Colligative Properties
• What are they?
– The word Colligative means “depending
on the collection”
– Change the physical properties of the
solvent.
– Depends on the number of particles of the
solute NOT which solute is used!
Colligative Properties
• Lowers the vapor pressure!
• Raises the boiling point!
• Lowers or depresses the freezing
point!
• Osmotic pressure
• Why?
Colligative Properties
• When a solute is dissolved in a solvent,
the vapor pressure of the solvent is
reduced.
• The reduction depends on the number
of solute particles in a given amount of
solvent.
• The French chemist, Raoult, first
discovered the vapor pressure
lowering relationship experimentally in
1882 which lead to…
Colligative Properties
• Raoult’s Law:
Law Any nonvolatile
solute at a specific concentration
lowers the vapor pressure of the
solvent by an amount that is
characteristic of that solvent.
Vapor Pressure Lowering
• The vapor pressure above a liquid is lowered
due to the attractive forces of the solvent on
the dissolved solute particles.
• Because of this, less solvent particles have
the energy to transition to the gaseous state
(evaporate), and therefore the vapor
pressure is lower.
• So… The greater the number of solute
particles in a solvent, the lower the VP
Pure Solvent
Solution
Beaker #1
Beaker #2
Which one has lower VP?
• #1 – solvent
has a large
surface area
to evaporate
from
• #2 – mixed
with solute =
fewer
solvent
particles at
surface
Boiling Point Elevation
• Similar factors (as with the vapor
pressure lowering), contribute to the
increase of the boiling point of a
solvent .
– The more solute particles the higher the
BP (the lower the VP)
• Practical application – adding salt to
water to increase the BP of water to
cook foods.
Boiling Point Elevation
Freezing Point Depression
• Freezing occurs when the particles no
longer have the energy to overcome their
interparticle attractive forces – they
organize and solidify (molecules slow way
down, loss of kinetic energy).
• Adding solute to a pure solvent lowers
the FP!
– WHY?
• Because the solute interferes with the
solvents interparticle attractions, therefore the
solid forms at cooler or lower temperature.
• So… the FP of a solution is always lower
than the FP of a pure solvent.
Freezing Point Depression
___ =
Pure Solvent
---- =
Solution
100oC
0oC
Osmotic Pressure
• What is osmosis?
• The amount of additional pressure
caused by the water molecules that
move into a concentrated solution is
called osmotic pressure. (The diffusion
of water)
• This pressure depends on the number
of solute particles in a given volume of
solution.
As water is moving
the pressure exerted by the additional water
molecules, osmotic pressure, is increasing on the left side of the
semipermeable membrane. Higher osmotic pressure on left,
lower osmotic pressure on right.
Colligative Properties (now the math)
• The change in the freezing and boiling
pts varies directly with the
concentration of particles.
• Molal freezing pt constant: 1.86C˚ for
water. Each mole of solute causes
the freezing pt of water to drop by
this much.
• Molal boiling pt constant: 0.512C˚ for
water. Each mole of solute causes
the boiling point to rise by this much.
Colligative Properties
These can be used to determine:
• The freezing point of the water
• The boiling point of the water
• The molecular mass of the solute from the
freezing point or the boiling point
• (see table 19-1 for other constants)
Colligative Properties
Ex. 6
Calculate the freezing point of a solution
containing 5.70 g of sugar, C12H22O11, in
50.0 g of water.(Molal freezing pt constant: 1.86C˚ for water. )
Convert grams of solute per gram of water to moles of solute per kg of
water (molality). Then multiply by the conversion ratio to obtain the
change in FP
5.70 g C12H22O11 103 g H2O
50.0 g H2O
1kg H2O
1 mol C12H22O11 1.86C˚
342 g C12H22O11 1 m
= 0.620C˚,
To determine the FP, subtract this from the FP of water
0 oC – 0.620 = - 0.620 oC
Calculating Molecular Mass Ex. 7
When 72.0 g of dextrose were dissolved in 100.0
g of water, the boiling point of the solution was
observed to be 102.05˚ C. What is the
molecular mass of dextrose?
Step 1: determine the molality of the
solution
100 oC - 102.05˚C = 2.05˚C determine the Tb
2.05 oC m
= 4.00 m
0.512 ˚C
molal boiling pt. constant for H2O
Step 2: determine the grams per mole
72.0 g dextrose
0.100 kg H2O
1 kg H2O = 180 g
4.00 mol
mol
One last thing: Colloids
• Colloids are not true solutions, but
special types of mixtures that behave
like solutions.
– There are two parts, the dispersed phase
and continuous phase.
• Dispersed phase has particles from 1 to 100
nm in size and remain dispersed by the
random motion of the molecules (kinetic
energy).
• Any particle larger than 100 nm will usually
settle out over time.