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Notes One Unit Seven– Chapter 13 Solutions

Notes One Unit Seven– Chapter 13 Solutions

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Notes One Unit Seven– Chapter 13 Solutions

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  1. Pages 466-486 Notes One Unit Seven– Chapter 13 Solutions Definitions Types of Mixtures Example Solutions Factors Affecting Solubility Like Dissolves Like Solubility of Solids Changes with Temperature Solubility of Gases Changes with Temperature Pressure Factor Molar Concentration Finding Molarity From Mass and Volume Finding Mass from Molarity and Volume Finding Volume from Molarity and Mass

  2. Definitions • Solutions are homogeneous mixtures. • Uniform throughout. • Solvent. • Determines the state of solution • Is the largest component. • Solute. • Other components dissolved in solvent.

  3. Common Mixtures SOLUTE EXAMPLE SOLVENT Type liquid mayonnaise liquid emulsion gas whipped cream liquid liquid foam solid dust in air gas aerosol liquid hair spray gas aerosol solid ruby glass solid solid liquid pearl solid emulsion gas Styrofoam solid solid foam

  4. Solution Types SOLUTE EXAMPLE SOLVENT PHASE gas a i r gas gas gas soda pop liquid liquid liquid antifreeze liquid liquid liquid filling solid solid solid seawater liquid liquid solid brass solid solid

  5. Factors Affecting Solubility • 1. Nature of Solute / Solvent. • 2. Temperature Increase • i) Solid/Liquid…increase • ii) gas…decreases • 3. Pressure Factor - • i) Solids/Liquids - Very little • ii) gas - Increases. • iii) squeezes gas into solution.

  6. Like Dissolves Like • Non-polar does not dissolve polar • Oil in H2O • Polar dissolves Polar • C2H5OH in H2O • Ionic compounds soluble in polar solvents • NaCl in H2O

  7. Solubility of solids Changes with Temperature • Does Solubility always increase for solids or gases? • How many grams potassium chromate will dissolve 100g water at 70oC? • 70g • How many grams lead(II) nitrate will precipitate from 250g water as it cools from 70oC to 50oC? solids gases 101g 82g 250g _____ 19gx =48g 100g

  8. Calculating Freezing Point Depression Mass • A solution containing 1.89 g of methanol in 51.96 g of water freezes at -3.4oC. Calculate the molecular weight of methanol . • 1.)Calculate Temperature Change • ΔTb = • 2.)Calculate moles per Kilograms • ΔTf = Kf x m  m = ΔTf /Kf • m =0.500m/kg • 3.)Calculate grams / kilograms • g = • g =23.0g/kg • fm= • 46.0g/m 0.000oC- 0.929oC= 0.929oC 0.929÷ 1.858oC/m= m 7.67 g ÷ 0.3330kg 23.0 g/ 0.500m

  9. Solubility of Gases Changes with Temperature • a) Why are fish stressed, if the temperature of the water increases? • How much does the solubility of oxygen change, for a 20oC to 60oC change? • 0.90-0.60=0.30mg 0.90mg 0.60mg

  10. Pressure Factor Greater pressure… more dissolved gas

  11. Pressure Factor

  12. Molar Concentration • M=n/V • MxV=n • V=n/M

  13. Finding Molarity From Mass and Volume • Calculate molarity for 25.5 g of NH3 in 600. mL solution. • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • 3) Calculate the Moles/Liters Ratio • M = n / V • M = 1.50 moles / 0.600 L • M = 2.52 mol/L Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m 25.5g ÷ 17.0g/m= 1.50m

  14. Finding Volume from Molarity and Mass • How many milliliters of 2.50M solution can be made using 25.5grams of NH3? • 1)Calculate formula mass: • 2)Calculate the moles of solute: • 3)Calculate Volume from Moles/Liters Ratio • V=1.50m/2.50M=0.600L • 600.mL solution Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m 25.5g ÷ 17.0g/m= 1.50m

  15. Finding Mass from Molarity and Volume • How many grams of NH3 are in 600. mL solution at 2.50M? • 1) Calculate formula mass: • 2) Calculate moles • M=n/V  n=MxV • n=1.50m • 3) Calculate mass • n=g/MW  g=MWxn • g=25.5g NH3 Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m n= 2.50M x 0.600L x g= (17.0g/m) (1.50m)

  16. Notes Two Unit Seven– Chapter 13 Solutions Pages 487-501 • Saturated versus Unsaturated • Colligative properties of water • Forming a Saturated Solution • How Does a Solution Form? • Colligative Properties • Vapor Pressure • Boiling and Freezing Point • BP Elevation and Freezing FP Depression • Calculating Freezing Point Depression Mass

  17. Characteristics of Saturated Solutions water precipitate precipitate dissolve dissolve dissolve Solid Unsaturated Saturated Unsaturated Dynamic Equilibrium Cooling causes precipitation. Warming causes dissolving.

  18. Solvation • As a solution forms, the solvent pulls solute particles apart and surrounds, or solvates, them.

  19. Colligative Properties • Colligative properties depend on moles dissolved particles. • Vapor pressure lowering • Boiling point elevation • Melting point depression • Osmotic pressure

  20. Vapor Pressure • vapor pressure of a solvent. • vapor pressure of a solution.

  21. Phase Diagram solid Critical point melting freezing 1 atm Liquid vaporizing Pressure Triple point condensing gas sublimation depostion NBP NFP Temperature 100.0oC 0.0oC

  22. One Molal Solution of Water solid 1 atm Liquid Pressure gas Kb Kf Temperature 0.512oC 1.858oC

  23. BP Elevation Constants (Kb)FP Depression Constants( Kf)

  24. Tb = Kb  m Kb=0.5120C/m Tf = Kf  m Kf=1.8580C/m BP Elevation and FP Depression

  25. Molarity versus Molality moles of solute ________________ Molality (m) = kilograms solvent moles of solute ________________ Molarity (M) = liters of solution

  26. Calculating Tf andTb • Calculate the freezing and boiling points of a solution made using 1000.g antifreeze (C2H6O2) in 4450g water. • 1) Calculate Moles • 2) Calculate molality • 3) Calculate Temperature Change • Δt=Kxm • ΔTf = • Tf = • ΔTb = • Tb = Mass E # 24.0 C 2x 12.0 = 1000.g ÷ 62.0g/mol = 16.1 moles 32.0 O 2x 16.0 = 6.0 H 6x 1.0 = 16.1 mole ÷ 4.45 Kg water = 62.0g/m 3.62m (1.858oC/m) (3.62 m) = 6.73oC 0.000oC- 6.73oC= -6.73oC (0.512oC/m) (3.62 m) = 1.96oC 100.000oC + 1.96oC = 101.96oC

  27. Calculating Boiling Point Elevation Mass • A solution containing 18.00 g of glucose in 150.0 g of water boils at 100.34oC. Calculate the molecular weight of glucose. • 1.)Calculate Temperature Change • ΔTb = • 2.)Calculate moles per Kilograms • ΔTb = Kb x m  m = ΔTb /Kb • m =0.67m/kg • 3.)Calculate grams / kilograms • g = • g =120 g/kg • MW=120 g/0.67m • 180g/m 100.34oC- 100.00oC= 0.34oC 0.34÷ 0.512oC/m= m 18.00 g ÷ 0.1500kg

  28. Notes Three Unit Seven • Ice-cream Lab A Calculating Freezing Point • Depression Mass • Colligative Properties of Electrolytes • Distillation • Osmotic Pressure • Dialysis Pages 487-501

  29. Ice-cream

  30. Calculating Freezing Point Depression Mass • 1.)Calculate Temperature Change • 2.)Calculate moles per Kilograms • 3.)Calculate grams / kilograms

  31. Colligative Properties of Electrolytes • Colligative properties depend on the number of particles dissolved. • NaClNa+1+Cl-1 CH3OHAl2(SO4)32Al+3 + 3SO4-2 C6H12O6

  32. Distillation

  33. Distillation

  34. Osmotic Pressure • Hypertonic • > 0.92% (9.g/L) • Crenation • Isotonic Saline • = 0.92% (9.g/L) • Hypotonic • < 0.92% (9.g/L) • Rupture

  35. Dialysis

  36. Kidney

  37. Dialysis