Ways of Expressing
Concentration
Mass Percentage, ppm, and ppb
• Definitions:
mass of componentin solution
mass % of component 
 100
total mass of solution
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Chapter 13
Example: How would you prepare 425 g of an aqueous
solution containing 2.40% by mass of sodium acetate,
NaC2H3O3?
Ans:
Mass of NaC2H3O3 = 10.2 g
Mass of H2O = mass of solution - mass of NaC2H3O3
= 415 g
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Chapter 13
Exercise:
Concentrated aqueous nitric acid has 69.0% by mass of
HNO3 and has a density of 1.41 gcm-3. What volume of
this solution contains 14.2 g of HNO3?
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Chapter 13
mass of component in solution
ppm of component 
10 6
total mass of solution
Also mgl-1
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Chapter 13
mass of componentin solution
ppb of component 
 109
total mass of solution
Also μgl-1
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Chapter 13
Exercise: Seawater contains 0.0064 g of dissolved
oxygen, O2, per litre. The density of seawater is 1.03
gcm-3. What is the concentration of oxygen, in ppm?
Ans: 6.2 ppm
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Chapter 13
Mole Fraction, Molarity, and Molality
moles of componentin solution
Mole fraction of component 
total moles of solution
moles solute
Molarity 
liters of solution
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Chapter 13
moles solute
Molality, m 
kg of solvent
• Converting between molarity (M) and molality (m) requires
density.
Exercise: 0.2 mol of ethylene glycol is dissolved in 2000 g of water.
Calculate the molality
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Chapter 13
Example: What is the molality of a solution containing 5.67 g of
glucose, C6H12O6 (Mr = 180.2 g), dissolved in 25.2 g of water?
(Calc. the mole fractions of the components as well).
Solution:
 Think about the solute!................glucose (express in moles)
 Think about the solvent!...............water (express in kilograms)
Ans: 1.25 m
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Chapter 13
Example: Converting molarity to molality
An aqueous solution is 0.907 M Pb(NO3)2. What is the
molality of lead nitrate, Pb(NO3)2, in this solution? The
density of the solution is 1.252 g/mL. (Molar mass of
Pb(NO3)2 = 331.2 g)
Solution:
 Mass of solution = density x volume
 Calculate mass of Pb(NO3)2 , ie, moles x Mr
 Mass of H2O = mass of solution – mass of Pb(NO3)2
 Molality = 0.953 m Pb(NO3)2
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Chapter 13
Colligative Properties
• Colligative properties - depend only on the
number of particles in solution and not on their
identity.
• So NaCl(aq) a +(aq) + Cl-(aq)
• K2SO4(aq)  2K+(aq) + SO42-(aq)
• C12H22O11(aq)  C12H22O11(aq)
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Chapter 13
• Examining the effect of adding a non-volatile
solute to a solvent on:
• 1. vapor pressure
• 2. boiling point
• 3. freezing point
• 4. osmosis
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Chapter 13
 Examples are: anti-freeze in the radiator water in a car
prevents freezing in winter and boiling in summer; snow
is melted by adding salt on sidewalks and streets
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Chapter 13
Lowering Vapor Pressure
• VP lowering depends on the amount of solute.
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Chapter 13
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Chapter 13
Lowering Vapor Pressure
• Raoult’s law:
Psoln = XsolventPosolvent
• Recall Dalton’s Law:
Ptotal = PA + PB + PC +….PN
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Chapter 13
• Ideal solution - obeys Raoult’s law
• Raoult’s law is to solutions what the ideal gas law is to gases
• Raoult’s law breaks down when the solvent-solvent and solutesolute intermolecular forces are greater than solute-solvent
intermolecular forces
• For liquid-liquid solutions where both components are volatile, a
modified form of Raoult’s law applies:
Ptotal = PA + PB = XAPoA + XBPoB
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Chapter 13
Example: Predict the vapour pressure of a solution
prepared by mixing 35 g solid Na2SO4 (Mr = 142
g/mol) with 175 g water at 25oC. The vapour pressure
of pure water at 25oC is 23.76 torr.
Ans: 22.1 torr
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Chapter 13
Exercise: The hydrocarbon limonene is the major
constituent of lemon oil. A solution of limonene in
78.0 g of benzene had a vapour pressure of 90.6 mm
Hg at 25oC, and the vapour pressure of pure benzene
at 25oC is 95.2 mm Hg. What is its mass and
molecular formula?
Ans: C10H16
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Chapter 13
 As with gases, ideal behaviour
for solutions is never perfectly
achieved
 Nearly ideal behaviour is
observed if solute-solute,
solvent-solvent and solutesolvent interactions are very
similar
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Chapter 13
Boiling-Point Elevation
Goal: interpret the phase diagram for a solution.
 Non-volatile solute lowers the vapor pressure
 Therefore the triple point - critical point curve is
lowered.
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Chapter 13
• Molal boiling-point-elevation constant, Kb, expresses
how much Tb changes with molality, m:
Tb  Kb m
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Chapter 13
Freezing Point Depression
T f  K f m
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Chapter 13
Colligative Properties
Freezing Point Depression
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Chapter 13
Example: How many grams of ethanol, C2H5OH, must be added
to 37.8 g of water to give a freezing point of -0.15oC?
Solution:
Water is the solvent and ethanol the solute
From table 13.4, Tf = 0.15oC; Kf for water is 1.86oC/m
T f  K f m
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Chapter 13
Colligative properties of ionic solutions
Tf = iKfm
where i is the no. of ions resulting from each formula unit
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Chapter 13
Example: Estimate the freezing point of a 0.010 m
aqueous solution of aluminium sulphate, Al2(SO4)3.
Assume the value of i based on the formula of the
compound.
Ans: -0.093oC
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Chapter 13
Osmosis
• Semipermeable membrane: permits passage of some
components of a solution. Example: cell membranes and
cellophane.
• Osmosis: the movement of a solvent from low solute
concentration to high solute concentration.
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Chapter 13
Osmosis
• Eventually the pressure difference between the arms
stops osmosis.
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Chapter 13
Osmosis
• Osmotic pressure, , is the pressure required to stop
osmosis:
V  nRT
n

    RT
V 
 MRT
• Isotonic solutions are solutions….?
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Chapter 13
•
•
•
•
Hypotonic solutions are solutions….?
Hypertonic solutions are solutions…?
Osmosis is spontaneous.
Red blood cells are surrounded by semipermeable
membranes.
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Chapter 13
Example: The formula for low-molecular weight
starch is (C6H10O5)n, where n averages 2x102. When
0.798 g of starch is dissolved in 100 mL of water
solution, what is the osmotic pressure at 25oC?
 = 0.006 atm
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Chapter 13
Exercise: Fish blood has an osmotic pressure equal to
that of seawater. If seawater freezes at -2.3oC, what is
the osmotic pressure of the blood at 25oC?
Ans: 30 atm
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Chapter 13
• Crenation:
– red blood cells placed in hypertonic solution
(relative to intracellular solution);
– The cell shrivels or swells up?
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Chapter 13
• Hemolysis:
there is a higher solute concentration in the cell;
What happens to the cell?
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Chapter 13
Osmosis
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Chapter 13
Hypertonic
solution
Hypotonic solution
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Chapter 13
To prevent crenation or hemolysis,
(intravenous) solutions must be isotonic.
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Chapter 13
IV
– Cucumber placed in NaCl solution loses water to shrivel up and
become a pickle.
– Limp carrot placed in water becomes firm because water enters
via osmosis.
– Salty food causes retention of water and swelling of tissues
(edema).
– Water moves into plants through osmosis.
– Salt added to meat or sugar to fruit prevents bacterial infection
(a bacterium placed on the salt will lose water through osmosis
and die).
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Chapter 13
• Active transport is the movement of nutrients and waste
material through a biological system.
• Active transport is not spontaneous.
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Chapter 13
Colloids
A colloid is a dispersion of particles of one substance
(the dispersed phase) throughout another substance or
solution (the continuous phase)
Examples….??
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Chapter 13
The particle sizes range from ~1 x 103 pm to 2 x 105
pm in size
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Chapter 13
Although a colloid appears to be homogeneous because
the dispersed particles are quite small, it can be
distinguished from a true solution by its ability to scatter
light
This is called the……effect?
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Chapter 13
Left: vessel containing colloid; Right: true solution
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Chapter 13
Aerosols – liquid droplets or solid particles dispersed in a
gas e.g. fog and smoke
Emulsion – liquid droplets dispersed throughout another
liquid e.g. butterfat in milk
Sol – solid particles dispersed in a liquid e.g. AgCl(s) in
H2O
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Chapter 13
Colloids in which the continuous phase in water can be
hydrophilic (e.g. protein molecules) or hydrophobic
colloids (Au particles in water).
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Chapter 13
How does soap stabilise oil in water?
And how do we digest fats in our digestive systems?
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Chapter 13
Removal of colloidal particles
• Because of their small size, colloidal particles tend to
be difficult to remove by processes such as filtration.
• Thus enlargement of particles by coagulation is
required
• Heating or
coagulation
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adding
an
Chapter 13
electrolyte
can
cause
• Heating increases collisions and hence particle size
• Electrolytes neutralise surface charges and reduce
repulsions e.g. Alum in water purification, clay deposits
in deltas
• Semi-permeable membranes can also be used to
separate ions from colloidal particles (dialysis e.g. waste
removal from blood by kidneys)
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Chapter 13