This document discusses different types of mixtures and solutions. It defines homogeneous and heterogeneous mixtures, and describes true solutions, suspensions, and colloids. It also defines key terms related to solutions such as solute, solvent, saturated, unsaturated and supersaturated solutions. Finally, it discusses various ways of expressing the concentration of solutions, including molarity, molality, mass percent, parts per million/billion and more. It provides examples and equations to calculate concentrations using these different units.

1. Solutions
IMRAN TARIQ
ASSISTANT PROFESSOR
UNIVERSITY COLLEGE OF PHARMACY,
UNIVERSITY OF THE PUNJAB

2. TYPES OF MIXTURES
HOMOGENOUS e.g. True solutions
HETEROGENOUS e.g. coarse and colloidal
dispersions
Homogeneity: something evenly distributed
Heterogeneity: something not distributed evenly in
space; a clump or cluster (not a single phase as in case
of colloidal and coarse dispersion).

3. Definitions
A solution is a homogeneous mixture of two or more
substances
OR
A solution is a homogenous mixture of solute and
solvent.
OR
 A solution is a homogenous mixture of two
substances but consisting of one phase.

4. A solute is dissolved in a solvent.
– solute is the substance being dissolved
– solvent is the liquid in which the solute is dissolved
– an aqueous solution has water as solvent
Binary solution: A homogenous mixture consisting
of one phase and containing only two components
i.e. one solute and one solvent e.g. Solution of Nacl
in water.

5. Dilute Solutions: A solution containing relatively
small quantity of solute as compared with the amount
of solvent.
Concentrated Solution: A solution containing large
amount of solute in the solution than that in dilute
solution.
Un-saturated solution: a solution in which more
solute can be dissolved at a given temperature is
called as an unsaturated solution.

6. A saturated solution is one where the concentration
is at a maximum - no more solute is able to dissolve
at a given temperature.
– A saturated solution represents an equilibrium: the rate of
dissolving is equal to the rate of crystallization. The salt
continues to dissolve, but crystallizes at the same rate or
under these condition the number of molecules leaving the
solute is equal to the number of molecule returning to the
solid phase i.e. solute.

7. Super Saturated Solution:
A solution that contains relatively larger amount of
solute than that required for saturation it is prepared
by heating and adding more and more solute.

8. Types of Solutions: Based on
physical states of solute and solvent:
Gas
Gas Solid Smoke in air
Gas
Gas
liquid Water vapors in air

9. Solutions
How does a solid dissolve
into a liquid?
What ‘drives’ the
dissolution process?
What are the energetics of
dissolution?

10. How Does a Solution Form?
1. Solvent molecules attracted to surface ions.
2. Each ion is surrounded by solvent molecules.
3. Enthalpy (∆H) changes with each interaction broken or
formed.
Ionic solid dissolving in water

11. How Does a Solution Form
The ions are solvated
(surrounded by
solvent).
If the solvent is water,
the ions are
hydrated.
The intermolecular
force here is ion-
dipole.

12. Dissolution vs reaction
Dissolution is a physical change—you can get back the
original solute by evaporating the solvent.
If you can’t, the substance didn’t dissolve, it reacted.

13. Degree of saturation
Saturated solution
Solvent holds as much
solute as is possible at
that temperature.
Undissolved solid
remains in flask.
Dissolved solute is in
dynamic equilibrium
with solid solute
particles.

14. Degree of saturation
Unsaturated Solution
Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.
No solid remains in
flask.

15. Degree of saturation
Supersaturated
Solvent holds more solute than is normally
possible at that temperature.
These solutions are unstable; crystallization can
often be stimulated by adding a “seed crystal” or
scratching the side of the flask.

16. 3 Stages of Solution Process
Separation of Solute
– must overcome IMF or ion-ion attractions in solute
– requires energy, ENDOTHERMIC ( + ∆H)
Separation of Solvent
– must overcome IMF of solvent particles
– requires energy, ENDOTHERMIC (+ ∆H)
Interaction of Solute & Solvent
– attractive bonds form between solute particles and solvent
particles
– “Solvation” or “Hydration” (where water = solvent)
– releases energy, EXOTHERMIC (- ∆H)

17. Dissolution at the molecular level?
Consider the dissolution of NaOH in H2O

18. TYPES OF HETEROGENEOUS
MIXTURES
SUSPENSIONS
A suspension is a mixture in
which particles of a material are
dispersed throughout a liquid or a
gas but are large enough that they
settle out.
The particles in a suspension are
large enough to scatter or block
light. A suspension can be
separated by passing it through a
filter.

19. SUSPENSIONS
Different components are in different phase, such as
solids in liquids or liquids in gases
It is necessary to shake the substance before using it

20. COLLOIDS
Is a mixture in which the particles are
spread throughout but are not large
enough to settle out.
The particles are not as small as
those of a solution, however are
smaller than those of a suspension

21. COLLOIDS
Particles in a colloid are large enough to scatter light.
A colloid cannot be separated by passing it through a
filter.

22. COLLOIDS
Consists of two separated
phases: Disperse phase (or
internal phase) and a continuos
phase (or dispersion medium).
May be solid, liquid or gas
Some are translucent because of
the Tyndall Effect (which is the
scattering of light)

23. Solutions Colloids Suspensions
Homogeneous Heterogeneous Heterogeneous
Solution in which dispersoid
or consists of small
molecules or ions.
(Evenly distributed)
Solution in which dispersoid
consists of single large
molecule or group of small
molecule.
Solution in which dispersoid
consists of group of large
molecules. ( not evenly
distributed at all)
Particle size: 0.01-1 nm;
can be atoms, ions,
molecules
Particle size: 1-1000 nm,
dispersed; can be combined
or large molecules
Particle size: over 1000 nm,
suspended; can be large
particles or combined
particles
Do not separate on standing Do not separate on standing Particles settle out
Do not scatter light Scatter light (Tyndall effect) May scatter light, but are
not transparent
They may pass through
ordinary as well as ultra
filters.
They may pass through
ordinary filter but not usually
through ultra filters
Thy cant pass through
ordinary as well as ultra
filters.
Examples:
1)Solution of Nacl in water.
2)Solution of glucose in
Examples:
1)Solution of satrch
2)Milk
Examples:
1)Pharmaceutical
suspension and emulsions

24. Ultra filtration is a separation process using
membranes with a pore sizes in the range of 0.1 to
0.001 micron. Typically, ultrafiltration will remove high
molecular weight substances, colloidal materials and
organic polymeric molecules. Low molecular weight
organics and ions such as sodium, calcium,
magnesium and sulfates are not removed.

25. HETEROGENEOUS MIXTURES
You can distinguish the two or more phases.

26. Gases in Solution
In general, the solubility of
gases in water increases
with increasing mass.
Larger molecules have
stronger dispersion
forces.

27. Gases in Solution
The solubility of
liquids and solids
does not change
appreciably with
pressure.
The solubility of a gas
in a liquid is directly
proportional to its
pressure.

28. Henry’s Law
Sg = kPg
where
Sg is the solubility of the
gas;
k is the Henry’s law
constant for that gas in that
solvent;
Pg is the partial pressure of
the gas above the liquid.

29. At a certain temperature, the Henry’s
law constant for N2 is 6.0 × 10−4
M/atm.
If N2 is present at 3.0 atm, what is the
solubility of N2?
1. 6.0 × 10−4
M
2. 1.8 × 10−3
M
3. 2.0 × 10−4
M
4. 5.0 × 10−5
M

30. Correct Answer:
Henry’s law, Sg
= kPg
Sg = (6.0 × 10−4
M/atm)(3.0 atm)
S = 1.8 × 10−3
M
1. 6.0 × 10−4
M
2. 1.8 × 10−3
M
3. 2.0 × 10−4
M
4. 5.0 × 10−5
M

31. Calculate the concentration of CO2 in a soft
drink that is bottled with a partial pressure
of CO2 of 4.0 atm over the liquid at 25°C.
The Henry’s law constant for CO2 in water
at this temperature is 3.1 × 10–2
mol/L-atm.
Solve:

32. Temperature
Generally, the
solubility of solid
solutes in liquid
solvents
increases with
increasing
temperature.

33. Temperature
The opposite is
true of gases:
Carbonated soft
drinks are more
“bubbly” if stored
in the refrigerator.
Warm lakes have
less O2 dissolved
in them than cool
lakes.

34. CONCENTRATION
EXPRESSION

35. 5 ways of expressing concentration5 ways of expressing concentration
–Mass percent: (mass solute / mass of solution) * 100
–Molarity(M): moles solute / Liter solution
–Molality* (m) - moles solute / Kg solvent
–Normality (N)- gram equivalent of solute/ liter solution
–Mole Fraction(χ
A) - moles solute / total moles solution
* Note that molality is the only concentration unit in which
denominator contains only solvent information rather than
solution.

36. Mass Percentage or Percentage
Expression
Mass % of A =
mass of A in solution
total mass of solution
× 100
% expression is an expression of parts of solute per 100 parts
Of the solution

37. Determine the mass percentage of
hexane in a solution containing
11 g of butane in 110 g of hexane.
1.9.0 %
2.10. %
3.90.%
4.91 %

38. Correct Answer:
Thus,
110 g
(110 g + 11 g)
100
solutionofmasstotal
solutionincomponentofmass
componentof%mass ×=
× 100 = 91%
1. 9.0 %
2. 10. %
3. 90.%
4. 91 %

39. % (w/w) =
% (w/v) =
% (v/v) =
% Concentration
100x
solutionmass
solutemass
100x
solutionvolume
solutemass
100x
solutionvolume
solutevolume

40. % w/w:
It expresses the no. of grams of the solute per 100
gram of the solution.
e.g. a 10 % w/w aqueous glycerine solution means 10 g
of glycerine dissolved in sufficient water to make
overall 100 gram of the solution.

41. % v/v:
It expresses the no. of milliliters of the solute per 100
milliliters of the solution.
e.g. a 10 % v/v aqueous ethanolic solution means 10 ml
of ethanol dissolved in sufficient water to make
overall 100 mls of the solution.

42. % w/v
It expresses the no. of grams of the solute per 100
mls of the solution.
e.g. a 10 % w/v aqueous Nacl solution means 10 g of
Nacl dissolved in sufficient water to make overall 100
mls of the solution.

43. % v/w
It expresses the no. of mls of the solute per 100 gram
of the solution.
e.g. a 10 % v/w aqueous glycerine solution means 10
ml of glycerine dissolved in sufficient water to make
overall 100 gram of the solution.

44. This is another way to expressing concentration,
particularly those of very dilute solutions. e.g. to
express the impurities of substances in water.
ppm denotes the amount of given substance in a total
amount of 1,000,000 of solution e.g. one milligram
per kilogram. 1 part in 10 6
ppb denotes the amount of given substance in a total
amount of 1,000,000,000 of solution e.g. 0.001
milligram per kilogram. 1 part in 10 9
Parts per Million and
Parts per Billion

45. Parts per Million and
Parts per Billion
ppm =
mass of A in solution
total mass of solution
× 106
Parts per Million (ppm)
Parts per Billion (ppb)
ppb =
mass of A in solution
total mass of solution
× 109

46. If 3.6 mg of Na+
is detected in a 200. g
sample of water from Lake Erie, what is
its concentration in ppm?
1. 7.2 ppm
2. 1.8 ppm
3. 18 ppm
4. 72 ppm

47. Correct Answer:
6
10
solutionofmasstotal
solutionincomponentofmass
componentofppm ×=
ppm1810
g200.
g0.0036
g200.
mg3.6 6
=×=
1. 7.2 ppm
2. 1.8 ppm
3. 18 ppm
4. 72 ppm

48. mole of solute
L of solution
M =
Molarity (M)
molarity= Given weight / molecular weight substance x
1/ volume of solution in liters
e.g.1 mole of Nacl= 58.5 gm of Nacl
No. of moles= Given weight / molecular weight substance
Because volume is temperature dependent, molarity can
change with temperature.
It expresses the no. of moles of solute
dissolved per liter of the solution.

49. Concentration: Molarity Example
If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of
solution, what is the molarity of KMnO4?
Now that the number of moles of substance is known, this can be
combined with the volume of solution — which must be in
liters — to give the molarity. Because 250. mL is equivalent to
0.250 L .
As is almost always the
case, the first step is to
convert the mass of
material to moles.
0.435 g KMnO4 • 1 mol KMnO4 = 0.00275 mol KMnO4
158.0 g KMnO4
Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M
0.250 L solution

50. moles of solute
kg of solvent
m =
Molality (m)
Because both moles and mass do not
change with temperature, molality
(unlike molarity) is not temperature
dependent.
It expresses the no. of moles of solute dissolved
per kg of the solvent.

51. Changing Molarity to Molality
If we know the
density of the
solution, we can
calculate the
molality from the
molarity, and
vice versa.

52. A solution is made by dissolving 4.35 g glucose (C6H12O6)
in 25.0 mL of water at 25°C. Calculate the molality of
glucose in the solution.
Solution molar mass of glucose, 180.2 g/mol
water has a density of 1.00 g/mL, so the mass of the solvent is

53. Normality (N)
It expresses the no. of gram equivalent of solute
dissolved per liter of the solution.
Normality= Given weight / equivalent weight x 1/
volume of solution in liters
No. gram equivalent of solute
L of solution
N =

54. EQUIVALENT WEIGHT
The no. of parts by weight of that substance that will
combine with one part by weight of replaceable H+ or
OH- or number of positive charge on elements

56. moles of A
total moles in solution
XA =
Mole Fraction (X)
The sum of the mole fractions of all the components
is always equal to unity i.e. 1
The ratio of the number of the moles of that
component to the total number of moles of all the
components of the solution

57. Colligative Properties
Colligative Properties are properties of a liquid that
change when a solute is added.
The magnitude of the change depends on the numbernumber
of solute particles in the solution, NOT on the naturenature
and sizeand size of the solute particles.
The colligative properties arise from the attractive
forces that are exerted by the solute on the solvent.
e.g. such attractive forces reduces the tendency of the
solvent molecules to escape from the liquids as
vapours and vapours pressure of the solvent is
therefore reduced by the presence of the solute.

58. Following are the colligative properties of the solution.
 Lowering of the vapour pressure
 Elevation of the boiling point
 Depression of the freezing point
 Osmotic pressure

59. Lowering of the vapour pressure

60. If any non-volatile and non- electrolyte solute is
dissolved in the solvent, the escaping tendency of the
solvent molecule is reduced and in turn vapour
pressure is also reduced. This is because solvent
molecules are hindered by the solute molecules and
so escaping tendency of the solvent molecules is
reduced. Therefore a less no. of solvent molecules
will be able to escape and so less no. of the vapours
will be formed and ultimately the vapour pressure of
the solvent is reduced. Greater the no. of the solute
particles, greater will be the hindrance thus greater is
the vapour pressure.

61. In case, when any non-volatile and electrolyte is
dissolved in the solvent, the lowering in the vapour
pressure will be abnormally lowered than that of the
non- electrolyte. This is because, the electrolyte
dissociate into ions in the solution, so no. of the
particle will increased and as a result, lowering in
vapour pressure will also high.

62. Elevation in boiling point

63. Greater the vapour pressure of a liquid, lower will be
its boiling point. Similarly, lower the vapour pressure
of the liquid, higher will be its boiling point. So If any
non-volatile and non- electrolyte solute is dissolved in
the solvent, the escaping tendency of the solvent
molecule is reduced and in turn vapour pressure is
also reduced and ultimately its boiling point will be
elevated.
In case, when any non-volatile and electrolyte is
dissolved in the solvent, the lowering in the vapour
pressure will be abnormally high than that of the non-
electrolyte. This is because, the electrolyte dissociate
into ions in the solution, so no. of the particle will
increased and as a result, lowering in vapour
pressure will also high. There fore elevation in boiling
point is also greater than that in case of non
electrolyte.

64. Normal Boiling Process
Extension of vapor pressure concept:
Normal Boiling Point: BP of Substance @ 1atm
When solute is added, BP > Normal BP
Boiling point is elevated when solute inhibits solvent from escaping.
Elevation of B. pt.
Express by BoilingExpress by Boiling
point Elevationpoint Elevation
equationequation

65. Depression in freezing point

66. The normal freezing point of solvent (pure liquid) is
defined as the temperature at which solid and liquids
forms of the solvent co- exist in equilibrium at a fixed
external pressure commonly 1 atm. ( i.e. 760 mm Hg)
Pure water has freezing point at 0˚C.
“At this equilibrium state the solid and liquid forms of the
solvent must have the same vapour pressure. If it is
not so, the form having the higher vapour pressure
would change into that form having lower vapour
pressure.”
( lowering in vapour pressure of the liquid after addition
of solute has been discussed earlier)

68. So the freezing point of the solution is the
temperature at which solid forms of the pure solvent
co- exist in equilibrium with the solution at a fixed
external pressure commonly 1 atm. ( i.e. 760 mm Hg)
As the vapour pressure of the solution is lower than
that of pure solvent so it is obvious that the solid and
solution cant be co exist in equilibrium ( this is
because at equilibrium both states must have same
vapour pressure). They can now co – exist at some
lower temperature ( new freezing point), where the
solid and solution have the same vapour pressure.
That’s why freezing point of the solvent will be lowered
upon addition of solute.
Greater no. of particles= lower of V.P= Depression in F.P

69. Osmotic pressure

70. Osmotic pressure
Osmosis is the spontaneous movement of water across a semi-
permeable membrane from an area of low solute concentration to an
area of high solute concentration
Osmotic Pressure - The Pressure that must be applied to stop osmosis
Π = i CRT
where P = osmotic pressure
i = van’t Hoff factor
C = molarity
R = ideal gas constant
T = Kelvin temperature

71. Osmosis and Blood Cells
(a) A cell placed in an isotonic solution. The net movement of
water in and out of the cell is zero because the concentration of
solutes inside and outside the cell is the same.
(b) In a hypertonic solution, the concentration of solutes outside
the cell is greater than that inside. There is a net flow of water
out of the cell, causing the cell to dehydrate, shrink, and
perhaps die.
(c) In a hypotonic solution, the concentration of solutes outside of
the cell is less than that inside. There is a net flow of water into
the cell, causing the cell to swell and perhaps to burst.

72. Vapor Pressure Lowering for a Solution
The diagram below shows how a phase diagram is affected by
dissolving a solute in a solvent.
The black curve represents the pure liquid and the blue curve
represents the solution.
Notice the changes in the freezing & boiling points.

73. Raoult’s Law
Describes vapor pressure lowering mathematically.
The lowering of the vapour pressure when a non-
volatile solute is dissolved in a volatile solvent (A) can
be described by Raoult’s Law:
PA = χAP°A
PA = vapour pressure of solvent A above the solution
cA = mole fraction of the solvent A in the solution
P°A = vapour pressure of pure solvent A
only the solvent (A) contributes to
the vapour pressure of the solution

74. What is the vapor pressure of water above a sucrose
(MW=342.3 g/mol) solution prepared by dissolving
158.0 g of sucrose in 641.6 g of water at 25 ºC?
The vapor pressure of pure water at 25 ºC is 23.76
mmHg.
mol sucrose = (158.0 g)/(342.3 g/mol) = 0.462 mol
mol water = (641.6 g)/(18 g/mol) = 35.6 mol
Xwater =
mol water
(mol water)+(mol sucrose)
=
35.6
35.6+0.462
= 0.987
Psol’n = Xwater P°water = (0.987)(23.76 mm Hg)
= 23.5 mm Hg

75. Mixtures of Volatile Liquids
Both liquids evaporate & contribute to the vapor pressure

76. Raoult’s Law: Mixing Two Volatile Liquids
Since BOTH liquids are volatile and contribute to the
vapour, the total vapor pressure can be represented
using Dalton’s Law:
PT = PA + PB
The vapor pressure from each component follows
Raoult’s Law:
PT = χAP°A + χBP°B
Also, χA + χB = 1 (since there are 2 components)

77. Benzene and Toluene
Consider a two solvent (volatile) system
– The vapor pressure from each component follows
Raoult's Law.
– Benzene - Toluene mixture:
• Recall that with only two components, χBz + χTol = 1
• Benzene: when χBz = 1, PBz = P°Bz = 384 torr &
when χBz = 0 , PBz = 0
• Toluene: when χTol = 1, PTol = P°Tol = 133 torr & when
χTol = 0, PBz = 0

78. 384 torr
133 torr
X Benzene
X Toluene
0 1
1 0
P (Total)
P (Benzene)
P (Toluene)
133 torr
384 torr