### Mixtures and Solutions

Types of mixtures
 Heterogeneous mixture:

Heterogeneous mixtures do not blend smoothly. Individual
substances can still be seen
 Suspension:


A suspension is a heterogeneous mixture where particles will
settle to the bottom if left undisturbed.
Example: sand in water.
 Colloids:


Colloids are heterogeneous mixtures where particles will not
settle to the bottom.
Example: Milk
Homogeneous Mixtures
 Homogeneous mixtures are mixtures where a
substance is dissolved in another substance. These are
also called solutions.
 Parts of a solution:
 Solute: The substance that is dissolved
 Solvent: The substance that does the dissolving
 When a solute readily dissolves in a solvent it is said to
be soluble.
 Two liquids that can be combined to make a solution
are said to be miscible.
 A solute that does not readily dissolve in a solvent is
said to be insoluble.
 Two liquids that do no mix to form a new solution are
said to be immiscible.
Solution Concentration
 Concentration is the measure of how much solute is dissolved in
the solution.
 There are many ways to measure concentration
 Percent by mass:
 (Mass of solute/Mass of solution) x 100
 Percent by volume:
 (Volume of solute/Volume of solution) x 100
 Molarity:
 Moles of solute/Liter of solution
 Molality:
 Moles of solute/Kilograms of solvent
 Mole Fraction:
 Moles of solute/Moles of solute + Moles of solvent
Percent by mass
 In order to maintain a sodium chloride concentration
similar to ocean water, an aquarium must contain 3.6g
NaCl per 100.0 g of water. What is the percent by mass
of NaCl in the solution?
 3.5%
 What is the percent by mass of NaHCO3 in a solution
containing 20.0 g of NaHCO3 dissolved in 600.0 mL of
H2O?
 3%
Percent by Volume
 What is the percent by volume of ethanol in a solution
that contains 35mL of ethanol dissolved in 155mL of
water?
 18%
Molarity
 A 100.5 mL intravenous solution contains 5.10 g of
glucose. What is the molarity of the solution? The
molar mass of glucoes is 180.6 g/mol
 0.282M
 If you have 1500 g of a bleach solution with a percent
by mass of NaOCl of 3.62 % how many grams of
NaCOl are in the solution?
 54.3 g
 What is the percent by volume of isopropyl alcohol in
a solution that contains 24 mL of isopropyl alcohol in
1.1 L of water?
 2.1 %
 Calculate the molarity of 1.6 L of a solution containing
1.55 g of KBr.
 8.13 x 10-3
Molality
 In the lab a student adds 4.5 g of NaCl to 100.0 g of
water. Calculate the molaity of the solution.
 0.77 mol/kg
Preparing Solutions
 How would you prepare 1 L of a 1.5 molar CuSO4
Solution?
 Diluting Solutions:
 Dilution equation M1V1 = M2V2
 Calculate the new volume if you wanted to dilute the
above solution to 0.5 molar.
Diluting Solutions












Remember
Molarity = (Moles of solute/Liters of solvent)
The Dilution Equation:
M1V1 = M2V2
Example:
What volume of a 2.00M CaCl2 solution would you use to make
0.50 L of a 0.300M solution?
M1 = 2.00M
V1 = ?
M2 = 0.300M
V2 = 0.5L
(2.00M)(V1) = (0.300M)(0.5L)
V1 = 0.075 L
 What volume of a 3.00M KI solution would you use to
make 0.300 L of a 1.25M KI solution?
 125mL
 How many milliliters of a 5.0M H2SO4 stock solution
would you need to prepare 100.0mL of 0.25M H2SO4?
 5.0mL
Mole Fraction
 To express concentration as a mole fraction we need to
know the number of moles of solute and the number
of moles of solvent.
 The mole fraction is calculated by dividing the number
of moles of either the solute or solvent by the total
number of moles in the solution.
Example
 If we have a solution that contains 36 g of HCl and 64 g
of H2O. What are the mole fractions of HCl and H2O?
 HCl:
 H2O:
Practice
 What is the mole fraction of NaOH in an aqueous
solution that contains 22.8% NaOH by mass?
 If the mole fraction of H2SO4 is an aqueous solution is
0.325, what is the percent by mass of H2SO4
Factors Affecting Solvation
 Many physical factors affect the solubility of a solute.
 Such as:
 Temperature
 Pressure
 Polarity
 Solvation is the process of surrounding solute particles
with solvent particles.
Aqueous ionic solutions
 Many ionic compounds
are soluble in water.
Why?
 Water is a polar
molecule.
 Why might an ionic
compound not be water
soluble?
Heat of solution
 The overall energy change that occurs during the
formation of a solution is called the heat of solution.
 certain solvation processes are exothermic while others
are endothermic.
Factors that affect solvation
 Agitation:
 Stirring or shaking moves solvated particles away from
the surface of the solid and allows more solute particles
to move into solution.
 Surface Area:
 Breaking the solute into small pieces increases the
available surface area and increase the rate of solvation.
 Temperature:
 As temperature increase so does the rate of solvation.
Types of solutions
 Unsaturated:
 A solution that more solute could be dissolved in.
 Saturated:
 In a saturated solution the rate of solvation and the rate
of crystallization are in equilibrium.
 Supersaturated solution:
 A super saturated solution is a solution that has more
solute dissolved in it that it would normally allow.
Solubility of Gases
 Unlike solid solutes the solubility of gases (CO2, or O2
for example) increases as the temperature of the
solvent goes down.
 The solubility of gases is also affected by pressure.
How?
Henry’s Law
 Henry’s Law states that at a constant temperature, the
solubility (S) of a gas in a liquid is directly proportional
to the pressure of the gas above the liquid.
 Henry’s Law
 S1/P1 = S2/P2
 Example:
 If 0.85g of a gas at 4.0 atm of pressure dissolves in 1.0 L
of water at 25o C, how much will dissolve in 1.0 L at 1.0
atm at the same temperature?
 0.21 g
Decompression Sickness
 Decompression sickness is a term used to describe the
effects of a drop in external pressure on a persons
body.
 One common symptom of decompression sickness is
AGE, Arterial Gas Embolism.
 The bubbles that form in the blood stream from gas
leaving solution can restrict blood flow.
Boiling Point Elevation
 Dissolving a solute in a solution raises the boiling
point of the solution.
 This is called boiling point elevation.
 We can calculate the difference in boiling point (Δ Tb)
by multiplying a solutions molality by a constant we
look up for our particular solvent (Kb).
 Δ Tb = Kbm
 What is the boiling point of a 0.625m aqueous solution
of a nonelectrolyte?
 What is the boiling point of a 0.4m aqueous solution
of Ca(OH)2 (A strong electrolyte) Kb = 0.512 oC/m
Molal Boiling Point Elevation
Constants.
Solvent
Boiling Point (oC)
Kb (oC/m)
Water
100.0
0.512
Benzene
80.1
2.53
Chloroform
61.7
3.63
Ethanol
78.5
1.22
Freezing Point Depression
 Adding a solute to a solvent also lowers the solvents
freezing point.
 The equation is very similar to boiling point elevation.
 ΔTf = Kf m
Solvent
Water
Freezing Point
(oC)
Kf (oC/m)
0.00
1.86
Benzene
5.5
5.12
Ethanol
-114.1
1.99
Example
 Sodium chloride is often used to prevent icy roads and
to freeze ice cream. What are the boiling point and
freezing point of a 0.029m AQUEOUS solution of
sodium chloride?
 BP: 100.03 0C
 FP: -0.11 oC
 Calculate the freezing and boiling point of a solution containing
6.42 g of sucrose (a nonelectrolyte) in 100.0 g of water.
 Tf = -0.350 oC
 Tb = 100.096 oC
 Calculate the freezing and boiling point of a solution containing
23.7 g of Copper (II) sulfate (A strong electrolyte) in 250 g of
water
 Tf = -2.02 oC
 Tb = 100.606 oC
 Calculate the freezing and boiling point of a solution containing
0.15 moles of the compound naphthalene, a nonelectrolyte in 175
g of benzene (Normal F.P. = 5.5, Normal B.P. = 80.1, Kf = 5.12, Kb
= 2.53)
 Tf = 1.1 oC
 Tb = 82.3 oC