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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 • Largest component • Solute. • 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 • ii) gas • 3. Pressure Factor - • i) Solids/Liquids - Very little • ii) gas • iii) squeezes gas into solution.

  6. Like Dissolves Like • Non-polar in Non-polar • Butter in Oil • Non-polar in polar • Oil in H2O • Polar in Polar • C2H5OH in H2O • Ionic compounds in polar solvents • NaCl in H2O

  7. Solubility of solids Changes with Temperature • How does the solubility Δ with temperature Increase? • How many grams of potassium chromate will dissolve in100g water at 70oC? • 70g • How many grams of lead(II) nitrate will precipitate from 250g water cooling from 70oC to 50oC? solids gases 101g 82g 250g _____ 19gx =48g 100g

  8. 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

  9. Pressure Factor Greater pressure… more dissolved gas

  10. Pressure Factor

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

  12. 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 E Mass # N 14.0 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m 25.5g ÷ 17.0g/m= 1.50m M = 1.50m 0.600L / 2.52 mol/L M =

  13. 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: • V= Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m 25.5g ÷ 17.0g/m= 1.50m V=n/M V= (1.50m) / (2.50M) 0.600L solution 600.mL

  14. 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 • n=1.50m • 3) Calculate mass • g=25.5g NH3 Mass E # 14.0 N 1x 14.0 = 3.0 H 3x 1.0 = 17.0g/m g ÷ fm= mol n = x L M x n= 2.50M 0.600L x g = n fm x g= (17.0g/m) (1.50m)

  15. 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

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

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

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

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

  20. 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

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

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

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

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

  25. 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

  26. 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

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

  28. Ice-cream

  29. Calculating Freezing Point Depression Mass • 1.)Calculate Temperature Change • ΔTf = • 2.)Calculate moles per Kilograms • ΔTf = Kfx m  m = ΔTf /Kf • m =1.83mol/kg • 3.)Calculate grams / kilograms • g = • g =36.4 g/kg • MW=36.4 g/1.83m • 19.9g/m 1.1oC- (-2.3oC)= 3.4oC 3.4oC÷ 1.858oC/m = m 0.05196kg 1.89 g ÷

  30. 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

  31. Distillation

  32. Distillation

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

  34. Dialysis

  35. Kidney

  36. Dialysis

  37. Final Quiz Notes

  38. Finding Molarity From Mass and Volume • Calculate molarity for 14.0 g of sodium peroxide(Na2O2) in 615 mL solution. • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • 3) Calculate the Moles/Liters Ratio Mass E # 46.0 Na 2x 23.0 = 32.0 O 2x 16.0 = 78.0g/m 14.0g ÷ 78.0g/m= 0.179m 0.289M 0. 179moles ÷ 0.615L = M = n/ v M =

  39. Finding Volume From Mass and Molarity • Find the volume for a solution having 14.0 g of sodium peroxide(Na2O2) in a 0.289M solution. • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • 3) Calculate the Liters. Mass E # 46.0 Na 2x 23.0 = 32.0 O 2x 16.0 = 78.0g/m 14.0g ÷ 78.0g/m= 0.179m 0.615L 0. 179moles ÷ 0.289M= v = n / M v =

  40. Finding Mass from concentration and Volume • How many grams of sodium peroxide(Na2O2) would be needed for a 0.289M solution of 615mL volume? • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • M=n/V  MxV=n • 3) Calculate the Grams. Mass E # 46.0 Na 2x 23.0 = 32.0 O 2x 16.0 = 78.0g/m 0.289mol/L X 0.615L = 0.179m 0. 179moles x 78.0g/m= g = n x fm g = 14.0g

  41. Calculating Freezing Point Depression Mass • A solution containing 7.67 g of ethanol in 333.0 g of water freezes at -0.929oC. Calculate the molecular weight of ethanol. • 1.)Calculate Temperature Change • ΔTf = • 2.)Calculate moles per Kilograms • ΔTf = Kf x m  m = ΔTf /Kf • m =0.500m/kg • 3.)Calculate grams / kilograms • g/kg = • g/Kg =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

  42. Solubility of solids Changes with Temperature 200g 182g • How many grams Cs2SO4 will precipitate from 267g water as it cools from 60oC to 25oC? 267g _____ 18gx =48g 100g

  43. 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

  44. end

  45. 80 70 60 50 Mg of gas per 100 grams of water 40 30 20 10 0 1.0 0 6.0 7.0 8.0 9.0 10.0 2.0 3.0 4.0 5.0 Gas pressure in atmospheres

  46. Calculating Tf andTb • Calculate the freezing and boiling points of a solution made using 36.0g glucose(C6H12O6) in 225.0g water. • 1) Calculate Moles • 2) Calculate molality of H2O • 3) Calculate Temperature Change • Δt= • ΔTf = • Tf = • ΔTb = • Tb = Mass E # 72.0 C 6x 12.0 = 36.0g ÷ 180.0g/mol = 0.200 moles 96.0 O 6x 16.0 = 12.0 H 12x 1.0 = 0.200 mol ÷ 0.2250 Kg= m = n/ Kg 180.0g/m 0.889m xm K (1.858oC/m) (0.889m) = 1.65oC 0.000oC- 1.65oC= -1.65oC (0.512oC/m) (0.889 m) = 0.455oC 100.000oC + 0.455oC = 100. 455oC

  47. Finding Molarity From Mass and Volume • Calculate molarity for 24.0 g of antifreeze(C2H6O2) in 445. mL solution. • 1) Calculate Formula Mass: • 2) Calculate the moles of solute: • 3) Calculate the Moles/Liters Ratio Mass E # 24.0 C 2x 12.0 = 32.0 O 2x 16.0 = 6.0 H 6x 1.0 = 62.0g/m 24.0g ÷ 62.0g/m= 0.387m 0.870M 0. 387moles ÷ 0.445L = M = n/ v M =

  48. Seven/Eight Rows

  49. 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

  50. Calculating Freezing Point Depression Mass • A solution containing 1.89 g of ethanol in 51.96 g of water freezes at -3.4oC. Calculate the molecular weight of ethanol . • 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