1. Physical Properties of Solutions
Chapter 12 1

2. 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 2

3. 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 3

4. Three types of interactions in the solution process:
solvent-solvent interaction
solute-solute interaction
solvent-solute interaction
Hsoln = H1 + H2 + H3
12.2 4

5. “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 5

6. Mass Percentage of solute
The Concentration of a solute can be quantitatively expressed in several ways:
Molarity
Mass Percentage of solute
Molality
Mole Fraction

7. Solve this?
A ml solution contains 19.6 g H2SO4. Calculate molarity and molality of the solution
Determine the number of equivalents of H2SO4 7

8. 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 8

9. Concentration Units Continued
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3 9

10. 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 10

11. Problem #1
A 3.0 g sample of ground water contains 3.5 ppm of Arsenic ion. Calculate;
(A) How many grams Arsenic ion are in this sample? What is the weight percentage of Arsenic ion?
(B) If 0.25 L aqueous solution with density of g/ml contains 13.7 g of pesticide, express the
(C) concentration in ppm and ppb. 11

12. Temperature and Solubility
Solid solubility and temperature
solubility increases with increasing temperature
solubility decreases with increasing temperature
12.4 12

13. 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 600C
Cool solution to 00C
All NaCl will stay in solution (s = 34.2g/100g)
78 g of PURE KNO3 will precipitate (s = 12 g/100g) g – 12 g = 78 g
12.4 13

14. Temperature and Solubility
O2 gas solubility and temperature
solubility usually decreases with increasing temperature
12.4 14

15. 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 for each gas (mol/L•atm) that depends only on temperature
low P
high P
low c
high c
12.5 15

16. Chemistry In Action: The Killer Lake
8/21/86
CO2 Cloud Released
1700 Casualties
Trigger?
earthquake
landslide
strong Winds
Lake Nyos, West Africa 16

17. 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 = P = X2
X2 = mole fraction of the solute
12.6 17

18. 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 18

19. < & > & 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 19

20. Solve this problem
. Dimethyl gloxime, DMG, is an organic molecule used to test for aqueous nickel (II) ions A solution prepared by dissolving 65.0 g of DMG in 375 g of ethanol boils at 80.3 oC.
What is the molar mass of DMG?
Kb = 1.22 oC/m, and boiling point of pure ethanol = 78.5 oC. 20

21. Problem
A solution is formed by dissolving 10.0 g of KCl in g of H2O.
(a) What is the vapor pressure of the solution at 25 oC if the vapor pressure of pure H2O is
23.8 mmHg at 25 oC?
(b) What is the boiling-point elevation? Kb for water is 0.52 oC/m.
(c) What is the freezing-point depression? Kf for water is 1.86 oC/m.
(d) What is the osmotic pressure of the solution? Assume that the volume of the solution is
0.500 L. 21

22. Fractional Distillation Apparatus
12.6 22

23. Boiling-Point Elevation
Tb = 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
Tb > 0
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
for a given solvent
12.6 23

24. Freezing-Point Depression
Tf = 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
Tf > 0
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
for a given solvent
12.6 24

25. 12.6 25

26. Tf = Kf m Kf water = 1.86 0C/m Tf = 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.
Tf = 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
Tf = Kf m
= C/m x 2.41 m = C
Tf = T f – Tf
Tf = T f – Tf
= C – C = C
12.6 26

27. Problem#2
An Antifreeze mixture is 11.0 molal ethylene glycol solution. Water is the solvent and it
has a mass of 800 g. What is the mass of the solution? The molar mass of ethylene glycol
is 62.0 g/mol 27

28. Osmotic Pressure ()
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 () is the pressure required to stop osmosis.
more
concentrated
dilute
12.6 28

29. Osmotic Pressure ()  = MRT High P Low P
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
12.6 29

30. A cell in an: isotonic hypotonic hypertonic
solution
hypotonic
hypertonic
12.6 30

31. 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
Boiling-Point Elevation
Tb = Kb m
Freezing-Point Depression
Tf = Kf m
Osmotic Pressure ()
 = MRT
12.6 31

32. 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
12.7 32

33. Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
Tb = i Kb m
Freezing-Point Depression
Tf = i Kf m
Osmotic Pressure ()
 = iMRT
12.7 33

34. 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 much larger than solute molecules
collodial suspension is not as homogeneous as a solution
12.8 34

35. The Cleansing Action of Soap
12.8 35

36. Chemistry In Action:
Desalination 36

37. Problem
What is the vapor pressure of a solution made by dissolving 24.6 g of camphor (C10H16O) in 98.5 g of
Benzene. Vapor pressure of benzene is mmHg. (Camphor is a low-volatility solid). 37