Uploaded by Hossain Rahman

Colligative Properties ppt Colligative Properties
Vapour pressure
Boiling point
Freezing point
Osmotic pressure
Learning objectives
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Describe meaning of colligative property
Use Raoult’s law to determine vapor pressure of
solutions
Describe physical basis for vapor pressure
lowering
Predict magnitude of vapor pressure lowering
based on chemical formula
Calculate osmotic pressure in solution and use to
determine molar mass of solute
Predict direction of deviation in non-ideal cases
based on intermolecular forces
Physical vs Chemical
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Mixing is physical process; chemical
properties don’t change
Properties of solutions are similar to those
of the pure substances
Addition of a foreign substance to water
alters the properties slightly
Colligative: particles are particles
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Colligative comes from colligate – to tie together
Colligative properties have common origin
Colligative properties depend on amount of
solute but do not depend on its chemical identity
Solute particles exert their effect merely by
being rather than doing
The effect is the same for all solutes
Colligative properties for
nonvolatile solutes:
Take it to the bank
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Vapour pressure is always lower
Boiling point is always higher
Freezing point is always lower
Osmotic pressure drives solvent from
lower concentration to higher
concentration
Non-volatile solutes and Raoult’s law
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Vapor pressure of solvent in solution containing nonvolatile solute is always lower than vapor pressure of
pure solvent at same T
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At equilibrium rate of vaporization = rate of condensation
Solute particles occupy volume reducing rate of evaporationthe
number of solvent molecules at the surface
The rate of evaporation decreases and so the vapor pressure
above the solution must decrease to recover the equilibrium
Molecular view of Raoult’s law:
Boiling point elevation
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In solution vapor
pressure is reduced
compared to pure
solvent
Liquid boils when
vapor pressure =
atmospheric pressure
Must increase T to
make vapor pressure
= atmospheric
Molecular view of Raoult’s law:
Freezing point depression
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Depends on the solute only being in the liquid phase
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Fewer water molecules at surface: rate of freezing drops
Ice turns into liquid
Lower temperature to regain balance
Depression of freezing point
Raoult’s Law
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Vapor pressure above solution is vapor
pressure of solvent times mole fraction of
solvent in solution
Pso ln  Psolv X solv
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Vapour pressure lowering follows:
Pso ln  Psolv X solute
Counting sheep (particles)
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The influence of the solute depends only on the
number of particles
Molecular and ionic compounds will produce
different numbers of particles per mole of
substance
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1 mole of a molecular solid → 1 mole of particles
1 mole of NaCl → 2 moles of particles
1 mole of CaCl2 → 3 moles of particles
Solution Deviants
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Like ideal gas law, Raoult’s Law works for an
ideal solution
Real solutions deviate from the ideal
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Concentration gets larger
Solute – solvent interactions are unequal
Solvent – solvent interactions are stronger than
the solute – solvent: Pvap is higher
Solvent – solute interactions are stronger than
solvent – solvent interactions: Pvap is lower
Incomplete dissociation
Not all ionic
substances dissociate
completely
 Van’t Hoff factor
accounts for this
Van’ t Hoff factor:
i = moles of particles in
soln/moles of solute
dissolved
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Riding high on a deep depression
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Blue curves are phase
boundaries for pure solvent
Red curves are phase
boundaries for solvent in
solution
Freezing point depression
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Pure solid separates out at
freezing – negative ΔTf
Boiling point elevation
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Vapour pressure in solution is
lower, so higher temperature
is required to reach
atmospheric – positive ΔTb
Magnitude of elevation
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Depends on the number of particles
present
Concentration is measured in molality
(independent of T) T  K m
b
b
Kb is the molal boiling point elevation
constant
Note: molality is calculated in terms of
particles
Magnitude of depression
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Analagous to boiling point, the freezing
point depression is proportional to the
molal concentration of solute particles
T f  K f m
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For solutes which are not completely
dissociated, the van’t Hoff factor is applied
to modify m:
T f  K f m  i
Osmosis: molecular discrimination
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A semi-permeable membrane
discriminates on the basis of molecular
type
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Solvent molecules pass through
Large molecules or ions are blocked
Solvent molecules will pass from a place of
lower solute concentration to higher
concentration to achieve equilibrium
Osmotic pressure
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Solvent passes into more conc solution
increasing volume
Passage of solvent can be prevented by applying
pressure
Pressure required to prevent transport equals
osmotic pressure
Calculating osmotic pressure
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The ideal gas law states
PV  nRT
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But n/V = M and so
  MRT
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Where M is the molar concentration of
particles and Π is the osmotic pressure
Note: molarity is used not molality
Osmotic pressure and molecular
mass
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Molar mass can be determined using any
of the colligative properties
Osmotic pressure provides the most
accurate determination because of the
magnitude of Π
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0.0200 M solution of glucose exerts osmotic
pressure of 374 mm Hg (0.5 atm) but freezing
point depression of only 0.02ºC
Determining molar mass
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A solution contains 20.0 mg insulin in
5.00 ml develops osmotic pressure of 12.5
mm Hg at 300 K
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M
12.5mmHg 1
RT
760mmHg
M
 6.68 104 M
L  atm
0.0821
300K
mol  K
Converting molarity to molar mass:
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Moles insulin = MxV = 3.34x10-6 mol
Molar mass = mass of insulin/moles of
insulin
= 0.0200 g/3.34x10-6 mol
= 5990 g/mol
Volatile solute: two liquids
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Total pressure is the sum of the pressures
of the two components
Ptotal  PA  PB
Ptotal  P X A  P X B
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A
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B
Ideal behaviour of liquid mixture
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Total pressure in a mixture of toluene (b.p. =
110.6ºC) and benzene (b.p. = 80.1ºC) equals
sum of vapor pressures of components
Ptotal  P X ben  P X tol
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ben

tol
Deviations from ideal
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Real solutions can deviate from the ideal:
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Positive (Pvap > ideal) solute-solvent interactions
weaker
Negative (Pvap < ideal) solute-solvent interactions
stronger
Fractional distillation: separation of
liquids with different boiling points
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The vapour above a liquid is richer in the
more volatile component
Boiling the mixture will give a distillate
more concentrated in the volatile
component
The residue will be richer in the less
volatile component
Purification in stages
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A 50:50 mixture produces a vapour rich in hexane
That mixture condensed is about 90:10 hexane
The 90:10 mixture produces vapour about 95:5
The practice of fractional distillation
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In practice, it is not necessary
to do the distillation in
individual steps
The vapour rising up the
column condenses and reevaporates continuously,
progressively becoming
enriched in the volatile
component higher up the tube
If the column is high enough,
pure liquid will be collected in