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Notes One Unit Seven– Chapter 13 Solutions

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1 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 Pages

2 Definitions Solutions are homogeneous mixtures. Uniform throughout.
Solvent. Determines the state of solution Largest component Solute. Dissolved in solvent

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

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

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 iii) squeezes gas into solution.

6 Non-polar in Non-polar Non-polar in polar Polar in Polar
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
200 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? 180 solids gases 160 140 grams solute /100g H2O 120 101g Pb(NO3)2 100 82g 80 K2CrO4 60 40 _____ 250g 20 19gx =48g 100g 20 30 40 50 60 70 80 90 100 Temperature 0C

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? =0.40mg 5.0 4.5 4.0 3.5 3.0 milligrams solute /100g H2O 2.5 2.0 1.5 1.00mg N2 1.0 O2 0.60mg 0.5 10 20 30 40 50 60 70 Temperature 0C

9 Pressure Factor Greater pressure… more dissolved gas

10 Pressure Factor grams solute /100g H2O 1.0 0.9 0.8 0.7 0.6 CO2 0.5 0.4
0.3 0.2 0.1 10 20 30 40 50 60 70

11 Molar Concentration and Molality
M=n/V n=MxV V=n/M Molality(m) = Moles Solute Kg of Solvent

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 1x 14.0 = 14.0 H 3x 1.0 = 3.0 17.0g/m 25.5g ÷ 17.0g/m= 1.50m M = 1.50m / 0.600L M = 2.52 mol/L

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= E # Mass N 1x 14.0 = 14.0 H 3x 1.0 = 3.0 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 E # Mass N 1x 14.0 = 14.0 H 3x 1.0 = 3.0 17.0g/m g ÷ fm= mol n = M x L n= 2.50M x 0.600L g = fm x n g= (17.0g/m) x (1.50m)

15 Molality Is defined as : MOLES OF SOLUTE Kg of SOLVENT

16 Notes Two Unit Seven– Chapter 13 Solutions
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 Pages

17 Solubility of solids Changes with Temperature
How many grams of cesium sulfate will precipitate from 330g water cooling from 50oC to 30oC? 195g 182g _____ 330g 13gx =43g 100g

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

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

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

21 Vapor Pressure vapor pressure of a solvent.
vapor pressure of a solution.

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

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

24 Boiling-Point Elevation and Freezing-Point Depression
Boiling-Point Elevation (∆Tb): The boiling point of the solution (Tb) minus the boiling point of the pure solvent (T°b): ∆Tb = Tb – T°b ∆Tb is proportional to concentration: ∆Tb = Kb m Kb = molal boiling-point elevation constant.

25 Boiling-Point Elevation and Freezing-Point Depression
Freezing-Point Depression (∆Tf): The freezing point of the pure solvent (T°f) minus the freezing point of the solution (Tf). ∆Tf = T°f – Tf ∆Tf is proportional to concentration: ∆Tf = Kf m Kf = molal freezing-point depression constant.

26 Boiling-Point Elevation and Freezing-Point Depression

27 Boiling-Point Elevation and Freezing-Point Depression

28 Boiling-Point Elevation and Freezing-Point Depression
The phase diagram shows a close-up of the liquid–vapor phase transition boundaries for pure chloroform. Estimate the boiling point of pure chloroform. Estimate the molal concentration of the nonvolatile solute.

29 Boiling-Point Elevation and Freezing-Point Depression
van’t Hoff Factor, i: This factor equals the number of ions produced from each molecule of a compound upon dissolving. i = 1 for CH3OH i = 3 for CaCl2 i = 2 for NaCl i = 5 for Ca3(PO4)2 For compounds that dissociate on dissolving, use: ∆Tb = iKb m ∆Tf = iKf m ∆P = ix2 P°1

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

31 Calculating Boiling Point Elevation Mass
A solution of18.00 g of glucose in g of water boils at oC. Calculate its formula mass. 1) Find Temp Δ 100.34oC - 100.00oC= 0.34oC ΔTb = 2) Find m / Kg m= 0.34oC ÷ 0.512oC/ m m= 0.67m/Kg 3)Find g / Kg 18.00 g ÷ 0.1500kg 120 g/Kg 4)Find fm Fm= 120 g / 0.67m =180g/m

32 Notes Three Unit Seven Ice-cream Lab A Calculating Freezing Point
Depression Mass Colligative Properties of Electrolytes Distillation Osmotic Pressure Dialysis Pages

33 Ice-cream

34 Calculating Freezing Point Depression Mass
1.1oC 38.61g 90.57g 92.46g -2.3oC 1) Find Temp Δ 1.1oC - (-2.3oC)= 3.4oC ΔTf = 2) Find m / Kg 92.46g -90.57g m= 3.4oC ÷ 1.858oC/ m 1.89g m= 1.83m/kg 90.57g 3)Find g/kg -38.61g 51.96g 1.89 g ÷ kg 36.4 g/kg E # Mass C 1x 12.0 = 12.0 4)Find fm O 1x 16.0 = 16.0 Fm= 36.4g ÷ 1.83m =19.9g/m H 4x 1.0 = 4.0 32.0g/m

35 Colligative Properties of Electrolytes
Colligative properties depend on the number of particles dissolved. NaClNa+1+Cl CH3OH Al2(SO4)32Al+3 + 3SO C6H12O6

36 Distillation

37 Distillation

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

39 Dialysis

40 Kidney

41 Dialysis

42 Final Quiz Notes

43 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 E # Mass Na 2x 23.0 = 46.0 O 2x 16.0 = 32.0 78.0g/m 14.0g ÷ 78.0g/m= 0.179m M = 0. 179moles ÷ 0.615L = 0.289M

44 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. E # Mass Na 2x 23.0 = 46.0 O 2x 16.0 = 32.0 78.0g/m 14.0g ÷ 78.0g/m= 0.179m v = n / M v = 0. 179moles ÷ 0.289M= 0.615L

45 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. E # Mass Na 2x 23.0 = 46.0 O 2x 16.0 = 32.0 78.0g/m 0.289mol/L X 0.615L = 0.179m g = n x fm g = 0. 179moles x 78.0g/m= 14.0g

46 Calculating Freezing Point Depression Mass
A solution of 7.67 g of ethanol in g of water freezes at oC. Find the formula mass of ethanol. 1) Find Temp Δ 0.000oC- (-0.929oC)= 0.929oC ΔTf = 2) Find m / Kg m= 0.929oC ÷ 1.858oC/ m m= 0.500m/kg 3)Find g / Kg 7.67 g ÷ 0.3330kg 23.0 g/Kg 4)Find fm Fm= 23.0g / 0.500m =46.0g/m

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

48 end

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

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

51 Calculating Tf andTb m = n/ Kg
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 = E # Mass C 6x 12.0 = 72.0 36.0g ÷ 180.0g/mol = 0.200 moles O 6x 16.0 = 96.0 H 12x 1.0 = 12.0 m = n/ Kg 0.200 mol ÷ Kg= 0.889m 180.0g/m K xm (1.858oC/m) (0.889m) = 1.65oC 0.000oC- 1.65oC= -1.65oC (0.512oC/m) (0.889 m) = 0.455oC oC + 0.455oC = oC

52 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 E # Mass C 2x 12.0 = 24.0 O 2x 16.0 = 32.0 H 6x 1.0 = 6.0 62.0g/m 24.0g ÷ 62.0g/m= 0.387m M = n/ v M = 0. 387moles ÷ 0.445L = 0.870M

53 Seven/Eight Rows

54 Calculating Freezing Point Depression Mass
A solution containing 1.89 g of methanol in 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

55 Calculating Freezing Point Depression Mass
A solution containing 1.89 g of ethanol in 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

56 Calculating Freezing Point Depression Mass
A solution containing 1.89 g of methanol in 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

57 BP Elevation Constants (Kb) FP Depression Constants( Kf)

58 Aqueous Solubility of CO2 at 101.3 kPa (1 atm) partial pressure[9]
Temperature ‡Dissolved CO2 volume per volume H2O grams CO2 per 100 ml H2O 0 °C 1.713 0.3346 1 °C 1.646 0.3213 2 °C 1.527 0.3091 3 °C 0.2978 4 °C 1.473 0.2871 5 °C 1.424 0.2774 6 °C 1.377 0.2681 7 °C 1.331 0.2589 8 °C 1.282 0.2492 9 °C 1.237 0.2403 10 °C 1.194 0.2318 11 °C 1.154 0.2239 12 °C 1.117 0.2165 13 °C 1.083 0.2098 14 °C 1.050 0.2032 15 °C 1.019 0.1970 16 °C 0.985 0.1903 17 °C 0.956 0.1845     18 °C 0.928 0.1789 19 °C 0.902 0.1737 20 °C 0.878 0.1688 21 °C 0.854 0.1640 22 °C 0.829 0.1590 23 °C 0.804 0.1540 24 °C 0.781 0.1493 25 °C 0.759 0.1449 26 °C 0.738 0.1406 27 °C 0.718 0.1366 28 °C 0.699 0.1327 29 °C 0.682 0.1292 30 °C 0.655 0.1257 35 °C 0.592 0.1105 40 °C 0.530 0.0973 45 °C 0.479 0.0860 50 °C 0.436 0.0761 60 °C 0.359 0.0576

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