This document discusses key concepts related to solution concentration including: - Solutions are homogeneous mixtures with a solvent as the major component and solute as the minor component. - Concentration can be expressed in various ways such as percentage by weight or volume, molarity, and parts per million or billion. - Dilute solutions have low concentrations while concentrated solutions have high concentrations. - Molarity calculations allow converting between moles, liters, and grams of solutes and solvents in solutions. - Dilution decreases concentration by adding more solvent while maintaining the same amount of solute.

1. Solution Concentration

2. Review
A solution is a homogeneous mixture.
The solvent is the major component of the
solution.
The solute is the minor component and
active ingredient.
A saturated solution holds the maximum
amount of solute that is theoretically
possible for a given temperature.

3. How would you describe this picture?

4. Solution Concentration
Is one glass of tea stronger than the other?
– What’s true about the “stronger” glass of tea?
– How much tea does it have in it compared to
the other glass?

5. Solution Concentration
Concentration – a ratio comparing the
amount of solute to the amount of solution.
Many ways of expressing concentration:
– % by weight (% w/w)
– % by volume (% v/v)
– parts per million (ppm) or parts per billion
(ppb) for very low concentrations
– molality (m)
– Molarity (M)

6. Concentrated vs. Dilute
The words “concentrated” and “dilute” are
opposites.
EX: The dark tea is more concentrated than
the light tea.
EX: The light tea is more dilute than the
dark tea.

7. Concentrated vs. Dilute
Concentrated
solution
Dilute
solution
= solute particles

8. Percent by Weight
% by weight (% w/w)
What is the % w/w of a solution if 3.00 grams of
NaCl are dissolved in 17.00 g of water?
– mass of solute = 3.00 g
– mass of solution = 3.00 g + 17.00 g = 20.00 g
– (3.00 g / 20.00 g) x 100% = 15.0% w/w
100%x
solutionofmasstotal
soluteofmass
%w/w =

9. Percent by Volume
% by volume (% v/v)
What is the % v/v of a solution if 20.0 mL of
alcohol are dissolved in 50.0 mL of solution?
– volume of solute = 20.0 mL
– volume of solution = 50.0 mL
– (20.0 mL / 50.0 mL) x 100% = 40.0%
100%x
solutionofvolumetotal
soluteofvolume
%v/v =

10. Molarity
Molarity (M)
– UNITS: mol/L or Molar (M)
– Example: 0.500 mol/L = 0.500 M
solutionofLiters
soluteofmoles
Molarity =

11. Molarity
 What is the Molar concentration of a sol’n if 20.0 grams
of KNO3 (MM = 101.11 g/mol) is dissolved in enough
water to make 800. mL?
– Convert g of KNO3 to mol of KNO3
– Convert mL to L
3
3
3
3 KNOmol0.198
KNOg101.11
KNOmol1
xKNOg20.0 =
L0.800
mL1
L0.001
xmL800. =

12. Molarity
What is the Molar concentration of a sol’n if
0.198 mol KNO3 is dissolved in enough water to
make 0.800 L?
M0.248
L0.800
KNOmol0.198
Molarity 3
==

13. Molarity
What is the Molar concentration of a sol’n if 10.5 grams of
glucose (MM = 180.18 g/mol) is dissolved in enough water
to make 20.0 mL of sol’n?
– Convert g of glucose to mol of glucose.
– Convert mL to L.
glucosemol0.0583
glucoseg180.18
glucosemol1
xglucoseg10.5 =
L0.0200
mL1
L0.001
xmL20.0 =

14. Molarity
What is the Molar concentration of a sol’n if 0.0583 mol of
glucose is dissolved in enough water to make 0.0200 L of
sol’n?
M2.92
L0.0200
glucosemol0.0583
Molarity ==

15. Calculating Grams
 How many grams of KI (MM = 166.00 g/mol) are needed to prepare
25.0 mL of a 0.750 M solution?
– Convert mL to L.
– Solve for moles.
– moles of KI = 0.750 M x 0.0250 L = 0.0188 mol KI
L0.0250
mL1
L0.001
xmL25.0 =
L0.0250
KIofmoles
xM0.750

16. Calculating Grams
 How many grams of KI (MM = 166.00 g/mol) are needed to prepare
25.0 mL of a 0.750 M solution?
– Convert 0.0188 mol KI to grams.
KIg3.12
KImol1
KIg166.00
xKImol0.0188 =

17. Calculating Grams
How many grams of HNO3 (MM = 63.02
g/mol) are present in 50.0 mL of a 1.50 M
sol’n?
– Convert mL to L.
• 50.0 mL = 0.0500 L
– Solve for moles:
• moles = (1.50 M)(0.0500 L) = 0.0750 mol HNO3
– Convert 0.0750 mol HNO3 to grams:
• 0.0750 mol HNO3 = 4.73 g HNO3

18. Dilution
Dilute (verb) - to add solvent to a solution.
– Decreases sol'n concentration.
– M1
V1
= M2
V2
• M1
= initial conc.
• V1
= initial volume
• M2
= final conc.
• V2
= final volume
– Assumes no solute is added.

19. Dilution
Stock Solution
Impractically
High Concentration
Usable
Solution
Add H2
O
Question for Consideration: Why do you
think chemical supply companies typically sell
acids (and other solutions) in extremely high
concentrations when it would be safer to ship
more dilute solutions?

20. Dilution
 To what volume should 40.0 mL of 18 M H2
SO4
be
diluted if a concentration of 3.0 M is desired?
– What do we want to know?
• V2
– What do we already know?
• M1
= 18 M
• V1
= 40.0 mL
• M2
= 3.0 M
– (18 M)(40.0 mL) = (3.0 M)V2
– 720 M*mL = (3.0 M)V2
– V = 240 mL

21. Dilution
 You are asked to prepare 500. mL of 0.250 M HCl,
starting with a 12.0 Molar stock sol'n. How much
stock should you use?
– What do we want to know?
• V1
– What do we already know?
• M1
= 12.0 M
• M2
= 0.250 M
• V2
= 500. mL
– (12.0 M) V1
= (0.250 M)(500. mL)
– (12.0 M) V1
= 125 M*mL

22.  To how much water should you add 20.0 mL of
5.00 M HNO3
to dilute it to 1.00 M?
– What do we want to know?
• How much water to add. (V2
- V1
)
– What do we already know?
• M1
= 5.00 M
• V1
= 20.0 mL
• M2
= 1.00 M
– (5.00 M)(20.0 mL) = (1.00 M) V2
– 100. M*mL = (1.00 M) V2
– V = 100. mL
Dilution

23. Colligative Properties
Colligative properties are properties of
solutions that are affected by the number of
particles but not the identity of the solute

24. Boiling Point Elevation: A Colligative
Property
Boiling point elevation is the temperature
difference between a solution and pure solvent
The value of the boiling point elevation is
directly proportional to molality, meaning the
greater the number of solute particles, the
greater the elevation

25. Freezing Point Depression: A Colligative
Property
Freezing point depression is the difference in
temperature between a solution and a pure
solvent
The value of freezing point depression is
directly proportional to molality, meaning the
greater the number of solute particles