new way chemistry for hong kong a-level book 11 the mole concept 2.1the mole 2.2 ideal gas equation...

New Way Chemistry for Hong Kong A-Level Book 11 The Mole Concept 2.1The Mole 2.2 Ideal Gas Equation 2.3 Determination of Relative Molecular Mass 2.4 The Faraday and the Mole Chapter 2

New Way Chemistry for Hong Kong A-Level Book 12 2.1 The Mole (SB p.28) What is Mole? ItemUnit used to countNo. of items per unit Shoespairs2 Eggsdozens12 Paperreams500 Particles in Chemistry moles6.02 x 10 23 for counting particles like atoms, ions, molecules for counting common objects

New Way Chemistry for Hong Kong A-Level Book 13 6.02 x 10 23 = 602000000000000000000000 Avogadro constant (the amount in 1 mole ) How large is the amount in 1 mole? 2.1 The Mole (SB p.28)

New Way Chemistry for Hong Kong A-Level Book 14 $ 6.02 x 10 23 All 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.28)

New Way Chemistry for Hong Kong A-Level Book 15 1 or number of particles = number of moles x (6.02 x 10 23 ) number of particles = number of moles x (6.02 x 10 23 ) How to find the number of moles? 2.1 The Mole (SB p.29) Number of moles =

New Way Chemistry for Hong Kong A-Level Book 16 Why defining 6.02 x 10 23 as the amount for one mole? 12 g carbon contains 6.02 x 10 23 12 C 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.29)

New Way Chemistry for Hong Kong A-Level Book 17 Relative mass 12 1 C atom . 6.02 x 10 23 H atom . 6.02 x 10 23 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.29)

New Way Chemistry for Hong Kong A-Level Book 18 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.29)

New Way Chemistry for Hong Kong A-Level Book 19 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) Number of moles = = 2 massmolar mass 2.1 The Mole (SB p.30)

New Way Chemistry for Hong Kong A-Level Book 110 2 or Mass = number of moles x molar mass Mass = number of moles x molar mass Number of moles = 2.1 The Mole (SB p.30)

New Way Chemistry for Hong Kong A-Level Book 111 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 CaCO 3. Molar mass of calcium carbonate = (40.1 + 12.0 + 3 x 16.0) g mol -1 = 100.1 g mol -1 Answer 2.1 The Mole (SB p.30)

New Way Chemistry for Hong Kong A-Level Book 112 Solution: (contd) Mass of calcium carbonate = Number of moles x Molar mass = 0.2 mol x 100.1 g mol -1 = 20.02g 2.1 The Mole (SB p.30)

New Way Chemistry for Hong Kong A-Level Book 113 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 10 23 mol -1 = 6.11 x 10 23 Answer 2.1 The Mole (SB p.31)

New Way Chemistry for Hong Kong A-Level Book 114 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: (a)Mass of water = Number of moles x Molar mass = 4 mol x 18.0 g mol -1 = 72.0 g Answer 2.1 The Mole (SB p.31)

New Way Chemistry for Hong Kong A-Level Book 115 Solution: (contd) (b) There are 4 moles of water molecules. Number of water molecules = Number of moles x Avogadro constant = 4 mol x 6.02 x 10 23 mol -1 = 2.408 x 10 24 2.1 The Mole (SB p.31)

New Way Chemistry for Hong Kong A-Level Book 116 Solution: (contd) (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 10 23 mol -1 = 7.224 x 10 24 2.1 The Mole (SB p.31)

New Way Chemistry for Hong Kong A-Level Book 117 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 2.1 The Mole (SB p.31) Solution: (a)The chemical formula of magnesium chloride is MgCl 2. Molar mass of MgCl 2 = (24.3 + 35.5 x2) g mol -1 = 95.3 g mol -1 Number of moles of MgCl 2 = = 0.105 mol

New Way Chemistry for Hong Kong A-Level Book 118 Solution: (contd) (b) 1 mole of MgCl 2 contains 1 mole of Mg 2+ and 2 moles of Cl -. Therefore, 0.105 mole of MgCl 2 contains 0.105 mol x 6.02 x 10 23 mol -1. Number of Mg 2+ ions = Number of moles of Mg 2+ x Avogadro constant = 0.105 mol x 6.02 x 10 23 mol -1 = 6.321 x 10 22 2.1 The Mole (SB p.31)

New Way Chemistry for Hong Kong A-Level Book 119 Solution: (contd) (c) 0.105 mole of MgCl 2 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 10 23 mol -1 = 1.264 x 10 23 (d) Total number of ions = 6.321 x 10 22 + 1.264 x 10 23 = 1.896 x 10 23 2.1 The Mole (SB p.31)

New Way Chemistry for Hong Kong A-Level Book 120 Example 2-5 What is the mass of carbon dioxide molecule? (R.a.m. : C = 12.0, O = 16.0) Answer 2.1 The Mole (SB p.32) Solution: The chemical formula of carbon dioxide is CO 2. Molar mass of CO 2 = (12.0 + 16.0 x 2) g mol -1 = 44.0 g mol -1 Number of mole = = = Mass of a CO 2 molecule= = 7.31 x 10 -23 g

New Way Chemistry for Hong Kong A-Level Book 121 2.1 The Mole (SB p.32) Check Point 2-1 (a)Find the mass in grams of 0.01 mole of zinc sulphide.(R.a.m. : S = 32.1, Zn = 65.4) (b)Find the number of ions in 5.61 g of calcium oxide. (R.a.m. : O = 16.0, Ca = 40.1) (c)Find the number of atoms in 32.05g of sulphur dioxide. (R.a.m. : O 16.0, S = 32.1) (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 hydrogen atoms, and (v) number of hydrogen atoms. Answer (a) Mass = No. of moles x Molar mass Mass of ZnS = 0.01 mol x (65.0 + 32.1) g mol -1 = 0.01 mol x 95.7 g mol -1 = 0.975g

New Way Chemistry for Hong Kong A-Level Book 122 2.1 The Mole (SB p.32) (b) No. of moles of CaO = = 0.1 mol 1 CaO formula unit contains 1 Ca 2+ ion and 1 O 2- ion. No. of moles of ions = 0.1 mole x 2 = 0.2 mol No of ions = 0.2 mol x 6.02 x 10 23 mol -1 = 1.204 x 10 23

New Way Chemistry for Hong Kong A-Level Book 123 2.1 The Mole (SB p.32) (c) No. of moles of SO 2 = = 0.5 mol 1 SO 2 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 10 23 mol -1 = 9.03 x 10 23

New Way Chemistry for Hong Kong A-Level Book 124 2.1 The Mole (SB p.32) (d) Molar mass of (NH 4 ) 2 CO 3 = 96.0 g mol- 1 (i)No. of mole of (NH 4 ) 2 CO 3 = (ii) 1 mole (NH 4 ) 2 CO 3 gives 2 moles NH 4+. No. of moles of (NH 4 ) 2 CO 3 = 0.05 mol x 2 = 0.1 mol

New Way Chemistry for Hong Kong A-Level Book 125 2.1 The Mole (SB p.32) (iii) 1 mole (NH 4 ) 2 CO 3 gives 1 mole CO 3 2-. No. of moles of CO 3 2- = 0.05 mol (iv) 1 (NH 4 ) 2 CO 3 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 10 23 mol -1 = 2.408 x 10 23

New Way Chemistry for Hong Kong A-Level Book 126 What is Molar Volume of Gases? at 25 0 C & 1 atm (Room temp & pressure / rtp) 2.1 The Mole (SB p.33)

New Way Chemistry for Hong Kong A-Level Book 127 at 0 0 C & 1 atm (Standard temp & pressure / stp) 22.4 2.1 The Mole (SB p.33) 22.4 dm 3

New Way Chemistry for Hong Kong A-Level Book 128 at any other fixed temp & pressure vv v 2.1 The Mole (SB p.33) V V V V

New Way Chemistry for Hong Kong A-Level Book 129 Avogadros 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 2.1 The Mole (SB p.33)

New Way Chemistry for Hong Kong A-Level Book 130 Answer 2.1 The Mole (SB p.34) 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 dm 3 mol -1 ; R.a.m. : Cl = 35.5) Solution: Molar mass of chloride gas (Cl 2 ) = 35.5 x 2 g mol -1 = 71.0 g mol -1 Number of moles of Cl 2 = = 0.05 mol Volume of Cl 2 = Number of moles of Cl 2 x Molar volume = 0.05 mol x 24.0 dm 3 mol -1 =1.2 dm 3

New Way Chemistry for Hong Kong A-Level Book 131 Example 2-7 Find the number of molecules in 4.48 cm 3 of carbon dioxide gas at standard temperature and pressure. (Molar volume of gas at S.T.P. = 22.4 dm 3 mol -1 ; Avogrado constant = 6.02 x 10 23 mol -1 ) Answer 2.1 The Mole (SB p.34) Solution: Molar volume of carbon dioxide gas at S.T.P. = 22.4 dm 3 mol -1 Number of moles of CO 2 = = 2 x 10 4 mol Number of CO 2 molecules = 2 x 10 -4 mol x 6.02 x 10 23 mol -1 =1.204 x 10 20

New Way Chemistry for Hong Kong A-Level Book 132 Example 2-8 The molar volume of nitrogen gas is found to be 24.0 dm 3 mol -1 at room temperature and pressure. Find the density of nitrogen gas. (R.a.m. : N = 14.0) 2.1 The Mole (SB p.35) Solution: Molar mass of nitrogen gas (N 2 ) = (14.0 + 14.0) g mol -1 = 28.0 g mol -1 Density = = Density of N 2 = = 1.167 g dm -3 Answer

New Way Chemistry for Hong Kong A-Level Book 133 Example 2-9 1.6 g of a gas occupies 1.2 dm 3 at room temperature and pressure. What is the relative molecular mass of the gas? (Molar Volume of gas at R.T.P. = 24.0 dm 3 mol -1 ) 2.1 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) Answer

New Way Chemistry for Hong Kong A-Level Book 134 2.1 The Mole (SB p.35) Check Point 2-2 (a)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 dm 3 mol -1) (b)Calculate the number of molecules in 4.48 dm 3 of hydrogen at standard temperature and pressure. ( Molar volume at S.T.P. = 22.4 dm 3 mol -1 ) (c)The molar volume of oxygen is 22.4 dm 3 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) (d)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) Answer (a)No. of moles of H 2 = = 0.3 mol Volume = No. of moles x Molar volume = 0.3 mol x 24.0 dm 3 mol -1 = 7.2 dm 3

New Way Chemistry for Hong Kong A-Level Book 135 2.1 The Mole (SB p.35) (b) No. of moles of H 2 = = 0.2 mol No of H 2 molecules = 0.2 mol x 6.02 x 10 23 mol -1 = 1.204 x 10 23

New Way Chemistry for Hong Kong A-Level Book 136 2.1 The Mole (SB p.35) ( c) Density = = Molar mass of O 2 = 16.0 x 2 g mol -1 =32.0 g mol -1 Molar volume of O 2 = 22..4 dm 3 mol -1 = 22 400 cm 3 mol -1 Density = = 1.43 x 10 -3 g cm -3

New Way Chemistry for Hong Kong A-Level Book 137 2.1 The Mole (SB p.35) (d) No. of moles of SO 2 = = 0.05 mol No of moles of O 2 = 0.05 mol Mass = No. of moles x Molar mass Mass of O 2 = 0.05 mol x 16.0g mol -1 = 1.6g

New Way Chemistry for Hong Kong A-Level Book 138 Gas Laws 2.2 Ideal Gas Equation (SB p.36) Boyles 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

New Way Chemistry for Hong Kong A-Level Book 139 2.2 Ideal Gas Equation (SB p.36)

New Way Chemistry for Hong Kong A-Level Book 140 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.

New Way Chemistry for Hong Kong A-Level Book 141 2.2 Ideal Gas Equation (SB p.37)

New Way Chemistry for Hong Kong A-Level Book 142 Combining: V n (Avogadros Law) V 1/P (Boyles Law) V T (Charles Law) V nT/P V = RnT/P where R is a constant (called the Universal Gas Constant) PV = nRT 2.2 Ideal Gas Equation (SB p.37) The Boyles law and Charles law gives the ideal gas equation

New Way Chemistry for Hong Kong A-Level Book 143 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 101 325 Nm -2 (Pa) V = 22.4 dm 3 mol -1 or 22.4 x 10 -3 m 3 mol -1 T = 0 o C 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 = = 8.314 JK -1 mol -1

New Way Chemistry for Hong Kong A-Level Book 144 Example 2-10 A 500 cm 3 sample of a gas in a sealed container at 700 mm Hg and 25 o C is heated to 100 o C. What is the final pressure of the gas? 2.2 Ideal gas equation (SB p.38) Solution: As the number of moles of the gas is fixed, PV/T should be a constant = P 2 = 876.17 mmHg. Note: All temperature values used in gas laws are on the Kelvin scale. Answer

New Way Chemistry for Hong Kong A-Level Book 145 Example 2-11 A reaction vessel of 500 cm 3 is filled with oxygen at 25 o C and the final pressure exerted on it is 101 325 Nm -2. How many moles of oxygen are there? (Ideal gas constant = 8.314 J K -1 mol -1 ) 2.2 Ideal gas equation (SB p.38) Solution: PV = nRT 101 325 Nm -2 x 500 x 10 -6 m 3 = n x 8.314 J K -1 mol -1 x (273 + 25) K n = 0.02 mol There is 0.02 mole of oxygen in the reaction vessel.. Answer

New Way Chemistry for Hong Kong A-Level Book 146 Example 2-12 A 5 dm 3 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 = 101 325 Nm -2, ideal gas constant = 8.314 J K -1 mol -1 ) 2.2 Ideal gas equation (SB p.39) Solution: Applying the equation, T = = = 1523.4 K The highest temperature it can be safely heated to is 1250.4 o C. Answer

New Way Chemistry for Hong Kong A-Level Book 147 2.2 Ideal gas equation (SB p.39) Check Point 2-3 (a)A reaction vessel is filled with a gas at 20 o C 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? (b)A balloon is filled with helium at 25 o C. The pressure exerted and the volume of balloon are found to be 1.5 atm and 450 cm 3 respectively. How many moles of helium have been introduced into the balloon? ( 1 atm = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) (c) 25.8 cm 3 sample of a gas has a pressure of 690 mm Hg and a temperature of 17 o C. What is the volume if the pressure is changed to 1.85 atm and the temperature to 345 K? ( 1 atm = 760 mmHg) Answer (a) T 2 = 586 K

New Way Chemistry for Hong Kong A-Level Book 148 (b) PV= nRT 1.5 x 101 325 Nm -2 x 450 x 10 -6 m 3 = n x 8.314 J K -1 mol -1 x (273+25) K n = 0.0276 mol 2.2 Ideal gas equation (SB p.39)

New Way Chemistry for Hong Kong A-Level Book 149 2.2 Ideal gas equation (SB p.39) (c ) V=15.06cm 3

New Way Chemistry for Hong Kong A-Level Book 150 2.3 Determination of Relative Molecular Mass (SB p.39) Mass of volatile liquid injected = 26.590 - 26.330 = 0.260 g

New Way Chemistry for Hong Kong A-Level Book 151 2.3 Determination of Relative Molecular Mass (SB p.39) Volume of trichloromethane vapour = 74.4 - 8.2 = 66.2 cm 3

New Way Chemistry for Hong Kong A-Level Book 152 2.3 Determination of Relative Molecular Mass (SB p.39) Temperature = 273 + 99 = 372 K

New Way Chemistry for Hong Kong A-Level Book 153 2.3 Determination of Relative Molecular Mass (SB p.39) Pressure = 101325 Nm -2

New Way Chemistry for Hong Kong A-Level Book 154 2.3 Determination of Relative Molecular Mass (SB p.40)

New Way Chemistry for Hong Kong A-Level Book 155 2.3 Determination of Relative Molecular Mass (SB p.40)

New Way Chemistry for Hong Kong A-Level Book 156 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

New Way Chemistry for Hong Kong A-Level Book 157 Example 2-13 A sample of gas occupying a volume of 50 cm 3 at 1 atm and 25 o C is found to have a mass of 0.0286 g. Find the relative molecular mass of the gas. ( Ideal gas constant = 8.314 J K -1 mol -1 ; 1 atm = 101 325 Nm -2 ) Answer 2.3 Determination of Relative Molecular Mass (SB p.41) Solution: PV = m/M RT 101 325 Nm -2 x 50 x 10 -6 m 3 = x 8.314 JK -1 mol -1 x (273 + 25) K M = 13.99 g mol -1 Therefore, the relative molecular mass of the gas is 13. 99.

New Way Chemistry for Hong Kong A-Level Book 158 Example 2-14 The density of a gas at 450 o C and 380 mmHg is 0.033 7 g dm -3. What is its relative molecular mass? ( 1 atm = 760 mmHg = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) Answer 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. 0.0337 g dm -3 = 0.0337 x 10 3 gm 3 = 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.

New Way Chemistry for Hong Kong A-Level Book 159 Check Point 2-4 (a)0.204 g of phosphorus vapour occupies a volume of 81.0 cm 3 at 327 o C and 1 atm. Determine the relative molecular mass of phosphorus. ( 1 atm = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) (b)A sample of gas has a mass of 12.0g and occupies a volume of 4.16dm 3 measured at 97 o C and 1.62 atm. Calculate the relative molecular mass of the gas. ( 1 atm = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) (c) A sample of 0.037g magnesium reacted with hydrochloric acid to give 38.2 cm 3 of hydrogen gas measured at 25 o C and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium. (1 atm = 760 mmHg = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) Answer (a) PV = m/M RT 101 325 Nm -2 x 81.0 x 10-6m 3 = (0.204 g/ M) x 8.314J K -1 mol -1 x (273 + 327) K M = 123.99 g mol -1 The relative molecular mass of phosphorus is 123.99. 2.3 Determination of Relative Molecular Mass (SB p.42)

New Way Chemistry for Hong Kong A-Level Book 160 (b) PV= (m/M)RT 1.62x 101 325 Nm -2 x 4.16 x 10 -3 m 3 = 12.0g/ M x 8.314 J K -1 mol -1 x (273+97)K M = 54.06 g mol -1 The relative molecular mass of the gas is 54.06. 2.3 Determination of Relative Molecular Mass (SB p.42)

New Way Chemistry for Hong Kong A-Level Book 161 (c) Mg(s) + 2 HCl(aq) MgCl 2 (aq) + H 2 (g) PV = nRT 740/760 x 101 325 Nm -2 x 38.2 x 10 -6 m 3 = n x 8.314 J K -1 mol -1 x (273 + 25 ) K n = 1.52 x 10 -3 mol No. of moles of H 2 produced = 1.52 x 10 -3. 2.3 Determination of Relative Molecular Mass (SB p.42)

New Way Chemistry for Hong Kong A-Level Book 162 No. of mole of Mg reacted = No. of moles of H 2 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. 2.3 Determination of Relative Molecular Mass (SB p.42)

New Way Chemistry for Hong Kong A-Level Book 163 2.4 Daltons 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). P T = P A + P B + P C Daltons Law of Partial Pressures

New Way Chemistry for Hong Kong A-Level Book 164 2.4 Daltons Law of Partial Pressures (SB p. 43) Consider a mixture of gases A, B and C occupying a volume V. It consists of n A, n B and n C moles of each gas. The total number of moles of gases in the mixture n total = n A + n B + n C If the equation is multiplied by RT/V, then n total (RT/V) = n A (RT/V) + n B (RT/V) + n C (RT/V) i.e. P total = P A + P B + P C (so Daltons Law is a direct consequence of the Ideal Gas Equation)

New Way Chemistry for Hong Kong A-Level Book 165 2.4 Daltons Law of Partial Pressures (SB p. 43) Besides, the partial pressure of each component gas can be calculated from the Ideal gas law. P A = n A (RT/V) and P total = n total (RT/V) i.e. P A = (nA/n total ) P total P A = x A P total

New Way Chemistry for Hong Kong A-Level Book 166 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 25 o C? ( 1 atm = 101 325 Nm -2 ) Answer 2.4 Daltons Law of Partial Pressures (SB p.43) Solution: Mole of fraction of N 2 = 80/100 Mole of fraction of O 2 = 20/100 Partial pressure of N 2 = 80/100 x 101 325 Nm -2 = 81 060 Nm -2 Partial pressure of O 2 = 20/100 x 101 325 Nm -2 = 20 265 Nm -2

New Way Chemistry for Hong Kong A-Level Book 167 Example 2-16 The value between a 5 dm 3 vessel containing gas A at a pressure of 15 atm and a 10 dm 3 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 in the vessel? (b) What are the mole fractions of gas A and gas B? Answer 2.4 Daltons Law of Partial Pressures (SB p.44) Solution: (a)Total volume of the system = (5 + 10) dm 3 = 15 dm 3 By Boyles Law: P 1 V 1 = P 2 V 2 Partial pressure of gas A = = 5 atm Partial pressure of gas B = = 8 atm

New Way Chemistry for Hong Kong A-Level Book 168 Solution: (contd) By Daltons law: P total = P A + P B Total pressure of the system= (5 + 8) atm = 13 atm (b) Mole fraction of gas A = P A /P total = 5 atm/13 atm = 0.385 Mole fraction of gas B = P B /P total = 8atm/13 atm = 0.615 2.4 Daltons Law of Partial Pressures (SB p.44)

New Way Chemistry for Hong Kong A-Level Book 169 Example 2-17 0.25 mole of nitrogen and 0.30 mole of oxygen are introduced into a vessel of 12 dm 3 at 50 o C. Calculate the partial pressures of nitrogen and oxygen and hence the total pressure exerted by the gases. ( 1 atm = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) Answer 2.4 Daltons Law of Partial Pressures (SB p. 44) Solution: Let the partial pressure of nitrogen be P A. Using the ideal gas equation PV = nRT, P A x 12 x 10 -3 m 3 = 0.25 mol x 8.314 JK -1 mol -1 x (273 + 50) K P A = 55 946 Nm -2 ( or 0.552 atm) Let the partial pressure of oxygen be P B. Using the ideal gas equation PV = nRT, P B = 0.30 mol x 8.314 JK -1 mol -1 x (273 + 50) K P B = 67 136 Nm -2 ( or 0.663 atm)

New Way Chemistry for Hong Kong A-Level Book 170 2.4 Daltons Law of Partial Pressures (SB p. 44) Solution: (contd) Total pressure = (55 946 + 67 136) Nm -2 = 123 082 Nm -2 Hence, the partial pressures of nitrogen and oxygen are 0.552 atm and 0.663 atm respectively, and the total pressure of the mixture is 1.215 atm.

New Way Chemistry for Hong Kong A-Level Book 171 Example 2-18 4.0 g oxygen and 6.0 g of nitrogen are introduced into a 5 dm 3 vessel at 27 o C. (a)What are the mole fractions of oxygen and nitrogen in the mixture? (b)What is the final pressure of the system? ( 1 atm = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ; R.a.m. : N = 14.0, O = 16.0) Answer 2.4 Daltons Law of Partial Pressures (SB p. 45) Solution: (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

New Way Chemistry for Hong Kong A-Level Book 172 2.4 Daltons Law of Partial Pressures (SB p. 45) Solution: (contd) 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 m 3 = 0.339 mol x 8.314 JK -1 mol -1 x (273 + 27)K P = 169 107 Nm -2 ( or 1.67 atm)

New Way Chemistry for Hong Kong A-Level Book 173 Check Point 2-5 (a)The valve between a 6 dm 3 vessel containing gas A at a pressire of 7 atm and an 8 dm 3 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 = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ) Answer 2.4 Daltons Law of Partial Pressures (SB p. 45) By Boyles law: P 1 V 1 = P 2 V 2 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 + 5.14 atm = 8.14 atm

New Way Chemistry for Hong Kong A-Level Book 174 Check Point 2-5 (b) 2 g of helium, 3 g of nitrogen and 4 g of argon are introduced into 15 dm 3 vessel at 100 o C. (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 = 101 325 Nm -2 ; ideal gas constant = 8.314 J K -1 mol -1 ; R.a.m. : He = 4.0, N = 14.0, Ar = 39.9) Answer 2.4 Daltons Law of Partial Pressures (SB p. 45) (i)No. of moles of He = = 0.05 mol No. of moles of N 2 = = 0.11 mol No. of moles of Ar = = 0.10 mol Total no. of moles of gases = 0.05 mol + 0.11 mol + 0.10 mol = 0.71 mol Mole faction of He = = 0.704 Mole fraction of N 2 = = 0.155 Mole fraction of Ar = = 0.141

New Way Chemistry for Hong Kong A-Level Book 175 2.4 Daltons Law of Partial Pressures (SB p. 45) (ii) Let the total pressure of the system be P. PV = nRT P x 15 x 10 -3 m 3 = 0.71 mol x 8.314 JK -1 mol -1 x ( 273 + 100 ) K P = 146 786 Nm -2 Partial pressure of He = 146 786 Nm -2 x 0.704 = 103 337 Nm -2 Partial pressure of N 2 = 146 786 Nm -2 x 0.155 = 22 752 Nm -2 Partial pressure of Ar = 146 786 Nm -2 x 0.141 = 20 697 Nm -2

New Way Chemistry for Hong Kong A-Level Book 176 The Faraday and the Mole? Quantity of electricity = current x time (coulombs, C) (amperes, A) (seconds, s) 2.5 The Faraday and the Mole (SB p. 46)

New Way Chemistry for Hong Kong A-Level Book 177 Faradays first law of electrolysis Faradays 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. Cu 2+ (aq) + 2e - Cu(s) vary directly 2.5 The Faraday and the Mole (SB p. 46)

New Way Chemistry for Hong Kong A-Level Book 178 Faradays second law of electrolysis 2.5 The Faraday and the Mole (SB p. 47)

New Way Chemistry for Hong Kong A-Level Book 179 Faradays 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. Cu 2+ (aq) + 2e - Cu(s) Ag + (aq) + e - Ag(s)2Ag + (aq) + 2e - 2Ag(s) 2.5 The Faraday and the Mole (SB p. 47)

New Way Chemistry for Hong Kong A-Level Book 180 2.5 The Faraday and the Mole (SB p. 48) Faraday Constant ElementElectrolyteNo. of coulombs to liberates 1 mole of metal atom Iron Copper Iron Zinc Silver Fe 2 (SO 4 ) 3 (aq) CuSO 4 (aq) Fe(NO 3 ) 2 (aq) ZnCl 2 (aq) AgNO 3 (aq) 289 500 193 000 96 500

New Way Chemistry for Hong Kong A-Level Book 181 2.5 The Faraday and the Mole (SB p. 48) Charge of 1 mole of electrons = 1 Faraday = 96 500 coulombs Number of moles of product formed = It / nF

New Way Chemistry for Hong Kong A-Level Book 182 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 = 96 500 C) Answer Solution: Q = It = 5A x ( 60 x 60) s = 18 000 C Number of moles of electrons = = 0.187 mol The number of Faradays passed is 0.187.

New Way Chemistry for Hong Kong A-Level Book 183 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? Answer 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

New Way Chemistry for Hong Kong A-Level Book 184 2.5 The Faraday and the Mole (SB p. 49) 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 Solution: Cu 2+ (aq) + 2e - Cu(s) To discharge 1 mole of Cu 2+, 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 = 0.296 g

New Way Chemistry for Hong Kong A-Level Book 185 2.5 The Faraday and the Mole (SB p. 50) Answer 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) 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 - H 2 (g) To give 1 mole of hydrogen gas, 2 moles of electrons (i.e. 2F) are required. Number of moles of H 2 (g) formed= It/nF = = 0.016 8 mol Mass of H 2 (g) formed = 0.016 8 mol x 1.0 x 2 g mol -1 = 0.033 6g

New Way Chemistry for Hong Kong A-Level Book 186 2.5 The Faraday and the Mole (SB p. 50) Solution: (contd). At anode: 4OH - (aq) O 2 (g) + 2H 2 O(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 O 2 (g) formed= It/nF = = 8.394 x 10 -3 mol Mass of O 2 (g) formed = 8.394 x 10 -3 mol x 16.0 x 2 g mol -1 = 0.268 6g

New Way Chemistry for Hong Kong A-Level Book 187 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 dm 3 of oxygen measured at room temperature and pressure? (Molar volume of gas at R.T.P. = 24.0 dm 3 mol -1 ; R.a.m. : O = 16.0, Cu = 63.5) Answer Solution: 4OH - (aq) O 2 (g) + 2H 2 O(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 O 2 given out = = 0.1 mol Number of moles of electrons given out = 0.1 mol x 4 = 0.4 mol

New Way Chemistry for Hong Kong A-Level Book 188 2.5 The Faraday and the Mole (SB p. 51) Solution: (contd) Cu 2+ (aq) + 2e - Cu(s) To discharge 1 mole of Cu 2+, 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

New Way Chemistry for Hong Kong A-Level Book 189 Check Point 2-6 (a)What current in amperes is required to deposit 6.35g of copper in 50 minutes from a copper(II) sulphate solution? (1 F = 96 500; 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 = 96 500C) 2.5 The Faraday and the Mole (SB p. 51) Answer (a) Q = It 96 500 C = 0.35 A x t t = 275 714 s (b) Cu 2+ (aq) + 2e - Cu(s) No. of moles of Cu formed = It/2F = I = 6.43 A

New Way Chemistry for Hong Kong A-Level Book 190 2.5 The Faraday and the Mole (SB p. 51) 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 = 96 500C; 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 (c) Al 3+ (l) + 3e - Al(s) No. of moles of Al(s) formed = It / 3F = = 0.622 mol Mass of Al formed = 0.622 mol x 27.0g mol -1 = 16.794

New Way Chemistry for Hong Kong A-Level Book 191 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 = 0.324 g

New Way Chemistry for Hong Kong A-Level Book 192 The END

new way chemistry for hong kong a-level book 11 the mole concept 2.1the mole 2.2 ideal gas equation...

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