Chapter 19: Molality and
Colligative Properties
Chapter 14 —Big Book p. 487 & 14.1 (p.
498-504)
HW Ch. 19 Blue Book: #1-17, 19
Molality… a little review to start
http://www.youtube.com/watch?v=WNr
SexmBDXU&feature=related
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!
• 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.
• Antifreeze, Electrolytes…:
– http://www.youtube.com/watch?v=n0W7Y2G
wi2E&feature=related
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
• BP elevation & FP depression
– http://www.youtube.com/watch?v=tjHaIDSzH
so&feature=related
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.
• MEMORIZE THESE 2 CONSTANTS!!
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 p. 166 in Blue Book 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