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Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties Colloids HW: Chpt 11 - pg. 531-538, #s 12, 14, 20, 24, 29, 34,

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Presentation on theme: "Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties Colloids HW: Chpt 11 - pg. 531-538, #s 12, 14, 20, 24, 29, 34,"— Presentation transcript:

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2 Chpt 11 - Solutions Concentrations Energy of solutions Solubility Colligative Properties Colloids HW: Chpt 11 - pg , #s 12, 14, 20, 24, 29, 34, 41, 42, 44, 46, 52, 66, 72, 77, 81 Due Mon Dec. 3

3 Various Types of Solutions Example State of Solution State of Solute State of Solvent Air, natural gasGas Vodka, antifreezeLiquid BrassSolid Carbonated water (soda)LiquidGasLiquid Seawater, sugar solutionLiquidSolidLiquid Hydrogen in platinumSolidGasSolid Solvent is majority component. Solute is minority component, usually the substance dissolved in the solvent (liquid).

4 Solution composition

5 Molarity You have 1.00 mol of sugar in mL of solution. Calculate the concentration in units of molarity M You have a 10.0 M sugar solution. What volume of this solution do you need to have 2.00 mol of sugar? L

6 Molarity (M) example Consider separate solutions of NaOH and KCl made by dissolving g of each solute in mL of solution. Calculate the concentration of each solution in units of molarity M NaOH 5.37 M KCl

7 Mass percent (%) What is the percent-by-mass concentration of glucose in a solution made my dissolving 5.5 g of glucose in 78.2 g of water? 6.6%

8 Mole fraction ( A ) A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in mL of water. Calculate the mole fraction of H 3 PO 4. (Assume water has a density of 1.00 g/mL.)

9 Molality (m) A solution of phosphoric acid was made by dissolving 8.00 g of H 3 PO 4 in mL of water. Calculate the molality of the solution. (Assume water has a density of 1.00 g/mL.) m

10 Solution Formation Schematic

11 Solution Formation Process 1.Separating the solute into its individual components (expanding the solute). 2.Overcoming intermolecular forces in the solvent to make room for the solute (expanding the solvent). 3.Allowing the solute and solvent to interact to form the solution.

12 Solution Formation Energies Steps 1 and 2 require energy, since forces must be overcome to expand the solute and solvent. Step 3 usually releases energy. Steps 1 and 2 are endothermic, and step 3 is often exothermic. Enthalpy change associated with the formation of the solution is the sum of the ΔH values for the steps: ΔH soln = ΔH 1 + ΔH 2 + ΔH 3 ΔH soln may have a positive sign (energy absorbed) or a negative sign (energy released).

13 Exo vs. Endo H soln Demo NH 4 NO 3 and NaOH examples

14 Explain why water and oil (a long chain hydrocarbon) do not mix. In your explanation, be sure to address how ΔH plays a role. H 1 H 2 H 3 H soln Outcome Polar solute, polar solvent Large Large, negative SmallSolution forms Nonpolar solute, polar solvent SmallLargeSmallLarge, positive No solution forms Nonpolar solute, nonpolar solvent Small Solution forms Polar solute, nonpolar solvent LargeSmall Large, positive No solution forms

15 Solubility Factors Structural Effects: Polarity (like dissolves like) Pressure Effects: Henrys law (for dissolved gases) Temperature Effects: Affecting aqueous solutions

16 Pressure effects Henrys law:C = kP C = concentration of dissolved gas k = constant P =partial pressure of gas solute above the solution Amount of gas dissolved in a solution is directly proportional to the pressure of the gas above the solution.

17 Gas solubility in liquid Soda pops carbonated water has the carbon dioxide forced into the solution under pressure. When the can is opened P atm is much lower than P can so CO 2 leaves -> pop goes flat.

18 Temperature effects Although the solubility of most solids in water increases with temperature, the solubilities of some substances decrease with increasing temperature. Predicting temperature dependence of solubility is very difficult. Solubility of a gas in solvent typically decreases with increasing temperature.

19 Temp solubility charts

20 Colligative Properties Depend only on the number, not on the identity, of the solute particles in an ideal solution: Vapor pressure lowering Boiling-point elevation Freezing-point depression Osmotic pressure

21 Vapor Pressure of solutions If the P vap of the solvent (water) > P vap of the solution, equilibrium is reached when the solvent evaporates and the solvent is absorbed by solution. It does this to lower the P vap towards its equilibrium value.

22 Raoults Law Nonvolatile solute lowers the vapor pressure of a solvent. Raoults Law: P soln =observed vapor pressure of solution solv =mole fraction of solvent P o solv =vapor pressure of pure solvent

23 Raoults Law - ideal solution Ideal solution occurs with a nonvolatile solute in solution Also the vapor pressure is then proportional to the mole fraction of the solvent using (total moles of ions of solute) in the solvent

24 Boiling Point elevation Nonvolatile solute elevates the boiling point of the solvent. ΔT = K b m solute ΔT = boiling-point elevation K b = molal boiling-point elevation constant m solute = molality of solute particles

25 Freezing Point depression When a solute is dissolved in a solvent, the freezing point of the solution is lower than that of the pure solvent. ΔT = K f m solute ΔT = freezing-point depression K f = molal freezing-point depression constant m solute = molality of solute particles

26 Phase Diagram of solutions

27 Boiling Point - Freezing Point explanation point_elevation point_depression

28 Boiling Pt Elev Problem A solution was prepared by dissolving g glucose in g water. The molar mass of glucose is g/mol. What is the boiling point of the resulting solution (in °C)? Glucose is a molecular solid that is present as individual molecules in solution. K b = 0.51 o C. kg/mol °C

29 Osmotic Pressure Osmosis – flow of solvent into the solution through a semipermeable membrane. (Kidney dialysis uses this Principle). = MRT =osmotic pressure (atm) M=molarity of the solution R= gas law constant T=temperature (Kelvin)

30 Osmotic Pressure graphic

31 Osmotic Pressure Problem When 33.4 mg of a compound is dissolved in 10.0 mL of water at 25°C, the solution has an osmotic pressure of 558 torr. Calculate the molar mass of this compound. Strategy: need Temp in K, Pressure in atm, use R with atm unit to get molarity. Then use vol get moles, then mass get molar mass. 111 g/mol

32 Colloids Intermediate mixture - a heterogeneous mixture with particle size between a suspension and a solution A suspension of tiny particles in some medium. Tyndall effect – scattering of light by particles. Suspended particles are single large molecules or aggregates of molecules or ions ranging in size from 1 to 1000 nm.

33 Types of colloids

34 Tyndall Effect graphic

35 Freezing Pt problem You take 20.0 g of a sucrose (C 12 H 22 O 11 ) and NaCl mixture and dissolve it in 1.0 L of water. The freezing point of this solution is found to be °C. K f = 1.86 o C. kg/mol Assuming ideal behavior, calculate the mass percent composition of the original mixture, and the mole fraction of sucrose in the original mixture. 72.8% sucrose and 27.2% sodium chloride; mole fraction of the sucrose is 0.313

36 Derivation of Colligative Properties Specific derivation of the partial derivatives and derivation for colligative properties are found on the website. 80a/480ants/colprop/colprop.html


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