2 The Mole Concept
2 1 The Mole Concept 2.1 The Mole 2.2 Molar Volume and Avogadro’s Law 2.3 Ideal Gas Equation 2.4 Determination of Molar Mass 2.5 Dalton’s Law of Partial Pressures 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 Avogadro constant (the amount in 1 mole) 4 New Way Chemistry for Hong Kong A-Level Book 1 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 = Avogadro consant 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 9 New Way Chemistry for Hong Kong A-Level Book 1 2.2 Molar Volume and Avogadro’s Law 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) 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. 12 New Way Chemistry for Hong Kong A-Level Book 1 2.2 Molar volume and Avogadro’s law (SB p.24) Avogadro’s Law So 1 mole of gases should have the same volume at the same temperature and pressure. Vn 13 where n is the no. of moles of gas New Way Chemistry for Hong Kong A-Level Book 1 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 New Way Chemistry for Hong Kong A-Level Book 1 2.3 15 Ideal Gas Equation New Way Chemistry for Hong Kong A-Level Book 1 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 New Way Chemistry for Hong Kong A-Level Book 1 2.3 Ideal gas equation (SB p.28) Schematic diagrams explaining Boyle’s law 17 New Way Chemistry for Hong Kong A-Level Book 1 2.3 Ideal gas equation (SB p.28) A graph of volume against the reciprocal of pressure for a gas at constant temperature 18 New Way Chemistry for Hong Kong A-Level Book 1 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 New Way Chemistry for Hong Kong A-Level Book 1 2.3 Ideal gas equation (SB p.28) Schematic diagrams explaining Charles’ law 20 New Way Chemistry for Hong Kong A-Level Book 1 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: Vn 1 V P VT V nT P V = RnT P (Avogadro’s Law) (Boyle’s Law) (Charles Law) where R is a constant (called the universal gas constant) PV = nRT 22 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 23 New Way Chemistry for Hong Kong A-Level Book 1 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 New Way Chemistry for Hong Kong A-Level Book 1 2.3 Ideal gas equation (SB p.29) Relationship between the ideal gas equation and the individual gas laws 25 New Way Chemistry for Hong Kong A-Level Book 1 2.3 Ideal gas equation (SB p.30) 26 Example 2-3A Example 2-3B Example 2-3C Check Point 2-3 New Way Chemistry for Hong Kong A-Level Book 1 2.4 Determination of Molar Mass 27 New Way Chemistry for Hong Kong A-Level Book 1 2.4 Determination of molar mass (SB p.32) Mass of volatile liquid injected = 26.590 - 26.330 = 0.260 g 28 New Way Chemistry for Hong Kong A-Level Book 1 2.4 Determination of molar mass (SB p.32) Volume of trichloromethane vapour = 74.4 - 8.2 = 66.2 cm3 29 New Way Chemistry for Hong Kong A-Level Book 1 2.4 Determination of molar mass (SB p.32) Temperature = 273 + 99 = 372 K 30 New Way Chemistry for Hong Kong A-Level Book 1 2.4 Determination of molar mass (SB p.32) Pressure = 101325 Nm-2 31 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 32 New Way Chemistry for Hong Kong A-Level Book 1 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 33 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 New Way Chemistry for Hong Kong A-Level Book 1 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 New Way Chemistry for Hong Kong A-Level Book 1 The END 39 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? Answer 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 40 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. Answer 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 41 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? 42 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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 43 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 44 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. 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 10 g Number of moles of MgCl2 = 95.3 g mol 1 = 0.105 mol 45 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. Answer (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 46 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. 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 47 New Way Chemistry for Hong Kong A-Level Book 1 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 48 New Way Chemistry for Hong Kong A-Level Book 1 Answer 2.1 The mole (SB p.23) Back What is the mass of a carbon dioxide molecule? Answer 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 Avogadro constant 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 49 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 50 New Way Chemistry for Hong Kong A-Level Book 1 Answer 2.1 The mole (SB p.23) (b) Find the number of ions in 5.61 g of calcium oxide. Answer 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 51 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. Answer 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 52 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. 53 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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 54 New Way Chemistry for Hong Kong A-Level Book 1 2.2 Molar volume and Avogadro’s law (SB p.24) What is the difference between a theory and a law? Answer A law tells what happens under a given set of circumstances while a theory attempts to explain why that behaviour occurs. Back 55 New Way Chemistry for Hong Kong A-Level Book 1 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) Answer 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 Answer 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. Answer 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 58 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 59 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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 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 60 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) Answer 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 61 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. Answer 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 62 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? Answer 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 63 New Way Chemistry for Hong Kong A-Level Book 1 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? Answer 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. 64 New Way Chemistry for Hong Kong A-Level Book 1 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) 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. 65 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? Answer 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. 66 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. Answer (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. 67 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? Answer (b) P1 P2 T1 = T2 10 atm 5 atm = T2 (273 20) K T2 = 586 K 68 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? 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 69 New Way Chemistry for Hong Kong A-Level Book 1 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? Answer (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 70 New Way Chemistry for Hong Kong A-Level Book 1 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) Answer 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. 71 New Way Chemistry for Hong Kong A-Level Book 1 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. Answer 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. 72 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. 73 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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. 74 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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) 75 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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. 76 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? Answer 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 77 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. Answer (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 79 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 80 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) Answer 81 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) 82 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. 83 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) 84 New Way Chemistry for Hong Kong A-Level Book 1 Answer 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 85 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 86 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. 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) 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 + … … 87 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? Answer 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 89 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) Answer 90 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 91 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 92 New Way Chemistry for Hong Kong A-Level Book 1
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