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
• The volume of a solution changes with a
change in temperature which alters the
molarity. (ex: what happens when you
boil 3 cups of water… do you still have 3
cups?)
• Masses, however, do not change with
temperature.
• So… we use molality (m)- the number of
moles of solute per kilogram of solvent
(mol/Kg, another concentration ratio).
Example #1
If 60.0 g of NaOH are dissolved in 1500g of water, what is
the concentration of this solution using Molality?
1. Analyze the problem – what info are we given?
mass of solute, 60.0 g NaOH
mass of solvent, 1500 g H2O
2. Solve for the unknown
Couple of ways to do this:
1. “Railroad-track” the whole thing all at once
OR
2. Break it up…
calculate number of moles of solute
Convert mass of water from g  kg
Plug in and solve for molality
=1.00m
Example #2
Calculate the mass of ethanol, C2H5OH, which must
be dissolved in 750.0 g water to make a 2.00 m
solution.
Multiply the mass of water by the concentration
(2.00 m), then convert to grams of ethanol by using
molecular mass of ethanol
750g H2O 2.00 mol C2H5OH 46.1g C2H5OH
1000 g H2O
1 mol C2H5OH
Answer: 69.2g C2H5OH
Example #3
Determine the mass of H2SO4 which
must be dissolved in 2500 g of H2O to
make a 4.00m solution.
Multiply the mass of H2O by the
concentration ratio. Then convert
from moles  grams of H2SO4 using
the formula mass of the acid.
2500g H2O 4.00 mol H2SO4
1000 g H2O
98.1 g H2SO4
1mol H2SO4
= 981 g H2SO4
Example # 4
What is the percentage by mass of Mg(NO3)2 in a
2.50 m solution?
The concentration involves 1000g H2O. Find the g
Mg(NO3)2 & add these two masses together for
total mass, then calculate percentage Mg(NO3)2 .
2.50 mol Mg(NO3)2 148.0 Mg(NO3)2
= 370 g Mg(NO3)2
1000 g H2O
1 mol Mg(NO3)2
1000 g H2O
Mass of soln = 370.0 g + 1000 g = 1370 g soln
% Mg(NO3)2 = 370.0
X 100 = 27.0 %
1370 g
Example # 5
How many molecules of ethanol must be dissolved
in 500.0 g of water to make a 1.00 m solution?
Multiply the mass of water by the concentration.
Then multiply by Avogadro’s # /mol.
500.0 g H2O 1.00 mol C2H5OH 6.02 x 1023 molecules
1000 g H2O
1 mol
Answer: 3.01 x 1023 molecules C2H5OH
Molality
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!
• 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: 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,
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 rTb
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.
Mark Rosengarten videos
• Antifreeze, Electrolytes…:
– http://www.youtube.com/watch?v=n0W7Y2G
wi2E&feature=related
• BP elevation & FP depression
– http://www.youtube.com/watch?v=tjHaIDSzH
so&feature=related
• Molality
– http://www.youtube.com/watch?v=WNrSexm
BDXU&feature=related