

35. 2.3 Ideal gas equation (SB p.28)
/ K
A graph of volume against absolute temperature for a gas at constant pressure

36. 2.3 Ideal gas equation (SB p.27)
Avogadro’s law
Boyle’s law
Charles’ law
R is the same for all gases
R is known as the universal gas constant
PV = nRT
Ideal gas equation

37. 2.3 Ideal gas equation (SB p.29)
Relationship between the ideal gas equation and the individual gas laws

38. At fixed n,
a constant
= a constant
Ideal gas behaviour is assumed in all gas laws

39. 2.3 Ideal gas equation (SB p.27)
Gas laws
Avogadro’s law
Boyle’s law
Charles’ law
PV = nRT
Ideal gas equation

40. What is the difference between a theory and a law?
2.2 Molar volume and Avogadro’s law (SB p.24)
Gas laws vs kinetic theory of gases
What is the difference between a theory and a law?
A law describes what happens under a given set of circumstances.
A theory attempts to explain why that behaviour occurs.

41. Ideal gas behaviour
Four assumptions as stated in kinetic theory of gases
Gas particles are in a state of constant and random motion in all directions, undergoing frequent collisions with one another and with walls of the container.
Gas particles are treated as point masses, i.e. they do not occupy volume.
Volume of a gas = capacity of the vessel

42. Ideal gas behaviour 3. There is no interaction among gas particles.
4. Collisions between gas particles are perfectly elastic, i.e. kinetic energy is conserved.

43. The ideal gas equation is obeyed by
real gases only at
(i) low pressure
(ii) high temperature
(less deviation from 24 dm3 at R.T.P.)

44. At low pressure, gas particles are so far apart that
(1) any interaction among them becomes negligible (assumption 3)
(2) the volume occupied by the gaseous molecules becomes negligible when compared with that of the container (assumption 2)

45. At high temperature,
gaseous molecules possess sufficient energy to overcome intermolecular interactions readily. (assumption 3)

46. 2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to?
T2 = 586 K

47. Check Point 2-3 Or n = 0.0276 mol n = 0.0276 mol
2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(c) A balloon is filled with helium at 25 oC. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How many moles of helium have been introduced into the balloon?
n = mol Or
n = mol

48. 2.3 Ideal gas equation (SB p.31)
Check Point 2-3
25.8 cm3 sample of a gas has a pressure of 690 mmHg and a temperature of 17 oC. What is the volume of the gas if the pressure is changed to 1.85 atm and the temperature to 345 K?
(1 atm = 760 mmHg)
V2 = 15.1 cm3

49. 2.3 Ideal gas equation (SB p.29)
Calculate the universal gas constant at S.T.P.
For one mole of an ideal gas at S.T.P.,
P = 1 atm or 101,325 Nm-2 (Pa)
V = 22.4 dm3 or m3
n = 1 mol
T = 273K

50. 2.3 Ideal gas equation (SB p.29)
atm dm3 K1 mol1
Or,
= Nm K1 mol1
= J K1 mol1

51. Q.9
PV = nRT
m = mass of the gas
M = molar mass of the gas

52. Determination of Molar Mass
2.4
Determination of Molar Mass

53. Determination of Molar Mass
1. Mass Spectrometry
2. Density Measurement

54. 2.4 Determination of molar mass (SB p.32)
(Mass of syringe + liquid) before injection (m1)
= g

55. 2.4 Determination of molar mass (SB p.32)
(Mass of syringe + liquid) after injection (m2)
= g

56. 2.4 Determination of molar mass (SB p.32)
Mass of liquid injected (m1 – m2)
= g – g = g

57. 2.4 Determination of molar mass (SB p.32)
Volume of air in syringe before injection (V1)
= 10.5 cm3

58. 2.4 Determination of molar mass (SB p.32)
Volume of air + vapour in syringe after injection (V2)
= cm3

59. 2.4 Determination of molar mass (SB p.32)
Volume of vapour in syringe (V2 – V1)
= cm cm3 = cm3

60. 2.4 Determination of molar mass (SB p.32)
Once m and V of the vapour are known,
density( )can be determined

61. 2.4 Determination of molar mass (SB p.32)
Temperature = = 338 K

62. 2.4 Determination of molar mass (SB p.32)
1 atm
Pressure = 1 atm

63. 2.4 Determination of molar mass (SB p.32)
Q.10
Molar mass
= 58.0 g mol1
Relative molecular mass = 58.0

64. 2.5 Dalton’s law of partial pressures (SB p.35)
Unit conversions : -
R = J K1mol1 = atm dm3 K1mol1
1m3 = 103 dm3 = 106 cm3
1 atm = 760 mmHg = Nm2 = Pa

65. 2.4 Determination of molar mass (SB p.34)
Check Point 2-4
0.204 g of phosphorus vapour occupies a volume of cm3 at 327 oC and 1.00 atm. Determine the molar mass of phosphorus.
= 124 g mol-1

66. 2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(b) A sample of gas has a mass of 12.0 g and occupies a volume of 4.16 dm3 measured at 97 oC and 1.62 atm. Calculate the molar mass of the gas.
(1 atm = Nm-2; ideal gas constant = J K-1 mol-1) Nm
= 54.1 g mol-1

67. 2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(c) A sample of g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium.
= 1.52103 mol

68. Check Point 2-4 Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g) 1.52103 mol
2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(c) A sample of g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium.
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
1.52103 mol
1.52103 mol

69. Dalton’s Law of Partial Pressures
2.5
Dalton’s Law of Partial Pressures

70. At fixed T & n, PV = constant (15 atm)(5 dm3) = (PA)(15 dm3)
Experiment 1
empty
Tap opened
Gas A
At fixed T & n,
PV = constant
(15 atm)(5 dm3) = (PA)(15 dm3)
PA = 5 atm

71. At fixed T & n PV = constant (12 atm)(10 dm3) = (PB)(15 dm3)
Experiment 2
Tap opened
Gas B
12 atm
empty
Gas B
At fixed T & n
PV = constant
(12 atm)(10 dm3) = (PB)(15 dm3)
PB = 8 atm

72. The total pressure PT = 13 atm = 5 atm + 8 atm = PA + PB
Experiment 3
Gas B
Tap opened
Gas A + Gas B
12 atm
The total pressure PT = 13 atm
= 5 atm + 8 atm
= PA + PB
Partial pressures of gases A & B

73. Gas B
Tap opened
12 atm
PA = 5 atm
PB = 8 atm
Partial pressure of a constituent gas in a mixture is the pressure that the gas would exert if it were present alone under the same conditions

74. 2.5 Dalton’s law of partial pressures (SB p.35)
In a mixture of gases which do not react chemically, the total pressure of the mixture is the sum of the partial pressures of the component gases (the sum of the pressure that each gas would exert if it were present alone under the same conditions).
PT = PA PB PC

75. 2.5 Dalton’s law of partial pressures (SB p.35)
Derivation from ideal gas equation
Consider a mixture of gases A, B and C at fixed T & V.
nA, nB and nC are the numbers of moles of each gas.
The total number of moles of gases in the mixture
nT = nA + nB + nC
Multiply by the constant RT/V
nT(RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V)
If gases A, B and C obey ideal gas behaviour
Ptotal = PA + PB + PC

76. Partial Pressures and Mole Fractions
Consider a mixture of two gases A and B in a container of capacity V at temperature T
Mole fractions of A & B
PA = PTXA
PB = PTXB

77. Consider a mixture of gases A, B, C, D,…
PA = PTXA
PB = PTXB
PC = PTXC
PD = PTXD …

78. Q.11 At fixed T & n, PV = constant For N2, P1V1 = P2V2
(0.20 Pa)(1.0 dm3) = P2(4.0 dm3)  P2 = 0.05 Pa
For O2, P1’V1’ = P2’V2’
(0.40 Pa)(2.0 dm3) = P2’(4.0 dm3)  P2’ = 0.2 Pa
By Dalton’s law of partial pressures
= 0.05 Pa Pa = 0.25 Pa

79. Q.12 At 200oC At –40oC, only N2 exists as a gas in the mixture
For a given amount of N2 at fixed V, P  T
At 200oC

80. At fixed T & V,

81. Q.13(a)
At fixed P & T, V  n
= 20%
= 1.96  104 Nm2

82. Q.13(a)
= 5.39  104 Nm2

83. Q.13(a)
= 2.45  104 Nm2

84. remain unchanged, but PT changes Q.13(b)
NH3 is removed
= 5.39  104 Nm  104 Nm2
= 7.84  104 Nm2
Note :
remain unchanged,
but PT changes

85. 2.5 Dalton’s law of partial pressures (SB p.39)
Check Point 2-5
The valve between a 6 dm3 vessel containing gas A at a pressure of 7 atm and an 8 dm3 vessel containing gas B at a pressure of 9 atm is opened. Assuming that the temperature of the system remains constant and there is no reaction between the gases, what is the final pressure of the system?

86. 2.5 Dalton’s law of partial pressures (SB p.39)
Check Point 2-5
Total pressure = PA + PB = ( ) atm = 8.1 atm

87. 2.5 Dalton’s law of partial pressures (SB p.39)
2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are introduced into a 15 dm3 vessel at 100 oC.
(i) What are the mole fractions of helium, nitrogen and argon in the system?
Total no. of moles = ( ) mol = 0.71 mol

88. 2.5 Dalton’s law of partial pressures (SB p.39)
2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are introduced into a 15 dm3 vessel at 100 oC.
(ii) Calculate the total pressure of the system, and hence the partial pressures of helium, nitrogen and argon.

89. The END

90. 2.1 The mole (SB p.20)
Back
Example 2-1A
What is the mass of 0.2 mol of calcium carbonate?
Answer
The chemical formula of calcium carbonate is CaCO3.
Molar mass of calcium carbonate = (  3) g mol-1
= g mol-1
Mass of calcium carbonate = Number of moles  Molar mass
= 0.2 mol  g mol-1
= g

91. Example 2-1B Answer Back投影片 14
2.1 The mole (SB p.21)
Back投影片 14
Example 2-1B
Calculate the number of gold atoms in a 20 g gold pendant.
Answer
Molar mass of gold = g mol-1
Number of moles =
= mol
Number of gold atoms
= mol  6.02  1023 mol-1
= 6.11  1022

92. 2.1 The mole (SB p.21)
Example 2-1C
It is given that the molar mass of water is 18.0 g mol-1.
What is the mass of 4 moles of water molecules?
How many molecules are there?
How many atoms are there?
Answer

93. Example 2-1C 2.1 The mole (SB p.21)
Mass of water = Number of moles  Molar mass
= 4 mol  18.0 g mol-1
= 72.0 g
There are 4 moles of water molecules.
Number of water molecules
= Number of moles  Avogadro constant
= 4 mol  6.02  1023 mol-1
=  1024

94. Example 2-1C Back 2.1 The mole (SB p.21)
1 water molecule has 3 atoms (i.e. 2 hydrogen atoms and 1 oxygen atom).
1 mole of water molecules has 3 moles of atoms.
Thus, 4 moles of water molecules have 12 moles of atoms.
Number of atoms = 12 mol  6.02  1023 mol-1
=  1024

95. 2.1 The mole (SB p.22)
Example 2-1D
A magnesium chloride solution contains 10 g of magnesium chloride solid.
Calculate the number of moles of magnesium chloride in the solution.
Answer
The chemical formula of magnesium chloride is MgCl2.
Molar mass of MgCl2 = (  2) g mol-1 = 95.3 g mol-1
Number of moles of MgCl2 =
= mol

96. 2.1 The mole (SB p.22)
Example 2-1D
(b) Calculate the number of magnesium ions in the solution.
Answer
1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Cl- ions.
Therefore, mol of MgCl2 contains mol of Mg2+ ions.
Number of Mg2+ ions
= Number of moles of Mg2+ ions  Avogadro constant
= mol  6.02  1023 mol-1
=  1022

97. 2.1 The mole (SB p.22)
Example 2-1D
(c) Calculate the number of chloride ions in the solution.
Answer
0.105 mol of MgCl2 contains 0.21 mol of Cl- ions.
Number of Cl- ions
= Number of moles of Cl- ions  Avogadro constant
= 0.21 mol  6.02  1023 mol-1
=  1023

98. Example 2-1D Answer Back投影片 14
2.1 The mole (SB p.22)
Back投影片 14
Example 2-1D
(d) Calculate the total number of ions in the solution.
Answer
Total number of ions
=   1023
=  1023

99. 2.1 The mole (SB p.23)
Example 2-1E
Back
What is the mass of a carbon dioxide molecule?
Answer
The chemical formula of carbon dioxide is CO2.
Molar mass of CO2 = (  2) g mol-1 = 44.0 g mol-1
Number of moles = = =
Mass of a CO2 molecule =
= 7.31  g

100. 2.1 The mole (SB p.23)
Check Point 2-1
Find the mass in grams of 0.01 mol of zinc sulphide.
Answer
Mass = No. of moles  Molar mass
Mass of ZnS = 0.01 mol  ( ) g mol-1
= 0.01 mol  97.5 g mol-1
= g

101. 2.1 The mole (SB p.23)
Check Point 2-1
(b) Find the number of ions in 5.61 g of calcium oxide.
Answer
No. of moles of CaO =
= 0.1 mol
1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion.
No. of moles of ions = 0.1 mol  2
= 0.2 mol
No. of ions = 0.2 mol  6.02  1023 mol-1
=  1023

102. 2.1 The mole (SB p.23)
Check Point 2-1
(c) Find the number of atoms in g of sulphur dioxide.
Answer
Number of moles of SO2 =
= 0.5 mol
1 SO2 molecule contains 1 S atom and 2 O atoms.
No. of moles of atoms = 0.5 mol  3
= 1.5 mol
No. of atoms = 1.5 mol  6.02  1023 mol-1
= 9.03  1023

103. Check Point 2-1 Answer There is 4.80 g of ammonium carbonate. Find the
2.1 The mole (SB p.23)
Check Point 2-1
There is 4.80 g of ammonium carbonate. Find the
(i) number of moles of the compound,
(ii) number of moles of ammonium ions,
(iii) number of moles of carbonate ions,
(iv) number of moles of hydrogen atoms, and
(v) number of hydrogen atoms.
Answer

104. Check Point 2-1 Back 2.1 The mole (SB p.23)
Molar mass of (NH4)2CO3 = 96.0 g mol-1
(i) No. of moles of (NH4)2CO3 = = 0.05 mol
(ii) 1 mole (NH4)2CO3 gives 2 moles of NH4+ ions.
No. of moles of NH4+ ions = 0.05 mol  2 = 0.1 mol
(iii) 1 mole (NH4)2CO3 gives 1 mole of CO32- ions.
No. of moles CO32- ions = 0.05 mol
(iv) 1 (NH4)2CO3 formula unit contains 8 H atoms.
No. of moles of H atoms = 0.05 mol  8
= 0.4 mol
(v) No. of H atoms = 0.4 mol  6.02  1023 mol-1 =  1023

105. What is the difference between a theory and a law?
2.2 Molar volume and Avogadro’s law (SB p.24)
Let's Think 1
What is the difference between a theory and a law?
Answer
A law tells what happens under a given set of circumstances while a theory attempts to explain why that behaviour occurs.
Back

106. 2.2 Molar volume and Avogadro’s law (SB p.25)
Example 2-2A
Find the volume occupied by 3.55 g of chlorine gas at room temperature and pressure.
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
Answer
Molar mass of chlorine gas (Cl2) = (35.5  2) g mol-1 = 71.0 g mol-1
Number of moles of Cl2 =
= 0.05 mol
Volume of Cl2 = Number of moles of Cl2  Molar volume
= 0.05 mol  24.0 dm3 mol-1
= 1.2 dm3
Back

107. 2.2 Molar volume and Avogadro’s law (SB p.25)
Example 2-2B
Find the number of molecules in 4.48 cm3 of carbon dioxide gas at standard temperature and pressure.
(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1; Avogadro constant = 6.02  1023 mol-1)
Answer
Molar volume of carbon dioxide at S.T.P. = 22.4 dm3 mol-1
= cm3 mol-1
Number of moles of CO2 =
= 2  10-4 mol
Number of CO2 molecules = 2  10-4 mol  6.02  1023 mol-1
=  1020
Back

108. 2.2 Molar volume and Avogadro’s law (SB p.26)
Example 2-2C
The molar volume of nitrogen gas is found to be dm3 mol-1 at room temperature and pressure. Find the density of nitrogen gas.
Answer
Molar mass of nitrogen gas (N2) = ( ) g mol-1 = 28.0 g mol-1
Density = =
Density of N2 =
=1.167 g dm-3
Back

109. 2.2 Molar volume and Avogadro’s law (SB p.26)
Example 2-2D
1.6 g of a gas occupies 1.2 dm3 at room temperature and pressure. What is the relative molecular mass of the gas?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
Answer
Number of moles of the gas =
= 0.05 mol
Molar mass of the gas =
= 32 g mol-1
Relative molecular mass of the gas = 32 (no unit)
Back

110. 2.2 Molar volume and Avogadro’s law (SB p.27)
Check Point 2-2
Find the volume occupied by 0.6 g of hydrogen gas at room temperature and pressure.
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1)
Answer
No. of moles of H2 = = 0.3 mol
Volume = No. of moles  Molar volume
= 0.3 mol  24.0 dm3 mol-1
= 7.2 dm3

111. 2.2 Molar volume and Avogadro’s law (SB p.27)
Check Point 2-2
(b) Calculate the number of molecules in 4.48 dm3 of hydrogen gas at standard temperature and pressure.
(Molar volume of gas at S.T.P. = 22.4 dm3 mol-1)
Answer
(b) No. of moles of H2 = = 0.2 mol
No. of H2 molecules = 0.2 mol  6.02  1023 mol-1
=  1023

112. 2.2 Molar volume and Avogadro’s law (SB p.27)
Check Point 2-2
The molar volume of oxygen gas is 22.4 dm3 mol-1 at standard temperature and pressure. Find the density of oxygen gas in g cm-3 at S.T.P.
Answer
Density = =
Molar mass of O2 = (16.0  2) g mol-1 = 32.0 g mol-1
Molar volume of O2 = 22.4 dm3 mol-1 = cm3 mol-1
Density =
= 1.43  10-3 g cm-3

113. 2.2 Molar volume and Avogadro’s law (SB p.27)
Check Point 2-2
(d) What mass of oxygen has the same number of moles as that in 3.2 g of sulphur dioxide?
Answer
No. of moles of SO2 =
No. of moles of O2 = 0.05 mol
Mass = No. of moles  Molar mass
Mass of O2 = 0.05 mol  (16.0  2) g mol-1
= 1.6 g
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114. 2.3 Ideal gas equation (SB p.30)
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Example 2-3A
A 500 cm3 sample of a gas in a sealed container at 700 mmHg and 25 oC is heater to 100 oC. What is the final pressure of the gas?
Answer
As the number of moles of the gas is fixed, should be a constant. =
P2 = mmHg
The final pressure of the gas at 100 oC is mmHg.
Note: All temperature values used in gas laws are on the Kelvin scale.

115. 2.3 Ideal gas equation (SB p.30)
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Example 2-3B
A reaction vessel of 500 cm3 is filled with oxygen gas at 25 oC and the final pressure exerted on it is Nm-2. How many moles of oxygen gas are there?
(Ideal gas constant = J K-1 mol-1)
Answer
PV = nRT
Nm-2  500  10-6 m3 = n  J K-1 mol-1  ( ) K
n = 0.02 mol
There is 0.02 mol of oxygen gas in the reaction vessel.

116. 2.3 Ideal gas equation (SB p.30)
Back
Example 2-3C
A 5 dm3 vessel can withstand a maximum internal pressure of 50 atm. If 2 moles of nitrogen gas are pumped into the vessel, what is the highest temperature it can be safely heated to?
Answer
Applying the equation,
T = = = K
The highest temperature it can be safely heated to is oC.

117. 2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(b) A reaction vessel is filled with a gas at 20 oC and 5 atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to?
T2 = 586 K

118. Check Point 2-3 Or n = 0.0276 mol n = 0.0276 mol
2.3 Ideal gas equation (SB p.31)
Check Point 2-3
(c) A balloon is filled with helium at 25 oC. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm3 respectively. How many moles of helium have been introduced into the balloon?
n = mol Or
n = mol

119. 2.3 Ideal gas equation (SB p.31)
Check Point 2-3
25.8 cm3 sample of a gas has a pressure of 690 mmHg and a temperature of 17 oC. What is the volume of the gas if the pressure is changed to 1.85 atm and the temperature to 345 K?
(1 atm = 760 mmHg)
V2 = 15.1 cm3

120. 2.4 Determination of molar mass (SB p.33)
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Example 2-4A
A sample of gas occupying a volume of 50 cm3 at 1 atm and 25 oC is found to have a mass of g. Find the molar mass of the gas.
(Ideal gas constant = J K-1 mol-1; 1 atm = Nm-2)
Answer
M = g mol-1
Therefore, the molar mass of the gas is g mol-1.

121. 2.4 Determination of molar mass (SB p.34)
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Example 2-4B
The density of a gas at 450 oC and 380 mmHg is g dm-3. What is its molar mass? (1 atm = 760 mmHg = Nm-2; ideal gas constant = J K-1 mol-1)
Answer
The unit of density of the gas has to be converted to g m-3 for the calculation.
g dm-3 =  103 g m-3 = 33.7 g m-3
PM = RT
M =
= = 4.0 g mol-1
Therefore, the molar mass of the gas is 4.0 g mol-1.

122. 2.4 Determination of molar mass (SB p.34)
Check Point 2-4
0.204 g of phosphorus vapour occupies a volume of cm3 at 327 oC and 1 atm. Determine the molar mass of phosphorus.
(1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
Answer
PV = RT
Nm-2  81.0  10-6 m3
=  J K-1 mol-1  ( ) K
M = g mol-1
The molar mass of phosphorus is g mol-1.

123. 2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(b) A sample of gas has a mass of 12.0 g and occupies a volume of 4.16 dm3 measured at 97 oC and 1.62 atm. Calculate the molar mass of the gas.
(1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
Answer
(b) PV = RT
1.62  Nm-2  4.16  10-3 m3
=  J K-1 mol-1  ( ) K
M = g mol-1
The molar mass of the gas is g mol-1.

124. 2.4 Determination of molar mass (SB p.34)
Check Point 2-4
(c) A sample of g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium.
(1 atm = 760 mmHg = Nm-2; ideal gas constant = J K-1 mol-1)
Answer

125. Check Point 2-4 Back 2.4 Determination of molar mass (SB p.34)
Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
PV = nRT
 Nm-2  38.2  10-6 m3
= n  J K-1 mol-1  ( ) K
n = 1.52  10-3 mol
No. of moles of H2 produced = 1.52  10-3 mol
No. of moles of Mg reacted = No. of moles of H2 produced
= 1.52  10-3 mol
Molar mass of Mg = = = g mol-1
The relative atomic mass of Mg is

126. 2.5 Dalton’s law of partial pressures (SB p.36)
Back
Example 2-5A
Air is composed of 80 % nitrogen and 20 % oxygen by volume. What are the partial pressures of nitrogen and oxygen in air at a pressure of 1 atm and a temperature of 25 oC?
Answer
Mole fraction of N2 =
Mole fraction of O2 =
Partial pressure of N2 =
= Nm-2
Partial pressure of O2 =
= Nm-2

127. 2.5 Dalton’s law of partial pressures (SB p.36)
Example 2-5B
The valve between a 5 dm3 vessel containing gas A at a pressure of 15 atm and a 10 dm3 vessel containing gas B at a pressure of 12 atm is opened.
(a) Assuming that the temperature of the system remains constant, what is the final pressure of the system?
(b) What are the mole fractions of gas A and gas B?
Answer

128. 2.5 Dalton’s law of partial pressures (SB p.36)
Example 2-5B
Total volume of the system = (5 + 10) dm3 = 15 dm
By Boyle’s law, P1V1 = P2V2
Partial pressure of gas A (PA) =
= 5 atm
Partial pressure of gas B (PB)
= 8 atm
By Dalton’s law of partial pressures, Ptotal = PA + PB
Final pressure of the system = (5 + 8) atm = 13 atm

129. 2.5 Dalton’s law of partial pressures (SB p.37)
Example 2-5B
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Mole fraction of gas A = =
= 0.385
Mole fraction of gas B =
= 0.615

130. 2.5 Dalton’s law of partial pressures (SB p.37)
Example 2-5C
0.25 mol of nitrogen and 0.30 mol of oxygen are introduced into a vessel of 12 dm3 at 50 oC. Calculate the partial pressures of nitrogen and oxygen and hence the total pressure exerted by the gases.
(1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
Answer

131. 2.5 Dalton’s law of partial pressures (SB p.37)
Example 2-5C
Let the partial pressure of nitrogen be PA.
Using the ideal gas equation PV = nRT,
PA  12  10-3 m3 = 0.25 mol  J K-1 mol-1  ( ) K
PA = Nm-2 (or atm)
Let the partial pressure of oxygen be PB.
PB  12  10-3 m3 = 0.30 mol  J K-1 mol-1  ( ) K
PB = Nm-2 (or atm)

132. 2.5 Dalton’s law of partial pressures (SB p.37)
Example 2-5C
Back
Total pressure of gases
= ( ) Nm-2
= Nm-2 Or
= ( ) atm
= atm
Hence, the partial pressures of nitrogen and oxygen are atm and atm respectively, and the total pressure exerted by the gases is atm.

133. 2.5 Dalton’s law of partial pressures (SB p.38)
Example 2-5D
4.0 g of oxygen and 6.0 g of nitrogen are introduced into a 5 dm3 vessel at 27 oC.
What are the mole fraction of oxygen and nitrogen in the gas mixture?
What is the final pressure of the system?
(1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
Answer

134. 2.5 Dalton’s law of partial pressures (SB p.38)
Example 2-5D
Number of moles of oxygen =
= mol
Number of moles of nitrogen =
= mol
Total number of moles of gases = ( ) mol
= mol
Mole fraction of oxygen = = 0.369
Mole fraction of nitrogen = = 0.631

135. 2.5 Dalton’s law of partial pressures (SB p.38)
Example 2-5D
Let P be the final pressure of the system.
Using the ideal gas equation PV = nRT,
P  5  10-3 m3 = mol  J K-1 mol-1  ( ) K
P = Nm-2 (or 1.67 atm)
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