1. Physical Properties of Solutions
Chapter 13
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2. Homework - pages
1-11, 14, 16, 18, 20, 22, 27,
31, 32, 34, 37, 38,
, 56, 58, 62, 64,
67, , 80, 85, 87

3. A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount
12.1

4. A saturated solution contains the maximum amount of a solute that will dissolve in a given solvent at a specific temperature.
An unsaturated solution contains less solute than the solvent has the capacity to dissolve at a specific temperature.
A supersaturated solution contains more solute than is present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
12.1

5. Temperature and Solubility
Solid solubility and temperature
12.4

6. PREVIEW
There are two factors that govern whether or not a process will occur spontaneously.
1. There is a tendency to decrease energy.
2. There is a tendency to increase disorder
(entropy).

7. “like dissolves like”
Two substances with similar intermolecular forces are likely to be soluble in each other.
non-polar molecules are soluble in non-polar solvents
CCl4 in C6H6
polar molecules are soluble in polar solvents
C2H5OH in H2O
ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
12.2

9. Like Dissolves Like: Solubility of Methanol in Water

10. Equilibrium in a Saturated Solution
Fig. 13.9

12. Predicting Relative Solubilities of Substances
Problem: Predict which solvent will dissolve more of the given solute.
(a) Sodium Chloride in methanol (CH3OH) or in propanol
(CH3CH2CH2OH).
(b) Ethylene glycol (HOCH2CH2OH) in water or in hexane
(CH3CH2CH2CH2CH2CH3).
(c) Diethyl ether (CH3CH2OCH2CH3) in ethanol (CH3CH2OH) or in
water.

13. The Cleansing Action of Soap
12.8

14. Test on chapters 12 & 13 – EH Assignment Due
Unit 15 Properties of Solutions
Mininum score
Sec 1 Solvents and Solutions 90
Sec 2 Concentration Calculations
Sec 3 Pressure and Temp Effects
Sec 4 Vapor Pressure of Solutions 60
Sec 5 FP Lowering and BP Rise
Unit 8 Section 2 only (Phase Changes and Heat) 80
Test on chapters 12 & 13 –

15. Concentration Units
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass
x 100%
mass of solute
mass of solute + mass of solvent
% by mass = =
x 100%
mass of solute
mass of solution
Mole Fraction (X)
XA =
moles of A
sum of moles of all components
12.3

16. Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3

17. Calculating Molality
Problem: What is the molality of a solution prepared by
dissolving 75.0 g Ba(NO3)2 (s) in to g of water at 250C.
Solution: molar mass of Ba(NO3)2 = g/mol
75.0 g Ba(NO3)2 x mole = mole g
mole
molality = = m = m kg

18. Expressing Concentration by Parts by Mass,
Parts by Volume, and Mole Fraction
Problem:
a) Calculate the PPB by mass of Iron in a 1.85 g Iron supplement
pill that contains g of Iron.
b) The label on a can of beer (340 mL) indicates “4.5% alcohol by
volume”. What is the volume in liters of alcohol it contains?
c) A sample of alcohol contains 118g of ethanol (C2H5OH), and
375.0 g of water. What are the mole fractions?

19. Converting Concentration Units
Problem: Commercial concentrated hydrochloric acid is 11.8 M HCl and has a density of 1.190g/ml. Calculate the (a) mass% HCl,
(b) molality and (c) mole fraction of HCl.
Calculate the molarity of a 1.74 m sucrose (C12H22O11) solution whose density is 1.12 g/mL

20. 5.86 moles ethanol = 270 g ethanol
What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is g/mL?
m =
moles of solute
mass of solvent (kg)
M =
moles of solute
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = kg
m =
moles of solute
mass of solvent (kg) =
5.86 moles C2H5OH
0.657 kg solvent
= 8.92 m
12.3

21. Temperature and Solubility
Solid solubility and temperature
solubility increases with increasing temperature
solubility decreases with increasing temperature
12.4

22. Fractional crystallization is the separation of a mixture of substances into pure components on the basis of their differing solubilities.
Suppose you have 90 g KNO3 contaminated with 10 g NaCl.
Fractional crystallization:
Dissolve sample in 100 mL of water at 60oC
Cool solution to 0oC
All NaCl will stay in solution (s = 34.2g/100g)
78 g of PURE KNO3 will precipitate (s = 12 g/100g). 90 g – 12 g = 78 g
12.4

23. Temperature and Solubility – O2
Gas solubility and temperature
solubility usually decreases with increasing temperature
12.4

24. Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).
c is the concentration (M) of the dissolved gas
c = kP
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only
on temperature
low P
high P
low c
high c
12.5

25. TA p414

26. Colligative Properties
Properties of solutions that depend on the number of solute
particles (atoms, molecules, ions) and not on the nature of the
solute
I ) Vapor Pressure Lowering - Raoult’s Law
II ) Boiling Point Elevation
III ) Freezing Point Depression
IV ) Osmotic Pressure

27. Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 1
P 1
= vapor pressure of pure solvent
X1 = mole fraction of the solvent
Raoult’s law
If the solution contains only one solute:
X1 = 1 – X2
P 1
- P1 = DP = X2
X2 = mole fraction of the solute
12.6

28. PA = XA P A Ideal Solution PB = XB P B PT = PA + PB PT = XA P A
Ideal Solution
PB = XB P B
PT = PA + PB
PT = XA P A
+ XB P B
12.6

29. < & > & PT is greater than predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
A-B
A-A
B-B
<
&
Force
A-B
A-A
B-B
>
&
12.6

30. Fractional Distillation Apparatus
12.6

31. Next Test
From 1999 – test #1
Spring 2001 – test #2

32. Boiling-Point Elevation
DTb = Tb – T b
T b is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
12.6

33. Freezing-Point Depression
DTf = T f – Tf
T f is the freezing point of
the pure solvent
T f is the freezing point of
the solution
T f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
12.6

34. Problem: Calculate the the boiling point and freezing point of a benzene solution if 257g of napthalene (C10H8, mothballs) is dissolved in g of benzene (C6H6).
napthalene = g/mol

35. DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m
What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g.
DTf = Kf m
Kf water = C/m =
3.202 kg solvent
478 g x
1 mol
62.01 g
m =
moles of solute
mass of solvent (kg)
= 2.41 m
DTf = Kf m
= C/m x 2.41 m = C Tf
= C
12.6

36. Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
more
concentrated
dilute
12.6

37. A cell in an: isotonic hypotonic hypertonic
solution
hypotonic
solution
(less conc.)
hypertonic
Solution
(more conc.)
12.6

38. Determining Molar Mass from Osmotic Pressure
Problem: A physician studying a type of hemoglobin formed during a
fatal disease dissolves 21.5 mg of the protein in water at 5.00C to make
1.5 ml of solution in order to measure its osmotic pressure. At
equilibrium, the solution has an osmotic pressure of 3.61 torr. What is
the molar mass of the hemoglobin?
Plan: We know the osmotic pressure (),R, and T. We convert  from
torr to atm and T from 0C to K and use the osmotic pressure equation to
solve for molarity (M). Then we calculate the moles of hemoglobin from
the known volume and use the known mass to find M.
Solution:
1 atm
760 torr
P = 3.61 torr x = atm
Temp = 5.00C = K

39. Determining Molar Mass from Freezing Point Depression
Problem: A 7.85 g sample of a compound with the empirical formula C5H4 is dissolved in 301 g of pure benzene. The freezing point is 4.50 °C. What are the
molar mass and molecular formula of this compound?

40. Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 1 o
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
p = MRT
12.6

41. Colligative Properties of Ionic Solutions
For ionic solutions we must take into account the number of ions present
i = van’t Hoff factor or the number of ions present
For vapor pressure lowering: P = i XsoluteP 0solvent
For boiling point elevation: Tb = i Kb m
For freezing point depression: Tf = i Kf m
For osmotic pressure:  = i MRT
im = conc. of particles

42. Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution
0.1 m Na+ ions & 0.1 m Cl- ions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution
0.2 m ions in solution
van’t Hoff factor (i) =
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes 1
NaCl 2
CaCl2 3
12.6

43. Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iMRT
12.7

44. Arrange the solutions in order of increasing FP
a. 0.1 m CaCl2, 0.1 m C12H22O11, 0.1m NaCl
b m HCl, 0.1m HCl, 0.1m HC2H3O2
What is the FP of m K2SO4 ?
The osmotic pressure of M KI is atm at 25 °C. What is the van’t Hoff factor at this concentration?

45. Colloid versus solution
A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.
Colloid versus solution
collodial particles are larger than solute molecules ( nm)
collodial suspension is not as homogeneous as a solution
12.8

46. Fig A