1.Types of Solution 2. Energy of Solution 3, Concentration of Solution

1. GENERAL CHEMISTRY II
WEEK 3-4

2. 1. Use different ways of expressing
concentration of solutions: percent by mass,
mole fraction, molarity, molality, percent by
volume, percent by mass, ppm.
2. Perform stoichiometric calculations for
reactions in solution.
3. Describe the effect of concentration on the
colligative properties of solutions .

3. 4. Differentiate colligative properties of
nonelectrolyte solutions and of electrolyte
solution.
5. Calculate the boiling point elevation and
freezing point depression from the
concentration of a solute in a solution

4. 6. Calculate molar mass from colligative
property data.
7. Describe laboratory procedures in
determining concentration of solution.

5. General Chemistry 2 – Senior High School (STEM)
EQ: What types of solutions are encountered
everyday?

6. Identify the solute and solvent in each of the
following solutions, explain your answers.
1. 25 grams of salt dissolved in 95 mL of water.

7. 2. 25 mL of water mixed with 75 mL of isopropyl
alcohol.

8. 2. Tincture of iodine prepared with 0.20 gram of
iodine and 20.0 ml of ethanol

9.  Solutions are mixtures of two or more
substances distributed throughout a single
phase.
 Atoms, ions, or molecules are thoroughly
mixed in a solution such that each part of the
mixture has uniform composition and
properties.

10.  A solution consists of a solute and a solvent.
 The Solute is the substance dissolved in a
solution and is usually present in smaller
amount; solvent is the dissolving medium.

11. SOLUTE SOLVENT SOLUTION EXAMPLE
GAS GAS GAS Oxygen in
Nitrogen
GAS LIQUID LIQUID Carbon Dioxide
in water
GAS SOLID SOLID Hydrogen in
Palladium
LIQUID LIQUID LIQUID Ethanol in
water
LIQUID SOLID SOLID Mercury in
silver
SOLID LIQUID LIQUID Salt in water
SOLID SOLID SOLID Copper in tin
(bronze)

12.  A solution can be classified as unsaturated,
saturated, or supersaturated depending on
the quantity of solute present in it.

13.  Unsaturated Solution contains less solute
than solvent’s capacity to dissolve. This
means that the solvent can still dissolve more
solute.
 Saturated Solution contains the maximum
amount of solute that the solvent can
dissolve at a certain temperature.

14.  Supersaturated solution contains more
dissolved solute than is present in a saturated
solution. This type of solution is unstable such
that the excess solute can crystallize in the
solution by adding a “seed crystal” (a process
called seeding) or by scratching the sides of
the container.

15. General Chemistry 2
EQ: How do liquid solution forms?

16.  For two substances to form a solution, they
must have the same nature in terms of
Polarity.
 The formation of liquid solution takes place in
three steps:
1. Overcoming the intermolecular forces in the
solvent to give room for the solute.

17. 2. Separating the solute into its individual
components.
3. Allowing the solute and solvent to interact
to form the solution.

18.  It is the enthalpy change associated with the
formation of the solution.
 It is equal to the sum of the H values for
the three steps.
A positive enthalpy of solution signifies an
endothermic process, while negative indicates
an exothermic process.

19.  The process is exothermic, if more energy is
released when new bonds form than is used
when bonds are broken.
 The process is endothermic, if more energy is
required (used) when bonds are broken than
is released when new bonds are formed.

20. General Chemistry 2
EQ: Why should solutions be expressed in a
correct concentration units?

21.  It is a measure of the amount of solute in a
given amount of solvent or solution.
 One type of solution may be prepared and
expressed in different concentrations.
Example:
 A cup of hot water with one teaspoon of
coffee is an altogether different solution than
a cup of water with a few teaspoon of coffee.

22. Example:
 Medications are also solutions of substances,
A 5 mL dose of a certain concentration of
medicine might cure a sick person, but a 5 mL
dose of a higher concentration might already
harmful to the person.

23. 1. Percent by Mass
2. Percent byVolume
3. Percent by Mass-Volume
4. Mole Fraction
5. Molality
6. Molarity
7. Parts per Million

25. 1. A saline solution with a mass of 355 g has
36.5 g of NaCl dissolved in it. What is the
mass percent concentration of the
solution?

26. Given:
Mass of Solute = 36.5 grams
Mass of Solution = 355 grams
Formula: Percent by mass =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
Percent by mass =
36.5 𝑔𝑟𝑎𝑚𝑠
355 𝑔𝑟𝑎𝑚𝑠
x 100
Percent by mass = 10.28%

27. 2. A wine contains 12% alcohol by volume.
Calculate the volume (in mL) of alcohol in
350 mL of the wine.

28. Given:
Percent byVolume = 12%
Volume of Solution = 350 mL
Volume of solute = x
Formula: % by volume =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
12 % =
𝑥
35𝑂 𝑚𝐿
x 100
12
100
(350 mL) = x
42 mL = x

29. Percent by Mass –Volume = mass of solute in grams
volume of solution in mL
1. What is the concentration in percent by
mass/volume of 150 mL of solution
containing 30 g of solute?
X 100

30. Given:
volume of solution (mL) : 150 ml
mass of solute (g) : 30 grams
Find:
% by mass-volume
Formula:
% by mass-volume=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝐿
x 100

31. % by mass-volume =
30 𝑔𝑟𝑎𝑚𝑠
150 𝑚𝐿
x 100
= (0.20) x 100
= 20 %

32. A 50 mL of 12% by mass-volume solution was
used in an experiment. How many grams of
solute does the solution contain?

33. A 50 mL of 12% by mass-volume solution was
used in an experiment. How many grams of
solute does the solution contain?
Given:
volume of solution in mL = 50 mL
% by mass-volume = 12%
Find:
Mass of solute in grams = ?

34. Formula:
% by mass-volume=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒 𝑖𝑛 𝑔
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝐿
x 100
12 % =
𝑥
50 𝑚𝑙
x 100
12
100
(50ml) = x
6 g = x

35.  The mole fraction (x) of a component in a
solution is equal to the number of moles of
the component divided by the total number
of moles of all the components present.
 The sum of all of the mole fractions of all
components in all solution will always be
equal to one.

37. 1. 25g of NaF is mixed with 200 g of H2O.
What is the mole fraction of NaF in the
solution?
Step 1:
Given:
solute: ______________
solvent: ______________
Find: _______________

38. 1. 25g of NaF is mixed with 200 g of H2O.
What is the mole fraction of NaF in the
solution?
Step 1:
Given:
solute: 25 g of NaF
solvent: 200 g of H2O
Find: mole fraction of NaF

40. Formula:
Step 2: Convert the given mass of solute and
solvent into moles using molar mass.
Given:_____________________

41. Mass to mole
- solute: 25 g of NaF
Molar Mass
Na : 1 x 23 = 23 g/mole
F : 1 x 19 = 19 g/mole
Total Molar Mass: 42 g/mole

42. Mass to mole
25 g of NaF X
1 𝑚𝑜𝑙𝑒
42 𝑔
= 0.60 moles of NaF
Moles of solute: 0.60 mole of NaF

43. Mass to mole
- solvent: 200 g of H2O
Molar Mass
H : 2 x 1 = 2 g/mole
O : 1 x 16 = 16 g/mole
Total Molar Mass: 18 g/mole

44. Mass to mole
200 g of H2O X
1 𝑚𝑜𝑙𝑒
18 𝑔
= 11.11 moles of H2O
moles of solvent: 11.11 moles of H2O

45. moles of solute + moles of solvent
=
Total Moles of Solution

46. Moles of solute = 0.60 moles of NaF
Moles of solvent = 11.11 moles of H2O
Total Moles of Solution = 11.71 moles

47. Moles of solute = 0.60 moles of NaF
Total Moles of Solution = 11.71 moles
Xsolute =
0.60 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝐹
11.71 𝑚𝑜𝑙𝑒𝑠
Xsolute = 0.05 (Final Answer)

48. Moles of solvent = 11.11 moles of H2O
Total Moles of Solution = 11.71 moles
Xsolvent =
11.11 𝑚𝑜𝑙𝑒𝑠
11.71 𝑚𝑜𝑙𝑒𝑠
Xsolvent = 0.95

49. Xsolute = 0.05
Xsolvent = 0.95
Total = 1

50. Calculate the mole fraction of each component
of a solution containing 65 g of ethanol (C2H6O)
in 350 g of water?

51.  The molality of a solution is the number of
moles of solute per kilogram of solvent.

52.  Calculate the molality of a solution
containing 16.5 g of dissolved naphthalene
(C10H8) in 0.0543 Kg benzene (C6H6).
STEP 1
Given:
Solute:_______________
Solvent: ______________
Find: _____________

53.  Calculate the molality of a solution
containing 16.5 g of dissolved naphthalene
(C10H8) in 0.0543 Kg benzene (C6H6).
Given:
Solute: 16.5 g C10H8
Solvent: 54.3 g C6H6
Find: molality

55. STEP 2: Convert the mass of solute into moles
using its molar mass.
Solute: 16.5 g C10H8
C : 10 x 12 = 120 g/mole
H : 8 x 1 = 8 g/mole
Total Molar Mass = 128 g/mole

56. STEP 2: Convert the mass of solute into moles
using its molar mass.
16.5 g C10H8 x
𝟏 𝒎𝒐𝒍𝒆
𝟏𝟐𝟖 𝒈
= 0.132 moles of C10H8

57. STEP 3: Calculate the Molality using the
formula.
m =
𝟎.𝟏𝟑𝟐 𝒎𝒐𝒍𝒆𝒔
0.0543 Kg
m = 2.49 moles/Kg

58. 10g of NaOH is dissolved in 500 g water . What
is the molality of solution?

59.  Molarity of solution is the number of moles of
solute per liter of solution. It is especially
useful in doing stoichiometric calculations
involving solution.

60.  Determine the molarity of a solution
containing 2.40 g of sodium chloride (NaCl) in
40.0 mL of solution.

61.  Determine the molarity of a solution
containing 2.40 g of sodium chloride (NaCl) in
40.0 mL of solution.
STEP 1
Given
Solute: __________________
Solution: _________________
Find: Molarity

62.  Determine the molarity of a solution
containing 2.40 g of sodium chloride (NaCl) in
40.0 mL of solution.
STEP 1
Given
Solute: 2.40 g of NaCl
Solution: 40 mL ------ 0.04 L
Find: Molarity

63. STEP 2: Convert the given mass of solute into
moles using molar mass.
Solute: 2.40 g of NaCl
Na : 1 x 23 = 23 g/mole
Cl : 1 x 35 = 35 g/mole
Total Molar Mass of solute: 58 g/mole

64. STEP 2: Convert the given mass of solute into
moles using molar mass.
2.40 g of NaCl x
1 𝑚𝑜𝑙𝑒
58 𝑔𝑟𝑎𝑚𝑠
= 0.04 moles

65. STEP 3: Calculate the molarity using the formula.
M =
𝟎.𝟎𝟒 𝒎𝒐𝒍𝒆𝒔
𝟎.𝟎𝟒 𝑳
M = 1 mole/L

66.  Determine the molar concentrations of a
solution that contains 25g of potassium
hydroxide (KOH) in 250 mL of solution.

67.  What is the molarity of 0.25 mol of NaCl in
300 mL of solution?

68.  Parts per million (ppm) expresses the number
of parts of solute per one million/billion parts
of solution.

69. 1. 25 grams of sodium chloride is dissolved in
100 grams of water, What is the
concentration of the sodium in parts per
million (ppm)?
STEP 1
Given Find: ppm
Solute : _______________
Solvent : _______________
Solution : ______________

70. 1. 25 grams of sodium chloride is dissolved in
100 grams of water, What is the
concentration of the sodium in parts per
million (ppm)?
STEP 1
Given Find: ppm
Solute : 25 grams NaCl
Solvent : 100 grams water
Solution : 125 grams

71. STEP 2: Calculate the ppm using the formula.
ppm =
𝟐𝟓 𝒈𝒓𝒂𝒎𝒔
𝟏𝟐𝟓 𝒈𝒓𝒂𝒎𝒔
𝒙 𝟏𝟎𝟔
ppm = 200,000