Solutions, solute and solvents combined to create solutions

1. General Chemistry 2
Presented by
Kristel Joy R. Somera
Subject teacher

3. Classification of Matter
Mixture
Pure substance

4. A solution is a homogenous mixture
made up of atoms, ions, or
molecules.
Example: Atmosphere (solution of
gases); seawater (solution of salt and
water; Soil (solution of elements and
minerals)

5. Two mediums of solutions
Solute – the dissolved medium
Solvent – the dissolving
medium

6. Types of Solutions
Type Solvent Solute Final
State
Solid Solid Liquid Solid
Liquid Liquid Liquid Liquid
Liquid Gas Liquid
Liquid Solid Liquid
Gas Gas Gas Gas
Gas Solid Gas

7. Solutions maybe classified;
1. Depending on the dissolution of the solute in the solvent
1.1. Saturated- a solution in which a solvent is not capable
of dissolving any more solute at a given temperature.
1.2. Supersaturated-comprises a large amount of solute at
a temperature wherein it will be reduced, as a result the
extra solute will crystallize quickly.
1.3. Unsaturated-a solution in which a solvent is capable of
dissolving any more solute at a given temperature.

8. Solutions maybe classified;
2. Depending on the solvent
2.1. Aqueous solution – When a solute is dissolved in
water the solution is called an aqueous solution. Eg, salt in
water, sugar in water and copper sulfate in water.
2.2. Non-aqueous solution – When a solute is dissolved in
a solvent other than water, it is called a non-aqueous
solution. Ex. iodine in carbon tetrachloride, sulfur in
carbon disulfide, phosphorus in ethyl alcohol.

9. Solutions maybe classified;
3. Depending on the amount of solute dissolve in solvent
3.1. dilute solution contains a small amount of
solute in a large amount of solvent
3.2. concentrated solution contains a large
amount of solute dissolved in a small amount of
solvent.

10. Concentrated solution
- contains greater
amount of solute
compared to the solvent
Diluted solution
- contains greater of solvent
compared to the solute

11. Task 2:
Objectives: To prepare different solutions according to their
classifications;
1. Make a plan on how to prepare the different solutions
according to their classification;
2. Make a step- by – step procedures in conducting the
activity;
3. Execute the procedure being made and label the
prepared solutions; and
4. Present the outputs in the class.

12. Identify the solution that is concentrated and
diluted in the given pair of solutions
A B A B

13. Solubility – refers to the amount of solute that can
dissolve in a given amount of solvent at room
temperature under given conditions.
Miscibility – refers to the property of two substances
to mix in all proportions, forming a homogenous
mixture.

14. Ways of Expressing Concentration
of Solution

15. I. Percent by mass
- This expresses the mass of solute per 100g
of solution
% by mass solute =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
Mass of the solution = mass of the solute + mass
of the solvent

16. I. Percent by mass
- if the solution is expressed in volume
% by volume solute =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
Volume of the solution = volume of the solute +
volume of the solvent

17. I. Percent by mass
Example: What is the % by mass of a 25 g CaCl2 and 120 g
H2O solution?
% by mass solute =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
Mass of the solution = mass of solute + mass of solvent
= 25 g and 120 g
= 145 g
% by mass solute =
25 𝑔
145 𝑔
x 100
=
17.24%

18. I. Percent by mass
Example: What is the % by volume of a 45 mL ethyl alcohol
and 105 mL H2O solution?
% by volume solute =
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 100
Volume of the solution =volume of solute + volume of solvent
= 45 Ml and 105 mL
= 150 mL
% by volume solute =
45 𝑚𝐿
150 𝑚𝐿
x 100
= 30%

19. II. Parts per million
This is used when the concentration of solute is very low. It is
being used to express the concentration of dissolved oxygen
in water or the amount of carbon dioxide in air.
Parts per million or ppm=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 1,000,000
Mass of the solution = mass of the solute + mass of the
solvent

20. II. Parts per million
Example: What is the concentration in ppm in the given
solution of 25 g CaCl2 and 120 g H2O
Mass of the solution = mass of solute + mass of solvent
= 25 g and 120 g
= 145 g
Parts per million or ppm=
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
x 1,000,000
Parts per million or ppm=
25 𝑔
145 𝑔
x 1,000,000
= 172,413 ppm

21. Activity : Solve the following problem
2. What is the % by volume of 10 mL isopropyl alcohol and
75 mL H2O?
3. What is the concentration in ppm of a solution compose
of 50 g NaCl and 200 g H2O?
1. What is the % by mass of 45 g of HCl in a 200 g solution?
4. What mass of NaCl is needed to prepare 30 % by mass of a
500 g solution of NaCl?

22. III. Mole Fraction
It is the ratio of the number of moles of one component to
the total number of moles in a solution.
Mole fraction of solute =
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Mole fraction of solvent =
𝑚𝑜𝑙𝑒 𝑜𝑓 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
Note: number of moles solution = number of moles solute + number of
moles solvent
Mole fraction of solute + mole fraction of solvent = 1

23. Sample Problem: (Note: you need a periodic table of element)
A solution is made by dissolving 1.25 g of Na2SO4+ in 65.0 g
of water. What is the mole fraction of solute and solvent?
Solution:
Step 1: Get the number of moles of solute (Na2SO4)
(Na2SO4) =
mass of Na2SO4
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 Na2SO4
=
1.25 g Na2SO4
142
𝑔
𝑚𝑜𝑙𝑒
Na2SO4
= 0.0088 mole

24. Solution:
Step 2: Get the number of moles of solvent (H2O)
H2O=
mass ofH2O
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓H2O
=
65 g H2O
18
𝑔
𝑚𝑜𝑙𝑒
H2O
= 3.61 mole
Step 3: Get the total number of moles
Number of moles = number of moles of solute + number of moles of
solvent
= 0. 0088 mole + 3.61 moles
= 3.62

25. Solution: Step 4: Get the mole fraction of solute and solvent
Mole fraction of solute =
0.0088 𝑚𝑜𝑙𝑒
3.62 𝑚𝑜𝑙𝑒𝑠
= 0.0024
Mole fraction of solvent =
3.61 𝑚𝑜𝑙𝑒𝑠
3.62 𝑚𝑜𝑙𝑒𝑠
= 0.997
To check your answer;
Mole fraction of solute + mole fraction of solvent = 1
0.0024 + 0.997 = 0.999 ≈ 1

26. Sample Problem: (Note: you need a periodic table of element)
A solution is made by dissolving 10 g of NaCl in 40 g of water.
What is the mole fraction of solute and solvent?
Solution:
Step 1: Get the number of moles of solute NaCl)
NaCl =
mass of Na𝐶𝑙
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 NaCl
=
10 g NaCl
58
𝑔
𝑚𝑜𝑙𝑒
NNaCl
= 0.17 moles

27. Solution: Step 2: Get the number of moles of solvent (H2O)
H2O=
mass ofH2O
𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓H2O
=
40 g H2O
18
𝑔
𝑚𝑜𝑙𝑒
H2O
= 2.22 mole
Step 3: Get the total number of moles
Number of moles = number of moles of solute + number of moles of
solvent
= 0.17mole +2.22moles
= 2.39

28. Solution: Step 4: Get the mole fraction of solute and solvent
Mole fraction of solute =
0.17 𝑚𝑜𝑙𝑒
2.39𝑚𝑜𝑙𝑒𝑠
=0.07
Mole fraction of solvent =
2.22 𝑚𝑜𝑙𝑒𝑠
2.39 𝑚𝑜𝑙𝑒𝑠
= 0.93
To check your answer;
Mole fraction of solute + mole fraction of solvent = 1
0.07 + 0.93 =1

29. Activity : Solve the following problem
5. A solution is prepared by dissolving 5.02 g of KNO3 in 95 g
of water. What is the mole fraction of the solute and the
solvent?

30. III. Molarity
It is the ratio of the number of moles of solute per liter of
solution.
Molarity=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑙𝑖𝑡𝑒𝑟 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛
A 3M (3 molar) NaOH means that there are three moles of
NaOH in one liter of solution.

31. Sample problem
2. What is the molarity of 5 gCaCl2 in 250 mL solution?
0.2 moles NaOH x
1
100𝑚𝐿
x
1000 𝑚𝐿
1𝐿
=
2 𝑚𝑜𝑙𝑒𝑠
𝐿
1. What is the molarity of 0.2 moles NaOH in 100 mL
solution?
5 g CaCl2 x
1 m
oleCaCl2
110 𝑔CaCl2
x
1
250 𝑚𝐿
x
1000𝑚𝐿
1 𝐿
=
0.182 𝑚𝑜𝑙𝑒𝑠
𝐿

32. Sample problem
500 mL x
1𝐿
1000𝑚𝐿
x
2 𝑚𝑜𝑙𝑒𝑠
1𝐿
x
36.35 𝑔 𝐻𝐶𝑙
1𝑚𝑜𝑙𝑒 𝐻𝐶𝑙
= 36.45 g HCl
3. Calculate the mass of solute needed to prepare 500 mL of
2M HCl (molar mass = 36.45 g). Note: 2M = 2 moles/liter and
the unknown is mass, therefore, the final answer is
expressed in g of HCl

33. Activity : Solve the following problem
6. What is the molarity of 236 g HCl in a liter of solution?

34. IV. Molality
It is the number of moles of solute per kilogram of solvent.
Molality=
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑠𝑜𝑙𝑢𝑡𝑒
𝑘𝑖𝑙𝑜𝑔𝑟𝑎𝑚 𝑠𝑜𝑙𝑣𝑒𝑛𝑡
A 3m (3 molal) solution means that there are three moles of
solute/ kilogram of solvent

35. Sample problem
2. What is the molality of a 0.40 g Na2CO3 dissolved in 150 g
grams of H2O? (molar mass of Na2CO3 = 106 g)
0.5 moles H3PO4 x
1
250 𝑔H2O
x
1000 𝑔
1𝑘𝑔
=
2 𝑚𝑜𝑙𝑒𝑠
𝑘𝑔
2m
1. What is the molality of 0.5 moles of H3PO4 in 250 g water?
Note that the final unit for molality is mole/kg.
0.40 g Na2CO3 x
1 m
oleNa2CO3
106 𝑔Na2CO3
x
1
150 𝑔H2O
x
1000𝑔
1 𝑘𝑔
=
0.025 𝑚𝑜𝑙𝑒𝑠
𝑘𝑔

36. Activity : Solve the following problem
7. What is the molality of 12.8 g Mg(OH)2 dissolved in 550 g
water? (molar mass of Mg(OH)2 = 58 g)

38. Electrical Properties of Solutions
Electrolytes are substances that dissociates or
form ions when dissolve in water (usually ionic
compounds and polar molecules)
Non - Electrolytes are substances that do not
ionize in a solution (usually non-polar)

39. Colligative Properties of Solutions
Colligative properties are properties that
depends solely on the number of particles of
solute present but not on the kind and nature
of solute.
Vapor pressure, osmotic pressure, freezing
point and boiling point

40. How do the amount of solute affect the
colligative properties?
A. Vapor Pressure Reduction
B. Boiling Point Elevation
C. Freezing Point Depression
C. Osmotic Pressure

41. Vaporization is the process where fast moving
molecules of a liquid overcome their attractive
forces and escape into the gaseous state.
A. Vapor Pressure Reduction