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Mol of Solute Molality ( ) = Kg Solvent M

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Mol of Solute Molality ( ) = Kg Solvent M Molality • Molality (m) is the number of moles of solute per kilogram of solvent. mol of solute Molality (m ) = kg solvent Copyright © Houghton Mifflin Company. All rights reserved. 11 | 51 Sample Problem Calculate the molality of a solution of 13.5g of KF dissolved in 250. g of water. mol of solute m = kg solvent 1 m ol KF ()13.5g 58.1 = 0.250 kg = 0.929 m Copyright © Houghton Mifflin Company. All rights reserved. 11 | 52 Mole Fraction • Mole fraction (χ) is the ratio of the number of moles of a substance over the total number of moles of substances in solution. number of moles of i χ = i total number of moles ni = nT Copyright © Houghton Mifflin Company. All rights reserved. 11 | 53 Sample Problem -Conversions between units- • ex) What is the molality of a 0.200 M aluminum nitrate solution (d = 1.012g/mL)? – Work with 1 liter of solution. mass = 1012 g – mass Al(NO3)3 = 0.200 mol × 213.01 g/mol = 42.6 g ; – mass water = 1012 g -43 g = 969 g 0.200mol Molality==0.206 mol / kg 0.969kg Copyright © Houghton Mifflin Company. All rights reserved. 11 | 54 Sample Problem Calculate the mole fraction of 10.0g of NaCl dissolved in 100. g of water. mol of NaCl χ=NaCl mol of NaCl + mol H2 O 1 mol NaCl ()10.0g 58.5g NaCl = 1 m ol NaCl 1 m ol H2 O ()10.0g+ () 100.g H2 O 58.5g NaCl 18.0g H2 O = 0.0299 Copyright © Houghton Mifflin Company. All rights reserved. 11 | 55 Conc. Based on mass g solute mass % or weight %=× 100 g solution gsolute ppm =×106 gsolution gsolute ppb =×109 g solution g solute ppt =×1012 gsolution Copyright © Houghton Mifflin Company. All rights reserved. 11 | 56 ppm, ppb, ppt in dilute aqueous solution 11 gsolute mgsolute ppm =≈ 106 g of solution L solution 11 g solute µ g solute ppb =≈ 109 g of solution L solution 11 g solute ng solute ppt =≈ 1012 g of solution L solution Copyright © Houghton Mifflin Company. All rights reserved. 11 | 57 11.8 Effect of Temperature on Solubility • The solubility of a gas decreases with temperature. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 58 Principles of Solubility • Nature of solute and solvent • Most nonelectrolytes that are appreciably soluble in water are hydrogen bonded (methanol, hydrogen peroxide, sugars). Other types of nonelectrolytes are generally more soluble in nonpolar or slightly polar solvents such as benzene or toluene. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 59 Principles of Solubility (cont.) • Effect of temperature a. Increase in T favors endothermic process: solid + water → solution ∆H usually positive, so solubility increases with T gas + water → solution ∆H usually negative, so solubility decreases with T Effect of pressure Henry’s Law Negligible, except for gases, where solubility is directly proportional to the partial pressure of the gas. Carbonated beverages. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 60 Temperature and pressure effects on gaseous solubility Copyright © Houghton Mifflin Company. All rights reserved. 11 | 61 Solubility of O2 Copyright © Houghton Mifflin Company. All rights reserved. 11 | 62 Effect of Temperature on Solubility • The solubility of an ionic solid generally increases with temperature. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 63 Effect of Pressure on Solubility Henry’s Law Pgas = kgasCgas –Pgas = pressure of the gas above the solution –Cgas = concentration of the gas –kgas = Henry’s law constant Henry’s law holds best for gases O2 and N2, does not hold HCl Copyright © Houghton Mifflin Company. All rights reserved. 11 | 64 Sample Problem A liter of water dissolves 0.0404 g of oxygen at 25oC at a pressure of 760. torr. What would be the concentration of oxygen (in g/L) if the pressure were increased to 1880 torr at the same temperature? PP 12= CC12 0.0404 g/L C = 2 760. torr 1880 torr C2 = 0.0999 g/L Copyright © Houghton Mifflin Company. All rights reserved. 11 | 65 11.9 Colligative Properties • Vapor pressure lowering – the vapor pressure of a solvent is lowered by the addition of a nonvolatile solute. cf) volatile solute Copyright © Houghton Mifflin Company. All rights reserved. 11 | 66 Colligative Properties of Nonelectrolytes • The properties of a solution differ considerably from those of the pure solvent. The solution properties depend primarily upon concentration of solute particles rather than type. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 67 Colligative Properties of Nonelectrolytes • Vapor pressure lowering • Boiling point elevation, freezing point lowering • Osmosis, osmotic pressure Copyright © Houghton Mifflin Company. All rights reserved. 11 | 68 Copyright © Houghton Mifflin Company. All rights reserved. 11 | 69 Raoult’s Law o Psolution= χ solventP solvent o P solvent = vapor pressure of the pure solvent It is independent of the nature of the solute but directly proportional to its concentration. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 70 Sample Problem What will be the vapor pressure of a solution made by dissolving 6.25g of glucose, C6H12O6 , in 50.0g of water at 25oC? How much was the vapor pressure of the pure water lowered? The vapor pressure of water at 25oC is 23.8 torr 1 mol mol glucose = 6.25 g = 0.0347 mol glucose 180. g 1 mol mol water = 50.0 g= 2.78 mol water 18.0g 2.78 χ == 0.988 water 0.0347+2.78 o Psoln== χ waterP water ()() 0.988 23.8 torr = 23.5 torr vapor pressure lowering = 23.8 torr - 23.5 torr = 0.2 torr Copyright © Houghton Mifflin Company. All rights reserved. 11 | 71 Vapor pressure lowering and freezing point depression Copyright © Houghton Mifflin Company. All rights reserved. 11 | 72 Colligative Properties Boiling point elevation – the change in the boiling point is: ∆Tb = iKbm – i = sum of the coefficients of the ions (i = 1 for molecular compounds) –Kb = boiling point elevation constant – m = molality Copyright © Houghton Mifflin Company. All rights reserved. 11 | 73 Colligative Properties Freezing point depression – the change in the freezing point is: ∆Tf = iKfm – i = sum of the coefficients of the ions (i = 1 for molecular compounds) –Kf = freezing point depression constant – m = molality Copyright © Houghton Mifflin Company. All rights reserved. 11 | 74 Colligative Properties Copyright © Houghton Mifflin Company. All rights reserved. 11 | 75 Boiling point elevation, freezing point lowering • Results from vapor pressure lowering. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 76 Antifreeze solution Copyright © Houghton Mifflin Company. All rights reserved. 11 | 77 Sample Problem Calculate the boiling point elevation and the freezing point depression of a solution made by dissolving 12.2g of KCl in 45.0g of water. Kb = o o 0.512 C/m and Kf = 1.86 C/m • i = 2 for KCl → K+ + Cl− 1 mol mol KCl = 12.2g= 0.164 mol 74.6g mol KCl 0.164 mol mm== =3.64 KCl kg water 0.045 kg oo ∆=Tbbim K =() 2() 0.512 C/ m() 3.64 m =3.73 C oo ∆=Tffim K =() 2() 1.86 C/ m() 3.64 m =13.5 C Copyright © Houghton Mifflin Company. All rights reserved. 11 | 78 Osmosis, osmotic pressure • Water moves through semi-permeable membrane from region of high vapor pressure (pure water) to region of low vapor pressure (solution) • π = nRT/V = MRT •1 M solution at 25°C has osmotic pressure of 24.5 atm Pickle, Fresh water, Nutrient solution Copyright © Houghton Mifflin Company. All rights reserved. 11 | 79 Colligative Properties Osmotic pressure П = iMRT – i = sum of the coefficients of the ions (i = 1 for molecular compounds) – M = molarity – R = gas constant (0.0821 L•atm/mol•K) – T = temperature in Kelvin Copyright © Houghton Mifflin Company. All rights reserved. 11 | 80 Osmosis Copyright © Houghton Mifflin Company. All rights reserved. 11 | 81 Osmosis Copyright © Houghton Mifflin Company. All rights reserved. 11 | 82 Osmosis Copyright © Houghton Mifflin Company. All rights reserved. 11 | 83 Electrolytes • Colligative effects are greater because of increased number of particles. • ∆Tf = 1.86°C × m × i a. where i is approximately equal to the number of moles of ions per mole of solute: (only in very dilute solution) • NaCl(s) → Na+(aq) + Cl-(aq) i = 2 • CaCl(s) → Ca2+(aq) + 2Cl-(aq) i = 3 a. Actually, i is usually less than predicted because of ionic atmosphere effects and ion pair. Copyright © Houghton Mifflin Company. All rights reserved. 11 | 84 Copyright © Houghton Mifflin Company. All rights reserved. 11 | 85 Sample Problem What is the osmotic pressure of a 100. mL solution containing 9.50 g of glucose, o C6H12O6, at 20.0 C? 1 mol mol glucose = 9.50 g =0.0528 mol 180. g 0.0528 mol M = = 0.528 mol/L glucose 0.100 L Latm⋅ π = (1)() 0.528 mol/L 0.08206 ()[20.0+273] K = 12.7 atm Kmol⋅ Copyright © Houghton Mifflin Company. All rights reserved. 11 | 86.
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