#### Transcript 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 2.1 The Mole 2 An undercover agent, a counterspy, a double agent Mole A burrowing mammal with fossorial forefeet A small congenital pigmented spot on the skin A breakwater 3 P. 3 / 66 A mole is the number of atoms in exactly 12.00 g ofpure 12 isotope. 6C This number, known as the Avogadro’s constant, can be determined by mass spectrometry. m B k e V 2 4 B is the magnetic field strength. V is the accelerating potential. k is a constant of the instrument. m B k e V 2 At fixed e, k, B and V m can be determined. 5 Q.5 Mass of one mole of C 12 6 atoms = 12.00 g mol1 = Mass of an Avogadro’s number of C 12 6 atoms = Avogadro’s number 1.992648 10-23 g 12.00 g mol1 Avogadro’s number 23 1.992648 10 g = 6.022 1023 mol1 6 2.1 The mole (SB p.18) What is “mole”? Item Unit used to count Shoes pairs Eggs dozens Paper reams No. of items per unit for 2 counting common 12 objects 500 Particles in Chemistry moles 6.022 1023 for counting particles like atoms, ions, molecules 7 sextillion quadrillion billion ~602,200,000,000,000,000,000,000 mol1 quintillion trillion million 千進制 1 mole ~ 602.2 sextillions 8 P. 8 / 66 The fastest supercomputer can count 1.7591015 atoms per second. Calculate the time taken for the superconductor to count 1 mole of carbon-12 atoms. 6.022 10 8 3.424 10 s 15 1.759 10 23 10.85 years 9 We can count the number of coins by weighing if the mass of one coin is known. Similarly, we can count the number of 12C by weighing if the mass of one 12C is known. no. of particles no. of moles 6.02 1023 mass molar mass 10 P. 10 / 66 Molar mass is the mass, in grams, of 1 mole of a substance 11 Q.6 12 13 C C Relative isotopic mass Relative intensity 12.000 100.00 13.003 1.12 Relative atomic mass 100.00 1.12 12.000 13.003 12.01 101.12 101.12 12 Q.6 12 13 C C Relative isotopic mass Relative intensity 12.000 100.00 13.003 1.12 Molar mass of carbon = 12.01 g mol1 13 Q.6 12 13 C C Relative isotopic mass Relative intensity 12.000 100.00 13.003 1.12 Relative isotopic mass is not exactly equal to mass number of the isotope 14 Number of moles of a substance number of particles = 6.022 1023 mol1 mass = molar mass Q.7 Number of moles of oxygen atoms = 15 number of oxygen atoms 6.022 1023 mol1 = 2g 16 g mol1 Q.7 Number of moles of oxygen atoms = number of oxygen atoms 6.022 1023 mol1 = 2g 16. g mol1 2 23 6 . 022 10 Number of oxygen atoms = 16 2 17 23 Number of atoms 6 . 022 10 0.04% O 16 = 3.011 1019 16 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 17 2.2 Molar Volume and Avogadro’s Law 18 2.2 Molar volume and Avogadro’s law (SB p.24) What is molar volume of gases? Volume occupied by one mole of molecules of a gas. 19 2.2 Molar volume and Avogadro’s law (SB p.24) What is molar volume of gases? Depends on T & P Two sets of conditions 20 2.2 Molar volume and Avogadro’s law (SB p.24) What is molar volume of gases? at 298 K & 1 atm (Room temp & pressure / R.T.P.) 21 2.2 Molar volume and Avogadro’s law (SB p.24) What is molar volume of gases? 22.4 dm3 22.4 dm3 22.4 dm3 at 273 K & 1 atm (Standard temp & pressure / S.T.P.) 22 Gas Molar mass/g Molar volume Molar volume at R.T.P./dm3 at S.T.P./dm3 O2 32 24.0 22.397 N2 28 24.0 22.402 H2 2 24.1 22.433 He 4 24.1 22.434 CO2 44 24.3 22.260 NH3 17 24.1 22.079 ~ 24 ~ 22.4 Not constant 23 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 moles of molecules. At fixed T & P, Vn If n = 1, V = molar volume 24 2.2 Molar volume and Avogadro’s law (SB p.24) Avogadro’s Law no. of moles of gas molecules volume of gas (dm ) molar volume (dm3 mol-1 ) 3 volume of gas (dm ) at R.T.P. 3 -1 24 dm mol 3 volume of gas (dm3 ) at S.T.P. 3 -1 22.4 dm mol 25 Vn V = Vm n 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 26 Example 2-2C Check Point 2-2 2.3 27 Ideal Gas Equation 2.3 Ideal gas equation (SB p.27) Boyle’s law At fixed n and T, PV = constant or 1 V P n = number of moles of gas molecules 28 2.3 Ideal gas equation (SB p.28) Schematic diagrams explaining Boyle’s law 29 1/P 30 2.3 Ideal gas equation (SB p.28) A graph of volume against the reciprocal of pressure for a gas at constant temperature 31 2.3 Ideal gas equation (SB p.28) Charles’ law At fixed n and P, V T T is the absolute temperature in Kelvin, K 32 2.3 Ideal gas equation (SB p.28) Schematic diagrams explaining Charles’ law 33 2.3 Ideal gas equation (SB p.28) Volume -273.15 oC 0o C Temperature / oC A graph of volume against temperature for a gas at constant pressure 34 2.3 Ideal gas equation (SB p.28) /K A graph of volume against absolute temperature for a gas at constant pressure 35 2.3 Ideal gas equation (SB p.27) Ideal gas equation Vn 1 V P Avogadro’s law VT Charles’ law RnT V P PV = nRT 36 Boyle’s law R is the same for all gases R is known as the universal gas constant Ideal gas equation 2.3 Ideal gas equation (SB p.29) Relationship between the ideal gas equation and the individual gas laws 37 At fixed n, PV nR a constant T P1V1 P2V2 P3V3 ...... = a constant T1 T2 T3 Ideal gas behaviour is assumed in all gas laws 38 2.3 Ideal gas equation (SB p.27) Gas laws Vn 1 V P Avogadro’s law VT Charles’ law PV = nRT 39 Boyle’s law Ideal gas equation 2.2 Molar volume and Avogadro’s law (SB p.24) Gas laws vs kinetic theory of gases What is the difference between a theory and a law? A law describes what happens under a given set of circumstances. A theory attempts to explain why that behaviour occurs. 40 Ideal gas behaviour Four assumptions as stated in kinetic theory of gases 1. Gas particles are in a state of constant and random motion in all directions, undergoing frequent collisions with one another and with walls of the container. 2. Gas particles are treated as point masses, i.e. they do not occupy volume. Volume of a gas = capacity of the vessel 41 Ideal gas behaviour 3. There is no interaction among gas particles. 4. Collisions between gas particles are perfectly elastic, i.e. kinetic energy is conserved. 42 The ideal gas equation is obeyed by real gases only at (i) low pressure (ii) high temperature (less deviation from 24 dm3 at R.T.P.) 43 (i) At low pressure, gas particles are so far apart that (1) any interaction among them becomes negligible (assumption 3) (2) the volume occupied by the gaseous molecules becomes negligible when compared with that of the container (assumption 2) 44 At high temperature, gaseous molecules possess sufficient energy to overcome intermolecular interactions readily. (assumption 3) 45 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? P1V P2V T1 T2 5 atm 10 atm (273 20)K T2 46 T2 = 586 K 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? PV nRT 1.5 atm 0.450dm3 n 0.082atm dm3 K -1 mol-1 298K n = 0.0276 mol Or 1.5 101325Nm-2 450 10-6 m3 n 8.314J K -1 mol-1 298K n = 0.0276 mol 47 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) P1V1 P2V2 T1 T2 690 mmHg 760 mmHg 25.8 cm3 (273 17) K 48 V2 = 15.1 cm3 1.85 atm V2 345K 2.3 Ideal gas equation (SB p.29) Q.8 Calculate the universal gas constant at S.T.P. For one mole of an ideal gas at S.T.P., P = 1 atm or 101,325 Nm-2 (Pa) V = 22.4 dm3 or 0.0224 m3 n = 1 mol T = 273K 49 2.3 Ideal gas equation (SB p.29) 3 PV 1 atm 22.4 dm R 0.082 atm dm3 K1 mol1 1 mol273K nT Or, 101325Nm 0.0224m R 2 1 mol273K = 8.314 Nm K1 mol1 = 8.314 J K1 mol1 50 3 Q.9 PV = nRT m = mass of the gas m PV RT M = molar mass of the gas M m RT P V M ρRT ρRT M M P 51 2.4 Determination of Molar Mass 52 Determination of Molar Mass 1. Mass Spectrometry 2. Density Measurement ρRT M P 53 m ρ V 2.4 Determination of molar mass (SB p.32) (Mass of syringe + liquid) before injection (m1) = 38.545 g 54 m ρ V 2.4 Determination of molar mass (SB p.32) (Mass of syringe + liquid) after injection (m2) = 38.260 g 55 m ρ V 2.4 Determination of molar mass (SB p.32) Mass of liquid injected (m1 – m2) = 38.545 g – 38.260 g = 0.285 g 56 m ρ V 2.4 Determination of molar mass (SB p.32) Volume of air in syringe before injection (V1) = 10.5 cm3 57 m ρ V 2.4 Determination of molar mass (SB p.32) Volume of air + vapour in syringe after injection (V2) = 146.6 cm3 58 m ρ V 2.4 Determination of molar mass (SB p.32) Volume of vapour in syringe (V2 – V1) = 146.6 cm3 - 10.5 cm3 = 136.1 cm3 59 2.4 Determination of molar mass (SB p.32) Once m and V of the vapour are known, m density( ρ )can be determined V 60 2.4 Determination of molar mass (SB p.32) Temperature = 273 + 65 = 338 K 61 2.4 Determination of molar mass (SB p.32) 1 atm Pressure = 1 atm 62 2.4 Determination of molar mass (SB p.32) Q.10 ρRT m RT m2 m1 RT Molar mass P V P V2 V1 P 0.285g 0.082atm dm K mol (338K) 3 3 1 atm 136.1 10 dm 3 1 = 58.0 g mol1 Relative molecular mass = 58.0 63 1 2.5 Dalton’s law of partial pressures (SB p.35) Unit conversions : R = 8.314 J K1mol1 = 0.082 atm dm3 K1mol1 1m3 = 103 dm3 = 106 cm3 1 atm = 760 mmHg = 101325 Nm2 = 101325 Pa 64 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 1.00 atm. Determine the molar mass of phosphorus. ρRT M P 0.204g 3 -1 -1 0.082 atm dm K mol (273 327)K 0.0810dm3 = 124 g mol-1 65 1.00 atm 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) ρRT M P 12.0 g 4.1610-3 m3 Nm 8.314 J K mol (273 97) K -1 1.62 101325Nm-2 = 54.1 g mol-1 66 -1 2.4 Determination of molar mass (SB p.34) (c) A sample of 0.037 g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium. PV n RT 740 mmHg 760 mmHg 0.082atm dm K mol (273 25) K = 1.52103 mol 67 0.0382dm 3 3 -1 -1 2.4 Determination of molar mass (SB p.34) (c) A sample of 0.037 g magnesium reacted with hydrochloric acid to give 38.2 cm3 of hydrogen gas measured at 25 oC and 740 mmHg. Use this information to calculate the relative atomic mass of magnesium. Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g) 1.52103 mol 1.52103 mol 0.037 g mass 1.52 10 mol molar mass of Mg molar mass of Mg 3 0.037 g -1 molar mass of Mg 24.3 g mol 1.52 10-3 mol 68 2.5 Dalton’s Law of Partial Pressures 69 Experiment 1 Tap opened empty Gas A At fixed T & n, PV = constant (15 atm)(5 dm3) = (PA)(15 dm3) PA = 5 atm 70 Experiment 2 Tap opened 12 atm empty Gas B Gas B At fixed T & n PV = constant (12 atm)(10 dm3) = (PB)(15 dm3) PB = 8 atm 71 Experiment 3 Tap opened 12 atm Gas B Gas A + Gas B The total pressure PT = 13 atm = 5 atm + 8 atm = PA + PB Partial pressures of gases A & B 72 Tap opened 12 atm Gas B PA = 5 atm PB = 8 atm Partial pressure of a constituent gas in a mixture is the pressure that the gas would exert if it were present alone under the same conditions 73 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 of the pressure that each gas would exert if it were present alone under the same conditions). PT 74 = PA + PB + PC 2.5 Dalton’s law of partial pressures (SB p.35) Derivation from ideal gas equation Consider a mixture of gases A, B and C at fixed T & V. nA, nB and nC are the numbers of moles of each gas. The total number of moles of gases in the mixture n T = nA + n B + n C Multiply by the constant RT/V nT(RT/V) = nA (RT/V) + nB (RT/V) + nC (RT/V) If gases A, B and C obey ideal gas behaviour Ptotal = PA + PB + PC RT nT RT (nA nB nC ) V V 75 Partial Pressures and Mole Fractions Consider a mixture of two gases A and B in a container of capacity V at temperature T PAV nART PBV nBRT PTV nTRT PAV nART PA nA XA PTV nT RT PT nA nB Mole fractions of A & B PB nB PBV nBRT XB PT nA nB PTV nTRT PA = PTXA 76 PB = PTXB XA XB 1 Consider a mixture of gases A, B, C, D,… XA XB XC XD ... 1 PA = PTXA PB = PTXB PC = PTXC PD = PTXD … 77 Q.11 At fixed T & n, PV = constant For N2, P1V1 = P2V2 (0.20 Pa)(1.0 dm3) = P2(4.0 dm3) P2 = 0.05 Pa For O2, P1’V1’ = P2’V2’ (0.40 Pa)(2.0 dm3) = P2’(4.0 dm3) P2’ = 0.2 Pa By Dalton’s law of partial pressures PT PN2 PO2 = 0.05 Pa + 0.2 Pa = 0.25 Pa 78 Q.12 At –40oC, only N2 exists as a gas in the mixture For a given amount of N2 at fixed V, P T P1 T1 P1 (273 200)K 1.50 atm (273 40) K P2 T2 At 200oC P1 3.05 atm PN2 PT 4.50 atm PN2 Ppropane Ppropane PT - PN2 (4.50- 3.05)atm 1.45 atm 79 At fixed T & V, Ppropane PTXpropane Xpropane 80 Ppropane PT 1.45 atm 0.322 4.50 atm Q.13(a) PT PNH3 PH2 PN2 9.810 Nm 4 2 At fixed P & T, V n XNH3 nNH3 nT VNH3 VT = 20% PNH3 PT XNH3 (9.810 Nm )(20%) 4 2 = 1.96 104 Nm2 81 Q.13(a) XH2 nH2 nT VH2 VT 55% PH2 PT XH2 (9.810 Nm )(55%) 4 2 = 5.39 104 Nm2 82 Q.13(a) XN2 nN2 nT VN2 VT 25% PN2 PT XN2 (9.810 Nm )(25%) 4 2 = 2.45 104 Nm2 83 Q.13(b) NH3 is removed PT PNH3 PH2 PN2 PH2 PN2 = 5.39 104 Nm2 + 2.45 104 Nm2 = 7.84 104 Nm2 Note : PH2 & PN2 remain unchanged, but PT changes 84 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? 85 2.5 Dalton’s law of partial pressures (SB p.39) P1V1 P2V2 P1V1 7 atm 6 dm3 Partialpressure of gas A P2 3 atm 3 V2 (6 8) dm P1V1 9 atm 8 dm3 Partialpressure of gas B P2 5.1 atm 3 V2 (6 8) dm Total pressure = PA + PB = (3 + 5.1) atm = 8.1 atm 86 2.5 Dalton’s law of partial pressures (SB p.39) (d) 2.0 g of helium, 3.0 g of nitrogen and 4.0 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? 2.0 g mass no. of moles of He 0.50 mol -1 molar mass 4.0 g mol 3.0 g mass no. of moles of N2 0.11 mol -1 molar mass 28.0 g mol 4.0 g mass no. of moles of Ar 0.10 mol -1 molar mass 39.9 g mol Total no. of moles = (0.50 + 0.11 + 0.10) mol = 0.71 mol XHe 87 0.50 0.70 0.71 XN2 0.11 0.15 0.71 0.10 XAr 0.14 0.71 2.5 Dalton’s law of partial pressures (SB p.39) (d) 2.0 g of helium, 3.0 g of nitrogen and 4.0 g of argon are introduced into a 15 dm3 vessel at 100 oC. (ii) Calculate the total pressure of the system, and hence the partial pressures of helium, nitrogen and argon. nT RT 0.71 mol 0.082atm dm3 K -1 mol-1 373K PT 1.45 atm 3 V 15 dm PHe PT XHe 1.45 atm PN2 PT XN2 0.50 1.0 atm 0.71 0.11 1.45 atm 0.22 atm 0.71 0.10 PAr PT XAr 1.45 atm 0.20 atm 0.71 88 The END 89 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 90 2.1 The mole (SB p.21) Back 投影 14gold pendant. Calculate the number of gold atoms in a 片 20 g 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 91 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? 92 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 93 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 94 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 95 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 96 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 97 2.1 The mole (SB p.22) Back 投影 片the 14solution. (d) Calculate the total number of ions in (d) Total number of ions = 6.321 1022 + 1.264 1023 = 1.896 1023 98 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 99 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 100 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 101 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 102 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. 103 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 104 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 105 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 106 Back 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 107 Back 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 108 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 109 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 110 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 111 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-3 112 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 113 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. 114 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. 115 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. 116 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? P1V P2V T1 T2 5 atm 10 atm (273 20)K T2 117 T2 = 586 K 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? PV nRT 1.5 atm 0.450dm3 n 0.082atm dm3 K -1 mol-1 298K n = 0.0276 mol Or 1.5 101325Nm-2 450 10-6 m3 n 8.314J K -1 mol-1 298K n = 0.0276 mol 118 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) P1V1 P2V2 T1 T2 690 mmHg 760 mmHg 25.8 cm3 (273 17) K 119 V2 = 15.1 cm3 1.85 atm V2 347 K 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. 120 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. 121 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. 122 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. 123 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) 124 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. 125 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-2 126 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) What are the mole fractions of gas A and gas B? 127 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 128 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 = 0.615 129 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 130 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) 131 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. 132 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) 133 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 = 0.631 134 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 135