2. Chapter 12 Topics
Physical Properties of Solutions
• Solution, solute, solvent, unsaturated, saturated, supersaturated,
crystallization
•Solution Process
•Concentrations (%mass, Mole fraction, molality, molarity, ppm)
•Factors affecting Solubility (T and P)
•Colligative properties of nonelectrolyte solution:
1)Vapour pressure lowering
2)B.P. Elevation
3) F.P. Depression
4) Osmotic Pressure (π )
•Using Colligative properties to determine molar mass
•Colligative properties of electrolyte solution
Dr. Ali Bumajdad
3. Solution, solute, solvent, unsaturated, saturated, supersaturated, crystallization
•Solution: homogenous mixture of 2 or more substances
•Solute: is(are) the substance(s) present in the smaller
amount(s)
•Solvent: is the substance present in the larger amount
∴Solution not necessarily liquid, it could be gas or solid
4. • Saturated solution: contains the maximum amount of a
solute that will dissolve in a given solvent at a specific
temperature.
• Unsaturated solution: contains less solute than the solvent
has the capacity to dissolve at a specific temperature.
• Supersaturated solution: contains more solute than is
present in a saturated solution at a specific temperature (not
stable).
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate
• Crystallization: dissolved solute comes out of solution and
forms crystals.
Solute + solvent
solvation
crystallization
Solution
5. Solution Process
Three types of interactions in the solution process:
• solvent-solvent interaction (endothermic)
• solute-solute interaction
(endothermic)
• solvent-solute interaction (endothermic or exothermic)
∆Hsoln>0, solution
may form and
maynot
∆Hsoln<0, solution
favourable
∆Hsoln = ∆H1 + ∆H2 + ∆H3
6. When solute-solute or solvent-solvent interaction are much
more than solute-solvent interaction
a solution will not form
“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)
7. •Miscible: two liquids completely soluble in each other in all
proportions
•Solvation: ion or molecules surrounded by solvent molecules
Q) Can we form a solution of octane (C8H18) in water?
No because:
- octane not polar while water is polar
-In this case ∆H3 is small while ∆H1 and ∆H2 large +ve
hence, ∆Hsoln is large positive
Sa Ex. :Predict whether each of the following substances is more likely to
dissolve in CCl4 or in Water: C7H16, Na2SO4, HCl, I2.
9. Concentration Units
•Concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
1) Percent by Mass
(1)
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
2) Mole Fraction (X)
moles of A
XA =
sum of moles of all components
(2)
10. Concentration Units Continued
3) Molarity (M)
M =
moles of solute
(3)
liters of solution
4) Molality (m)
m =
moles of solute
mass of solvent (kg)
(4)
14. Sa. Ex.:A solution is made by dissolving 13.5 g glucose (C6H12O6) in
0.100 kg of water. What is the mass percentage of solute in this solution?
11.9%
. Ex.
Calculate the mass percentage of NaCl in a solution containing 1.50 g of
NaCl in 50. g of water.
A commercial bleaching solution contains 3.62 mass% sodium hypochlorite,
NaOCl. What is the mass of NaOCl in a bottle containing 2500 g of
bleaching solution?
2.91%
90.5g
15. Sa. Ex.
a) A solution is made by dissolving 4.35 g glucose (C6H12O6) in
25.0 ml of water. Calculate the molality of glucose in the solution.
0.966 m
b) What is the molality of a solution made by dissolving 36.5 g
of naphthalene (C10H8) in 425 g of toluene (C7H8)
0.670 m
c) A solution of hydrochloric acid contains 36% HCl by mass
(i) Calculate the mole fraction of HCl in the solution.
(ii) Calculate the molality of HCl in the solution
XHCl= 0.22
15 m
16. Q) What is the molality of a 5.86 M ethanol (C2H5OH) solution
whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
(1) mass of solute: (n×Mm=mass)
5.86 moles ethanol contains 270 g ethanol
(2) mass of solution: solution density ×solution volume
927 g of solution
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
17. Sa. Ex. A solution contains 5.0 g of toluene (C7H8) and 225 g
of benzene and has a density of 0.876 g/ml. Calculate the molarity
of the solution.
0.21 M
Sa. Ex. A solution containing equal masses of glycerol (C3H8O3,
Mm=92gmol-1) and water has a density of 1.10 g/ml. Calculate:
10.9 m
(a) The molality of glycerol
0.163
(b) The mole fraction of glycerol
5.97 M
(c) The molarity of glycerol in the solution.
18.
19.
20. •Factors affecting Solubility (T and P, and nature of
solute and solvent)
(1) Temperature
Like dissolve like
a) Solid solubility and temperature
Solubility increases with
increasing temperature
Solubility decreases with
increasing temperature
Usually s
αT
For solid
21. •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 600C
2. Cool solution to 00C
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
22. (1) Temperature
(b) Gas solubility and temperature
Solubility usually
decreases with
increasing temperature
Usually s
α1
T
For gas
23. (2) Pressure
a) Gas solubility and Pressure
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
Henry’s law
c = kP
(5)
(there are some exceptions
when the gas react with water)
c is the concentration (M) of the dissolved gas
P is the pressure of the gas over the solution (vapor or partial pressure)
k is a constant for each gas (mol/L•atm) that depends only on temperature
Vapour Pressure:
pressure of gas above the
liquid
low P
high P
low c
high c
24.
25. Sample Ex. :Calculate the concentration of CO2 in soft drink (partial pressure
of CO2 = 4.0 atm, and the Henery's law constant for CO 2 in water
is 3.1 × 10-2 mol/L atm.
c = kP
(5)
=0.12 mol/l
26. Colligative Properties of Nonelectrolyte Solutions
•Colligative properties: properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
(e.g. adding salt and ethylene glycol reduce F.P.)
1) Vapor-Pressure Lowering
Adding nonvolatile solute to a solvent always lower the
vapour pressure
0
P1 = X1 P 1
(6)
For nonvolatile solute (no measurable
vapour pressure)
Raoult’s law
Where:
P 1 = vapor pressure of solution after adding solute
0
P 1 = vapor pressure of pure solvent
X = mole fraction of the solvent
27. If the solution contains only one solute:
X1 = 1 – X2
0
P 1 = (1 - X2) P 1
P 1 - P1 = ∆P = X2 P 1
0
0
(7)
X2 = mole fraction of the solute
The decrease in vapour pressure (∆P) directly proportion to
solute concentration (not to the nature of solute)
28. When both components are volatile:1) Ideal behaviour
Ideal Solution
0
PA = XA P A
0
PB = XB P B
PT = PA + PB
0
PT = XA P A + XB P 0
B
29. When both components are volatile: 2) Nonideal behaviour
PT is greater than
predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
Force
Force
< A-A & B-B
A-B
Force
Force
Force
> A-A & B-B
A-B
31. Vapour pressure of water at 30 °C = 31.82mmHg
0
P1 = X1 P 1
(6)
Raoult’s Law
P 1 - P1 = ∆P = X2 P 1
0
X1 =
X2=1-X1
0
(2)
n1
n1+n2
(7)
32. Sa. Ex. Glycerin (C3H8O3) is a nonvolatile nanelectrolyte with a density
of 1.26 g/ml at 25 °C. Calculate the vapor pressure at 25 °C of a solution made
by adding 50.0 ml of glycerin to 500.0 ml of water. The vapor pressure of pure
water at 25 °C is 23.8 torr.
23.2 torr
Sa. Ex. The vapor pressure of pure water at 110 °C is 1070 torr. A solution
of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 °C.
Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glyco
in the solution?
0.290
33. 2) Boiling-Point Elevation
By the addition of nanvolatile solute
0
∆Tb = Tb – T b
0
T b is the boiling point of
the pure solvent
T b is the boiling point of
the solution
0
Tb > T b
∆Tb > 0
∆Tb = Kb m
(8)
m is the molality of the solution
Tb0
Tb
Kb is the molal boiling-point
elevation constant (0C/m)
for a given solvent
34. 3) Freezing-Point Depression
By the addition solute
∆Tf = T 0 – Tf
f
T
0
is the freezing point of
the pure solvent
T f is the freezing point of
the solution
f
T 0 > Tf
f
∆Tf > 0
∆Tf = Kf m
(9)
m is the molality of the solution
Tf
Tf0
Kf is the molal freezing-point
depression constant (0C/m)
for a given solvent
(depend only on solvent)
35. The decrease in Tf upon addition of solute is due to the more
energy needs to be removed to form order solution than to form
order solvent (solution is more disorder)
The solid that separate when a solution freeze is the pure solvent only
that why the V.P. of solid is the same as that for pure liquid (see the
phase diagram)
37. Q) 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.(Kf water = 1.86 0C/m)
∆Tf = Kf m
m =
(9)
moles of solute
mass of solvent (kg)
=
478 g
62.01 g
3.202 kg solvent
= 2.41 m
∆Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
∆Tf = T 0 – Tf
f
Tf = T 0 – ∆Tf = 0.00 0C – 4.48 0C = -4.48 0C
f
38. Sa. Ex. Automotive antifreeze consists of ethylene glycol (C2H6O2),
a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point
of a 25.0 mass% solution of ethylene glycol in water.
(Kb water=0.52 and Kf water = 1.86 °C/m)
∆Tb = Kb m
Tb = Tb0 +∆Tb
102.8 °C
(8)
∆Tf = Kf m
(9)
Tf = Tf0 -∆Tb
-10.0 °C
Sa. Ex. Calculate the freezing point of a solution containing 0.600 kg of
CHCl3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found
in the leaves of eucalyptus trees.
39. 4) Osmotic Pressure (π)
•Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution to a more concentrated one.
•Semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
• Osmotic pressure (π ) is the pressure required to stop osmosis.
Cucumber in concentrated brine lose water
dilute
more
concentrated
Pickle
40. High
P
Low
P
Osmotic pressure
in unit atm
π = MRT
Colligative property = depend on concentration only
M is the molarity of the solution
R is the gas constant (0.0821 L atm/mol K)
T is the temperature (in K)
(10)
41. • When the concentrations of the two solution are equal and, hence, have the
same osmotic pressure, the solutions called isotonic
•Isotonic: the two solutions are of equal concentration and, hence,
they have the same osmotic pressure
•When concentrations of the two solution are different, the lower concentration
called hypotonic and the higher concentration called hypertonic
•Hypotonic: the solution of lower concentration
•Hypertonic: the solution of higher concentration
42. A cell in an:
A cell in
isotonic
solution
the two solution are
equal in concentration
A cell in
hypotonic
solution
A cell in
hypertonic
solution
the outer solution is less the outer solution is more
concentrated than the
concentrated than the
solution in the cell
solution in the cell
44. Sa. Ex.:The average osmotic pressure of blood is 7.7 atm at 25 °C.
what concentration of glucose (C6H12O6) will be isotonic with
blood?
0.31 M
Sa. Ex.:What is the osmotic pressure at 20 °C of a 0.0020 M
sucrose (C12H22O11) solution?
0.048 atm
45. •Using Colligative Properties to determine molar mass
B.P. Elevation
∆Tb = Kb m
F.P. Depression
(9)
(8)
∆Tf = Kf m
m = ∆Tb
m = ∆Tf
Kb
Kf
m=
n
m=
mass of solvent in Kg
m=
mass of solvent in Kg
mass in g
m=
M.m. (mass of solvent in Kg)
M.m =
mass in g
n
(11)
m (mass of solvent in Kg)
mass in g
M.m. (mass of solvent in Kg)
M.m =
mass in g
(11)
m (mass of solvent in Kg)
46. Osmotic Pressure
(10)
π = MRT
M= π
RT
M=
n
VL
M=
mass in g
M.m. (VL)
M.m =
mass in g
(12)
M (VL)
M.m =
(mass in g) R T
(VL) π
(13)
47. Kf of benzene = 5.12 °C/m
m = ∆Tf
Kf
M.m =
mass in g
(11)
m (mass of solvent in Kg)
127 g/mol
C10H8
49. Sa. Ex.: Coamphor (C10H16O) melts 1t 179.8 °C, and it has
a particularly large freezing-point-depression constant Kf = 40.0
°C/m. When 0.186 g of an organic substance of unknown molar
mass is dissolved in 22.01 g of liquid camphor, the freezing point
of the mixture is found to be 176.7 °C. what is the molar mass of
the solute?
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.
Vapor-Pressure Lowering
0
P1 = X1 P 1
(7)
Boiling-Point Elevation
∆Tb = Kb m
(8)
Freezing-Point Depression
∆Tf = Kf m
(9)
π = MRT
(10)
Osmotic Pressure (π)
51. 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
van’t Hoff factor (i) =
0.2 m ions in solution
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes
NaCl
CaCl2
1
2
3
52. Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
∆Tb = i Kb m (8b)
Freezing-Point Depression
∆Tf = i Kf m
Osmotic Pressure (π)
(9b)
π = iMRT (10b)
54. Sa. Ex.: List the following solutions in the order of their expected
freezing points: 0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl;
0.050 m HC2H3O2; 0.10 m C12H22O11.
Sa. Ex: Which of the following solutes will produce the largest
increase in boiling point upon addition to 1 Kg of water:
1 mol Co(NO3)2, 2 mol of KCl, 3 mol of ethylene glycol (C2H6O2)?
