The document provides information on different ways of expressing the concentration of solutions, including percent by mass, mole fraction, molarity, molality, percent by volume, and parts per million (ppm). It discusses an activity where students are asked to mix a substance with water and observe whether the mixture is uniform or non-uniform. Finally, it defines key terms related to solution concentration such as solution, solute, solvent, concentration, solubility, miscible, and immiscible.

2. Use different ways of expressing
concentration of solutions: percent by mass,
mole fraction, molarity, molality, percent by
volume, percent by
mass, ppm
(STEM_GC11PPIIId-f-111)
Learning Competency

3. MIX IT
UP!

4. Directions: Each group will be given
1 plastic cup to be filled with water.
One or two members of the group
will get any material/substance
within the school premises and mix it
up with water.

5. Questions:
1. Does the water and the substance/
material mix? Why do you think so?
2. Does your mixture has uniform or
non-uniform appearance?

6. 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?

7. A B

10. 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. 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

13. Miscible liquids can easily dissolve in one another.
Immiscible liquids are not soluble in each other.
Chemistry-Borders

14. 1 mol = 6.02214 × 1023 particles.
A mole is the amount (10) of material containing
6.02214 × 1023 particles. 1 mol = 6.02214 × 1023 particles.
This number is also called Avogadro's number.
Mole

15. Mole Conversion

16. Molarity
(M)

17. 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 (mol)
Liters of solution (L)

18. Example 1: What is the molarity of a solution
that has 2.3 moles of sodium chloride in 0.45
liters of solution?
Molarity

19. 2.3 moles NaCl = 5.1M NaCl
0.45 L
Molarity
Example 1: What is the molarity of a solution
that has 2.3 moles of sodium chloride in 0.45
liters of solution?

20. 3.45 moles of sodium chloride is
added in 0.56 millimeters of
solution. What is the Molarity of
the solution?

21. Example 2: How many moles of Na2CO3 are
there in 10.0 L of 2.0 M solution?
Molarity

22. Example 2: How many moles of Na2CO3 are
there in 10.0 L of 2.0 M solution?
Molarity
10.0 L 2.0 mol Na2CO3
1 1 L
= 20.0 moles Na2CO3

23. How many moles of HCl are there
in 15.0 L of 3.0 M solution?

24. Example 3: How many moles of
Na2CO3 are needed to make 450 mL of
1.5 molar solution?
Molarity

25. Example 3: How many moles of KNO3
are needed to make 450 mL of 1.5
molar solution?
Molarity
450 mL 1L 1.5 mol KNO3
1 1000mL 1L
= 0.675 moles KNO3

26. How many moles of Na2CO3 are
needed to make 679 cL of 4.8
molar solution?

27. Example 4: How many grams of NaCl are
needed to make 3.0 L of 1.5 M solution?
Molarity

28. Example 4: How many grams of NaCl are
needed to make 3.0 L of 1.5 M solution?
Molarity
3.0 L 1.5 mol NaCl 58.44 g NaCl
1 1L 1 mol NaCl
= 260 g NaCl

29. How many grams of HCl are
needed to make 4.0 L of 3.5 M
solution?

30. Example 5: How many L of 4.0 M solution
can be made with 132g of NaCl ?
Molarity

31. Example 5: How many L of 4.0 M solution
can be made with 132g of NaCl ?
Molarity
132 g NaCl 1 mol NaCl 1 L
1 58.44 g NaCl 4.0 mol NaCl
= 0.565 L

32. How many L of 5.0 M solution can
be made with 132g of Mg(CN)2 ?

33. 1.How many grams of magnesium cyanide
Mg(CN)2 are needed to make 275 mL of a 0.075
M solution?
2.How many grams of hydrochloric acid are
present in 3.0 L of a 0.750 M solution?
3.What is the molarity of a solution in which 0.850
grams of ammonium nitrate (NH₄NO₃)are
dissolved in 345 mL of solution?

34. 1.15 mg of NaBr is dissolved in 400 mL of
solution. What is the Molarity of the solution?
2.What is the molarity of the solution in which
58g of NaCl are dissolved in 20 dL of solution?
3.How many grams of Ca(OH)2 are needed to
produce 500 ml of 1.66 M Ca(OH)2 solution?
4. What is the molarity of 650 ml of solution
containing 63 grams of NaCl?

35. Mass
Percent

36. Mass Percent
Solutions can also be represented as percent of solute in a
specific mass of solution.
For a solid dissolved in water, you use percent by mass which is
Mass Percent.
% by mass = mass solute (g) x 100
mass of solution (g)
**Mass of solution = solute mass + solvent mass

37. If a solution that has a mass of 800.0 grams contains 20.0
grams of NaCl, what is the concentration using Percent by
Mass?
Example 1

38. If a solution that has a mass of 800.0 grams contains 20.0 grams of
NaCl, what is the concentration using Percent by Mass?
% by mass = mass solute x 100
mass of solution
% by mass = 20.0g NaCl x 100
800.0g solution
= 2.50% NaCl
Example 1

39. The mass of the solution is 7800 mg
containing 10.0 grams of Mg(CN)2 , what
is the concentration of the solution using
Percent by Mass?

40. If 10.0 grams of NaCl is dissolved in 90.0 grams of water,
what is the concentration using Percent by Mass?
Example 2

41. If 10.0 grams of NaCl is dissolved in 90.0 grams of water,
what is the concentration using Percent by Mass?
% by mass = mass solute x 100
mass of solution
% by mass = 10.0g NaCl x 100 = 10.0%NaCl
100.0g solution
Example 2

42. What is the concentration using
Percent by Mass if 2800 mg of HCl is
dissolved in 8900 mg of methanol?

43. How many grams of sodium bromide are in 200.0g
of solution that is 15.0% sodium bromide by mass?
Example 3

44. How many grams of sodium bromide are in 200.0g of
solution that is 15.0% sodium bromide by mass?
% by mass = mass solute x 100
mass of solution
% by mass = g NaBr x 100 = 15.0%NaBr
200.0g solution
g NaBr = 200.0 x 15.0
100
= 30 g NaBr
Example 3

45. How many grams of sodium
chloride are in 500.0g of solution
that is 25.0% sodium chloride by
mass?

46. 1) 15 g of NaCl was dissolved in 225g of water. What is the mass
percent of NaCl in the solution?
2) What is the percent by mass of 5.0 g of iron (II) sulfate
dissolved in 75.0 g of water?
3) A solution is made by dissolving 125 g of sodium chloride in
1.5 kg of water. What is the percent by mass?
4) 25 mL of Methanol (CH3OH) with a density of d = 0.792
g/cm*3 is mixed with 150mL of water (d= 1g/cm*3). What is
the mass percent of methanol?

47. 1.What is the percent by volume of a solution
formed by added 15 L of acetone to 28 L of
water?
2. An experiment requires a solution that is 80%
methyl alcohol by volume. What volume of
methyl alcohol should be added to 200 mL of
water to make this solution?
3. What is the percent by volume of a solution
formed by mixing 25 mL of isopropanol with
45 mL of water?

49. 1) Calculate the mole fraction of a solution made by 20g of NaOH and 900 g
of water.
2) A tank is charged with a mixture of 1.0 x 103 mol of oxygen and 4.5 x
103 mol of helium. Calculate the mole fraction of each gas in the mixture.
3) 0.100 mole of NaCl is dissolved into 100.0 grams of pure H2O. What is the
mole fraction of NaCl?
4) A solution contains 10.0 g pentane (C5H12), 10.0 g hexane (C6H14) and
10.0 g benzene (C6H6). What is the mole fraction of hexane?
5) How many moles of water must be used to dissolve 100.0 grams of
sucrose (C12H22O11) to prepare a 0.020 mole fraction of sucrose in the
solution?

50. Solution Stoichiometry
Chemistry-Borders IPC-SOLUTIONS-BORDERS
reactants products

51. 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

52. 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

53. 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

54. 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 m 1 mol NaOH
= 450. g NaOH

56. 1) 25 mL of Methanol (CH3OH) with a density of d=0.792 g/cm3
is mixed with 150mL of water (d= 1g/cm*3). What is the
volume and mass percent of methanol?
2) 2 mol of KCl is dissolved in 8 moles of water. What is the mol
fraction of KCl? Of water?
3) 10 g of NaOH is dissolved in 500 g of water. What is the
molality of the solution.

58. Dilutions and Molarity
Use this formula to make a more dilute solution
from a concentrated solution
Molarity1 x Volume1 = Molarity2 x Volume2
(Concentrated) (Dilute)
(before) = (after)
M1V1 = M2V2

59. Example 1
How many liters of 2.5 M HCl are required to make 1.5 L of 1.0 M
HCl?
M1V1 = M2V2
M1 = 2.5 M M2 = 1.0 M
V1 = ? V2 = 1.5 L

60. Example 1
How many liters of 2.5 M HCl are required to make 1.5 L of 1.0 M
HCl?
M1V1 = M2V2
M1 = 2.5 M M2 = 1.0 M
V1 = ? V2 = 1.5 L
(2.5M) V1 = (1.0M) (1.5 L)
2.5M
2.5M
= 0.60L

61. M1 = 2.5M V1 = 0.60L M2 = 1.0 M V2= 1.5 L
.
How much water should you add to the volume of 2.5M HCl you
calculated above to make the solution? (draw this in your notes)
1st add .60L of HCl to
measuring device
Example 1

62. M1 = 2.5M V1 = 0.60L M2 = 1.0 M V2= 1.5 L
.
How much water should you add to the volume of 2.5M HCl you
calculated above to make the solution? (draw this in your notes)
Example 1
Then add enough water to get to 1.5L
of solution
V2 – V1 = Amount of water
1.5L – 0.60L = 0.90L water

63. M1 = 2.5M V1 = 0.60L M2 = 1.0 M V2= 1.5 L
.
How much water should you add to the volume of 2.5M HCl you
calculated above to make the solution? (draw this in your notes)
Example 1
Final solution is 1.5L of 1.0M
HCl

64. Example 2
250.0 mL of a 0.500 M HCl solution needs to be made from
concentrated HCl. What volume of the concentrated solution
is needed if its molarity is 12.0 M?
M1V1 = M2V2
M1 =
V1 =
M2 =
V2 =

65. Example 2
250.0 mL of a 0.500 M HCl solution needs to be made from
concentrated HCl. What volume of the concentrated solution is
needed if its molarity is 12.0 M?
M1V1 = M2V2
M1 = 12.0M
V1 = 10.4mL
M2 = 0.500M
V2 = 250.0mL
How much water would you
add to make the final
solution?
250.0mL - 10.4mL = 239.6mL