Physical properties of solutions Chemistry Solutions colligatice properties Example questions

1. CHE 106 CHEMISTRY
Bu dersin amacı, öğrencilerde kimyanın temel kavram ve prensiplerine
ilişkin sağlam bir temel oluşturmak, kimyanın günlük yaşantımızda
önemli bir rol oynadığını örneklerle göstererek öğrencilere kimyayı
sevdirmektir.
Chang R., Chemistry , Mc Graw Hill Inc., 4-13 Baskılar
Yardımcı Kitaplar:
1. Kenneth W. Raymond, General Organic and Biochemical
Chemistry, John Wiley & Sons,
2. Olmsted, J., &Williams, G. M., Chemistry, John Wiley& Sons,
3. Baskı
3. Blackman, A., Bottle, S.E., Schmid, S., Mocerino, M., Wille, U.,
Chemistry, John Wiley & Sons
4. Petrucci, R.H., Harwood, W.S., Herring, F.G., Madura, J.D.,
General Chemistry, Pearson Education International.

3. Physical
Properties
of Solutions

4. Content

5. 12.1
A solution is a homogenous mixture of two 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.

6. Solubility is a measure of how much solute
will dissolve at a specific temperature.
Substances are referred as:
•soluble
•slightly soluble
•insoluble
A substance is said to be soluble if a fair amount of it visibly
dissolves when added to water/solvent.
All ionic compounds are strong electrolytes, but they are not
equally soluble.

7. 7
What happens when you mix acetone and
polystyrene?
Acetone is a solvent that can dissolve
or weaken polystyrene. If you mix it
with polystyrene, the acetone will
begin to dissolve the polystyrene and
can cause warping or dissolution of
the plastic.

8. Examples of Insoluble Compounds
8
CdS PbS Ni(OH)2 Al(OH)3

9. 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.
Crystalization is the process in which a dissolved solute comes out of solution
and forms cristals.
12.1
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.

10. The ease of dissolving a
solute in a solvent is
governed by
intermolecular forces.
Energy and disorder that
results when molecules of
the solute and solvent mix to
form a solution are the
forces driving the solution
process.

11. Molecular view of solution
process
During the solution process, solvent molecules interact with
solute molecules.
If the solvent-solute interactions are stronger than the
solvent-solvent and solute-solute interactions, then the
solute is dissolved in the solvent (Energy-driven process).
Another driving force of the solution process is an increase
in disorder in the final system.
If the disorder of the solution is bigger than the disorder
of the solvent and disorder of the solute, the solute is
dissolved in the solvent (Entropy-driven process).

12. Three steps of solution process
(energy driven)
• Heat of separation of solute molecules, DH1
• Heat of separation of solvent molecules, DH2
• Heat of mixing of solute and solvent molecules, DH3

13. 12.2
Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
DHsoln = DH1 + DH2 + DH3
Solvent molecules
are separated
(endothermic)
Solute molecules
are separated
(endothermic)
Solvent and solute
molecules mix
(exothermic or
endothermic)

15. “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

17. Dissolution of NaCl in Water
https://youtu.be/betbKEJf7jQ

19. Example
Predict the solubilities in the following
cases:
(a) Br2 in benzene (C6H6, µ=0 D) and in water
(H2O, µ=1.87 D)
(b) KCl in carbon tetrachloride (CCl4, µ=0
D) and in liquid ammonia (NH3, µ=1.46 D)
(c) Formaldehyde (CH2O) in carbon disulfide
(CS2, µ=0 D) and in water.

20. H2O
CCl4 C6H6
NH3
Br2
KC
l
CH2O
CS2
Non-
polar

22. Miscible liquids
Two liquids are said to be miscible if they are
completely soluble in each other in all proportions.

23. Solvation
The process in which a solute molecule (ion) is
surrounded by solvent molecules

24. Concentration units
• Percent by mass (units: %), is independent of temperature
and does not require calculations of molar masses
• Mole fraction (no units); is used in calculating partial
pressures of gases
• Molarity (units: mol/L); used practically in laboratories
because the volume of a solution is easy to measure but is
dependent of temperature
• Molality (units: mol/kg); is independent of temperature
because the concentration is expressed in the solute number of
moles and mass of solvent

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

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

27. What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 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 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
m =
moles of solute
mass of solvent (kg)
=
5.86 moles C2H5OH
0.657 kg solvent
= 8.92 m
12.3

28. Example: What is the percent by mass of 0.892 g KCl
dissolved in 54.6 g water?
% by mass = x 100%
0.892 g
0.892 g+ 54.6 g
= 1.61 %

29. (no units)
Data
-mass of solute = 12.5 g
-mass of solvent = 80.5 g
Calculation of moles:
nwater = 4.46 mol
Number of water moles:
1 mol = 18.02 g
n mol = 80.5 g
n = 1 mol x 80.5 g / 18.02 g
= 4.46 mol
Example: What is the mole fraction of 12.5 g sodium chloride (NaCl)
dissolved in 80.5 g water?
XA =
moles of A
sum of moles of all components
Mole Fraction (X),

31. nA = 0.21 mol 1 mol = 58.44 g
n mol = 12.5 g
n = 1 mol x 12.5 g / 58.44 g
= 0.21 mol
Calculation of mole fraction:
XA = 0.04
Number of NaCl moles:
XA =
0.21
0.21 mol +4.46 mol

32. GCh12-16
Example: Calculate molarity of a 1.2 L of HNO3 solution containing 24.4 g of HNO3.
Data mass of solute = 24.4 g
volume of solution = 1.2 L
M =
moles of solute
liters of solution
Molarity (M)

33. GCh12-17
Solution
Step 1. Calculation of moles
nsolute = 0.38 mol
Step 2. Calculation of molarity
Number of HNO3 moles
1 mol = 63.02 g
n mol = 24.4 g
n = 1 mol x 24.4 g / 63.02 g
= 0.38 mol
M =
0.38 mol
1.2 L
= 0.31 mol/L or 0.31 M

34. Example: Calculate the molality of H2SO4 solution, containing 24.4 g
of H2SO4 in 198 g of water.
Data
-mass of solute = 24.4 g
-mass of solvent = 198 g = 0.198 kg
Molality (m)
m =
moles of solute
mass of solvent (kg)

35. GCh12-19
Solution
Step 1. Calculation of moles:
nsolute = 0.24 mol
Step 2. Calculation of molality:
= 1.26 mol / kg
Number of H2SO4 moles
1 mol = 98.10 g
n mol = 24.4 g
n = 1 mol x 24.4 g / 98.10 g = 0.248 mol
m =
0.248 mol
0.198 kg

36. GCh12-
Density of solution
mass of solution (msn)
d =
(units: g / mL)
volume of solution (Vsn)
Examples
msn
Vsn
-mass of solution = 128 g
-volume of solution = 1.45 L = 1450 mL
d = 128 g / 1450 mL = 0.08 g/mL

37. GCh12-21
Problem
The molarity if an aqueous solution of methanol is 2.45 M. The density of
this solution is 0.976 g/mL. What is the molality of the solution?
Data
-density of solution = 0.976 g/mL
-molarity of solution = 2.45 M
Solution
moles of solute
M =
volume of solution
mass of solution
d =
volume of solution
moles of solute
m =
mass of solvent

38. GCh12-22
Step 1. Calculation of mass of solute
2.45 moles - 1 liter of solution
79.49 g - 1 liter of solution
Mass of CH3OH:
1 mol = 32.04 g
2.45 mol = n g
Step 2. Calculation of mass of water
0.976 g of solution - 1 mliter of solution
976 g of solution - 1 liter of solution
mass of water = 976 g - 76.49 g
= 898 g
n = 2.45 mol x 32.04 g / 1mol= 79.49 g
Step 3. Calculation of molality:
m = 2.45 mol / 0.898 kg = 2.73 m

39. GCh12-
Problem
The density of an aqueous solution containing 10 % of ethanol (C2H2OH) by
mass, is 0.984 g/mL. Calculate the molality of this solution. Calculate the
molarity.
moles of solute
M =
volume of solution
mass of solution
d =
volume of solution
moles of solute
m =
mass of solvent
Data
- density of solution = 0.984 g/mL
- percent by mass of solution = 10%

40. Step 1. Calculation of mass of water in 100 g of
solution
10g (Ethanol, E) exist in 100g (Solution, S)
Mass (Water, W) = 100 g (S) - 10 g(E)
= 90 g = 0.09 kg
Step 2. Calculation of ethanol moles in 100 g of solution
10g (E) (1 mol) / (46.07 g) = 0.217 mol (E)
Step 3. Calculation of molality of 100 g of solution
m = (0.217 mol) / (0.09 kg)=2.41 mol/kg
Moles of ethanol:
1 mol = 46.07 g
n mol = 10 g
n = 10 g x 1 mol / 46.07 g = 0.217 mol
moles of solute
m =
mass of solvent

41. GCh12
Step 4. Calculation of a volume of 100 g of solution
100 g (1 mL ) / (0.984 g) = 102 mL = 0.102 L
Step 5. Calculation of molarity of 100 g of solution
M = (0.217 mol) / (0.102 L) = 2.13 mol/Lof
Volume of solution:
1 mL = 0.984 g
V mL = 100 g
V = 100 g x 1 mL / 0.984 g
= 102 mL

42. Dilution
M1V1 = M2V2

43. GCh12-27
Effect of temperature on solubility
• Solubility of solid solutes
• Solubility of gas solutes
Solid solubility
Generally, the solubility of solids is bigger at the higher
temperature
There is no correlation between the solubility of a solute at
a particular temperature, and its heat of solution DHsol
Solubility
• Maximal concentration of a solute in a solvent
• CaCl2
• NH4NO3
exothermic
endothermic
Solubility of both compounds increases with the temperature

44. Temperature and Solubility
solubility increases
with increasing
temperature
solubility decreases
with increasing
temperature
12.4

45. 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:
1. Dissolve sample in 100 mL of water at 60°C
2. Cool solution to 0°C
3. All NaCl will stay in solution (s = 34.2g/100g)
4. 78 g of PURE KNO3 will precipitate (s = 12 g/100g).
90 g – 12 g = 78 g
12.4
If we have a solution of a mixture of two solutes, by using
fractional crystalization we can separate those two solutes.

46. O2 gas solubility and temperature
solubility usually
decreases with
increasing temperature
12.4
Fish prefer to be deeper in water,
where the temperature is lower, and
the oxygen concentration is bigger
Global warming decreases the
concentration of oxygen in water,
which is harmful to its living animals

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

49. Dynamic equilibration
A number of molecules going from
the gas to the liquid is the same
as the number of molecules going
from the liquid to the gas at the
same time
For all practical purposes, external pressure has no influence on the
solubility of liquids and solids, but it does greatly affect the solubility of
gases.

50. 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.
The colligative properties are:
 Vapor pressure lowering.
 Boiling point elevation.
 Melting point depression.
 Osmotic pressure.

51. Colligative Properties of Nonelectrolyte Solutions
Vapor-Pressure Lowering
Raoult’s law
If the solution contains only one solute:
X1 = 1 – X2
P 1
0
- P1 = DP = X2 P 1
0
P 1
° = vapor pressure of pure solvent
X1 = mole fraction of the solvent
X2 = mole fraction of the solute
12.6
P1 = X1 P 1
°
The vapor pressure of its solution is always smaller than that of the pure solvent

55. Problem
At 25°C the vapor pressure of pure water is 23.76 mmHg and that of an aqueous urea solution is 22.98 mmHg.
Estimate the molality of the solution.
Moles of
solvent, n1
Moles of
solute, n2

56. Molality (m) m =
moles of solute
mass of solvent (kg)
Mass of water=(0.976mol)×(18gmol−1)=17.406g
17.405 g of water contain solute = 0.033 mole
1000 g of water contain solute =(0.033mol)(17.406g)×1000g=1.896mol
∴Molality of solution=1.896 m.

58. Boiling-Point Elevation
DTb = Tb – T b
0
Tb > T b
0
DTb > 0
T b is the boiling point of
the pure solvent
0
T b is the boiling point of
the solution
DTb = 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
T b
T b°

60. Freezing-Point Depression
DTf = T f – Tf
0
T f > Tf
0
DTf > 0
T f is the freezing point of
the pure solvent
0
T f is the freezing point of
the solution
DTf = 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

61. 12.6

63. 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 62.01 g/mol.
DTf = Kf m
m =
moles of solute
mass of solvent (kg)
= 2.41 m
=
3.202 kg solvent
478 g x
1 mol
62.01 g
Kf water = 1.86 0C/m
DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
DTf = T f – Tf
0
Tf = T f – DTf
0
= 0.00 0C – 4.48 0C = -4.48 0C
12.6

66. Osmotic Pressure (p)
12.6
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.
dilute
more
concentrated

67. Semipermeable Membrane in Osmosis

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

69. A cell in an:
isotonic
solution
hypotonic
solution
hypertonic
solution
If two solutions are of equal
concentration and, hence, have the
same osmotic pressure, they are
said to be isotonic.
If two solutions are of unequal
osmotic pressures, the more
concentrated solution is said to be
hypertonic and the more dilute
solution is described as hypotonic.

70. Summary - 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.
12.6
Vapor-Pressure Lowering P1 = X1 P 1
0
Boiling-Point Elevation DTb = Kb m
Freezing-Point Depression DTf = Kf m
Osmotic Pressure (p) p = MRT

71. Colligative Properties of Electrolyte Solutions
12.7
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl– ions
Colligative properties are properties that depend (ideally) 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
nonelectrolytes
NaCl
CaCl2
i should be
1
2
3

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

73. A colloid is a dispersion of actual particles of one substance
throughout a dispersing medium of another substance. The
particles are small enough to stay suspended but large enough
to scatter light (the Tyndall effect).
Colloid versus solution
• collodial particles are much larger than solute molecules
• collodial suspension is not as homogeneous as a solution
12.8

74. The Cleansing Action of Soap
12.8