2. Mixtures – a review
• Mixture: a combination of two or more
substances that do not combine
chemically, but remain the same
individual substances; can be separated
by physical means.
• Two types:
– Heterogeneous
– Homogeneous
3. Heterogeneous Mixture
• “Hetero” means “different”
• Consists of visibly different substances or
phases (solid, liquid, gas)
• Can be separated by filtering
• Example:
4. Homogeneous Mixture
• “Homo” means the same
• has the same uniform appearance and
composition throughout; maintain one phase
(solid, liquid, gas)
• Commonly referred to as solutions
• Example:
Salt Water
5. IDENTIFY WHAT TYPE OF MIXTURE
Heterogeneous or Homogeneous
1. Pizza
2. Sandwich
3. Tea
4. Salad
5. Fruit juice
6. Oil and water
7. Wine
8. Air
9. Paella
10. Salt and pepper
11. Chocolate
12. ice cubes in a drink
13. Coffee
14. Vinegar
15. Concrete
6.
7. Separating mixtures
• There are many ways to separate
mixtures into their components.
distillation
Magnetic
separation
filtering
chromatography
8. Solution
• Solution: a mixture of two or more substances
that is identical throughout (homogeneous)
• can be physically separated
• composed of solutes and solvents
the substance being dissolved
the substance that dissolves the solute
Iced Tea Mix
(solute)
Water
(solvent)
Iced Tea
(solution)
Salt water is
considered a
solution. How can it
be physically
separated?
9. Solution
• The solvent is the largest part of the solution
and the solute is the smallest part of the
solution
S O L V E N T
S O L U T E
10. Types of Solutions
Gaseous solutions – air = Oxygen + Nitrogen
Liquid solutions – drinks = mix + water
Solid solutions – alloys = steel, brass, etc
11. Concentration
• the amount of solute dissolved in a
solvent at a given temperature
•described as dilute if it has
a low concentration of solute
dissolved
•described as concentrated
if it has a high concentration
of solute dissolved
12.
13. Concentration
•Unsaturated - has a less than the
maximum concentration of solute
dissolved
•Saturated - has the maximum
concentration of solute dissolved
(can see solid in bottom of
solution)
•Supersaturated -contains more
dissolved solute than normally
possible (usually requires an
increase in temperature followed
by cooling)
14.
15. Solubility
• the amount of solute that
dissolves in a certain amount of a
solvent at a given temperature
and pressure to produce a
saturated solution
16. Conductivity in Solutions
• Conductivity is a measure of water’s
ability to conduct electrical current.
Measurements of conductivity provide a
general indication of water quality.
17. Electrolytes
What are electrolytes?
• Electrolytes are substances that become ions in
solution and acquire the capacity to conduct
electricity.
• Non electrolyte is a substance that does not
dissociate into ions and so in solution it is a
nonconductor of electricity.
18. • Tap water conducts electricity:
• You should avoid water when you are using
electricity. Unlike distilled water that does
not conduct enough electricity to light a bulb
in the conductivity apparatus tap water
contains various ions from dissolved minerals.
Water sources like water from a well has
more concentration of salts than the surface
water such as lakes and rivers.
19. • While swimming in the pool, get out of
the pool if a thunderstorm strikes.
Chlorinated water conducts electricity.
20. Factors affecting solubility of solids
Temperaturee
increased temperature causes
solids to dissolve faster
Shaking
Note: Increasing the amount of solute
DOES NOT increase the rate of dissolving
Shaking (agitation) causes
solids to dissolve faster
Smaller particles dissolve
Faster because they have
more surface area
Particle Sizee
21. Polarity and Dissolving
• Chemists use the saying
“like dissolves like”:
➢Polar solutes tend to
dissolve in polar
solvents.
➢Nonpolar solutes tend to
dissolve in nonpolar
solvents.
Oil is nonpolar while water is
polar. They are immiscible.
22. • Miscible liquids can easily dissolve in one
another.
• Immiscible liquids are not soluble in
each other.
Chemistry-Borders
23. • The amount of solute in a
solution can be expressed in
several ways. These includes
percentage by mass, volume, or
mass- volume, mole fraction,
molality, molarity, and parts per
million, among others
24. PERCENT BY MASS, BY VOLUME, AND BY
MASS- VOLUME
• Percent by mass (or percent by weight,
%w/w) expresses the mass of solute per 100 g
solution. Mass of solution is equal to the mass
of solute plus the mass of solvent. The formula
for percent by mass is:
Chemistry-Borders IPC-Solutions-Borders
25. Sample Problem 1:
If 28.5 g of calcium hydroxide Ca (OH)2 is
dissolved in enough water to make 185 g of
solution, what is the percent by mass of Ca
(OH)2 in the solution?
Solution:
26. Sample Problem 2:
If mass of solution = 25.0g
sugar + 100.0g water = 125.0
g, What is its percent by
mass?
27. Sample Problem 3:
A saline solution with a mass of 355 g has
36.5 g of NaCl dissolved in it. What is the
mass/mass percent concentration of the
solution?
We can substitute the quantities given in the
equation for mass/mass percent:
28. If the solution involves a solute and a
solvent that are both liquids, the
percent by volume (%v/v) is used
instead of by mass with the following
formula:
31. Sample Problem #2
What is the % volume of NaOH if 20 ml
of NaOH are added to 300 ml of water?
Percent volume = volume of part (solute) x 100
volume of whole (solution)
% volume = 20 ml x 100
(20 ml + 300 ml)
% volume = (20 / 320) x 100
= 6.25% solution
32. Sample Problem 3:
A wine contains 12% alcohol by volume.
Calculate the volume (in mL) of alcohol in 350
mL of the wine.
Solution: % by volume =
volume solute in mL
volume solution in mL
× 100
volume solute in mL =
% by volume 𝑥 volume solution in mL
100
33. Percent by mass- volume (or percent by weight- volume, %
w/v) of a solution is given by:
34. • A mole is the SI unit of number of particles and
can be used as an expression of the molecular
weight of a substance.
Units of Concentration
The formula weight of an
element is expressed as
grams/mole
35. • The molar mass of a compound can be
calculated by adding the molar mass of the
individual elements.
Units of Concentration
22.99 + 35.45 = 58.44 g/mol
36. MOLE FRACTION
Mole fraction is the ratio of the number of
moles of one component to the total number of
moles in a solution. It is represented by a capital
letter X.
37. Sample problem 1:
Calculate the mole fraction of sulfuric
acid (H2SO4) in 8% (% w/w) aqueous H2SO4
solution. (molar masses: H2SO4 98 g/ mol;
H2O = 18 g/ mol)
Chemistry-Borders IPC-Solutions-Borders
38. MOLE FRACTION
1. Calculate the mole fraction of each
component of a solution containing 65 g of
ethanol 𝐶2𝐻6O in 350 g of water.
2. A solution is made by dissolving 1.25g
𝑁𝑎2S𝑂4 in 65.0g water. Calculate the mole of
fraction of the solute and the solvent.
40. mol of solute
kg of solvent
m =
Molality (m)
Because both moles and mass do not
change with temperature, molality
(unlike molarity) is not temperature
dependent.
41. •Molality is define as the
number of moles of solute per
kilogram of solvent. It can be
mathematically expressed as
molality =
moles of solute
mass of solvent (in kg)
42. mol of solute
L of solution
M =
Molarity (M)
• Because volume is temperature
dependent, molarity can change
with temperature.
43. Molarity
• Molarity is the concentration of a
solution expressed in moles of solute
per Liter of solution.
• Molarity is a conversion factor for
calculations
Molarity (M) = moles of solute
Liters of solution
44. Example#1 What is the molarity of a
solution of NaOH if there are 4 moles of
NaOH dissolved water to make 1 liter
of solution?
Chemistry-Borders IPC-Solutions-Borders
4 moles = 4 M
1 liter
45. Molarity
M = mol (solute)
L (solution)
• Example 2: What is the molarity of a solution
that has 2.3 moles of sodium chloride in 0.45
liters of solution?
2.3 moles NaCl = 5.1M NaCl
0.45 L
46. Molarity Problems
What is the molarity of a solution of NaOH if
there are 4 moles of NaOH dissolved
water to make 2 liters of solution?
4 moles = 2 M
2 liter
47. Molarity Problems
What is the molarity of a solution of NaOH if
there are 2 moles of NaOH dissolved
water to make 6 liters of solution?
2 moles = .33 M
6 liters
48. Molarity Problems
What is the molarity of a solution of NaOH if
there are 2 moles of NaOH dissolved
water to make .5 liters of solution?
2 moles = 4 M
.5 liters
49. Molarity Problems
What is the molarity of a solution of NaOH if
there are 40 grams of NaOH dissolved
water to make 1 liters of solution?
First, convert grams to moles:
Na = 23 g/mol
O = 16 g/mol so 40 g = 1 mole
H = 1 g/mol
40 g/mole GFM
50. Molarity
M = mol (solute)
L (solution)
• Example 2: How many moles of KNO3 are needed
to make 450. mL of 1.5 molar solution?
450. mL 1L 1.5 mol KNO3
1 1000mL 1L
= .675 moles KNO3
51. Molarity
M = mol (solute)
L (solution)
• Example 3: How many grams of NaCl are needed
to make 3.0 L of 1.5 M solution?
3.0 L 1.5 mol NaCl 58.44 g NaCl
1 1 L 1 mol NaCl
= 260 g NaCl
52. Molarity
M = mol (solute)
L (solution)
• Example 4: How many L of 4.0 M solution can be
made with 132g of NaCl ?
132 g NaCl 1 mol NaCl 1 L
1 58.44 g NaCl 4.0 mol NaCl
= .565 L
54. Solution Stoichiometry
• When we previously did stoichiometry
for a reaction to determine theoretical
yield, we only worked with GRAMS and
MOLES
• Ex/ How many MOLES of HCl are
required to react with 13 GRAMS of
zinc?
Zn + 2 HCl → ZnCl2 + H2
55. Solution Stoichiometry
• But we may be given something OTHER than
grams and moles
• We can use stoichiometry to solve for ANY
unit. We just need to make sure units cancel
out and we end up with the unit we are trying
to solve for!
• The mole ratio using coefficients from the
balanced chemical equation is the key to
switching between compounds
56. Solution Stoichiometry
Ex/ How many LITERS of 12 M HCl are
required to react with 13.0 GRAMS of zinc?
Zn + 2 HCl → ZnCl2 + H2
13.0g Zn 1 mole Zn 2 mol HCl 1L HCl
1 65.38g Zn 1 mol Zn 12 mol HCl
Remember – Molarity (M) is a conversion Factor
= 0.0331 L HCl
57. Solution Stoichiometry
• Ex/ How many grams of NaOH would be
required to react with 1.50 L of 3.75M
sulfuric acid?
H2SO4 + NaOH → Na2SO4 + H2O
1.50L 1 H2SO4 3.75 mole H2SO4 2 mol NaOH 40.00g NaOH
1 1 L H2SO4 1 mole H2SO4 1 mol NaOH
= 450. g NaOH
