1. Chapter 2 The Mole Concept 2.1 The Mole 2.2 Ideal Gas Equation
Determination of Relative Molecular Mass
The Faraday and the Mole

2. What is “Mole”? Item Unit used to count No. of items per unit Shoes
2.1 The Mole (SB p.28)
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 x 1023
for counting common objects
for counting particles like atoms, ions, molecules

3. Avogadro constant (the amount in
The Mole (SB p.28)
How large is the amount in 1 mole?
6.02 x 1023 =
Avogadro constant (the amount in
1 mole)

4. ? $ 6.02 x 1023 All the people in the world so that each get:
The Mole (SB p.28)
All the people in the world
$ 6.02 x 1023
so that each get:
$ 1000 note
count at a rate of 2 notes/sec ?
2000 years

5. How to find the number of moles?
2.1 The Mole (SB p.29)
How to find the number of moles?
Number of moles = 1 or
number of particles
= number of moles x (6.02 x 1023)

6. Why defining 6.02 x 1023 as the amount for one mole?
The Mole (SB p.29)
Why defining 6.02 x 1023 as the amount for one mole?
12 g carbon contains 6.02 x C atoms
The mole is the amount of substance containing as many particles as the number of atoms in 12 g of carbon-12.

7. Relative atomic masses
The Mole (SB p.29)
Molar mass
Relative mass
C atom
…….
6.02 x 1023 12
Relative atomic masses
H atom
…….
6.02 x 1023 1
1 g
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. Molar mass is the same as the relative atomic mass in grams.
The Mole (SB p.29)
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.

9. Calculations using molar mass
The Mole (SB p.30)
Calculations using molar mass
Question
We have a sample of 2 g of hydrogen atoms. What is the number of moles? (Relative atomic mass of H = 1)
…….
1 g (1 mole)
…….
1 g (1 mole)
mass
Number of moles =
molar mass
= 2

10. or Number of moles = Mass = number of moles x molar mass 2
The Mole (SB p.30)
Number of moles = 2 or
Mass
= number of moles x molar mass

11. The Mole (SB p.30)
Example 2-1
What is the mass of 0.2 mole of calcium carbonate? (R.a.m.* : C = 12.0, O = 16.0, Ca = 40.1)
Solution:
The chemical formula of calcium carbonate is CaCO3.
Molar mass of calcium carbonate
= ( x 16.0) g mol-1
= g mol-1
Answer

12. Mass of calcium carbonate = Number of moles x Molar mass
The Mole (SB p.30)
Solution: (cont’d)
Mass of calcium carbonate
= Number of moles x Molar mass
= 0.2 mol x g mol-1
= 20.02g

13. The Mole (SB p.31)
Example 2-2
Calculate the number of gold atoms in 20g of gold atom. (R.a.m. : Au = 197.0)
Solution:
Number of gold atoms in 20g of gold coin
= x 6.02 x 1023 mol-1
= 6.11 x 1023
Answer

14. Example 2-3 It is given that the molar mass of water is 18.0g mol-1.
The Mole (SB p.31)
Example 2-3
It is given that the molar mass of water is 18.0g mol-1.
(a)What is the mass of 4 moles of water molecule?
(b) How many molecules are there?
(c) How many atoms are there?
Solution:
Mass of water
= Number of moles x Molar mass
= 4 mol x 18.0 g mol-1
= 72.0 g
Answer

15. (b) There are 4 moles of water molecules. Number of water molecules
The Mole (SB p.31)
Solution: (cont’d)
(b) There are 4 moles of water molecules.
Number of water molecules
= Number of moles x Avogadro constant
= 4 mol x 6.02 x 1023 mol-1
= x 1024

16. 1 water molecule has 3 atoms (including
The Mole (SB p.31)
Solution: (cont’d)
(c)
1 water molecule has 3 atoms (including
2 hydrogen atoms and 1 oxygen atoms).
1 mole of water molecule has 3 moles of atoms.
Thus, 4 moles of water molecules have 12 moles
of atoms.
Number of atoms
= 12 mol x 6.02 x 1023 mol-1
= x 1024

17. The Mole (SB p.31)
Solution:
The chemical formula of magnesium chloride is MgCl2.
Molar mass of MgCl2
= ( x2) g mol-1
= 95.3 g mol-1
Number of moles of MgCl2 =
= mol
Example 2-4
A magnesium chloride solution contains 10 g of magnesium chloride solid
(a) Calculate the number of moles of magnesium chloride in the solution.
(b) Calculate the number of magnesium ions in the solution.
(c) Calculate the number of chloride ions in the solution.
(d) Calculate the total number of ions in the solution.
(R.a.m.: Mg = 24.3, Cl = 35.5)
Answer

18. = Number of moles of Mg2+ x Avogadro constant
The Mole (SB p.31)
Solution: (cont’d)
(b) 1 mole of MgCl2 contains 1 mole of Mg2+ and 2 moles of Cl- . Therefore, mole of MgCl2 contains mol x 6.02 x 1023 mol-1.
Number of Mg2+ ions
= Number of moles of Mg2+ x Avogadro constant
= mol x 6.02 x 1023 mol-1
= x 1022

19. (c) 0.105 mole of MgCl2 contains 0.21 mole of Cl- . Number of Cl- ions
The Mole (SB p.31)
Solution: (cont’d)
(c) mole of MgCl2 contains 0.21 mole of Cl- .
Number of Cl- ions
= Number of moles of Cl- x Avogadro constant
= 0.21 mol x 6.02 x 1023 mol-1
= x 1023
(d) Total number of ions
= x x 1023
= x 1023

20. Example 2-5 Solution: The chemical formula of carbon dioxide is CO2.
The Mole (SB p.32)
Solution:
The chemical formula of carbon dioxide is CO2.
Molar mass of CO2
= ( x 2) g mol-1
= 44.0 g mol-1
∵ Number of mole
= = ∴ =
Mass of a CO2 molecule=
= 7.31 x g
Example 2-5
What is the mass of carbon dioxide molecule?
(R.a.m. : C = 12.0, O = 16.0)
Answer

21. The Mole (SB p.32)
Check Point 2-1
Find the mass in grams of 0.01 mole of zinc sulphide.(R.a.m. : S = 32.1, Zn = 65.4)
Find the number of ions in 5.61 g of calcium oxide. (R.a.m. : O = 16.0, Ca = 40.1)
Find the number of atoms in 32.05g of sulphur dioxide. (R.a.m. : O 16.0, S = 32.1)
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 hydrogen atoms, and
(v) number of hydrogen atoms.
(a) Mass = No. of moles x Molar mass
Mass of ZnS
= 0.01 mol x ( ) g mol-1
= 0.01 mol x 95.7 g mol-1
= 0.975g
Answer

22. 1 CaO formula unit contains 1 Ca2+ ion and 1 O2- ion.
The Mole (SB p.32)
(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 mole x 2 = 0.2 mol
No of ions = 0.2 mol x 6.02 x 1023 mol-1
= x 1023

23. 1 SO2 molecule contains 1 S atom and 2O atoms.
The Mole (SB p.32)
(c) No. of moles of SO2 =
= 0.5 mol
1 SO2 molecule contains 1 S atom and 2O atoms.
No. of mole of atoms = 0.5 mole x 3
= 1.5 mol
No of atoms = 1.5 mol x 6.02 x 1023 mol-1
= 9.03 x 1023

24. (d) Molar mass of (NH4)2CO3 = 96.0 g mol-1 No. of mole of (NH4)2CO3 =
The Mole (SB p.32)
(d) Molar mass of (NH4)2CO3 = 96.0 g mol-1
No. of mole of (NH4)2CO3 =
∵ 1 mole (NH4)2CO3 gives 2 moles NH4+.
No. of moles of (NH4)2CO3 = 0.05 mol x 2
= 0.1 mol

25. (iii) ∵ 1 mole (NH4)2CO3 gives 1 mole CO32- .
The Mole (SB p.32)
(iii) ∵ 1 mole (NH4)2CO3 gives 1 mole CO32- .
No. of moles of CO32- = 0.05 mol
(iv) 1 (NH4)2CO3 formula unit contains 8H atoms.
No. of moles of H atoms = 0.05 mol x 8
= 0.4 mol
(v) No. of H atoms
= 0.4mol x 6.02 x 1023 mol-1 = x 1023

26. What is Molar Volume of Gases?
The Mole (SB p.33)
What is Molar Volume of Gases?
at 250C & 1 atm
(Room temp & pressure / rtp)

27. (Standard temp & pressure / stp)
The Mole (SB p.33)
22.4 dm3
22.4
22.4
22.4
at 00C & 1 atm
(Standard temp & pressure / stp)

28. at any other fixed temp & pressure
The Mole (SB p.33) V V v v v
at any other fixed temp & pressure

29. The Mole (SB p.33)
Avogadro’s Law
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.
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

30. The Mole (SB p.34)
Solution:
Molar mass of chloride gas (Cl2) = 35.5 x 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 x Molar volume
= 0.05 mol x 24.0 dm3mol-1
=1.2 dm3
Example 2-6
Find the volume occupied by 3.55 g of chloride gas at room temperature and pressure (Molar volume of gas at R.T.P. = 24.0 dm3 mol-1; R.a.m. : Cl = 35.5)
Answer

31. Example 2-7 Solution: Molar volume of carbon dioxide gas at S.T.P.
The Mole (SB p.34)
Solution:
Molar volume of carbon dioxide gas at S.T.P.
= 22.4 dm3 mol-1
Number of moles of CO2 =
= 2 x 10 –4 mol
Number of CO2 molecules
= 2 x 10-4 mol x 6.02 x 1023 mol-1
=1.204 x 1020
Example 2-7
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; Avogrado constant = 6.02 x 1023 mol-1)
Answer

32. Example 2-8 Solution: Molar mass of nitrogen gas (N2)
The Mole (SB p.35)
Solution:
Molar mass of nitrogen gas (N2)
= ( ) g mol-1
= 28.0 g mol-1
∵ Density = =
∴ Density of N2 =
= g dm-3
Example 2-8
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. (R.a.m. : N = 14.0)
Answer

33. Example 2-9 Solution: Number of moles of the gas
The Mole (SB p.35)
Solution:
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)
Example 2-9
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

34. The Mole (SB p.35)
Check Point 2-2
Find the volume of 0.6 g of hydrogen gas at room temperature and pressure. (R.a.m. : H = 1.0; molar volume at R.T.P. = 24.0 dm3 mol-1)
Calculate the number of molecules in 4.48 dm3 of hydrogen at standard temperature and pressure. ( Molar volume at S.T.P. = 22.4 dm3 mol-1)
The molar volume of oxygen is 22.4 dm3 mol-1 at standard temperature and pressure. Find the density of oxygen in g cm-3 at S.T.P.. (R.a.m. : O = 16.0)
What mass of oxygen has the same number of moles as that in 3.2 g of sulphur dioxide? (R.a.m. : O = 16.0, S = 32.1)
No. of moles of H2 =
= 0.3 mol
Volume = No. of moles x Molar volume
= 0.3 mol x 24.0 dm3 mol-1
= 7.2 dm3
Answer

35. No of H2 molecules = 0.2 mol x 6.02 x 1023 mol-1 = 1.204 x 1023
The Mole (SB p.35)
(b) No. of moles of H2 =
= 0.2 mol
No of H2 molecules = 0.2 mol x 6.02 x 1023 mol-1
= x 1023

36. Molar volume of O2 = 22..4 dm3 mol-1 = 22 400 cm3 mol-1 Density =
The Mole (SB p.35)
( c) Density = =
Molar mass of O2 = 16.0 x 2 g mol-1
=32.0 g mol-1
Molar volume of O2 = dm3 mol-1
= cm3 mol-1
Density =
= 1.43 x 10-3 g cm-3

37. Mass = No. of moles x Molar mass Mass of O2 = 0.05 mol x 16.0g mol-1
The Mole (SB p.35)
(d) No. of moles of SO2 =
= 0.05 mol
No of moles of O2 = 0.05 mol
Mass = No. of moles x Molar mass
Mass of O2 = 0.05 mol x 16.0g mol-1
= 1.6g

38. 2.2 Ideal Gas Equation (SB p.36)
Gas Laws
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
PV = constant

39. 2.2 Ideal Gas Equation (SB p.36)

40. 2.2 Ideal Gas Equation (SB p.37)
Charles’ Law states that:
At a constant pressure, the volume of a given mass of a gas is directly proportional to the absolute temperature.

41. 2.2 Ideal Gas Equation (SB p.37)

42. 2.2 Ideal Gas Equation (SB p.37)
The Boyle’s law and Charles’ law
gives the ideal gas equation
Combining:
V  n (Avogadro’s Law)
V  1/P (Boyle’s Law)
V  T (Charles Law)
V  nT/P
V = RnT/P where R is a constant (called the Universal Gas Constant)
PV = nRT

43. 2.2 Ideal Gas Equation (SB p.37)
For one mole of an ideal gas at standard temperature and pressure,
P = 760 mmHg, 1 atm or Nm-2 (Pa)
V = 22.4 dm3 mol-1 or 22.4 x 10-3 m3 mol-1
T = 0 oC or 273K
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 = PV/T =
= JK-1mol-1

44. 2.2 Ideal gas equation (SB p.38)
Example 2-10
A 500 cm3 sample of a gas in a sealed container at 700 mm Hg and 25oC is heated to 100oC. What is the final pressure of the gas?
Solution:
As the number of moles of the gas is fixed, PV/T should be a constant =
P2 = mmHg.
Note: All temperature values used in gas laws are on the Kelvin scale.
Answer

45. 2.2 Ideal gas equation (SB p.38)
Example 2-11
A reaction vessel of 500 cm3 is filled with oxygen at 25oC and the final pressure exerted on it is Nm-2. How many moles of oxygen are there? (Ideal gas constant = J K-1 mol-1)
Solution:
PV = nRT
Nm-2 x 500 x 10-6 m3
= n x J K-1 mol-1 x ( ) K
n = 0.02 mol
There is 0.02 mole of oxygen in the reaction vessel. .
Answer

46. 2.2 Ideal gas equation (SB p.39)
Example 2-12
A 5 dm3 vessel can withstand a maximum internal pressure of 50 atm. If 2 moles of nitrogen gas is pumped into the vessel, what is the highest temperature it can be safely heated to? (1 atm = Nm-2, ideal gas constant = J K-1 mol-1)
Solution:
Applying the equation,
T = =
= K
The highest temperature it can be safely heated to is oC.
Answer

47. 2.2 Ideal gas equation (SB p.39)
Check Point 2-3
A reaction vessel is filled with a gas at 20oC and 5atm. If the vessel can withstand a maximum internal pressure of 10 atm, what is the highest temperature it can be safely heated to?
A balloon is filled with helium at 25oC. 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? ( 1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
(c) 25.8 cm3 sample of a gas has a pressure of 690 mm Hg and a temperature of 17oC. What is the volume if the pressure is changed to 1.85 atm and the temperature to 345 K? ( 1 atm = 760 mmHg)
T2 = 586 K
Answer

48. 2.2 Ideal gas equation (SB p.39)
(b) PV= nRT
1.5 x Nm-2 x 450 x 10-6 m3
= n x J K-1 mol-1 x (273+25) K
n = mol

49. 2.2 Ideal gas equation (SB p.39)
(c )
V=15.06cm3

50. 2.3 Determination of Relative Molecular Mass (SB p.39)
Mass of volatile liquid injected = = g

51. 2.3 Determination of Relative Molecular Mass (SB p.39)
Volume of trichloromethane vapour = = 66.2 cm3

52. 2.3 Determination of Relative Molecular Mass (SB p.39)
Temperature = = 372 K

53. 2.3 Determination of Relative Molecular Mass (SB p.39)
Pressure = Nm-2

54. 2.3 Determination of Relative Molecular Mass (SB p.40)

55. 2.3 Determination of Relative Molecular Mass (SB p.40)

56. 2.3 Determination of Relative Molecular Mass (SB p.40)
PV = nRT………..(1)
n = m / M………(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 = (m/M) RT
M = (mRT)/PV

57. 2.3 Determination of Relative Molecular Mass (SB p.41)
Example 2-13
A sample of gas occupying a volume of 50 cm3 at 1 atm and 25oC is found to have a mass of g. Find the relative molecular mass of the gas. ( Ideal gas constant = J K-1mol-1; 1 atm = Nm-2)
Solution:
PV = m/M RT
Nm-2 x 50 x 10-6 m3
= x JK-1 mol-1 x ( ) K
∴ M = g mol-1
Therefore, the relative molecular mass of the gas is
Answer

58. 2.3 Determination of Relative Molecular Mass (SB p.42)
Solution:
The unit of density of the gas has to be converted to g m-3 for the calculation.
g dm-3 = x 103 gm3 = 33.7 g m-3
PM = RT
M = RT/ P =
= 4.0 g mol-1
Therefore, the relative molecular mass of the gas
is 4.0.
Example 2-14
The density of a gas at 450 oC and 380 mmHg is g dm-3. What is its relative molecular mass? ( 1 atm = 760 mmHg = Nm-2; ideal gas constant = J K-1mol-1)
Answer

59. 2.3 Determination of Relative Molecular Mass (SB p.42)
Check Point 2-4
0.204 g of phosphorus vapour occupies a volume of 81.0 cm3 at 327oC and 1 atm. Determine the relative molecular mass of phosphorus. ( 1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
A sample of gas has a mass of 12.0g and occupies a volume of 4.16dm3 measured at 97oC and 1.62 atm. Calculate the relative molecular mass of the gas.
( 1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
(c) A sample of 0.037g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25oC 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)
PV = m/M RT
Nm-2 x 81.0 x 10-6m3
= (0.204 g/ M) x 8.314J K-1 mol-1 x
( ) K
M = g mol-1
∴ The relative molecular mass of phosphorus is
Answer

60. 2.3 Determination of Relative Molecular Mass (SB p.42)
(b) PV= (m/M)RT
1.62x Nm-2 x 4.16 x 10-3m3
= 12.0g/ M x J K-1 mol-1 x (273+97)K
M = g mol-1
∴ The relative molecular mass of the gas is

61. 2.3 Determination of Relative Molecular Mass (SB p.42)
(c) Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g)
PV = nRT
740/760 x Nm-2 x 38.2 x 10-6 m3
= n x J K-1 mol-1 x ( ) K
n = 1.52 x 10-3 mol
∴ No. of moles of H2 produced = 1.52 x 10-3.

62. 2.3 Determination of Relative Molecular Mass (SB p.42)
No. of mole of Mg reacted
= No. of moles of H2 produced = 1.52 x 10-3 mol
Molar mass of Mg = Mass/ No. of moles
= 0.037g / 1.52 x 10-3 mol-1= 24.3 g mol-1
∴ The relative atomic mass of Mg is 24.3.

63. 2.4 Dalton’s Law of Partial Pressures (SB p. 43)
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

64. 2.4 Dalton’s Law of Partial Pressures (SB p. 43)
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)

65. 2.4 Dalton’s Law of Partial Pressures (SB p. 43)
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

66. 2.4 Dalton’s Law of Partial Pressures (SB p.43)
Example 2-15
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 25oC? ( 1 atm = Nm-2)
Solution:
Mole of fraction of N2 = 80/100
Mole of fraction of O2 = 20/100
Partial pressure of N2 = 80/100 x Nm-2
= Nm-2
Partial pressure of O2 = 20/100 x Nm-2
= Nm-2
Answer

67. 2.4 Dalton’s Law of Partial Pressures (SB p.44)
Solution:
Total volume of the system
= (5 + 10) dm3 = 15 dm3
By Boyle’s Law: P1V1 = P2V2
Partial pressure of gas A
= = 5 atm
Partial pressure of gas B
= = 8 atm
Example 2-16
The value 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.
Assuming that the temperature of the system remains constant, what is the final pressure in the vessel?
What are the mole fractions of gas A and gas B?
Answer

68. 2.4 Dalton’s Law of Partial Pressures (SB p.44)
Solution: (cont’d)
By Dalton’s law: Ptotal = PA + PB
Total pressure of the system= (5 + 8) atm = 13 atm
(b) Mole fraction of gas A
= PA/Ptotal = 5 atm/13 atm = 0.385
Mole fraction of gas B
= PB/Ptotal = 8atm/13 atm = 0.615

69. 2.4 Dalton’s Law of Partial Pressures (SB p. 44)
Solution:
Let the partial pressure of nitrogen be PA.
Using the ideal gas equation PV = nRT,
PA x 12 x 10-3 m3
= 0.25 mol x JK-1 mol-1 x ( ) K
PA = Nm-2 ( or atm)
Let the partial pressure of oxygen be PB.
PB = mol x JK-1 mol-1 x ( ) K
PB = Nm-2 ( or atm)
Example 2-17
0.25 mole of nitrogen and 0.30 mole 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

70. 2.4 Dalton’s Law of Partial Pressures (SB p. 44)
Solution: (cont’d)
Total pressure
= ( ) Nm-2 = Nm-2
Hence, the partial pressures of nitrogen and oxygen are atm and atm respectively, and the total pressure of the mixture is atm.

71. 2.4 Dalton’s Law of Partial Pressures (SB p. 45)
Solution:
(a) Number of moles of oxygen =
= mol
Number of moles of nitrogen
= mol
Total number of moles of gases
= ( ) mol = mol
Example 2-18
4.0 g oxygen and 6.0 g of nitrogen are introduced into a 5 dm3 vessel at 27oC.
What are the mole fractions of oxygen and nitrogen in the mixture?
What is the final pressure of the system? ( 1 atm = Nm-2; ideal gas constant = J K-1 mol-1; R.a.m. : N = 14.0, O = 16.0)
Answer

72. 2.4 Dalton’s Law of Partial Pressures (SB p. 45)
Solution: (cont’d)
Mole fraction of oxygen = = 0.369
Mole fraction of nitrogen = = 0.631
(b) Let P be the total pressure of the system.
By ideal gas equation PV = nRT,
P x 5 x 10-3 m3
= mol x JK-1mol-1 x ( )K
P = Nm-2 ( or 1.67 atm)

73. 2.4 Dalton’s Law of Partial Pressures (SB p. 45)
By Boyle’s law: P1V1 = P2V2
Partial pressure of gas A
= 7 atm x
= 3.00 atm
Partial pressure of gas B
= 9 atm x
= 5.14 atm
Final pressure = 3.00 atm atm
= 8.14 atm
Check Point 2-5
The valve between a 6 dm3 vessel containing gas A at a pressire 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? ( 1 atm = Nm-2; ideal gas constant = J K-1 mol-1)
Answer

74. 2.4 Dalton’s Law of Partial Pressures (SB p. 45)
(i)No. of moles of He = = 0.05 mol
No. of moles of N2 = = 0.11 mol
No. of moles of Ar = = 0.10 mol
Total no. of moles of gases
= 0.05 mol mol mol = 0.71 mol
Mole faction of He = = 0.704
Mole fraction of N2 = = 0.155
Mole fraction of Ar = = 0.141
Check Point 2-5
(b) 2 g of helium, 3 g of nitrogen and 4 g of argon are introduced into 15 dm3 vessel at 100 oC.
What are the mole fractions of helium, nitrogen and argon in the system?
Calculate the total pressure of the system, and hence the partial pressures of helium, nitrogen and argon.
(1 atm = Nm-2; ideal gas constant
= J K-1mol-1 ; R.a.m. : He = 4.0, N = 14.0, Ar = 39.9)
Answer

75. 2.4 Dalton’s Law of Partial Pressures (SB p. 45)
(ii) Let the total pressure of the system be P.
PV = nRT
P x 15 x 10-3m3
= 0.71 mol x JK-1mol-1 x ( ) K
P = Nm-2
Partial pressure of He
= Nm-2 x = Nm-2
Partial pressure of N2
= Nm-2 x = Nm-2
Partial pressure of Ar
= Nm-2 x = Nm-2

76. The Faraday and the Mole?
2.5 The Faraday and the Mole (SB p. 46)
The Faraday and the Mole?
Quantity of electricity = current x time (coulombs, C) (amperes, A) (seconds, s)

77. Faraday’s first law of electrolysis
2.5 The Faraday and the Mole (SB p. 46)
Faraday’s first law of electrolysis
Faraday’s First Law of Electrolysis
The mass of a substance liberated at or dissolved from an electrode during electrolysis is directly proportional to the quantity of electricity passing through the electrolyte.
Cu2+(aq) e-  Cu(s)
vary directly

78. Faraday’s second law of electrolysis
2.5 The Faraday and the Mole (SB p. 47)
Faraday’s second law of electrolysis

79. 2.5 The Faraday and the Mole (SB p. 47)
Faraday’s Second Law of Electrolysis
The no. of moles of different ions discharged by the same quantity of electricity is inversely proportional to their respective charge.
2Ag+(aq) e-  2Ag(s)
Ag+(aq) e-  Ag(s)
Cu2+(aq) e-  Cu(s)

80. Faraday Constant Element Electrolyte
2.5 The Faraday and the Mole (SB p. 48)
Faraday Constant
Element
Electrolyte
No. of coulombs to liberates 1 mole of metal atom
Iron
Copper
Zinc
Silver
Fe2(SO4)3(aq)
CuSO4(aq)
Fe(NO3)2(aq)
ZnCl2(aq)
AgNO3(aq)
96 500

81. 2.5 The Faraday and the Mole (SB p. 48)
Charge of 1 mole of electrons
= 1 Faraday
= coulombs
Number of moles of product formed
= It / nF

82. 2.5 The Faraday and the Mole (SB p. 48)
Example 2-19
How many Faradays have been pssing through a resistance in a circuit carrying a current of 5 A for 1 hour? ( 1 F = C)
Solution:
Q = It = 5A x ( 60 x 60) s = C
Number of moles of electrons =
= mol
The number of Faradays passed is
Answer

83. 2.5 The Faraday and the Mole (SB p. 49)
Example 2-20
What is the number of moles of silver formed when a current of 0.3 A is passed through a silver nitrate solution for 30 minutes?
Solution:
Ag+(aq) + e Ag(s)
To form 1 mole of Ag, 1 mole of electrons (i.e. 1 F) is required.
Number of moles of Ag formed = It / nF =
= 5.60 x 10-3 mol
Answer

84. 2.5 The Faraday and the Mole (SB p. 49)
Solution:
Cu2+(aq) + 2e Cu(s)
To discharge 1 mole of Cu2+, 2 moles of electrons
(i.e. 2F) are required.
Number of moles of Cu formed = It/nF =
= 4.66 x 10-3 mol
Mass of Cu formed = 4.66 x 10-3 mol x 63.5g mol-1
= g
Example 2-21
What is the mass of copper formed at the cathode when a current of 0.25 A is passed through a copper(II) sulphate solution for 1 hour ( R.a.m. : Cu = 63.5)?
Answer

85. 2.5 The Faraday and the Mole (SB p. 50)
Solution:
When a dilute sulphuric acid is electrolysed, hydrogen is formed at the cathode and oxygen is formed at the anode of the electrolytic cell.
At cathode: 2H+(aq) + 2e H2(g)
To give 1 mole of hydrogen gas, 2 moles of electrons (i.e. 2F) are required.
Number of moles of H2 (g) formed= It/nF =
= mol
Mass of H2(g) formed
= mol x 1.0 x 2 g mol-1 = g
Example 2-22
Find the masses of products formed when a dilute sulphuric acid solution is electrolysed with a current of 0.6 A for 90 minutes.
(R.a.m. : H = 1.0, O = 16.0)
Answer

86. 2.5 The Faraday and the Mole (SB p. 50)
Solution: (cont’d).
At anode: 4OH-(aq) O2(g) + 2H2O(l) + 4e-
To give 1 mole of oxygen gas, 4 moles of electrons (i.e. 4F) are given out by the hydroxide ions.
Number of moles of O2 (g) formed= It/nF =
= x 10-3 mol
Mass of O2(g) formed
= x 10-3 mol x 16.0 x 2 g mol-1 = g

87. 2.5 The Faraday and the Mole (SB p. 50)
Example 2-23
What mass of copper would be desposited by the quantity of electricity that liberates 2.4 dm3 of oxygen measured at room temperature and pressure?
(Molar volume of gas at R.T.P. = 24.0 dm3 mol-1;
R.a.m. : O = 16.0, Cu = 63.5)
Solution:
4OH-(aq) O2(g) + 2H2O(l) + 4e-
To give 1 mole of oxygen gas, 4 mole of electrons (i.e. 4 F)are given out by the hydroxide ions.
Number of moles of O2 given out
= = 0.1 mol
Number of moles of electrons given out
= 0.1 mol x 4 = 0.4 mol
Answer

88. 2.5 The Faraday and the Mole (SB p. 51)
Solution: (cont’d)
Cu2+(aq) + 2e Cu(s)
To discharge 1 mole of Cu2+, 2 moles of electrons (i.e. 2F) are required.
Number of moles of Cu formed = 0.4 mol / 2
= 0.2 mol
Mass of Cu deposited = 0.2 mol x 63.5 g mol-1
= 12.7 g

89. 2.5 The Faraday and the Mole (SB p. 51)
(a) Q = It
C = 0.35 A x t
t = s
(b) Cu2+(aq) + 2e Cu(s)
No. of moles of Cu formed = It/2F =
I = 6.43 A
Check Point 2-6
What current in amperes is required to deposit 6.35g of copper in 50 minutes from a copper(II) sulphate solution? (1 F = ; R.a.m. : Cu = 63.5)
(b) What is the time required to pass 1 Faraday of electricity through an electrolyte with a current of 0.35A? ( 1F = C)
Answer

90. 2.5 The Faraday and the Mole (SB p. 51)
(c)
Al3+(l) + 3e Al(s)
No. of moles of Al(s) formed = It / 3F =
= mol
Mass of Al formed
= mol x 27.0g mol-1 =
Check Point 2-6
(c) Calculate the mass of aluminium that would be deposited during the electrolysis of a molten aluminum salt by a current of 10 A for 5 hours. ( 1F = C; R.a.m. : Al = 27.0)
(d) A current of 0.37A flowing for 15 minutes through an electrolyte liberates 0.20 g of metal X. what mass of X would be liberated by a current of 0.30 A for minutes?
Answer

91. 2.5 The Faraday and the Mole (SB p. 51)
(d) No. of moles of X formed =
Since molar mass, n and F are constants,
It /( Mass of X) is a constant. =
Mass of X = g

92. The END