# 2 The Mole Concept

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1
The Mole Concept
2.1
The Mole
2.2
2.3
Ideal Gas Equation
2.4
Determination of Molar Mass
2.5
Dalton’s Law of Partial Pressures
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The Mole
2
New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.18)
What is “mole”?
Item
Shoes
Unit used to
count
pairs
No. of items
per unit
for
counting
2
Eggs
dozens
Paper
reams
500
Particles in
Chemistry
moles
6.02  1023
common
12
objects
for counting particles like atoms, ions,
molecules
3
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The mole (SB p.19)
How large is the amount in 1 mole?
6.02  1023
= 602 000 000 000 000 000 000 000
(the amount in 1 mole)
4
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2.1
The mole (SB p.19)
\$ 6.02  1023
All the people in the
world
so that each
get:
\$ 1000 note
?
2000 years
5
count at a
rate of 2
notes/sec
New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.19)
How to find the number of moles?
Number of particles
Number of moles =
or
number of particles
= number of moles  (6.02  1023)
6
New Way Chemistry for Hong Kong A-Level Book 1
1
2.1
The mole (SB p.19)
Why defining 6.02 x 1023 as the
amount for one mole?
12 g carbon contains 6.02  1023
12C atoms
The mole is the amount of substance
containing as many particles as the number
of atoms in 12 g of carbon-12.
7
New Way Chemistry for Hong Kong A-Level Book 1
2.1
The mole (SB p.19)
Relative
mass
…….
12
C atom
Molar
mass
6.02  1023
…….
1
H atom
1g
6.02  1023
Relative atomic
masses
Molar mass is the mass, in grams, of 1 mole of a
substance, e.g. the molar mass of H atom is 1 g.
8
New Way Chemistry for Hong Kong A-Level Book 1
2.1
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
Example 2-1D
Example 2-1E
Check Point 2-1
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2.2
Molar Volume and
10
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Molar volume and Avogadro’s law (SB p.24)
What is molar volume of gases?
at 25oC & 1 atm
(Room temp & pressure / R.T.P.)
11
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Molar volume and Avogadro’s law (SB p.24)
Equal volumes of all gases at the same
temperature and pressure contain the
same number of molecules.
Equal volumes of all gases at the same
temperature and pressure contain the
same number of moles of gases.
12
New Way Chemistry for Hong Kong A-Level Book 1
2.2
Molar volume and Avogadro’s law (SB p.24)
So 1 mole of gases should have the same
volume at the same temperature and
pressure.
Vn
13
where n is the no. of moles of gas
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2.2
Molar volume and Avogadro’s law (SB p.24)
Interconversions involving number
of moles
Example 2-2A
Example 2-2B
Example 2-2D
14
Example 2-2C
Check Point 2-2
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2.3
15
Ideal Gas
Equation
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2.3 Ideal gas equation (SB p.27)
Boyle’s law
At constant temperature,
the volume of a given
mass of a gas is
inversely proportional to
the pressure exerted on
it
PV = constant
16
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2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Boyle’s law
17
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2.3 Ideal gas equation (SB p.28)
A graph of volume against the reciprocal of
pressure for a gas at constant temperature
18
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2.3 Ideal gas equation (SB p.28)
Charles’ law
At a constant pressure, the volume of a given
mass of a gas is directly proportional to the
absolute temperature.
19
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2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Charles’ law
20
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2.3 Ideal gas equation (SB p.28)
A graph of volume against absolute temperature
for a gas at constant pressure
21
New Way Chemistry for Hong Kong A-Level Book 1
2.3 Ideal gas equation (SB p.27)
Ideal gas equation
Combining:
Vn
1
V
P
VT
V  nT
P
V = RnT
P
(Boyle’s Law)
(Charles Law)
where R is a constant
(called the universal gas constant)
PV = nRT
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New Way Chemistry for Hong Kong A-Level Book 1
2.3 Ideal gas equation (SB p.29)
For one mole of an ideal gas at standard
temperature and pressure,
P = 760 mmHg, 1 atm or 101 325 Nm-2 (Pa)
V = 22.4 dm3 mol-1 or 22.4  10-3 m3 mol-1
T = 0 oC or 273K
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2.3 Ideal gas equation (SB p.29)
By substituting the values of P, V and T in S.I. Units
into the equation, the value of ideal gas constant can
be found.
PV
R=
T
2  22.4  10-3 m3mol 1
101325
Nm
=
273K
= 8.314 J K-1mol-1
24
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2.3 Ideal gas equation (SB p.29)
Relationship between the ideal gas equation and the
individual gas laws
25
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2.3 Ideal gas equation (SB p.30)
26
Example 2-3A
Example 2-3B
Example 2-3C
Check Point 2-3
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2.4
Determination
of Molar Mass
27
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2.4 Determination of molar mass (SB p.32)
Mass of volatile liquid injected
= 26.590 - 26.330 = 0.260 g
28
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2.4 Determination of molar mass (SB p.32)
Volume of trichloromethane vapour
= 74.4 - 8.2 = 66.2 cm3
29
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2.4 Determination of molar mass (SB p.32)
Temperature = 273 + 99 = 372 K
30
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2.4 Determination of molar mass (SB p.32)
Pressure = 101325 Nm-2
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New Way Chemistry for Hong Kong A-Level Book 1
2.4 Determination of molar mass (SB p.32)
PV = nRT
101325 Nm-2  66.2  10-6 m3
= n  8.314 J K-1 mol-1  372 K
n = 2.169  10-3 mol
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2.4 Determination of molar mass (SB p.32)
Molar mass
Mass of trichlorom ethane
= Number of moles of trichlorom ethane
0.260 g
=
2.169 10-3 mol
= 119.87 g mol-1
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New Way Chemistry for Hong Kong A-Level Book 1
2.4 Determination of molar mass (SB p.32)
PV = nRT………..(1)
n = m ………..(2)
M
Where
m is the mass of the volatile substance
M is the molar mass of the volatile
substance
Combing (1) and (2), we obtain
m
PV = M RT
mRT
M = PV
34
Example 2-4A
Example 2-4B
Check Point 2-4
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2.5
Dalton’s Law of
Partial Pressures
35
New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.35)
Dalton’s Law of Partial Pressures
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
that each gas would exert as if it is present alone
under the same conditions).
PT
36
=
PA
+
PB
New Way Chemistry for Hong Kong A-Level Book 1
+
PC
2.5 Dalton’s law of partial pressures (SB p.35)
Consider a mixture of gases A, B and C occupying a
volume V. It consists of nA, nB and nC moles of each
gas.
The total number of moles of gases in the mixture
ntotal = nA + nB + nC
If the equation is multiplied by RT/V, then
ntotal (RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V)
i.e. Ptotal = PA + PB + PC
(so Dalton’s Law is a direct consequence
of the Ideal Gas Equation)
37
New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.35)
Besides, the partial pressure of each
component gas can be calculated from the
Ideal gas law.
PA = nA(RT/V) and Ptotal = ntotal(RT/V)
i.e. PA= (nA/ntotal) Ptotal
PA = xA Ptotal
Example 2-5A
Example 2-5B
Example 2-5D
38
Example 2-5C
Check Point 2-5
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The END
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.20)
Back
What is the mass of 0.2 mol of calcium carbonate?
The chemical formula of calcium carbonate is CaCO3.
Molar mass of calcium carbonate = (40.1 + 12.0 + 16.0  3) g mol-1
= 100.1 g mol-1
Mass of calcium carbonate = Number of moles  Molar mass
= 0.2 mol  100.1 g mol-1
= 20.02 g
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.21)
Back
Calculate the number of gold atoms in a 20 g gold pendant.
Molar mass of gold = 197.0 g mol-1
20 g
Number of moles =
197.0 g mol 1
= 0.1015 mol
Number of gold atoms
= 0.1015 mol  6.02  1023 mol-1
= 6.11  1022
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.21)
It is given that the molar mass of water is 18.0 g mol-1.
(a) What is the mass of 4 moles of water molecules?
(b) How many molecules are there?
(c) How many atoms are there?
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.21)
(a) Mass of water = Number of moles  Molar mass
= 4 mol  18.0 g mol-1
= 72.0 g
(b) There are 4 moles of water molecules.
Number of water molecules
= Number of moles  Avogadro constant
= 4 mol  6.02  1023 mol-1
= 2.408  1024
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.21)
Back
(c) 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
= 7.224  1024
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.22)
A magnesium chloride solution contains 10 g of magnesium
chloride solid.
(a) Calculate the number of moles of magnesium chloride
in the solution.
(a) The chemical formula of magnesium chloride is MgCl2.
Molar mass of MgCl2 = (24.3 + 35.5  2) g mol-1 = 95.3 g mol-1
10 g
Number of moles of MgCl2 =
95.3 g mol 1
= 0.105 mol
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.22)
(b) Calculate the number of magnesium ions in the solution.
(b) 1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Clions.
Therefore, 0.105 mol of MgCl2 contains 0.105 mol of Mg2+ ions.
Number of Mg2+ ions
= Number of moles of Mg2+ ions  Avogadro constant
= 0.105 mol  6.02  1023 mol-1
= 6.321  1022
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.22)
(c) Calculate the number of chloride ions in the solution.
(c) 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
= 1.264  1023
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2.1 The mole (SB p.22)
Back
(d) Calculate the total number of ions in the solution.
(d) Total number of ions
= 6.321  1022 + 1.264  1023
= 1.896  1023
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2.1 The mole (SB p.23)
Back
What is the mass of a carbon dioxide molecule?
The chemical formula of carbon dioxide is CO2.
Molar mass of CO2 = (12.0 + 16.0  2) g mol-1 = 44.0 g mol-1
Mass
Number of molecules
Number of moles =
=
Molar mass
Mass of a CO2 molecule
1
=
44.0 g mol -1
6.02  10 23 mol -1
44.0 g mol -1
Mass of a CO2 molecule =
6.02  10 23 mol -1
= 7.31  10-23 g
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.23)
(a) Find the mass in grams of 0.01 mol of zinc sulphide.
(a) Mass = No. of moles  Molar mass
Mass of ZnS = 0.01 mol  (65.4 + 32.1) g mol-1
= 0.01 mol  97.5 g mol-1
= 0.975 g
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.23)
(b) Find the number of ions in 5.61 g of calcium oxide.
5.61 g
(b) No. of moles of CaO =
(40.1  16.0) g mol 1
= 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
= 1.204  1023
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.23)
(c) Find the number of atoms in 32.05 g of sulphur dioxide.
32.05 g
(c) Number of moles of SO2 =
(32.1  16.0  2) g mol -1
= 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
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New Way Chemistry for Hong Kong A-Level Book 1
2.1 The mole (SB p.23)
(d) 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.
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2.1 The mole (SB p.23)
Back
(d) Molar mass of (NH4)2CO3 = 96.0 g mol-1
4.80 g
(i) No. of moles of (NH4)2CO3 =
= 0.05 mol
96.0 g mol 1
(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 = 2.408  1023
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2.2 Molar volume and Avogadro’s law (SB p.24)
What is the difference between a theory and a law?
A law tells what happens under a given set of
circumstances while a theory attempts to
explain why that behaviour occurs.
Back
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2.2 Molar volume and Avogadro’s law (SB p.25)
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)
Molar mass of chlorine gas (Cl2) = (35.5  2) g mol-1 = 71.0 g mol-1
3.55 g
Number of moles of Cl2 =
71.0 g mol 1
= 0.05 mol
Volume of Cl2 = Number of moles of Cl2  Molar volume
= 0.05 mol  24.0 dm3 mol-1
= 1.2 dm3
56
Back
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.25)
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)
Molar volume of carbon dioxide at S.T.P. = 22.4 dm3 mol-1
= 22400 cm3 mol-1
4.48 cm3
Number of moles of CO2 =
22400 cm3 mol 1
= 2  10-4 mol
Number of CO2 molecules = 2  10-4 mol  6.02  1023 mol-1
= 1.204  1020
57
Back
New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.26)
The molar volume of nitrogen gas is found to be
24.0 dm3 mol-1 at room temperature and pressure. Find the
density of nitrogen gas.
Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1 = 28.0 g mol-1
Molar mass
Mass
Density =
=
Molar volume
Volume
Density of N2
28.0 g mol -1
=
24.0 dm3 mol -1
=1.167 g dm-3
Back
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New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.26)
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)
1.2 dm3
Number of moles of the gas =
24.0 dm3 mol 1
= 0.05 mol
1.6 g
Molar mass of the gas =
0.05 mol
= 32 g mol-1
Relative molecular mass of the gas = 32 (no unit)
Back
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New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.27)
(a) 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)
0.6 g
(a) No. of moles of H2 =
= 0.3 mol
(1.0  2) g mol 1
Volume = No. of moles  Molar volume
= 0.3 mol  24.0 dm3 mol-1
= 7.2 dm3
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New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.27)
(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)
4.48 dm3
(b) No. of moles of H2 =
= 0.2 mol
3
1
22.4 dm mol
No. of H2 molecules = 0.2 mol  6.02  1023 mol-1
= 1.204  1023
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New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.27)
(c) 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.
Mass
Molar mass
(c) Density =
=
Volume
Molar volume
Molar mass of O2 = (16.0  2) g mol-1 = 32.0 g mol-1
Molar volume of O2 = 22.4 dm3 mol-1 = 22400 cm3 mol-1
32.0 g mol -1
Density =
22400 cm3 mol 1
= 1.43  10-3 g cm-3
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New Way Chemistry for Hong Kong A-Level Book 1
2.2 Molar volume and Avogadro’s law (SB p.27)
(d) What mass of oxygen has the same number of moles as
that in 3.2 g of sulphur dioxide?
3.2 g
(d) No. of moles of SO2 =
(32.1  16.0  2) g mol 1
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
Back
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2.3 Ideal gas equation (SB p.30)
Back
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?
As the number of moles of the gas is fixed,
P1V1 P2 V2
=
T2
T1
PV should be a constant.
T
700 mmHg  500 cm3
P2  500 cm3
=
(273  25) K
(273  100) K
P2 = 876.17 mmHg
The final pressure of the gas at 100 oC is 876.17 mmHg.
Note: All temperature values used in gas laws are on the Kelvin scale.
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2.3 Ideal gas equation (SB p.30)
Back
A reaction vessel of 500 cm3 is filled with oxygen gas at 25 oC
and the final pressure exerted on it is 101 325 Nm-2. How
many moles of oxygen gas are there?
(Ideal gas constant = 8.314 J K-1 mol-1)
PV = nRT
101325 Nm-2  500  10-6 m3 = n  8.314 J K-1 mol-1  (273 + 25) K
n = 0.02 mol
There is 0.02 mol of oxygen gas in the reaction vessel.
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New Way Chemistry for Hong Kong A-Level Book 1
2.3 Ideal gas equation (SB p.30)
Back
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?
Applying the equation,
PV
50  101325 Nm -2  5  10 -3 m3
T=
=
= 1523.4 K
-1
-1
nR
2 mol  8.314 J K mol
The highest temperature it can be safely heated to is 1250.4 oC.
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New Way Chemistry for Hong Kong A-Level Book 1
2.3 Ideal gas equation (SB p.31)
(a) State Boyle’s law, Charles’ law and ideal gas equation.
(a) Boyle’s law states that at constant temperature, the volume of a
given mass of a gas is inversely proportional to the pressure
exerted on it.
Charles’ law states that at constant pressure, the volume of a given
mass of a gas is directly proportional to the absolute temperature.
The ideal gas equation, PV = nRT, shows the relationship between
the pressure, volume, temperature and number of moles of a gas.
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New Way Chemistry for Hong Kong A-Level Book 1
2.3 Ideal gas equation (SB p.31)
(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?
(b)
P1
P2
T1 = T2
10 atm
5 atm
=
T2
(273  20) K
T2 = 586 K
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New Way Chemistry for Hong Kong A-Level Book 1
2.3 Ideal gas equation (SB p.31)
(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?
(c) PV = nRT
1.5  101325 Nm-2  450  10-6 m3
= n  8.314 J K-1 mol-1  (273 + 25) K
n = 0.0276 mol
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2.3 Ideal gas equation (SB p.31)
Back
(d) 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)
(d)
P1V1 P2 V2

T1
T2
690
atm  25.8 cm3
1.85 atm  V2
760

(273  17) K
345 K
V2 = 15.06 cm3
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2.4 Determination of molar mass (SB p.33)
Back
A sample of gas occupying a volume of 50 cm3 at 1 atm and
25 oC is found to have a mass of 0.0286 g. Find the molar mass
of the gas.
(Ideal gas constant = 8.314 J K-1 mol-1; 1 atm = 101325 Nm-2)
PV 
m
RT
M
0.0286 g
 8.314 J K 1 mol 1  (273  25) K
M
M = 13.99 g mol-1
101325 Nm - 2  50  10  6 m3 
Therefore, the molar mass of the gas is 13.99 g mol-1.
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2.4 Determination of molar mass (SB p.34)
Back
The density of a gas at 450 oC and 380 mmHg is
0.0337 g dm-3. What is its molar mass? (1 atm = 760 mmHg =
101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
The unit of density of the gas has to be converted to g
calculation.
for the
m-3
0.0337 g dm-3 = 0.0337  103 g m-3 = 33.7 g m-3
PM = RT
M = RT
P
33.7 g m -3  8.314 J K -1 mol -1  (273  450) K
=
= 4.0 g mol-1
380
 101325 Nm  2
760
Therefore, the molar mass of the gas is 4.0 g mol-1.
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New Way Chemistry for Hong Kong A-Level Book 1
2.4 Determination of molar mass (SB p.34)
(a) 0.204 g of phosphorus vapour occupies a volume of
81.0 cm3 at 327 oC and 1 atm. Determine the molar mass of
phosphorus.
(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
(a) PV = m RT
M
101325 Nm-2  81.0  10-6 m3
0.204 g
=
 8.314 J K-1 mol-1  (273 + 327) K
M
M = 123.99 g mol-1
The molar mass of phosphorus is 123.99 g mol-1.
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2.4 Determination of molar mass (SB p.34)
(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 = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
(b) PV = m RT
M
1.62  101325 Nm-2  4.16  10-3 m3
12.0 g
=
 8.314 J K-1 mol-1  (273 + 97) K
M
M = 54.06 g mol-1
The molar mass of the gas is 54.06 g mol-1.
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New Way Chemistry for Hong Kong A-Level Book 1
2.4 Determination of molar mass (SB p.34)
(c) A sample of 0.037 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 = 101325 Nm-2;
ideal gas constant = 8.314 J K-1 mol-1)
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New Way Chemistry for Hong Kong A-Level Book 1
2.4 Determination of molar mass (SB p.34)
Back
(c) Mg(s) + 2HCl(aq)  MgCl2(aq) + H2(g)
PV = nRT
740
 101325 Nm-2  38.2  10-6 m3
760
= n  8.314 J K-1 mol-1  (273 + 25) 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
0.037 g
Mass
Molar mass of Mg =
=
= 24.34 g mol-1
-3
No. of moles 1.52  10 mol
The relative atomic mass of Mg is 24.34.
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.36)
Back
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?
80
Mole fraction of N2 =
100
Mole fraction of O2 = 20
100 80
Partial pressure of N2 =
 101325 Nm  2
100
= 81060 Nm-2
20
 101325 Nm  2
Partial pressure of O2 =
100
= 20265 Nm-2
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.36)
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)78 What are the
mole fractions of gas A and gas B?
New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.36)
(a) Total volume of the system = (5 + 10) dm3 = 15 dm
By Boyle’s law, P1V1 = P2V2
Partial pressure of gas A (PA)
15 atm  5 dm3
=
15 dm3
= 5 atm
Partial pressure of gas B (PB)
12 atm  10 dm3
=
15 dm3
= 8 atm
By Dalton’s law of partial pressures, Ptotal = PA + PB
Final pressure of the system = (5 + 8) atm = 13 atm
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.37)
Back
PA
(b) Mole fraction of gas A =
Ptotal
5 atm
=
13 atm
= 0.385
P
Mole fraction of gas B = B
Ptotal
8 atm
=
13 atm
= 0.615
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.37)
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 = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.37)
Let the partial pressure of nitrogen be PA.
Using the ideal gas equation PV = nRT,
PA  12  10-3 m3 = 0.25 mol  8.314 J K-1 mol-1  (273 + 50) K
PA = 55946 Nm-2 (or 0.552 atm)
Let the partial pressure of oxygen be PB.
Using the ideal gas equation PV = nRT,
PB  12  10-3 m3 = 0.30 mol  8.314 J K-1 mol-1  (273 + 50) K
PB = 67136 Nm-2 (or 0.663 atm)
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.37)
Back
Total pressure of gases
= (55946 + 67136) Nm-2
= 123082 Nm-2
Or
Total pressure of gases
= (0.552 + 0.663) atm
= 1.215 atm
Hence, the partial pressures of nitrogen and oxygen are 0.552 atm and
0.663 atm respectively, and the total pressure exerted by the gases is
1.215 atm.
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.38)
4.0 g of oxygen and 6.0 g of nitrogen are introduced into a
5 dm3 vessel at 27 oC.
(a)
What are the mole fraction of oxygen and nitrogen in
the gas mixture?
(b) What is the final pressure of the system?
(1 atm = 101325 Nm-2;
ideal gas constant = 8.314 J K-1 mol-1)
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.38)
4.0 g
(a) Number of moles of oxygen =
32.0 gmol 1
= 0.125 mol
6.0 g
Number of moles of nitrogen =
28.0 gmol 1
= 0.214 mol
Total number of moles of gases = (0.125 + 0.214) mol
= 0.339 mol
0.125 mol
Mole fraction of oxygen =
= 0.369
0.339 mol
0.214 mol
Mole fraction of nitrogen = 0.339 mol = 0.631
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.38)
(b) Let P be the final pressure of the system.
Using the ideal gas equation PV = nRT,
P  5  10-3 m3 = 0.339 mol  8.314 J K-1 mol-1  (273 + 27) K
P = 169 107 Nm-2 (or 1.67 atm)
Back
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.39)
(a) Define mole fraction.
(a) Mole fraction of a substance is the ratio of the number of moles of
that substance to the total number of moles in the mixture.
(b) State Dalton’s law of partial pressures.
(b) Dalton’s law of partial pressures states that in a mixture of gases
which do not react with each other, the total pressure exerted is the
sum of the pressure that each gas would exert as if it it present
alone under the same conditions.
Ptotal = PA + PB + PC + … …
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.39)
(c) 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?
88
New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.39)
(c) By Boyle’s law: P1V1 = P2V2
6 dm3
Partial pressure of gas A = 7 atm 
(6  8) dm3
= 3.00 atm
8 dm3
Partial pressure of gas B = 9 atm 
(6  8) dm3
= 5.14 atm
By Dalton’s law: Ptotal = PA + PB
Final pressure of the system = 3.00 atm + 5.14 atm = 8.14 atm
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.39)
(d) 2 g of helium, 3 g of nitrogen and 4 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?
(ii) Calculate the total pressure of the system, and hence
the partial pressures of helium, nitrogen and argon.
(1 atm = 101325 Nm-2; ideal gas constant = 8.314 J K-1 mol-1)
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.39)
2g
4.0 g mol -1
= 0.50 mol
3g
No. of moles of N2 =
(14.0  2) g mol -1
= 0.11 mol
4g
No. of moles of Ar =
39.9 g mol -1
= 0.10 mol
(d) (i) No. of moles of He =
Total no. of moles of gases = 0.50 mol + 0.11 mol + 0.10 mol
= 0.71 mol
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New Way Chemistry for Hong Kong A-Level Book 1
2.5 Dalton’s law of partial pressures (SB p.39)
Back
Mole fraction of He = 0.50 mol
= 0.704
0.71mol
0.11
mol
Mole fraction of N2 =
= 0.155
0.71mol
Mole fraction of Ar = 0.10 mol
= 0.141
0.71
mol be P.
(ii) Let the total pressure of the
system
PV = nRT
P  15  10-3 m3 = 0.71 mol  8.314 J K-1 mol-1  (273 + 100) K
P = 146786 Nm-2
Partial pressure of He = 146786 Nm-2  0.704 = 103337 Nm-2
Partial pressure of N2 = 146786 Nm-2  0.155 = 22752 Nm-2
Partial pressure of Ar = 146786 Nm-2  0.141 = 20697 Nm-2
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