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.
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.
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 =
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.
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
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