Chem. 1: Notes Unit 12: Mixtures - PowerPoint PPT Presentation

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Chem. 1: Notes Unit 12: Mixtures

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  1. Chem. 1: Notes Unit 12: Mixtures 4 Sections: Types of Mixtures Solution process Solubility Colligative properties

  2. I. Types of Mixtures – combos. of 2 or more pure substances that are NOT chemically combined Type: SUSPENSION Properties: - large suspended particles (> 1000 nm) - visible particles that settle w/ gravity - heterogeneous Examples: granite, trail mix, soups, salt/sand/wood chips Methods of separation: - filter - decant

  3. Type: Colloid Properties: - medium sized particles (1 to 1000 nm) - cloudy appearance, Tyndall effect (background image) - heterogeneous to homogeneous - dispersed phase and dispersing medium

  4. Types of Colloids

  5. Methods of separation of colloids: - coagulation (heat + electrolyte) - dialysis (semi permeable membrane)

  6. Type: Solution Properties: - small particles (< 1 nm) - clear - homogeneous - solute and solvent solute can be – electrolyte (ions that conduct) - nonelectrolyte (molecules that are nonconductors

  7. Solutions exist in all these states: • Gas-Gas Solutions - Gaseous mixtures are usually homogeneous and all gases mixtures are gas-gas solutions. The air is a natural gas solution, but its water and carbon dioxide contents may vary depends on the temperature and places. • Liquid Solutions -When molecules of gas, solid or liquid are dispersed and mixed with those of liquid, the homogeneous (uniform) states are called liquid solutions. Solid, liquid and gas dissolve in liquid to form liquid solutions. In general, the terms solution and liquid solution are synonymous. • Solid Solutions -Many alloys, ceramics, and polymer blends are solid solutions. Within certain range, copper and zinc dissolve in each other and harden to give solid solutions called brass. Silver, gold, and copper form many different alloys with unique colors and appearances

  8. Methods of Separation for Solutions: • Evaporation • chromatography

  9. II. Solution Process – solvation (hydration when solvent = H2O) = solute is surrounded and broken down by solvent solvation solute + solvent solution crystalization (dynamic equilibrium) A. Factors Affecting Rate of Dissolving 1. stirring– increases rate at which solute and solvent mix 2. heating– increased T = increased KE = faster molecules = faster mixing 3. surface area – small particles = large surface area = more opportunity for solute/solvent mixing

  10. Solution Formation – involves 3 interactions 1. break solute-solute attractions – requires energy (endothermic) 2. break solvent-solvent attractions - requires energy (endothermic) 3. form solute-solvent attractions – releases energy (exothermic) if (1 + 2) < 3 solvation is exothermic (- heat of solution) • solute + solvent <==> solution + heat (so, heat is given off) • (increase in temp. causes rxn to shift towards opposite side (left) = gas dissolved in liquid) • if (1 + 2) > 3 solvation is endothermic (+ heat of solution) • solute + solvent + heat <==> solution • (increase in temp. causes rxn to shift towards opposite side (right) = solid dissolved in liquid)

  11. III. Solubility – amount of solute that can dissolve in a certain amount of solvent A. Factors Affecting Solubility 1. types of solute-solvent – “like dissolves like” (like = same polarity, IMF…) a. miscible – substance that will mix together b. immiscible – substances that will not mix together 2. temperature a. solids – increased T = increased solubility b. gases – increased T = decreased solubility 3. pressure – Henry’s law – increased P = increased solubility for a gas in a liquid S1/P1 = S2/P2 (S = solubility in mol/L) so, what type of relationship is this?

  12. B. Units of Concentration – measurements of solubility 1. Terms a. concentrated – large solute : solvent (this means large solute to solvent ratio) b. dilute – small solute : solvent c. saturated – maximum amount of solute dissolved in solvent d. unsaturated – less than max. amt. of solute dissolved in solvent e. supersaturated – more than max. amt. of solute dissolved in solvent

  13. 2. Units a. Molarity (M) Equation: M = moles solute L solution* b. Molality (m) m = moles solute kg solvent c. Mole Fraction (X) X = moles solute moles solution* d. Percent Weight (%) % Wt. = g solute x 100 g solution* * solution = solute + solvent

  14. Example: Molarity 80.5 g C6H8O6 are dissolved in 210 mL of H2O. The total volume of the solution once the solute is added is 238 mL. • Calculate the molarity of the solution M = moles solute L solution moles C6H8O6:80.5 g C6H8O6 1 mol C6H8O6 = 0.457 mol C6H8O6 176 g C6H8O6 L solution: 238mL = 0.238 L M = 0.457 mol C6H8O6 = 1.92 M 0.238 L

  15. 80.5 g C6H8O6 are dissolved in 210 mL of H2O. Solution density = 1.22 g/mL at 55oC. 2. Calculate the molality m = moles solute = 0.457 mol C6H8O6 = 2.18 m kg solvent 0.210 kg H2O • Calculate the mole fraction X = moles solute = 0.457 mol C6H8O6 moles solution 12.157 mol

  16. Calculate the Percent weight % Wt. = g solute x 100 = 80.5 g C6H8O6 x 100 = 27.2 % g solution 80.5 g + 210 g IV. Colligative Properties – depend on amount, NOT type, of solute Vant’ hoff factor: number of solute particles when a substance dissolves(how many pieces does the solute break up into) • for molecular compounds i = 1 • for ionic compounds you have to look at the compound Ex. NaCl___> Na+1 + Cl-1 }2 ions so i = 2 BaCl2___> Ba+2 + 2Cl-1 }3 ions so i = 3

  17. A. Vapor Pressure Lowering – vapor P of a liquid = P caused by evaporated molecules Adding solute = decrease in evap. = lower vapor P B. Boiling Point Elevation - ∆Tb Increased T = increases vapor P boiling occurs when liquid vapor P = PATM So, when you add solute = lower liquid vapor P As a result, a higher temp is needed to get liquid vapor P = PATM ∆Tb = kb x m x i (kb = boiling pt. constant) -for water, kbH2O = 0.52 oC/m; m = molality C. Freezing Point Depression- ∆ Tf Freezing occurs when liquid vapor P (evap.) = solid vapor P (sublimation) ∆Tf = kf x m x i (kf = freezing pt. constant ) -for water, kfH2O = -1.86 oC/m; m = molality

  18. Practice Problems -what would be the new boiling point of a solution containing 47.5g of MgCl2 in 200mL of H2O? ∆Tb = kb x m x I first, we need to find molality m = moles of solute/kg of solvent 47.5g MgCl2 1mol MgCl2 = 0.5 mol MgCl2 95g MgCl2 m = 0.5mol/0.200kg = 2.5m i for = 3, 1Mg+ and 2 Cl_ ions ∆Tb = 0.52 oC/m x 2.5m x 3 = 3.9oC temperature elevation So, the new boiling point of the solution would be 103.9oC

  19. Practice Problems -what would be the new freezing point of a solution containing 5.0g of LiCl in 50mL of H2O? ∆Tf = kf x m x i first, we need to find molality m = moles of solute/kg of solvent 5.0g LiCl 1mol LiCl = 0.118 mol LiCl 42.4g LiCl m = 0.118mol/0.05kg m = 1.18m i for LiCl = 2 ∆Tf = -1.86 oC/m x 2.36m x 2 = -8.8oC So, the new freezing point of the solution would be -8.8oC