Download - The Mole Concept
New Way Chemistry for Hong Kong A-Level Book 11
The Mole ConceptThe Mole Concept
2.12.1 The MoleThe Mole
2.22.2 Molar Volume and Avogadro’s LawMolar Volume and Avogadro’s Law
2.32.3 Ideal Gas EquationIdeal Gas Equation
2.4 2.4 Determination of Molar MassDetermination of Molar Mass
2.52.5 Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures
22
New Way Chemistry for Hong Kong A-Level Book 12
2.2.11 The MoleThe Mole
New Way Chemistry for Hong Kong A-Level Book 13
2.1 The mole (SB p.18)
What is “mole”?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 6.02 1023
for counting particles like atoms, ions, molecules
for counting common objects
New Way Chemistry for Hong Kong A-Level Book 14
6.02 1023
= 602 000 000 000 000 000 000 000
Avogadro constant(the amount in 1 mole)
How large is the amount in 1 mole?How large is the amount in 1 mole?
2.1 The mole (SB p.19)
New Way Chemistry for Hong Kong A-Level Book 15
$ 6.02 1023All the people in the world
so that each get:
$ 1000 note
count at a rate of 2 notes/sec
?2000 years
2.1 The mole (SB p.19)
New Way Chemistry for Hong Kong A-Level Book 16
1
or
number of particles
= number of moles (6.02 1023)
number of particles
= number of moles (6.02 1023)
How to find the number of How to find the number of moles?moles?
2.1 The mole (SB p.19)
Number of moles = consant Avogadroparticles of Number
New Way Chemistry for Hong Kong A-Level Book 17
Why defining 6.02 x 10Why defining 6.02 x 102323 as the as the amount for one mole?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.
2.1 The mole (SB p.19)
New Way Chemistry for Hong Kong A-Level Book 18
Relative mass12
1
C atom
…….
6.02 1023
H atom
…….
6.02 1023
1 g
Molar mass
Molar mass is the mass, in grams, of 1 mole of a substance, e.g. the molar mass of H atom is 1 g.
Relative atomic masses
2.1 The mole (SB p.19)
New Way Chemistry for Hong Kong A-Level Book 19
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.
2.1 The mole (SB p.20)
Example 2-1AExample 2-1A Example 2-1BExample 2-1B Example 2-1CExample 2-1C
Example 2-1DExample 2-1D Example 2-1EExample 2-1E Check Point 2-1Check Point 2-1
New Way Chemistry for Hong Kong A-Level Book 110
2.2.22Molar Volume anMolar Volume and Avogadro’s Lad Avogadro’s La
ww
New Way Chemistry for Hong Kong A-Level Book 111
What is molar volume of What is molar volume of gases?gases?
at 25oC & 1 atm
(Room temp & pressure / R.T.P.)
2.2 Molar volume and Avogadro’s law (SB p.24)
New Way Chemistry for Hong Kong A-Level Book 112
Avogadro’s LawAvogadro’s Law2.2 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 molecules.
Equal volumes of all gases at the same temperature and pressure contain the
same number of moles of gases.
Equal volumes of all gases at the same temperature and pressure contain the
same number of moles of gases.
New Way Chemistry for Hong Kong A-Level Book 113
Avogadro’s LawAvogadro’s Law
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 where n is the no. of moles of gas V n where n is the no. of moles of gas
New Way Chemistry for Hong Kong A-Level Book 114
2.2 Molar volume and Avogadro’s law (SB p.24)
Interconversions involving Interconversions involving number of molesnumber of moles
Example 2-2AExample 2-2A Example 2-2BExample 2-2B Example 2-2CExample 2-2C
Example 2-2DExample 2-2D Check Point 2-2Check Point 2-2
New Way Chemistry for Hong Kong A-Level Book 115
2.2.33 Ideal Gas Ideal Gas
EquationEquation
New Way Chemistry for Hong Kong A-Level Book 116
Boyle’s lawBoyle’s law
2.3 Ideal gas equation (SB p.27)
At constant temperature, the volume of a given mass of a gas is inversely proportional to the pressure exerted on it
PV = constant
New Way Chemistry for Hong Kong A-Level Book 117
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Boyle’s law
New Way Chemistry for Hong Kong A-Level Book 118
2.3 Ideal gas equation (SB p.28)
A graph of volume against the reciprocal of pressure for a gas at constant temperature
New Way Chemistry for Hong Kong A-Level Book 119
At a constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.
2.3 Ideal gas equation (SB p.28)
Charles’ lawCharles’ law
New Way Chemistry for Hong Kong A-Level Book 120
2.3 Ideal gas equation (SB p.28)
Schematic diagrams explaining Charles’ law
New Way Chemistry for Hong Kong A-Level Book 121
2.3 Ideal gas equation (SB p.28)
A graph of volume against absolute temperature for a gas at constant pressure
New Way Chemistry for Hong Kong A-Level Book 122
PV = nRTPV = nRT
2.3 Ideal gas equation (SB p.27)
Ideal gas equationIdeal gas equation
P1
Combining:
V n (Avogadro’s Law)
V (Boyle’s Law)
V T (Charles Law)
V
V = where R is a constant (called the universal gas constant)
PnT
PRnT
New Way Chemistry for Hong Kong A-Level Book 123
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
2.3 Ideal gas equation (SB p.29)
New Way Chemistry for Hong Kong A-Level Book 124
2.3 Ideal gas equation (SB p.29)
273K
1mol3m3-1022.42Nm 101325
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.
R =
=
= 8.314 J K-1mol-1
TPV
New Way Chemistry for Hong Kong A-Level Book 125
2.3 Ideal gas equation (SB p.29)
Relationship between the ideal gas equation and the individual gas laws
New Way Chemistry for Hong Kong A-Level Book 126
2.3 Ideal gas equation (SB p.30)
Example 2-3AExample 2-3A Example 2-3BExample 2-3B
Example 2-3CExample 2-3C Check Point 2-3Check Point 2-3
New Way Chemistry for Hong Kong A-Level Book 127
2.2.44DeterminatioDeterminatio
n of Molar n of Molar MassMass
New Way Chemistry for Hong Kong A-Level Book 128
Mass of volatile liquid injected
= 26.590 - 26.330 = 0.260 g
2.4 Determination of molar mass (SB p.32)
New Way Chemistry for Hong Kong A-Level Book 129
Volume of trichloromethane vapour
= 74.4 - 8.2 = 66.2 cm3
2.4 Determination of molar mass (SB p.32)
New Way Chemistry for Hong Kong A-Level Book 130
Temperature = 273 + 99 = 372 K
2.4 Determination of molar mass (SB p.32)
New Way Chemistry for Hong Kong A-Level Book 131
Pressure = 101325 Nm-2
2.4 Determination of molar mass (SB p.32)
New Way Chemistry for Hong Kong A-Level Book 132
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
New Way Chemistry for Hong Kong A-Level Book 133
Molar mass
=
=
= 119.87 g mol-1
ethanetrichlorom of moles of Numberethanetrichlorom of Mass
mol 3-102.169
g 0.260
2.4 Determination of molar mass (SB p.32)
New Way Chemistry for Hong Kong A-Level Book 134
PV = nRT………..(1)
n = ………..(2)
Where
m is the mass of the volatile substance
M is the molar mass of the volatile substance
Combing (1) and (2), we obtain
PV = RT
M =
2.4 Determination of molar mass (SB p.32)
Mm
Mm
PVmRT
Example 2-4AExample 2-4A
Example 2-4BExample 2-4B
Check Point 2-4Check Point 2-4
New Way Chemistry for Hong Kong A-Level Book 135
2.2.55 Dalton’s Law Dalton’s Law
of Partial of Partial PressuresPressures
New Way Chemistry for Hong Kong A-Level Book 136
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).
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 = PA + PB + PC
Dalton’s Law of Partial PressuresDalton’s Law of Partial Pressures
2.5 Dalton’s law of partial pressures (SB p.35)
New Way Chemistry for Hong Kong A-Level Book 137
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)
2.5 Dalton’s law of partial pressures (SB p.35)
New Way Chemistry for Hong Kong A-Level Book 138
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
2.5 Dalton’s law of partial pressures (SB p.35)
Check Point 2-5Check Point 2-5
Example 2-5AExample 2-5A Example 2-5BExample 2-5B Example 2-5CExample 2-5C
Example 2-5DExample 2-5D
New Way Chemistry for Hong Kong A-Level Book 139
The END
New Way Chemistry for Hong Kong A-Level Book 140
What is the mass of 0.2 mol of calcium carbonate?
2.1 The mole (SB p.20)
Back
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
Answer
New Way Chemistry for Hong Kong A-Level Book 141
Calculate the number of gold atoms in a 20 g gold pendant.
2.1 The mole (SB p.21)
Back
AnswerMolar mass of gold = 197.0 g mol-1
Number of moles =
= 0.1015 mol
Number of gold atoms
= 0.1015 mol 6.02 1023 mol-1
= 6.11 1022
1mol g 197.0g 20
New Way Chemistry for Hong Kong A-Level Book 142
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?
2.1 The mole (SB p.21)
Answer
New Way Chemistry for Hong Kong A-Level Book 143
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
New Way Chemistry for Hong Kong A-Level Book 144
Back
2.1 The mole (SB p.21)
(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
New Way Chemistry for Hong Kong A-Level Book 145
A magnesium chloride solution contains 10 g of magnesium chloride solid.
(a) Calculate the number of moles of magnesium chloride in the solution.
2.1 The mole (SB p.22)
Answer
(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
Number of moles of MgCl2 =
= 0.105 mol
1mol g 95.3g 10
New Way Chemistry for Hong Kong A-Level Book 146
(b) Calculate the number of magnesium ions in the solution.
2.1 The mole (SB p.22)
Answer
(b) 1 mole of MgCl2 contains 1 mole of Mg2+ ions and 2 moles of Cl- ions.
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
New Way Chemistry for Hong Kong A-Level Book 147
(c) Calculate the number of chloride ions in the solution.
2.1 The mole (SB p.22)
Answer
(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
New Way Chemistry for Hong Kong A-Level Book 148
(d) Calculate the total number of ions in the solution.
2.1 The mole (SB p.22)
Answer(d) Total number of ions
= 6.321 1022 + 1.264 1023
= 1.896 1023
Back
New Way Chemistry for Hong Kong A-Level Book 149
What is the mass of a carbon dioxide molecule?
2.1 The mole (SB p.23)
AnswerThe chemical formula of carbon dioxide is CO2.
Molar mass of CO2 = (12.0 + 16.0 2) g mol-1 = 44.0 g mol-1
Number of moles = =
=
Mass of a CO2 molecule =
= 7.31 10-23 g
constant Avogadromolecules of Number
mass MolarMass
1-
2
mol g 44.0
molecule CO a of Mass1-23 mol 10 6.02
1
1-23
-1
mol 10 6.02mol g 44.0
Back
New Way Chemistry for Hong Kong A-Level Book 150
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
Answer
New Way Chemistry for Hong Kong A-Level Book 151
2.1 The mole (SB p.23)
(b)Find the number of ions in 5.61 g of calcium oxide.
Answer
(b) 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
= 1.204 1023
1mol g 16.0)(40.1g 5.61
New Way Chemistry for Hong Kong A-Level Book 152
2.1 The mole (SB p.23)
(c) Find the number of atoms in 32.05 g of sulphur dioxide.
Answer(c) 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
1-mol g 2) 16.0 2.13(g 32.05
New Way Chemistry for Hong Kong A-Level Book 153
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. Answer
New Way Chemistry for Hong Kong A-Level Book 154
2.1 The mole (SB p.23)
(d) 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 = 2.408 1023
1mol g 96.0g 4.80
Back
New Way Chemistry for Hong Kong A-Level Book 155
What is the difference between a theory and a law?
Back
2.2 Molar volume and Avogadro’s law (SB p.24)
A law tells what happens under a given set of circumstances while a theory attempts to explain why that behaviour occurs.
Answer
New Way Chemistry for Hong Kong A-Level Book 156
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
1mol g 71.0g 3.55
2.2 Molar volume and Avogadro’s law (SB p.25)
Back
New Way Chemistry for Hong Kong A-Level Book 157
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
= 22400 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
= 1.204 1020
13
3
mol cm 22400cm 4.48
2.2 Molar volume and Avogadro’s law (SB p.25)
Back
New Way Chemistry for Hong Kong A-Level Book 158
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.
Answer
2.2 Molar volume and Avogadro’s law (SB p.26)
Back
Molar mass of nitrogen gas (N2) = (14.0 + 14.0) g mol-1 = 28.0 g mol-1
Density = =
Density of N2
=
=1.167 g dm-3
VolumeMass
volume Molarmass Molar
1-3
-1
mol dm 24.0mol g 28.0
New Way Chemistry for Hong Kong A-Level Book 159
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
2.2 Molar volume and Avogadro’s law (SB p.26)
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)
13
3
mol dm 24.0dm 1.2
mol 0.05g 1.6
Back
New Way Chemistry for Hong Kong A-Level Book 160
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)
Answer
(a) 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
1mol g 2)(1.0g 0.6
New Way Chemistry for Hong Kong A-Level Book 161
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)Answer
(b) No. of moles of H2 = = 0.2 mol
No. of H2 molecules = 0.2 mol 6.02 1023 mol-1
= 1.204 1023
13
3
mol dm 2.42dm 4.48
New Way Chemistry for Hong Kong A-Level Book 162
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.
Answer
(c) 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 = 22400 cm3 mol-1
Density =
= 1.43 10-3 g cm-3
VolumeMass
volume Molarmass Molar
13
-1
mol cm 22400mol g 32.0
New Way Chemistry for Hong Kong A-Level Book 163
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?
Answer
(d) 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
1mol g 2)16.0(32.1g 3.2
Back
New Way Chemistry for Hong Kong A-Level Book 164
2.3 Ideal gas equation (SB p.30)
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 = 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.
TPV
1
11
T
VP
2
22
T
VP
K 25)(273cm 500 mmHg 700 3
K )001(273
cm 500 P 3
2
Back
New Way Chemistry for Hong Kong A-Level Book 165
2.3 Ideal gas equation (SB p.30)
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)Answer
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.
Back
New Way Chemistry for Hong Kong A-Level Book 166
2.3 Ideal gas equation (SB p.30)
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
Back
Applying the equation,
T = = = 1523.4 K
The highest temperature it can be safely heated to is 1250.4 oC.
1-1-
3-3-2
mol K J 8.314 mol 2m 105 Nm 10132550
nRPV
New Way Chemistry for Hong Kong A-Level Book 167
2.3 Ideal gas equation (SB p.31)
(a)State Boyle’s law, Charles’ law and ideal gas equation.
Answer(a) Boyle’s law states that at constant temperature, the volume of a giv
en 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.
New Way Chemistry for Hong Kong A-Level Book 168
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? Answer
(b)
=
=
T2 = 586 K
1
1
T
P
2
2
T
P
K 20) (273atm 5 2T
atm 10
New Way Chemistry for Hong Kong A-Level Book 169
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?Answer
(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
New Way Chemistry for Hong Kong A-Level Book 170
2.3 Ideal gas equation (SB p.31)
(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)Answer
(d)
V2 = 15.06 cm3
2
22
1
11
T
VP
T
VP
K 345
V atm 1.85
K 17) (273
cm 25.8atm 760690
2
3
Back
New Way Chemistry for Hong Kong A-Level Book 171
2.4 Determination of molar mass (SB p.33)
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) Answer
Back
M = 13.99 g mol-1
Therefore, the molar mass of the gas is 13.99 g mol-1.
RTMm
PV
K 25) (273 mol K J 8.314M
g 0.0286m 10 50 Nm 101325 11362-
New Way Chemistry for Hong Kong A-Level Book 172
2.4 Determination of molar mass (SB p.34)
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)
Answer
Back
The unit of density of the gas has to be converted to g m-3 for the calculation.
0.0337 g dm-3 = 0.0337 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.
PRT
2
-1-1-3
Nm 101325760380
K 450) (273 mol K J 8.314 m g 33.7
New Way Chemistry for Hong Kong A-Level Book 173
(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)
Answer
2.4 Determination of molar mass (SB p.34)
(a) PV = RT
101325 Nm-2 81.0 10-6 m3
= 8.314 J K-1 mol-1 (273 + 327) K
M = 123.99 g mol-1
The molar mass of phosphorus is 123.99 g mol-1.
Mm
Mg 0.204
New Way Chemistry for Hong Kong A-Level Book 174
(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)
Answer
2.4 Determination of molar mass (SB p.34)
(b) PV = RT
1.62 101325 Nm-2 4.16 10-3 m3
= 8.314 J K-1 mol-1 (273 + 97) K
M = 54.06 g mol-1
The molar mass of the gas is 54.06 g mol-1.
Mm
Mg 12.0
New Way Chemistry for Hong Kong A-Level Book 175
(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)
Answer
2.4 Determination of molar mass (SB p.34)
New Way Chemistry for Hong Kong A-Level Book 176
2.4 Determination of molar mass (SB p.34)
(c) Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)
PV = nRT
101325 Nm-2 38.2 10-6 m3
= 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
Molar mass of Mg = = = 24.34 g mol-1
The relative atomic mass of Mg is 24.34.
moles of No.Mass
mol 10 1.52g 0.037
3-
760740
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New Way Chemistry for Hong Kong A-Level Book 177
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
Back2.5 Dalton’s law of partial pressures (SB p.3
6)
Mole fraction of N2 =
Mole fraction of O2 =
Partial pressure of N2 =
= 81060 Nm-2
Partial pressure of O2 =
= 20265 Nm-2
10080
10020
2Nm 10132510080
2Nm 10132510020
New Way Chemistry for Hong Kong A-Level Book 178
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
2.5 Dalton’s law of partial pressures (SB p.36)
New Way Chemistry for Hong Kong A-Level Book 179
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)
=
= 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
3
3
dm 15dm 5 atm 15
3
3
dm 15dm 10 atm 12
New Way Chemistry for Hong Kong A-Level Book 180
2.5 Dalton’s law of partial pressures (SB p.37)
(b) Mole fraction of gas A =
=
= 0.385
Mole fraction of gas B =
=
= 0.615
total
A
P
P
atm 13atm 5
total
B
P
P
atm 13atm 8
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New Way Chemistry for Hong Kong A-Level Book 181
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) Answer
2.5 Dalton’s law of partial pressures (SB p.37)
New Way Chemistry for Hong Kong A-Level Book 182
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)
New Way Chemistry for Hong Kong A-Level Book 183
2.5 Dalton’s law of partial pressures (SB p.37)
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 184
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) Answer
2.5 Dalton’s law of partial pressures (SB p.38)
New Way Chemistry for Hong Kong A-Level Book 185
2.5 Dalton’s law of partial pressures (SB p.38)
(a) Number of moles of oxygen =
= 0.125 mol
Number of moles of nitrogen =
= 0.214 mol
Total number of moles of gases = (0.125 + 0.214) mol
= 0.339 mol
Mole fraction of oxygen = = 0.369
Mole fraction of nitrogen = = 0.631
1gmol 32.0g 4.0
1gmol 28.0g 6.0
mol 0.339mol 0.125
mol 0.339mol 0.214
New Way Chemistry for Hong Kong A-Level Book 186
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)
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New Way Chemistry for Hong Kong A-Level Book 187
2.5 Dalton’s law of partial pressures (SB p.39)
(a)Define mole fraction.
(b)State Dalton’s law of partial pressures.
Answer(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) 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 + … …
New Way Chemistry for Hong Kong A-Level Book 188
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?
Answer
New Way Chemistry for Hong Kong A-Level Book 189
2.5 Dalton’s law of partial pressures (SB p.39)
(c) By Boyle’s law: P1V1 = P2V2
Partial pressure of gas A =
= 3.00 atm
Partial pressure of gas B =
= 5.14 atm
By Dalton’s law: Ptotal = PA + PB
Final pressure of the system = 3.00 atm + 5.14 atm = 8.14 atm
3
3
dm 8) (6dm 6
atm 7
3
3
dm 8) (6dm 8
atm 9
New Way Chemistry for Hong Kong A-Level Book 190
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) Answer
New Way Chemistry for Hong Kong A-Level Book 191
2.5 Dalton’s law of partial pressures (SB p.39)
(d) (i) No. of moles of He =
= 0.50 mol
No. of moles of N2 =
= 0.11 mol
No. of moles of Ar =
= 0.10 mol
Total no. of moles of gases = 0.50 mol + 0.11 mol + 0.10 mol
= 0.71 mol
1-mol g 4.0g 2
1-mol g 2) (14.0g 3
1-mol g 39.9g 4
New Way Chemistry for Hong Kong A-Level Book 192
2.5 Dalton’s law of partial pressures (SB p.39)
Mole fraction of He = = 0.704
Mole fraction of N2 = = 0.155
Mole fraction of Ar = = 0.141
(ii) Let the total pressure of the system be P.
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
mol 0.71mol 0.50
mol 0.71mol 0.11
mol 0.71mol 0.10
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