Solutions & Colligative
Properties 
CHAPTER 16
Solutions
 Particles less than 1 nm in size.
 Homogeneous mixtures
 Particles do not settle and cannot be separated by
filtration.
 Do not exhibit Tyndall effect (don’t scatter light)
Components of a Solution
 Solutions consist of two parts:
Solute = substance being dissolved
 Solvent = dissolving medium

 The solutes and solvents in a solution may exist as
gases, liquids, or solids.
Solute
Solvent
Example
Gas
Gas
Gas
Liquid
CO2 in water
Liquid
Liquid
Alcohol in H2O
Solid
Liquid
Sugar in H2O
Solid
Solid
Brass, 14 k gold
Air
Solutions
The intermolecular forces
between solute and
solvent particles must be
strong enough to
compete with those
between solute particles
and those between
solvent particles.
Woah wait, what does this
mean?!?!?
How Does a Solution Form?
As a solution forms, the solvent pulls solute
particles apart and surrounds, or solvates,
them.
How Fast is a Solution Formed?
Influenced by 3 Factors:
1. Stirring (agitation)
2. Temperature
3. Surface Area of the Dissolving Particles
Solubility
 There is a limit to the amount of solute that can be
dissolved in a solvent.
 The point at which this limit is reached for any
solute-solvent combination depends on
the nature of the solute (what’s being
dissolved in what solvent)
 the amount of the solvent
 the temperature

Saturated Versus Unsaturated Solutions
 A solution that contains the maximum amount of
dissolved solute is described as a saturated
solution.
 If
more solute is added to a saturated
solution, it falls to the bottom of the container
and does not dissolve.
 A solution that contains less solute than a
saturated solution under the same
conditions is an unsaturated solution.
Supersaturated Solutions
PUT VIDEO HERE
 Usually when a saturated solution is cooled, the
excess solute usually comes out of solution, leaving
the solution saturated at the lower temperature.
 But sometimes the excess solute does not separate
and forms a supersaturated solution.
 solution
that contains more dissolved solute than a
saturated solution contains under the same
conditions.

application: rock candy
Supersaturated Solutions
STOP! Do POGIL
Solubility Values
 The solubility of a substance is the amount of that
substance required to form a saturated solution
with a specific amount of solvent at a specified
temperature.
 example: The solubility of sodium chloride is
36 g per 100 g of water at 20°C.
 Solubilities vary widely and are determined
experimentally.
 They can be found in chemical handbooks and
are usually given as grams of solute per 100 g
of solvent at a given temperature.
Temperature and Solubility
1.
Increasing the temperature usually increases
solubility of solids in liquids.

The effect of temperature on solubility for a given
solute is difficult to predict.

The solubilities of some solutes vary greatly over
different temperatures, and those for other solutes
hardly change at all.

A few solid solutes are actually less soluble at higher
temperatures.
2. Increasing the temperature usually
decreases the solubility of a gas in
liquid.
16.2 Concentration of a Solution
 The concentration of a solution is a measure of
the amount of solute in a given amount of solvent
or solution.
 Chemists commonly use several different ways to
quantitatively express the concentration of a
solution. These include:




Percent Composition by mass
Parts per million / Part per billion
Molarity (M)
Molality (m)
Percent Composition by Mass
 Sometimes the amount of solute present in
the solution is expressed as a percent by
mass.
grams of solute
 % Composition = ------------------------x
100
grams of solution
Remember: solution is the solute and solvent
together
Percent Composition Sample Problems
 What is the percent composition of a
solution that contains 115 g NaCl dissolved
in 500 g of H2O?
 How many grams of KMnO4 are needed to
make 500 g of a 12.0% KMnO4 solution?
Parts per Million (ppm) and
Parts per Billion (ppb)
Parts per Million (ppm)
ppm =
mass of solute in solution
 106
total mass of solution
Parts per Billion (ppb)
mass of solute in solution
 109
ppb =
total mass of solution
What Do You Think?
 A 900.0 g sample of sea water is found to
contain 0.0067 g Zn. Express this
concentration in ppm.
Molarity (M)
 Molarity (M) is the number of moles of solute
per liter of solution.
moles of solute (mol)
 Molarity (M) = -------------------------------liters of solution (L)
Molarity Sample Problems
 What is the molarity of a 0.500 L sample
of solution that contains 60.0 g of sodium
hydroxide (NaOH)?
 How many grams of sodium chloride
(NaCl) are required to prepare 250.0 mL
of a 3.00 M solution?
MAKING DILUTIONS!
 Pay attention future lab workers, this one is for you!
 Make dilutions from a more concentrated solution:
M1V1 = M2V2
(sometimes also seen as C1V1=C2V2)
Dilutions practice worksheet
16.3 Colligative Properties & 16.4 Calculations w/
Colligative Properties
Another Way to Measure
Concentration: Molality (m)
 Molality is the concentration of a solution
expressed in moles of solute per kilogram of
solvent.
moles of solute (mol)
 Molality (m) = ----------------------------------kilogram of solvent (kg)
16.4
Molality and Mole Fraction

To make a 0.500m solution of
NaCl, use a balance to measure
1.000 kg of water and add
0.500 mol (29.3 g) of NaCl.
16.6
16.6
16.6
16.6
Molality Sample Problem
 A solution was prepared by dissolving 17.1 g
of glucose, C6H12O6, in 275 g of water. What
is the molality (m) of this solution?
16.4
Molality and Mole Fraction

The mole fraction of a solute in a solution is the ratio of the
moles of that solute to the total number of moles of solvent and
solute.
16.4
Molality and Mole Fraction

In a solution containing nA mol of solute A and nB mol of solvent B
(XB), the mole fraction of solute A (XA) and the mole fraction of
solvent B (XB) can be expressed as follows.
16.7
16.7
16.7
16.7
Sample Problem 16.7
Colligative Properties
 These are the effects that a solute has on a solvent.
 When water has something dissolved in it, its
physical properties change.
 It will no longer boil at 100oC and it will no longer
freeze at 0oC like pure water.
Three Main Effects
 Lowers the vapor pressure of a solvent.
 Lower vapor pressure means that fewer water molecules can
escape from the liquid phase into the gas phase at given
temperature. Remember, a lower vapor pressure means a
higher boiling point!
 2) Raises the boiling point of a solvent.
 3) Lowers the freezing point of a solvent.
Freezing Point Depression
Boiling Point Elevation
Applications
 salting
icy roads
 making ice cream
 antifreeze
cars (-64°C to 136°C)
fish & insects
Colligative Properties
-THESE DEPEND ONLY ON
THE NUMBER OF
DISSOLVED PARTICLES
-NOT ON WHAT KIND OF
PARTICLE
GENERAL RULE: THE MORE SOLUTE PARTICLES
THAT ARE PRESENT IN A SOLVENT, THE GREATER
THE EFFECT.
# of Particles
 Nonelectrolytes
(covalent)
remain intact when dissolved
1 particle
 Electrolytes
(ionic)
dissociate into ions when dissolved
2 or more particles

Electrolytes have a stronger affect in lowering the
freezing point and elevating the boiling point because it puts
more particles into the solution.
 Remember the rule, the more particles, the
greater the effect!
 The Dissociation Factor (d.f.) for an electrolyte
is the number of ions a compound dissociates
into.
 NaCl gives Na+ ions and Cl- ions, which is 2
particles, therefore d.f. = 2
 What is “d.f. “ for Al(NO3)3 ?
Al(NO3)3  Al3+ + 3 NO3-
= 4 particles
Non-electrolytes
 A compound that does not conduct electricity
when dissolved in water.
 Examples are glucose or any other sugars and
alcohols, such as ethanol (CH3CH2OH)
 The d.f. = 1 for any non-electrolyte.
 Why?
 Covalent molecules do not break apart when
they become solvated by water molecules.
Calculations
t = m · d.f. · K
t: change in temperature (°C)
K: constant based on the solvent (°C·kg/mol)
Use Kb for boiling point elevation and Kf for freezing
point depression. Each solvent has it’s own unique
factors!
m:molality (m) = moles solute/Kg solvent
d.f.:# of particles
Example 1:
If 48 moles of ethylene glycol, C2H4(OH)2, is
dissolved in 5.0kg of water. What is the boiling
point of the solution?




∆Tbp = m  d.f.  Kb
d.f. = 1 ( it is a non-eletrolyte)
Kb = 0.52oC/m (a constant for water)
Molality = m = moles solute/kg solvent
∆Tbp = (48 moles) (1) (0.52oC/molal)
(5.0 Kg)
∆Tbp = 5.0 oC
New BP = 100 + 5 = 105 oC
Example 2:
55.5 grams of CaCl2 are dissolved in 2.0kg of water.
What is the freezing point of this solution?




∆Tfp = m  d.f.  Kf
d.f. = 3 (CaCl2 produces 3 particles)
CaCl2 Ca2+ + 2 ClKf = 1.86oC/m (a constant for water)
Must calculate molality
Molality = mass/molar mass = (55.5 gram/111g/mole)
= 0.25m
2.0 kg
Kg Solvent
∆Tfp = (0.25molal) (3) (1.86oC/molal)
New FP = -1.4 oC
= 1.4 oC