SOLUTIONS AND THEIR PHYSICAL
PROPERTIES
TYPES OF SOLUTIONS
Solution: A homogenous mixture whos properties are uniform and which contains two or
more substances that can be varied
Solvent: The component that is present in the greatest quantity or that determines the state
of matter in which the solution exists.
Solute: A solution component that is present in a solution in lesser quantity than the
solvent.
Concentrated solution: Solutions containing a relatively high concentrations of solute.
Diluted solution: Solution containing a relatively low concentrations of solute.
Solubility: The max. amount of the solute that will dissolve in a definite amount of the
solvent and produce a stable system at a specified temperature
Saturated solution: When dissolving and precipitation occur at the same rate, the
quantity of dissolved solute remains constant with time and the solution is said to be
saturated.
Supersaturated solution: The solution in which the concentration of solute is higher than
that of a saturated solution.
Alloy: Solid solutions with a metal as the solvent ; e.g. brass (Cu-Zn), solder (Sn-Pb)
Some Common Solutions
SOLUTE
SOLUTION
Gaseous
SOLVENT
Liquid
Solid
Gaseous
Air(N2+O2+several
others)
Liquid
Mineral water(
CO2 in water)
Alcohol(Ethan
ol)-water,
Polyhydrocarb
on-gasoline
NaCl in water, Au in
Hg
Solid
Pt/ H2
Mercury in
Gold,
Hexane in
paraffin
Bronze(Cu-Sn)
Brass(Cu-Zn)
Solution Concentration
A measure of the quantity of the solute present in a given quantity of solvent is called
concentration denir. Concentration can be explained in many ways using expressions such
as saturated, unsaturated, supersaturated, diluted or concentrated solution. Besides, it can be
expressed also with some terms of quantity such as, molarity, ppt, ppm ve ppb, molality,
normality. In addition, some other definition such as mole fraction, mole percentage can be
used.
Mass Percent, Volume Percent, and Mass/Volume Percent
Mass percentage % (m/m) = msolute/msolution
If we dissolve 5,00 g NaCl, in 95 g water , we get 100 g of a solution that is % 5,00 NaCl .
% Volume (V/V) = Vsolute/Vsolution
25,0 mL methyl alcohol and 75,0 mL water is an antifreeze solution that is % 25,0 CH3OH by
volume.
Mass/Volume % (m/V) = msolute/Vsolution
An aqueous solution with 0,9 g NaCl in 100 mL of solution is said to be % 0,9
(mass/volume).
Solution Concentration
Mole Fraction and Mole Percent
Mol fraction
ni
i 
nt
Amount of component (in moles)
Total
amount
of
all solution
components(
in moles)
ni
Mole percent %  i 
 100
nt
Molarity ve Molality
Molarity
Molality
nsolute
M 
Vsolution
Amount of solute (in moles)
nsolute
M
msolvent
Amount of solute (in moles)
Volume of solution (L)
Mass of solvent ( Kg )
Illustrative Examples
The concentration of a solution in different units:
An ethanol-water solution is prepared by dissolving 10,00 mL of C2H5OH
(d=0,789 g/mL) in a sufficient volume of water to produce 100 mL of a
solution with a density 0,982 g/mL. What is the concentration of ethanol
in this solution as (a) volume percent (b) mass percent (c) Mass/Volume
percent (d) mole fraction(e) mole percent(f) molarity(g) molality?
10,00 mL ethanol
a) % ethanol (volume)  100,0 mL solution 100  %10,00
0,789 g ethanol
b) m
 7,89 g ethanol
ethanol  10,00 mL 
1 mL ethanol
0,982 g solution
msolution  100,0 mL solution 
 98,2 g solution
1 mL solution
7,89 g ethanol
% ethanol(m/ m) 
100  % 8,03
98,2 g solution
Illustrative Examples
c)% ethanol (m/V) 
7,89 g ethanol
100  % 7,89
100,0 mL solution
d) Convert the mass of ethanol from part b) to an amount in moles
nethanol  7,89 g C2 H 5OH 
1mol C2 H 5OH
 0,171mol C2 H 5OH
46,07 g C2 H 5OH
If 100,0 mL solution weighs 98,2 g ; 98,2 - 7,89 =
90,3g water.
nwater  90,3 g H 2O 
1 mol H 2O
 5,01mol H 2O
18,02 g H 2O
Mole Fraction of Ethanol
0,171 mol C2 H 5OH
 ethanol 
 0,0330
0,171 mol C2 H 5OH  5,01 mol H 2O
SOLUTION CONCENTRATION
e) Mole Percent of Ethanol =
f)
ethanol 100  % 3,30
0,171 mol C2 H 5OH
M
 1,71 M C2 H 5OH
0,1 L H 2O
g) mwater(kg)  90,3 g H 2O 
1kg H 2O
 0,0903 kg H 2O
1000 g H 2O
“Molality of Ethanol”
0,171 mol C2 H 5OH
M
 0,189 m C2 H 5OH
0,0903 kg H 2O
m: molality (mol/kg)
INTERMOLECULAR FORCES AND THE SOLUTION
PROCESS
In this section we focus on the behaviour of molecules in solution,
specifically on intermolecular forces
II. aşama
Enthalpy
III. aşama
I. aşama
Pure solvent
Pure solute
ΔHsoln= 0
seperated solvent molecules ΔHa >0
seperated solute molecules ΔHb >0
Seperated solvent and solid molecules
ΔHsoln= ΔHa+ ΔHb + ΔHc
solution ΔHc <0
INTERMOLECULAR FORCES IN MIXTURES
• The values ΔHa, ΔHb, and ΔHc depend on the
strengths of intermolecular forces of attraction
• If all intermolecular forces of attraction are of
about equal strength, a random of intermingling of
molecules occurs. A homogenous mixture or
solution results and this is called as ideal solutions.
• ΔHa + ΔHb ~ -ΔHc ΔHsoln ~ 0,
Intermolecular forces between:
ΔHa = solvent molecules A
ΔHb = solute molecules B ,
ΔHc = solvent molecules A and
solute molecules B (A-B)
IDEAL SOLUTIONS
Substances with similar molecular structures have similar
intermolecular forces of attraction. Example: The solution of the liquid
hydrocarbons (Benzene- Toluene mixture). ΔHsoln ~ 0 .
NONIDEAL SOLUTIONS
• If forces of attraction between same
molecules, A-A or B-B (molecules between
solvent or solute), are SMALLER, than
forces of attraction of unlike molecules A-B,
(forces between solute and solvent), a solution
is formed but its properties can not be
predicted. These are identified as non-ideal
solutions. Interactions between solute and
solvent(ΔHc) release more heat than the heat
absorbed to seperate the solvent and solute
molecules (ΔHa + ΔHb). The work of
solubility is an exothermic reaction (ΔHsoln <
0).
A-A < A-B
OR
B-B < A-B
CHCl3 Chloroform -- (CH3)2CO Acetone
NONIDEAL SOLUTIONS
• If forces of attraction between same
molecules, A-A or B-B (molecules between
solvent or solute), are ONLY A LITTLE
GREATER, than forces of attraction of
unlike molecules A-B, (forces between solute
and solvent), a NON-IDEAL solution is
formed as a complete mixture. The solution
has a higher enthalpy(ΔHc) than the pure
componentsΔHa+ ΔHb); so the solution
process is endothermic (ΔHsoln > 0).
A-A > A-B
veya
B-B > A-B
(CH3)2CO Acetone ----- CS2
Carbon disulfide(Polar-Non
polar solution)
NONIDEAL SOLUTIONS
• If forces of attraction between same
molecules, A-A or B-B (molecules between
solvent or solute), are MUCH GREATER,
than forces of attraction of unlike molecules
A-B, (forces between solute and solvent),
dissolving does not occur to any siginificant
extent. The components remain segregated in
a heterogenous mixture. In a mixture of
water and octane( a constituent of gasoline)
strong hydrogen bonds hold water molecules
together in clusters. The non-polar octane
molecules are not able to exert a strong
attractive force on the water molecules, and
two liquids do not mix.
A-A >> A-B
or
B-B >> A-B
Water molecules
Strong H bonds
Octane molecules
IONIC SOLUTIONS
KCl(s)
K+(g) + Cl-(g)
K+(g) + Cl-(g)
ΔH1=(- lattice energy of KCl)=701,2 kJ
K+(aq) + Cl-(aq) ΔH2= - 684,1 kJ
ΔHf= ΔH1 + ΔH2= 17,1 kJ
• If the ion-dipole forces of
attraction are strong enough to
overcome the interionic forces
of attraction in the crystal,
dissolving will occur. An ion
surrounded by a cluster of water
molecules
is said to be
hydrated.
SOLUTION FORMATION AND EQUILIBRIUM
• When the solute and the solvent are
mixed up;
• a) First dissolving occur,
• b) Then precipitation
increases with time
starts
and
• c) After a while the dissociation and
precipitation rates become equal.The
quantity of dissolved solid remains
constant with time
a)Diluted
b)Concentrated c)Saturated
• These solutions are called as saturated solutions
• The concentration of the saturated solution is called the solubility of the
solute in the given solvent . It can be expressed as Molarity or Mass Percent
• Solubility varies with temperature.
SOLUTION FORMATION AND EQUILIBRIUM
• If in preparing a solution we start with
less solute than would be present in the
saturated solution, the solution is
unsaturated.
• The precipitation occurs when a
saturated solution is cooled.
• Occasionally all the solute may remain
in the solution(as at that temperature the
solvent dissolves a greater amount of
solute than that in the saturated solution)
We
identify
these solutions
as
supersaturated solutions.
• KNO3 dissolves in 100 g water and
30oC at an amount of ~32g .This solution
is saturated. At the same temp., if 30g
KNO3 are dissolved, it is unsaturated, 35g
KNO3 are dissolved, it is supersaturated.
Solubility Curve for various substances
SOLUTION FORMATION AND EQUILIBRIUM
• The solubilities of ionic substances
increase with increasing temperature
Exceptions to this generalization
are:SO32-, SO42-, SeO42-, AsO43-, PO43- .
• When
Hsoln > 0, raising the
temperature increases the solubility.
• When, Hsoln < 0, the solubility
decreases with increasing temperature.
• We prepare a concentrated solution at a
high temperature. Then we let the solution
cool. At lower temperatures the solution
becomes saturated in the desired
compound. The excess compound
crystallizes from solution and the
impurities remain in solution. This
method of purifying
a solid called
fractional crystallisation.
SOLUBILITY OF GASES
Solubility: The amount of gas which is able to dissolve at 0oC and 1 atm pressure
in 1 L water.
• Effect of Temperature
In a gas, the molecules are much farther apart than they will
be in the solution(solid-liquid mixtures). Hence, the gas
must condense to a liquid before it dissolves in another
liquid. Condensation is an exothermic reaction and “The
solubilities of gases decrease with increased
temperature”. Ex: Beverages which contain CO2 are
consumed cold. Fish require cold watersince there is not
enough dissolved air(oxygen) in hot water
• Effect of Pressure
C  kP
The solubility of a gas increases as the
gas pressure is increased.This is called
gas
Henry’s law.
If the gas pressure increases, the
C = solubility of gas mL/L
solubility rises
k = proportionality constant
Henry’s Law fails for gases at high pressures and it also fails if the gas
ionizes in water or reacts with water.
We only expect Henry’s law to apply to equilibrium between molecules
of a gas and the same molecules in solution
SOLUBILITY OF GASES
Example: At 0oC and 1 atm pressure N2 gas has the solubility of
23,54 mL N2/L. in order to reach a solubility of 100,0 mL N2 per liter,
what pressure must be exerted at 0oC?
k
C
23,54 mL N 2 / L 100,0 mL N 2 / L



Pgaz
1,00 atm
PN 2
PN 2  4,25 atm
Example: At 0oC and O2 pressure of 1,00 atm, the aqueous solubility
of O2(g) is 48,9 mL O2/L. What is the molarity of O2 in a saturated
water solution when the O2 is under its normal partial pressure in air
(0,2095 atm)?
Molarity of O2 at
and 1 atm
0oC
M O2 
1mol O2
22,4 L O2 ( NK )
 2,18 103 M O2
1 L Solution
0,0489 L O2
C
2,18 10 3 M O2
C
k


 C  4,57 10  4 M O2
Pgaz
1,00 atm
0,2095 atm
VAPOR PRESSURES OF SOLUTIONS
We find the vapor pressures of solutions to be important
when we want devise a method of seperating volatile
liquid mixtures by distillation. Also, boiling point and
osmotic pressure play an important role in the fractional
distillation. In 1880, the French chemist Raoult found
that a dissolved solid lowers the vapor pressure of the
solvent.
The partial pressure exerted by solvent vapor above an
ideal solution, PA, is the vapor pressure of the pure
solvent at the given temperature PAo, multiplied by the
mole fraction of the solvent in the solution, XA
Fractional Distillation
Raoult’s Law:
PA   A  P
o
A
VAPOR PRESSURES OF SOLUTIONS
Predicting Vapor Pressures of Ideal Solutions: The vapor pressures of
pure benzene and toluene at 25 oC are 95,1 ve 28,4 mmHg. A solution
is prepared in which the mole fractions of benzene and toluene are both
0,500. What are the partial pressures of benzene and toluene in this
solution? What is the total pressure? What is the composition of the
vapor of benzene-toluene at equilibrium ?
0
Pbenz   benz  Pbenz
 0,500  95,1 mmHg  47,6 mmHg
Ptol   tol  Ptol0  0,500  28,4 mmHg  14,2 mmHg
Ptop  47,6 mmHg  14,2 mmHg
Composition of
solution
Ptop  61,8 mmHg
 benz 
 tol
Pbenz 47,6 mmHg

 0,770
Ptop 61,8 mmHg
Ptol 14,2 mmHg


 0,230
Ptop 61,8 mmHg
Composition of vapor
LIQUID-VAPOR EQUILIBRIUM: IDEAL SOLUTIONS
ben  0,500
The line combining the 3 - 4 points is
called “tie line”. The vapor ends of these
ties lines can be joined by the green curve.
From the relative placement of the liquidvapor curves we see for ideal solutions of
two components, the vapor phase is richer
in the more volatile component than is the
liquid phase.
Composition of Solution
ben  0,500
 tol  0,500
Comp. of Vapor
ben  0,770
 tol  0,230
When this solution is again vaporized;
Comp. of Vapor
ben  0,920
 tol  0,080
The
process
of
revaporisation
and
condensation
can
be
continiued. This is called
fractional distillation.
LIQUID- VAPOR EQUILIBRIUM:NONIDEAL
SOLUTIONS
LIQUID-VAPOR CURVES IN
AZEOTROPE MIXTURES
In nonideal solutions, if the departures
from ideal solution behaviour are
sufficiently great, certain solutions
may vaporize to produce a vapor that
has the same composition as the liquid.
These solutions, called azeotropes
boil at a constant temperature and
because the liquid and vapor have the
same composition, they can not be
seperated by fractional distillation
Propanol-Water Azeotrope
Mixture contains 71,69%
propanol and 28,31% water.
Ethyl alcohol-water azetrope
mixture has the boling point of
78,2˚C with the composition
95.6% ethyl alc. and 4.4% water
Freezing Point Depression and Boiling Point
Elevation of Nonelectrolyte Solutions
The properties such as Vapor pressure lowering, freezing point depression, boiling point
elevation, and osmotic pressure whose values depend only on the concentration of solute
particles in solution and not on what the solute is, are called colligative properties.
Freezing point depression
TFP  KFP  m
Pressure
TFP  TFP0  TFP
Boiling point elevation
ΔTFP
fp fp0
ΔTBP
bp0 bp
Temperature
TBP  KBP  m
TBP  TBP  TBP0
Freezing Point Depression and Boiling Point
Elevation of Nonelectrolyte Solutions
Establishing a molecular formula with freezing point data: Nicotine,
extracted from tobacco leaves, is a liquid completely miscible with water
at temperatures below 60 oC . (a)What is the molality of nicotine in an
aqueous solution that starts to freeze at -0,450 oC ? (b) If the solution is
obtained by dissolving 1,921 g of nicotine in 48,92 g of water, what must
be the molar mass of nicotine? KFP=1,86 oC/m
a)
b)
TFP 0 C  (0,450 C )
m

 0,242 m

C
K FP
1,86 m
m
n
mwater
1,921 g / MAni cot ine
mol

 0,242
0,04892 kg water
kg water
MAni cot ine 
1,921 g
 162 g / mol
(0,04892  0,242) mol
OSMOTIC PRESSURE
An aqueous sucrose(sugar) solution in a long glass tube is
seperated from pure water by a semipermeable
membrane(permeable to water only). Water molecules can
pass through the membrane in either direction. But because
the concentration of water molecules is greater in the pure
water than in the solution, there is a net flow from the pure
water into the solution. This net flow called osmosis, causes
the solution to rise up the tube.
Applying a pressure to the sucrose solution slows down the net flow of water into the solution.
The necesssary pressure to stop osmotic flow is called the osmotic pressure and represented by
the symbol p. This pressure is 15 atm for a 20% sucrose solution. Osmotic pressure is a
colligative property because its magnitude depends only on the number of solute particles per unit
volume of solution
Osmotic Pressure (p)
R(gas constant) = 0,08206 L atm/(mol K)
n
p
 R T
V
p  c  R T
c = Molarity of the solution
OSMOTIC PRESSURE
Calculating osmotic pressure: What is the osmotic pressure at 25 oC of
an aqueous solution that is 0,0010 M C12H22O11 (sucrose)?
0,0010 mol  0,08206 L atm /( mol K )  298 K
p
 0,024 atm (18 mmHg )
1L
Establishing a molar mass from a measurement of osmotic pressure: A
50,00 mL sample of an aqueous solution is prepared containing 1,08 g.
of a blood plasma protein, human serum albumin. The solution has an
osmotic pressure of 5,85 mmHg at 298 K. What is the molar mass of the
albumin?
1 atm
3
Palb  5,85 mmHg 
p
(m / MA) RT
mRT
 MA 
V
pV
760 mmHg
 7,70 10 atm
1,08 g  0,08206 L atm /( mol K )  298K
4
MAalb 

6
,
86

10
g / mol
3
7,70 10  0,050 L
Freezing Point Depression and Boiling Point
Elevation of Electrolyte Solutions
Some solutes produce a greater effect on colligative properties. For example consider a 0,0100 m
aqueous solution. The predicted freezing point depression of this solution is
TFP  K FP  m  1,86 C / m  0,0100 m  0,0186 C
If the 0,0100 m solution is urea, the measured freezing point is just -0,0186, if the solution is
0,0100 m NaCl (electrolyte), the measured freezing point is -0,0361. The ratio of the measured
value of a colligative property to the expected value, if the solute were electrolyte is called Van’t
Hoff factor (i) .
(TFP ) measured 0,0361 C
Van’t Hoff
i

 1,92  2

factor
(TFP)exp ected 0,0186 C
Equations of Colligative Properties for some electrolytes
For a non-electrolyte i = 1
NaCl i = 2
MgCl2 i = 3
TFP  i  KFP  m
TBP  i  KBP  m
p  i  c  R T
Freezing Point Depression and Boiling Point
Elevation of Nonelectrolyte Solutions
Predicting Colligative Properties for electrolyte solutions: Predict the freezing point of
aqueous 0,00145 m MgCl2 . KFP = 1,86 oC/m
MgCl2 (aq)  Mg2+(aq) + 2Cl-(aq)
Van’t Hoff factor, i = 3
TFP  i  K FP  m
TFP  3 1,86 C / m  0,00145 m

TFP  0,0081 C
REVERSE OSMOSIS-DESALINATION
In this case, the membran is only permeable for the water molecules. Under normal
conditions in the event of OSMOSIS , there is a net flow from the pure
water to the salty water. When a pressure is applied on side B (P> Pos), a
net flow of water from salty water to the pure water occurs. This is called
REVERSE OSMOSIS . By means of DESALINATION, pure water can
be obtained from seawater and other waste water can be reused by this
way .
Practical Applications -FP lowering
Salt can be used to deice roads.
Automobile antifreeze ethylene glycol
(C2H4(OH)2)