1. Notes One Unit Six– 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
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?
=0.30mg
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 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 x
g = fm n g=
(17.0g/m) x
(1.50m)

15. Notes Two Unit Six– 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

16. Characteristics of Saturated Solutions
water
precipitate
precipitate
dissolve
dissolve
dissolve
Solid
Unsaturated
Unsaturated
Saturated
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. How do you get from this…

20. …to this?

21. Add an ionic compound!

22. Vapor Pressure

23. Vapor Pressure Lowering
The particles of solute are surrounded by and attracted to particles of solvent.
Now the solvent particles have less kinetic energy and tend less to escape into the space above the liquid.
So the vapor pressure is less.

24. Ionic vs Molecular Solutes
Ionic solutes produce two or more ion particles in solution.
They affect the colligative properties proportionately more than molecular solutes (that do not ionize).
The effect is proportional to the number of particles of the solute in the solution.

25. How many particles do each of the following give upon solvation?
NaCl
CaCl2
Glucose

26. Freezing Point Depression

27. Example
Salt is added to melt ice by reducing the freezing point of water.

28. Boiling Point Elevation

29. Example
Addition of ethylene glycol C2H6O2 (antifreeze) to car radiators.

30. Freezing Point Depression and Boiling Point Elevation
∆Tb =imkb (for water kb=0.51 oC/m)
Freezing Point Depression
∆Tf=imkf (for water kf=1.86 oC/m)
Note: m is the molality of the particles, so if the solute is ionic, multiply by the #of particles it dissociates to. (i is # of particles called van hoff factor)

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

32. Which is more effective for lowering the freezing point of water?
NaCl or CaCl2

33. Example 1:
Find the new freezing point of 3m NaCl in water.

34. Example 2:
Find the new boiling point of 3m NaCl in water.

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

36. 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 = E #
Mass C 2x
12.0 =
24.0
1000.g ÷
62.0g/mol =
16.1 moles O 2x
16.0 =
32.0 H 6x
1.0 =
6.0
16.1 mole ÷
4.45 Kg water =
3.62m
62.0g/m
(1.858oC/m)
(3.62 m) =
6.73oC
0.000oC-
6.73oC=
-6.73oC
(0.512oC/m)
(3.62 m) =
1.96oC
oC +
1.96oC =
101.96oC

37. Calculating Boiling Point Elevation Mass
A solution containing g of glucose in g of water boils at oC. 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

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

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

40. Ice-cream

41. 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
1.89 g ÷ kg

42. 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

43. Distillation

44. Distillation

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

46. Dialysis

47. Kidney

48. Dialysis

49. Final Quiz Notes

50. 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 = n/ v
M =
0. 179moles ÷
0.615L =
0.289M

51. 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

52. 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

53. Calculating Freezing Point Depression Mass
A solution containing 7.67 g of ethanol in g of water freezes at oC. 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

54. 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

55. 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

56. end

57. 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

58. 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

59. 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

60. Seven/Eight Rows

61. 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

62. 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

63. 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