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Molality = Moles of Solute/Kg of Solvent (M) Moles of Solute/Kg of Solvent (M)

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Molality = Moles of Solute/Kg of Solvent (M) Moles of Solute/Kg of Solvent (M) 5/2/2010 Properties of Solutions Copyright © Houghton Mifflin Company. All rights reserved. 17a–2 Figure 17.2: (a) Enthalpy of solution Hsoln has a negative sign (the process is Figure 17.1: The formation of a liquid exothermic) if Step 3 releases more energy than is required by Steps 1 and 2. (b) Hsoln has a positive sign (the process is endothermic) if Stpes 1 and 2 require solution can be divided into three steps more energy than is released in Step 3. Copyright © Houghton Mifflin Company. All rights reserved. 17a–3 Copyright © Houghton Mifflin Company. All rights reserved. 17a–4 Concentration Molarity = Moles of solute/Liters of Solution (M) Entropy Usually Favors Making Solutions Molality = Because There Are More Moles of solute/Kg of Solvent (m) Possible Arrangements of the Atoms or Mole Fraction= Molecules Moles solute/total number of moles Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved. 17a–5 Copyright © Houghton Mifflin Company. All rights reserved. 17a–6 1 5/2/2010 Concentration Concentration Molarity = Molarity = Moles of solute/Liters of Solution (M) Moles of solute/Liters of Solution (M) Molality = Molality = Moles of solute/Kg of Solvent (m) Moles of solute/Kg of Solvent (m) Mole Fraction= Mole Fraction= Moles solute/total number of moles Moles solute/total number of moles Mass %= Mass %= Mass solute/total mass x 100 Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved. 17a–7 Copyright © Houghton Mifflin Company. All rights reserved. 17a–8 A sample of NaNO3 weighing 8.5 grams is placed in a 500 ml volumetric flask and distilled water was added to the mark on the Molarity = neck of the flask. Calculate the Molarity of the resulting solution. Moles of solute/Liters of Solution (M) Convert the given grams of solute to moles of solute : 1 mole NaNO3 8.5 g NaNO3 0.1 mole NaNO3 85 g NaNO3 Molality = Convert given ml of solution to liters Moles of solute/Kg of Solvent (m) 1lit1 liter 500 ml 0.5 liter 1000 ml Mole Fraction= Apply the definition for Molarity: Molarity = moles NaNO3 / volume of the solution in liters Moles solute/total number of moles M = 0.1 mole / .500 liters = 0.200 Molar NaNO3 Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved. 17a–9 Copyright © Houghton Mifflin Company. All rights reserved. 17a–10 Determine the mole fraction of KCl in 3000 grams of Molarity = aqueous solution containing 37.3 grams of Potassium Chloride KCl. Moles of solute/Liters of Solution (M) 1. Convert grams KCl to moles KCl using the molecular weight of KCl Molality = 1 mole KCl 37.3 g KCl 0.5 mole KCl Moles of solute/Kg of Solvent (m) 74.6 g KCl 2. Determine the grams of pure solvent water from the Mole Fraction= given grams of solution and solute Total grams = 3000 grams = Mass of solute + Mass of water Moles solute/total number of moles Mass of pure solvent = (3000 - 37.3) gram = 2962.7 gram Mass %= Mass solute/total mass x 100 Copyright © Houghton Mifflin Company. All rights reserved. 17a–11 Copyright © Houghton Mifflin Company. All rights reserved. 17a–12 2 5/2/2010 Determine the mole fraction of KCl in 3000 grams of Assuming the density of water to be 1 g/mL we approximate the aqueous solution containing 37.3 grams of Potassium density of a dilute aqueous solution to be 1 g/mL Chloride KCl. 3. Convert grams of solvent H O to mols 1g 2 1 ppm = 1 mol 2962.7 grams water 164.6 mols H O 1 g 18.0 grams 2 1g 1 g 1g 4. Apply the definition for mole fraction mole fraction = moles of KCl / Total mols of KCl and water = 1 g11 ml ml 0.5 / (0.5 + 164.6) = 0.5 / 165.1 = 0.00303 1 ppm = 1 μg/mL = 1 mg/L 1 ppb = 1 ng/mL = 1 μg/L Copyright © Houghton Mifflin Company. All rights reserved. 17a–13 Copyright © Houghton Mifflin Company. All rights reserved. 17a–14 3 5/2/2010 Factors Affecting Solubility Factors Affecting Solubility SoluteSolute--SolventSolvent Interactions SoluteSolute--SolventSolvent Interactions • Polar liquids tend to dissolve in polar solvents. • Miscible liquids: mix in any proportions. • Immiscible liquids: do not mix. • Intermolecular forces are important: water and ethanol are miscible because the broken hydrogen bonds in both pure liquids are re-established in the mixture. • The number of carbon atoms in a chain affect solubility: the more C atoms the less soluble in water. Copyright © Houghton Mifflin Company. All rights reserved. 17a–19 Copyright © Houghton Mifflin Company. All rights reserved. 17a–20 Factors Affecting Solubility Factors Affecting Solubility SoluteSolute--SolventSolvent Interactions SoluteSolute--SolventSolvent Interactions • The number of -OH groups within a molecule • Generalization: “like dissolves like”. increases solubility in water. • The more polar bonds in the molecule, the better it dissolves in a polar solvent. • The less polar the molecule the less it dissolves in a polar solvent and the better is dissolves in a non- polar solvent. • Network solids do not dissolve because the strong intermolecular forces in the solid are not re- established in any solution. Copyright © Houghton Mifflin Company. All rights reserved. 17a–21 Copyright © Houghton Mifflin Company. All rights reserved. 17a–22 Factors Affecting Solubility Factors Affecting Solubility Gas – solvent: Pressure Effects Gas – solvent: Pressure Effects • Solubility of a gas in a liquid is a function of the pressure of the gas. • The higher the pressure, the more molecules of gas are close to the solvent and the greater the chance of a gas molecule striking the surface and entering the solution. – Therefore, the higher the pressure, the greater the solubility. – The lower the pressure, the fewer molecules of gas are close to the solvent and the lower the solubility. Copyright © Houghton Mifflin Company. All rights reserved. 17a–23 Copyright © Houghton Mifflin Company. All rights reserved. 17a–24 4 5/2/2010 Factors Affecting Solubility Gas – solvent: Pressure Effects Gas – solvent: Pressure Effects Henry’s Law: Cg kPg Cg is the solubility of gas, Pg the partial pressure, k = Henry’s law constant. Carbonated beverages are bottled under > 1 atm. As the bottle is opened, Pg decreases and the solubility of CO2 decreases. Therefore, bubbles of CO2 escape from solution. Copyright © Houghton Mifflin Company. All rights reserved. 17a–25 Copyright © Houghton Mifflin Company. All rights reserved. 17a–26 Gas – solvent: Pressure Effects dissolution of SO2 and NOx ACID RAIN + 2+ CaCO3(s) + H CO2(g) + H2O + Ca Copyright © Houghton Mifflin Company. All rights reserved. 17a–27 Copyright © Houghton Mifflin Company. All rights reserved. 17a–28 Factors Affecting Solubility Temperature Effects Experience tells us that sugar dissolves better in warm water than cold, but Coca Cola goes flat when warm. • As temperature increases, solubility of solids generally increases. • As temperature increases, solubility of gasses generally decreases. Copyright © Houghton Mifflin Company. All rights reserved. 17a–29 Copyright © Houghton Mifflin Company. All rights reserved. 17a–30 5 5/2/2010 Factors Affecting Solubility Factors Affecting Solubility Temperature Effects: solids Temperature Effects: gasses Copyright © Houghton Mifflin Company. All rights reserved. 17a–31 Copyright © Houghton Mifflin Company. All rights reserved. 17a–32 Thermodynamics of Solubility Lake Nyos in Cameroon ∆G = ∆H - T∆S Solubility increases if ∆G becomes more negative . When ∆S is positive, ∆G becomes more negative when T increases. Solubility will increase when T increases. When ∆S is negative, ∆G becomes more positive when T increases. Solubility will decrease when T increases. Compare KCl(s) → K+ (aq) + Cl- (aq) ∆S positive - ∆G ↓ gas → liquid ∆S negative - ∆G ↑ Source: Corbis Copyright © Houghton Mifflin Company. All rights reserved. 17a–33 Copyright © Houghton Mifflin Company. All rights reserved. 17a–34 Thermodynamics of Solubility Temperature Dependence of Gas Solubility • ∆G = ∆H - T∆S •gas→solution • ∆S is negative • ∆H is negative •-T∆S is positive • ∆G becomes more positive when T increases • Therefore, solubility Cg kPg decreases Henry’s Law Constant Depends on T Copyright © Houghton Mifflin Company. All rights reserved. 17a–35 Copyright © Houghton Mifflin Company. All rights reserved. 17a–36 6 5/2/2010 Factors Affecting Solubility Thermodynamics of Phase Changes Temperature Effects • Gases are less soluble at higher temperatures. B A • Thermal pollution: if lakes get too warm, CO2 and O2 become less soluble and are not available for plants or animals. • In global warming modeling, CO2 may be released from the oceans Why does a liquid at A form a solid when the temperature is lowered to B Copyright © Houghton Mifflin Company. All rights reserved. 17a–37 Copyright © Houghton Mifflin Company. All rights reserved. 17a–38 Thermodynamics for Phase Change Gases: ∆G = ∆H - T∆S Large Negative for Negative for Positive for Entropy spontaneous process liquid to solid liquid to solid • liquid→solid • ∆H is negative (stronger intramolecular forces) Liquid: Smaller • ∆S is negative (more order) Entropy • -T∆S iiiis positive • As T decreases, -T∆S becomes smaller Solids: • ∆G goes to zero when ∆H = T∆S (at T = Tfusion) Smallest Entropy • For T less than Tfusion, ∆G is negative, solid is stable. Copyright © Houghton Mifflin Company. All rights reserved. 17a–39 Copyright © Houghton Mifflin Company. All rights reserved. 17a–40 Colligative Properties • Colligative properties depend on quantity of solute molecules. (E.g. freezing point depression and boiling point elevation.) . NB: quantity is what? Mass? Volume? Number? Copyright © Houghton Mifflin Company. All rights reserved. 17a–41 Copyright © Houghton Mifflin Company. All rights reserved. 17a–42 7 5/2/2010 Colligative Properties Question of the Day Lowering the Vapor Pressure What happens? • Non-volatile solvents reduce the ability of the surface solvent molecules to escape the liquid.
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