Ch 12.1
Types of Mixtures
Chapter 12.1


Standard 6.a.: Students know the definitions
of solute and solvent.
Objectives:
 We will distinguish between heterogeneous
and homogeneous mixtures
 We will compare the properties of
suspensions, colloids, and solutions.
Heterogeneous vs. Homogeneous
Mixtures


Heterogeneous Mixture: mixture does not
have a uniform composition.
 Ex: Milk and soil
Homogeneous Mixture: entire mixture has
the same or uniform composition.
 Ex: Salt water
Solutions




Soluble: capable of being dissolved.
 Ex. Sugar is soluble in water.
Sugar and water create a solution, or a
homogeneous mixture of two or more
substances in a single phase.
Solvent: the thing that does the dissolving.
Solute: the thing that is being dissolved.

Solutions may exist as gases, liquids, or
solids, and may also be combinations.
Solute State Solvent State
Example
Gas
Gas
Oxygen in Nitrogen
Gas
Liquid
CO2 in Water
Liquid
Liquid
Alcohol in Water
Liquid
Solid
Mercury in Silver & Tin
Solid
Liquid
Sugar in Water
Solid
Solid
Copper in Nickel (alloy)
Suspensions


Suspension: When the particles in a
solvent are so large that they settle out
unless the mixture is constantly agitated.
 Ex: Muddy water
The particles in a suspension can be
separated by passing the mixture through
a filter.
Colloids
Particles that are intermediate in size
between those in solutions and suspensions
form mixtures called colloids.
 These are also known as emulsions and
foams and cannot be separated using a
filter.
 Ex. Mayonnaise and Milk
 Tyndall Effect: when light is scattered by the
particles in a colloid.

Solutes: Electrolytes vs.
Nonelectrolytes


Electrolyte: a substance that dissolves in
water to give a solution that conducts an
electric current.
Nonelectrolyte: a substance that dissolves
in water to give a solution that doesn’t
conduct an electric current.
Chapter 12.1


Standard 6.a.: Students know the definitions
of solute and solvent.
Objectives:
 We will distinguish between heterogeneous
and homogeneous mixtures
 We will compare the properties of
suspensions, colloids, and solutions.
Homework
Ch 12.1
pg 406 #1, 2, 6
and pg 426 #3-5
Ch 12.2
The Solution Process
Chapter 12.2


Standard 6.c.: Students know temperature,
pressure, and surface area affect the
dissolving process.
Objectives:
 We will list the 3 factors that affect that
rate of dissolution.
 We will compare the effects of temperature
and pressure on solubility.
Factors Affecting Dissolution Rate


The compositions of the solvent and the
solute determine whether a substance will
dissolve.
Three factors that affect dissolving rate:



Stirring (agitation)
Temperature
Surface area of the dissolving particles.
Solubility


Solution Equilibrium: the physical state in
which the opposing processes of dissolution
and crystallization of a solute occur at equal
rates.
Solubility tells us how much solute can
dissolve in a certain amount of solvent at a
particular temperature and pressure to make
a saturated solution.

Expressed in grams of solute per 100 grams of
solvent



Saturated Solution: the solution cannot hold
any more solute.
Unsaturated Solution: the solution could still
dissolve more solute.
Supersaturated Solution: the solution is
holding more than it should at the given
temperature, and if you messed with the
solution by shaking it or adding even one
more crystal of solute, the whole thing would
crystallize rapidly.


Solubility Values: amount of substance
required to form a saturated solution with
a specific amount of solvent at a specified
temperature.
Solubility of sugar is 204 grams per 100
grams of water at 20°C.
Solute-Solvent Interactions



“Like dissolves Like”
Polar will dissolve other polar molecules
and Nonpolar dissolves other nonpolar.
Hydration: when water is used to dissolve
an ionic solution.
Liquid Solutes and Solvents


Miscible: two liquids that can dissolve in
each other.
Immiscible: the liquids don’t mix.

Ex. Oil and vinegar
Factors Affecting Solubility

Temperature affects the solubility of:




Solid Solutes
Liquid Solutes
Gaseous Solutes
Pressure affects the solubility of:

Gaseous Solutes
Temperature

Gas dissolved in a Liquid: as the temperature
increases, the solubility decreases.


Example: Warm soda loses its carbonation.
Solid dissolved in a Liquid: as the
temperature increases, the solubility
increases.

Example: Sugar in hot tea versus iced tea.
Pressure

Gas dissolved in Liquid: As pressure
increases, solubility increases.


Example: Soda is carbonated under high pressure.
Solid dissolved in Liquid: As pressure
increases, solubility does not change!

Since you cannot compress solids and liquids,
pressure has no effect on solubility.
Henry’s Law

Henry’s Law states that at a given
temperature, the solubility (S) of a gas in a
liquid is directly proportional to the pressure
(P) of the gas above the liquid.

So, as the pressure of the gas above the liquid
increases, the solubility of the gas increases.
S1
P1
S2
P2
Calculating Solubility of a Gas

If the solubility of a gas in water is 0.77 g/L at 3.5
atm of pressure, what is its solubility (g/L) at 1.0
atm of pressure and a constant temperature?




P1 = 3.5 atm
S1 = 0.77 g/L
P2 = 1.0 atm
S2 = ? g/L
0.77 g/L = S2
3.5 atm 1.0 atm
S2 = 0.22 g/L
Enthalpies of Solution



Solvated: when a solute particle is
surrounded by solvent molecules.
The formation of a solution is accompanied
by an energy change, it can be released or
absorbed.
Enthalpy of solution: the net amount of
energy absorbed as heat by the solution
when a specific amount of solute dissolves in
a solvent.
Chapter 12.2


Standard 6.c.: Students know temperature,
pressure, and surface area affect the
dissolving process.
Objectives:
 We will list the 3 factors that affect that
rate of dissolution.
 We will compare the effects of temperature
and pressure on solubility.
Homework

Ch 12.2 pg 426 #7-12
Ch 12.3
Concentrations of
Solutions
Chapter 12.3


Standard 6.d.: Students know how to calculate
the concentration of a solute in terms of grams
per liter, molarity, ppm, and percent
composition.
Objective: We will calculate the concentration
of a solute, the amount of solute in a given
amount of solution, and the amount of solution
that contains a given amount of solute.
Concentrations of Solutions


Concentration of a solution: a measure of
the amount of solute that is dissolved in a
given quantity of solvent.
Solutions can be referred to as dilute or
concentrated, but these are not very
definite terms.
Molarity

Molarity (M): the number of moles of solute
dissolved in one liter of solution.
moles of solute
Molarity (M) 
liters of solution
Note: it is the total volume in liters of solution, not
the liters of solvent.
Calculating Molarity of a Solution




IV Saline Solutions are 0.90 g NaCl in exactly
100 mL of solution. What is the molarity of
the solution?
Step 1: convert mL to L (divide by 1000)
Step 2: convert the grams of NaCl to moles
of NaCl using molar mass.
Step 3: put moles of NaCl and L of solution
into the molarity equation and divide.
Finding Moles of Solute



Household bleach is a solution of sodium
hypochlorite (NaClO). How many moles of
solute are present in 1.5L of 0.70M NaClO?
Moles Solute = M x L = mol/L x L
Multiply the given volume in L by the molarity
expressed in mol/L.
Molality

Another way to express solution concentration
is Molality (m)


NOT THE SAME AS MOLARITY!
Molality (m) is the concentration of a solution
expressed in moles of solute per kilogram
solvent.
moles of solute
m (molality ) 
1 kg solvent
Calculating Molality of a Solution
•
Calculate the molality of a solution
prepared by dissolving 10.0g of NaCl
in 600.g of water.
10.0g NaCl  0.171 mol NaCl
600.0 g  0.600 kg
m = mol of solute = 0.171 mol of NaCl = 0.285 m NaCl
kg of solvent
0.600 kg of water
Finding Moles of Solute using
molality.

How many moles of sodium fluoride are
needed to prepare a 0.40m NaF solution that
contains 750.0g of water?
m = mol of solute
kg of solvent
mol solute = m x kg of solvent
mol NaF= 0.40 mol x 0.75 kg = 0.30 mol
kg
Making Dilutions

Diluting a solution reduces the number of
moles of solute per unit volume, but the total
number of moles of solute in solution does
not change.
M1 x V1 = M2 x V2
Preparing a Dilute Solution



How many mL of 2.00M MgSO4 solution must
be diluted with water to prepare 100.0mL of
0.400M MgSO4?
Use the dilution formula and plug in the
known values and then solve for the
unknown.
Volume can be in any unit, as long as they are
both the same. (Just like gas laws).
0.400 M MgSO4 x 100.0 mL = 2.00 M MgSO4 x V2
V2 = 20.0 mL
Chapter 12.3


Standard 6.d.: Students know how to calculate
the concentration of a solute in terms of grams
per liter, molarity, ppm, and percent
composition.
Objective: We will calculate the concentration
of a solute, the amount of solute in a given
amount of solution, and the amount of solution
that contains a given amount of solute.
Homework

Ch 12.3 pg 427 #21-23 and 27-29