Ch15.1 – Water and Aqueous Solutions
Ex1) Draw H2O structure
shape:
bond angle:
polar:
Ch15.1 – Water and Aqueous Solutions
O
shape bent
H
H
bond 105˚
O
O
polar yes
H
H H
H
Acts like a skin at surface
(call this surface tension)
H20 has high surface tension, low vapor pressure, high boiling
point (compared to other covalent molecules)
High surface ttension makes
it stick to its self, defies gravity
coin
Add soap (surfactants-surface active agents)
interfere with hydrogen bonding, bead
spreads out.
coin
Ch15.1 – Water and Aqueous Solutions
O
shape bent
H
H
bond 105˚
O
O
polar yes
H
H H
H
Acts like a skin at surface
(call this surface tension)
H20 has high surface tension, low vapor pressure, high boiling
point (compared to other covalent molecules)
High s.t. makes it stick
to its self, defies gravity
coin
Add soap (surfactants-surface active agents)
interfere with hydrogen bonding, bead
spreads out.
coin
All liquids have some amount of surface tension.
Heat Capacity
Water has a high specific heat capacity(ability to absorb
energy without changing temp)
J
C p  4.18 
g C
- Instead of energy going in and increasing
average K.E. (molecular motion),
causes bonds to bend, twist, wiggle, stretch.
Evaporation
- water has a large heat of vaporization - 2260J/g
Strong dipole forces, require lots of energy to break free.
Ex2) How much heat energy is required to vaporize 2.0g of water?
Evaporation
- water has a large heat of vaporization - 2260J/g
Strong dipole forces, require lots of energy to break free.
Ex) How much heat energy is required to vaporize 2.0g of water?
2.0g
2260J
= 4520J
1g
Ammonia – HV = 1370 J/g
Methane – HV = 500 J/g
They vaporize much easier than water.
- since N and C aren’t as electronegative,
don’t have as strong hydrogen bonding.
They have higher vapor pressure than water
Evaporation
- water has a large heat of vaporization - 2260J/g
Strong dipole forces, require lots of energy to break free.
Ex) How much heat energy is required to vaporize 2.0g of water?
2.0g
2260J
= 4520J
1g
Ammonia – HV = 1370 J/g
Methane – HV = 500 J/g
They vaporize much easier than water.
- since N and C aren’t as electronegative,
don’t have as strong hydrogen bonding.
They have higher vapor pressure than water
Water has low vapor pressure
(only a little of it becomes a gas)
Ice
Solid H20 is less dense than liquid (ice floats).
Ice
Water
- water has its greatest density at 4˚C
(liquid, slow moving)
- important for life on earth.
- Most substances, solid is more dense than liquid
Exceptions besides H2O:
Bismuth, Antimony, some iron alloys
Ch15 HW#1 1-6
Lab15.1 – Adhesion/Cohesion
- due tomorrow
- Ch15 HW#1 due at beginning of period
Ch15 HW#1 1-6
1. What is surface tension? Why do particles at the surface of the liquid
behave differently from those in the bulk of the liquid?
2. Describe the origin of the vapor pressure of water.
3. Why is it easier to wash a car with soapy water than just water alone?
4. Why does water have a relatively high boiling point?
5. How many kilojoules are required to vaporize 5.0g of water at its
boiling point?
6. What unique characteristic of ice distinguishes it from other solids?
5. How many kilojoules are required to vaporize 5.0g of water at its
boiling point?
5.0 gH 2O 2260 J

 11,300 J or 11 .3kJ
1
1.0 gH 2O
6. What unique characteristic of ice distinguishes it from other solids?
5. How many kilojoules are required to vaporize 5.0g of water at its
boiling point?
5.0 gH 2O 2260 J

 11,300 J or 11 .3kJ
1
1.0 gH 2O
6. What unique characteristic of ice distinguishes it from other solids?
Floats in its own liquid
Ch15.2 – Aqueous Solutions
- Pure water doesn’t exist in nature because water dissolves
so many substances.
When water dissolves a substance, the solution is called an aqueous
solution. With any other solvent its is just called a solution.
Solvent – dissolving medium (substance doing the dissolving)
Solute – the dissolved particles in the solvent.
- water will dissolve any ionic compound (to some degree) or any polar
molecular compounds.
- it wont dissolve oil because oil is nonpolar.
“Like dissolves like”
Ch15.2 – Aqueous Solutions
- Pure water doesn’t exist in nature because water dissolves
so many substances.
When water dissolves a substance, the solution is called an aqueous
solution. With any other solvent its is just called a solution.
Solvent – dissolving medium (substance doing the dissolving)
Solute – the dissolved particles in the solvent.
- water will dissolve any ionic compound (to some degree) or any polar
molecular compounds.
- it wont dissolve oil because oil is nonpolar.
“Like dissolves like”
In polar solutions: ions or polar molecules are attracted to polarity
of solvent. (some don’t dissolve if attractions between ions
is stronger than attractions to solvent.)
In nonpolar solutions: substance just mix
If 2 solutions mix, the one in greater amount is the solvent.
Electrolytes and Nonelectrolytes
Electrolytes – ionic compounds conduct electricity when their ions are mobile.
Conduct as a liquid (molten), or when aqueous (dissolved in solution)
Do not conduct in the solid state - ions cant move around.
Exs: NaCl
conducts when dissolved in water, or when heated until it
becomes molten. Doesn’t conduct as solid. (strong electrolyte)
BaSO4
conducts when molten, poor conductor when dissolved
in water because it is insoluble – doesn’t dissolve well.
(weak electrolyte)
Nonelectrolytes – covalent compounds. Never conduct electricity.
Metals – conduct as solids or liquid
Ch15 HW#2 + Density Rev
Density Review
1) What is the density of a copper slug that a lab group masses at 180.01g
and finds the volume to be 20.5cm3?
2) If the actual density is 8.96 g/cm3, what is the % error?
3) Glycerin:
1. Mass of dry grad. cylinder = 12.36g
2. Mass of glycerin & cylinder = 24.74g
3. Volume of glycerin = 9.8mL
4) Salt water solution has density 1.13g/cm3.
Volume determined to be 8.0cm3. Find mass:
5) Mass a piece of aluminum at 90.20g. Its density is given as 2.80 g/cm3.
Find the volume:
Density Review
1) What is the density of a copper slug that a lab group masses at 180.01g
and finds the volume to be 20.5cm3?
D=m/v = 180.01g/20.5cm3
= 8.78g/cm3
2) If the actual density is 8.96 g/cm3, what is the % error?
% error = l actual – exp l / actual x 100%
= l 8.96 - 8.78 l / 8.96
x 100% = 2.0%
3) Glycerin:
1. Mass of dry grad. cylinder = 12.36g
2. Mass of glycerin & cylinder = 24.74g
3. Volume of glycerin = 9.8mL
Mass of glycerin = 12.38g
D = 12.38/9.8 = 1.26g/mL
4) Salt water solution has density 1.13g/cm3.
Volume determined to be 8.0cm3. Find Mass:
D = m/V  m = D.V =
1.13g 8.0 cm3
1cm3
= 9.04g
5) Mass a piece of aluminum at 90.20g. Its density is given as 2.80 g/cm3.
Find the volume:
D = m/V => 2.80 g/cm3 = 90.20g / V
V = 90.20g / 2.80g/cm3
V= 90.20g 1 cm3 = 32.3 cm3
2.80g
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
8) Tablespoon of NaCl in water. ID solute & solvent.
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
f) CaCO3
g) Gasoline
11) Electrolyte
and
nonelectrolyte
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solvent
Aqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
f) CaCO3
g) Gasoline
11) Electrolyte
and
nonelectrolyte
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solvent
Aqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.
Solute – NaCl
Solvent – water
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
f) CaCO3
g) Gasoline
11) Electrolyte
and
nonelectrolyte
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solvent
Aqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.
Solute – NaCl
Solvent – water
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
f) CaCO3
g) Gasoline
11) Electrolyte
and
nonelectrolyte
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solvent
Aqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.
Solute – NaCl
Solvent – water
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
polar cov
ionic
polar cov
ionic
nonpolar
yes
yes
yes
yes
no
f) CaCO3
g) Gasoline
ionic
nonpolar
(not very soluble)
no
11) Electrolyte
and
nonelectrolyte
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solvent
Aqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.
Solute – NaCl
Solvent – water
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
polar cov
ionic
polar cov
ionic
nonpolar
yes
yes
yes
yes
no
f) CaCO3
g) Gasoline
ionic
nonpolar
(not very soluble)
no
11) Electrolyte
and
nonelectrolyte
ions in solution
no ions in solution
conducts electricity
does not conduct electricity
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 12
7) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solvent
Aqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.
Solute – NaCl
Solvent – water
9) Describe how an ionic compound dissolves in water.
(picture from yesterday)
10) Which dissolves in water?
a) HCl
b) NaI c) NH3
d) MgSO4
e) CH4
polar cov
ionic
polar cov
ionic
nonpolar
yes
yes
yes
yes
no
f) CaCO3
g) Gasoline
ionic
nonpolar
(not very soluble)
no
11) Electrolyte
and
nonelectrolyte
ions in solution
no ions in solution
conducts electricity
does not conduct electricity
12) Equation for how calcium chloride dissociates in water.
CaCl2(s)
Ca(aq)+2 + 2Cl(aq)–
Ch15.3 – Suspensions, Colloids, Mixtures
Solutions Homogenous mixture
Small particle size (0.1 – 1nm)
Does not scatter light
Particles don’t separate
Exs: NaCl + water
Particles cant be filtered
KCl + water
Suspensions - Heterogeneous mixture
Large particle size (over 100nm)
Scatters light (Tyndall effect)
Particles separate (sediment forms)
Particles can be filtered Ex: Muddy water
Colloids -
Mixture is in-between
Medium particle size
Scatters light
Particles don’t separate from liquid
Particles cant be filtered
Exs: smoke, milk, marshmallows, gelatin, paint,
aerosol sprays, whipped cream
Emulsions -
colloidal dispersions of liquid in liquids
Ex: soapy water (one end of soap molecules is polar,
attracts to H2O. Other end is nonpolar,
will dissolve nonpolar oils.)
Hydrated Crystals
-water molecules attach to a salt crystal
Ex1) Chemical name for Na2B4O7.10H2O
HW#16a) Name Na2CO8.10H2O
Ex 2) Formula for calcium chloride dihydrate
HW#15a) Formula for sodium sulfate decahydrate
Ex 3) Find the percent water by mass of CaCl2.2H2O
Hydrated Crystals
-water molecules attach to a salt crystal
Ex1) Chemical name for Na2B4O7.10H2O
sodium borate decahydrate
HW#16a) Name Na2CO8.10H2O
sodium carbonate decahydrate
Ex 2) Formula for calcium chloride dihydrate
Ca+2
Cl-1
2 H2O
CaCl2.2H2O
HW#15a) Formula for sodium sulfate decahydrate
Ex 3) Find the percent water by mass of CaCl2.2H2O
Hydrated Crystals
-water molecules attach to a salt crystal
Ex1) Chemical name for Na2B4O7.10H2O
sodium borate decahydrate
HW#16a) Name Na2CO8.10H2O
sodium carbonate decahydrate
Ex 2) Formula for calcium chloride dihydrate
Ca+2
Cl-1
2 H2O
CaCl2.2H2O
HW#15a) Formula for sodium sulfate decahydrate
Na+1
SO4-2 10 H2O
Na2SO4.10 H2O
Ex 3) Find the percent water by mass of CaCl2.2H2O
Hydrated Crystals
-water molecules attach to a salt crystal
Ex1) Chemical name for Na2B4O7.10H2O
sodium borate decahydrate
HW#16a) Name Na2CO8.10H2O
sodium carbonate decahydrate
Ex 2) Formula for calcium chloride dihydrate
Ca+2
Cl-1
2 H2O
CaCl2.2H2O
HW#15a) Formula for sodium sulfate decahydrate
Na+1
SO4-2 10 H2O
Na2SO4.10 H2O
Ex 3) Find the percent water by mass of CaCl2.2H2O
1 Ca @ 40.1 = 40.1
2 Cl @ 35.5 = 71.0
2 H20 @ 18.0 = 36.0
147.1
% water: 36.0
147.1
x 100% =
Hydroscopic – substance that pulls water out of the air
(desiccant packs that come with your shoes and beef jerky)
Ch15 HW#3 13 – 17
Lab15.2 – Water of Hydration
- due in 2 days
- Ch15 HW#3 due at beginning of period
Ch15 HW#3 13 – 17
13. 2 ways to distinguish a suspension from colloid.
14. Why don’t solutions exhibit Tyndall effect?
15b) Magnesium sulfate heptahydrate c) barium hydroxide octahydrate
Mg+2 S04-2 . 7H20
16b) Magnesium carbonate trihydrate
d) Calcium nitrate trihydrate
Ba+2 OH-1 . 8H20
c) Mg3(P04)2.4 H20
e) CoCl2 .2H2O
Ch15 HW#3 13 – 17
13. 2 ways to distinguish a suspension from colloid.
particle size, filtration, ability to stay suspended.
14. Why don’t solutions exhibit Tyndall effect?
15b) Magnesium sulfate heptahydrate c) barium hydroxide octahydrate
Mg+2 S04-2 . 7H20
16b) Magnesium carbonate trihydrate
d) Calcium nitrate trihydrate
Ba+2 OH-1 . 8H20
c) Mg3(P04)2.4 H20
e) CoCl2 .2H2O
Ch15 HW#3 13 – 17
13. 2 ways to distinguish a suspension from colloid.
particle size, filtration, ability to stay suspended.
14. Why don’t solutions exhibit Tyndall effect?
Particles too small to scatter light.
15b) Magnesium sulfate heptahydrate c) barium hydroxide octahydrate
Mg+2 S04-2 . 7H20
16b) Magnesium carbonate trihydrate
d) Calcium nitrate trihydrate
Ba+2 OH-1 . 8H20
c) Mg3(P04)2.4 H20
e) CoCl2 .2H2O
Ch15 HW#3 13 – 17
13. 2 ways to distinguish a suspension from colloid.
particle size, filtration, ability to stay suspended.
14. Why don’t solutions exhibit Tyndall effect?
Particles too small to scatter light.
15b) Magnesium sulfate heptahydrate c) barium hydroxide octahydrate
Mg+2 S04-2 . 7H20
MgSO4.7H20
16b) Magnesium carbonate trihydrate
d) Calcium nitrate trihydrate
Ba+2 OH-1 . 8H20
Ba(OH)2.8H20
c) Mg3(P04)2.4 H20
e) CoCl2 .2H2O
Ch15 HW#3 13 – 17
13. 2 ways to distinguish a suspension from colloid.
particle size, filtration, ability to stay suspended.
14. Why don’t solutions exhibit Tyndall effect?
Particles too small to scatter light.
15b) Magnesium sulfate heptahydrate c) barium hydroxide octahydrate
Mg+2 S04-2 . 7H20
MgSO4.7H20
Ba+2 OH-1 . 8H20
Ba(OH)2.8H20
16b) Magnesium carbonate trihydrate
c) Mg3(P04)2.4 H20
magnesium phosphate
tetrahydrate
Mg+2 C03-2 3H20
MgCO3.3H20
d) Calcium nitrate trihydrate
e) CoCl2 .2H2O
Ch15 HW#3 13 – 17
13. 2 ways to distinguish a suspension from colloid.
particle size, filtration, ability to stay suspended.
14. Why don’t solutions exhibit Tyndall effect?
Particles too small to scatter light.
15b) Magnesium sulfate heptahydrate c) barium hydroxide octahydrate
Mg+2 S04-2 . 7H20
MgSO4.7H20
Ba+2 OH-1 . 8H20
Ba(OH)2.8H20
16b) Magnesium carbonate trihydrate
c) Mg3(P04)2.4 H20
magnesium phosphate
tetrahydrate
Mg+2 C03-2 3H20
MgCO3.3H20
d) Calcium nitrate trihydrate
Ca+2 NO3-1 3H2O
Ca(NO3)2.3H2O
e) CoCl2 .2H2O
Cobalt Chloride dihydrate
Ch16.1 – Properties of Solutions
Solution Formation
The nature of the solute and solvent determines if solute dissolves.
These affect how fast:
1) Amount of surface are in contact.
(only affects rate, not how much dissolves.)
2) Temperatures – faster when hotter.
(increase freq of collisions)
3) Stir it – puts fresh solute in contact with fresh solvent.
(agitation)
Solubility - amount of a substance that will dissolve at a given temp.
36.2g of NaCl dissolve in 100g of H2O @ 25oC. NO MORE!
- put more in, settles at bottom.
Dynamic Equilibrium:
NaCl  Na+(aq) + Cl-(aq)
Unsaturated solution – can dissolve more solute
Saturated solution – contains the max amount of solute
Supersaturated solution – “tricked” into dissolving more than max
(pressure, temp)
Mix 2 liquids:
If they dissolve in each other  miscible
If they are insoluble  immiscible
Factors Affecting Solubility
Solids and Liquids – solubility usually increases with temp
(some exceptions)
Gases dissolved in a liquid (like soda)
– solubility decreases as temp increases
– solubility increases as pressure above solution increases.
Ex1) What is the
solubility of:
a) KBr @ 70˚C: _____g
b) NaClO3 @ 100˚C: _____g
c) NaCl @ 40˚C: _____g
Henry’s Law:
solubility
pressure
S1 S 2

P1 P2
Ex2) If the solubility of a gas is 0.77 g/L at 350 kPa,
what is its solubility at 100 kPa? (Temp held const)
Ch16 HW#1 1 – 8
Lab16.1 – Solubility
- due in 3 days
- Ch16 HW#1 due at beginning of period
Ch16 HW#1
1. 2 components of solution:
- gets dissolved (lesser amount)
- does the dissolving (greater amount)
2. Name 3 factors that influence rate at which solute dissolve
3. What do to:
a) make a saturated soln unsaturated Add more
b) make an unsaturated soln saturated Add more
4. Explain miscible and immiscible
Ch16 HW#1
1. 2 components of solution:
Solute - gets dissolved (lesser amount)
Solvent - does the dissolving (greater amount)
2. Name 3 factors that influence rate at which solute dissolve
3. What do to:
a) make a saturated soln unsaturated Add more
b) make an unsaturated soln saturated Add more
4. Explain miscible and immiscible
Ch16 HW#1
1. 2 components of solution:
Solute - gets dissolved (lesser amount)
Solvent - does the dissolving (greater amount)
2. Name 3 factors that influence rate at which solute dissolve
surface area
heat it
stir it
3. What do to:
a) make a saturated soln unsaturated Add more
b) make an unsaturated soln saturated Add more
4. Explain miscible and immiscible
Ch16 HW#1
1. 2 components of solution:
Solute - gets dissolved (lesser amount)
Solvent - does the dissolving (greater amount)
2. Name 3 factors that influence rate at which solute dissolve
surface area
heat it
stir it
3. What do to:
a) make a saturated soln unsaturated Add more solvent
b) make an unsaturated soln saturated Add more solute
4. Explain miscible and immiscible
Ch16 HW#1
1. 2 components of solution:
Solute - gets dissolved (lesser amount)
Solvent - does the dissolving (greater amount)
2. Name 3 factors that influence rate at which solute dissolve
surface area
heat it
stir it
3. What do to:
a) make a saturated soln unsaturated Add more solvent
b) make an unsaturated soln saturated Add more solute
4. Explain miscible and immiscible
Mix
(likes dissolve likes)
Don’t mix
(polar and nonpolar)
5. How much NaCl can be dissolved in 750g H20 at 25˚c.
36.2g NaCl = ___g NaCl
100g H20
750g H20
6. Based on solids in table 16.4 (& lab) general statement about solubility
& temp.
As temp goes up, solubility …
7. Can a solution with undissolved solute be supersaturated?
8. A gas has solubility in water at 0˚C of 3.6g/L at 100kPa.
What pressure is needed for solubility of 9.5g/L at 0˚C?
S1 = S2
P1 P2
3.6g/L = 9.5g/L
100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.
36.2g NaCl = 271g NaCl
100g H20
750g H20
6. Based on solids in table 16.4 (& lab) general statement about solubility
& temp.
As temp goes up, solubility …
7. Can a solution with undissolved solute be supersaturated?
8. A gas has solubility in water at 0˚C of 3.6g/L at 100kPa.
What pressure is needed for solubility of 9.5g/L at 0˚C?
S1 = S2
P1 P2
3.6g/L = 9.5g/L
100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.
36.2g NaCl = 271g NaCl
100g H20
750g H20
6. Based on solids in table 16.4 (& lab) general statement about solubility
& temp.
As temp goes up, solubility goes up
7. Can a solution with undissolved solute be supersaturated?
8. A gas has solubility in water at 0˚C of 3.6g/L at 100kPa.
What pressure is needed for solubility of 9.5g/L at 0˚C?
S1 = S2
P1 P2
3.6g/L = 9.5g/L
100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.
36.2g NaCl = 271g NaCl
100g H20
750g H20
6. Based on solids in table 16.4 (& lab) general statement about solubility
& temp.
As temp goes up, solubility goes up
7. Can a solution with undissolved solute be supersaturated?
No. The solid solute present will draw the excess dissolved solute
out of the solution.
8. A gas has solubility in water at 0˚C of 3.6g/L at 100kPa.
What pressure is needed for solubility of 9.5g/L at 0˚C?
S1 = S2
P1 P2
3.6g/L = 9.5g/L
100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.
36.2g NaCl = 271g NaCl
100g H20
750g H20
6. Based on solids in table 16.4 (& lab) general statement about solubility
& temp.
As temp goes up, solubility goes up
7. Can a solution with undissolved solute be supersaturated?
No. The solid solute present will draw the excess dissolved solute
out of the solution.
8. A gas has solubility in water at 0˚C of 3.6g/L at 100kPa.
What pressure is needed for solubility of 9.5g/L at 0˚C?
S1 = S2
P1 P2
3.6g/L = 9.5g/L
100kPa P2
P2= 264kPa
Ch16.2 – Molarity
(Give Lab results, graph tomorrow)
A general measure of the concentration of a solution:
Dilute - contains only a small amount of solute
Concentrated -contains a large amount of solute
The best unit to give exact concentration is molarity
Molarity( M ) 
moles of solute
Liters of solution
U nits :
Ex1) Calculate the molarity of a sugar solution if 2 mol of glucose is
added to enough water to give 5 L of solution.
Ex2) A saline solution contains .90 g of NaCl per 100.0 mL of
solution. What is its molarity?
1 Na @ 23.0 = 23.0
1 Cl @ 35.5 = 35.5
58.5 g/mol
Ch16.2 – Molarity (Give Lab results, graph tomorrow)
A general measure of the concentration of a solution:
Dilute - contains only a small amount of solute
Concentrated -contains a large amount of solute
The best unit to give exact concentration is molarity
Molarity( M ) 
moles of solute
Liters of solution
U nits :
Ex1) Calculate the molarity of a sugar solution if 2 mol of glucose is
added to enough water to give 5 L of solution.
2mol glucose
 0.4M
5 L so ln
Ex2) A saline solution contains .90 g of NaCl per 100.0 mL of
solution. What is its molarity?
0.90 gNaCl 1mol NaCl

 0.15 mol L
0.100 L 58 .5 gNaCl
1 Na @ 23.0 = 23.0
1 Cl @ 35.5 = 35.5
58.5 g/mol
Ex3) How many grams of solute are added to make 1.5 L
of 0.2 M Na2SO4 ?
2 Na @ 23.0 = 46.0
1 S @ 32.1 = 32.1
4 O @ 16.0 = 64.0
142.1 g/mol
Ex3) How many grams of solute are added to make 1.5 L
of 0.2 M Na2SO4 ?
1.5L 0.2mol Na 2 SO4 142.1g Na 2 SO4


 43g Na 2 SO4
1L so ln
1mol Na 2 SO4
2 Na @ 23.0 = 46.0
1 S @ 32.1 = 32.1
4 O @ 16.0 = 64.0
142.1 g/mol
HW#11)
a. 1.0 mol of KCl in 750mL of soln.
b. 0.50 mol of MgCl2 in 1.5L of soln.
c. 400g of CuSO4 in 4.00L of soln.
1 Cu @ 63.5 =
1 S @ 32.1 =
4 O @ 16.0 =
g/mol
HW#11)
a. 1.0 mol of KCl in 750mL of soln.
1.0mol KCl
 1 . 3M
0.750 L so ln
b. 0.50 mol of MgCl2 in 1.5L of soln.
c. 400g of CuSO4 in 4.00L of soln.
1 Cu @ 63.5 =
1 S @ 32.1 =
4 O @ 16.0 =
g/mol
HW#11)
a. 1.0 mol of KCl in 750mL of soln.
1.0mol KCl
 1 . 3M
0.750 L so ln
b. 0.50 mol of MgCl2 in 1.5L of soln.
0.5mol MgCl 2
 0.33 mol L
1.5L so ln
c. 400g of CuSO4 in 4.00L of soln.
1 Cu @ 63.5 =
1 S @ 32.1 =
4 O @ 16.0 =
g/mol
HW#11)
a. 1.0 mol of KCl in 750mL of soln.
1.0mol KCl
 1 . 3M
0.750 L so ln
b. 0.50 mol of MgCl2 in 1.5L of soln.
0.5mol MgCl 2
 0.33 mol L
1.5L so ln
c. 400g of CuSO4 in 4.00L of soln.
400 gCuSO4 1molCuSO4

 0.63M
4L
159 .6 gCuSO4
1 Cu @ 63.5 = 63.5
1 S @ 32.1 = 32.1
4 O @ 16.0 = 64.0
159.6 g/mol
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
1.0 L 0.5mol NaCl 58.5g NaCl


 29 g NaCl
1L so ln
1mol NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
1.0 L 0.5mol NaCl 58.5g NaCl


 29 g NaCl
1L so ln
1mol NaCl
b) 500mL of 2.0M KNO3
0.500 L 2mol KNO3 101.1g KNO3


 101 .1g KNO3
1L so ln
1mol KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
1.0 L 0.5mol NaCl 58.5g NaCl


 29 g NaCl
1L so ln
1mol NaCl
b) 500mL of 2.0M KNO3
0.500 L 2mol KNO3 101.1g KNO3


 101 .1g KNO3
1L so ln
1mol KNO3
c) 250mL of 0.10M CaCl2
0.250 L 0.1mol CaCl 2 111.1g CaCl 2


 2.8 g CaCl 2
1L so ln
1mol CaCl 2
Ch16 HW#2 9 – 14
9. Molarity is more meaningful than the words dilute or concentrated. Why?
10. A salt has a volume of 250mL and contains 0.70mol of NaCl. Molarity?
12. Aqueous soln that has a volume of 2.0L and contains 36.0g of glucose.
If the gfm of glucose is 180.0g, molarity?
13. How many moles of solute are in 250mL of 2.0M CaCl2? Grams?
14. Calc moles & grams
c) 250mL of 0.10M CaCl2
Ch16 HW#2 9 – 14
9. Molarity is more meaningful than the words dilute or concentrated. Why?
10. A salt has a volume of 250mL and contains 0.70mol of NaCl. Molarity?
0.70 mol NaCl
 2. 8 M
0.250 L so ln
12. Aqueous soln that has a volume of 2.0L and contains 36.0g of glucose.
If the gfm of glucose is 180.0g, molarity?
13. How many moles of solute are in 250mL of 2.0M CaCl2? Grams?
14. Calc moles & grams
c) 250mL of 0.10M CaCl2
Ch16 HW#2 9 – 14
9. Molarity is more meaningful than the words dilute or concentrated. Why?
10. A salt has a volume of 250mL and contains 0.70mol of NaCl. Molarity?
0.70 mol NaCl
 2. 8 M
0.250 L so ln
12. Aqueous soln that has a volume of 2.0L and contains 36.0g of glucose.
If the gfm of glucose is 180.0g, molarity?
36 .0 g glucose 1mol glucose

 0.10 mol L
2.0 L
180 .0 g glucose
13. How many moles of solute are in 250mL of 2.0M CaCl2? Grams?
14. Calc moles & grams
c) 250mL of 0.10M CaCl2
Ch16 HW#2 9 – 14
9. Molarity is more meaningful than the words dilute or concentrated. Why?
10. A salt has a volume of 250mL and contains 0.70mol of NaCl. Molarity?
0.70 mol NaCl
 2. 8 M
0.250 L so ln
12. Aqueous soln that has a volume of 2.0L and contains 36.0g of glucose.
If the gfm of glucose is 180.0g, molarity?
36 .0 g glucose 1mol glucose

 0.10 mol L
2.0 L
180 .0 g glucose
13. How many moles of solute are in 250mL of 2.0M CaCl2? Grams?
2.0mol CaCl 2 .250 mol

 0.50 mol CaCl 2
1.0 L
0.50 mol CaCl 2 111 .1g CaCl 2
 56 g
1mol CaCl 2
14. Calc moles & grams
c) 250mL of 0.10M CaCl2
HW#14) (IC?) Calc moles and grams
a) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
1.0 L 0.5mol NaCl 58.5g NaCl


 29 g NaCl
1L so ln
1mol NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
1.0 L 0.5mol NaCl 58.5g NaCl


 29 g NaCl
1L so ln
1mol NaCl
b) 500mL of 2.0M KNO3
0.500 L 2mol KNO3 101.1g KNO3


 101 .1g KNO3
1L so ln
1mol KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and grams
a) 1.0L of 0.50M NaCl
1.0 L 0.5mol NaCl 58.5g NaCl


 29 g NaCl
1L so ln
1mol NaCl
b) 500mL of 2.0M KNO3
0.500 L 2mol KNO3 101.1g KNO3


 101 .1g KNO3
1L so ln
1mol KNO3
c) 250mL of 0.10M CaCl2
0.250 L 0.1mol CaCl 2 111.1g CaCl 2


 2.8 g CaCl 2
1L so ln
1mol CaCl 2
Ch16.2 – Molarity cont
Ex1) 2.23 moles of KCl are dissolved in 20.1 mLs of water.
What is its molarity?
Ex2) 36.2 of NaCl can be dissolved in 100.0 g of water when
at 25˚C. What is its molarity?
Ex3) In lab, we need a 6.0 M NaOH solution. To fill the dropper bottles,
I make up 250 mL. How many grams of NaOH are needed?
Lab16.1 – Solubility
Solubility g/100g H2O
100
90
80
70
60
50
40
30
20
10
10
20
30
Temp
Actual Temp
Solubility
Previous Yrs
40
20
50 60
Temp (˚C)
30
40
70
80
90
100
50
60
70
80
90
100
Ch16 HW#3 15 – 20
15) A solution of Na3Po4 contains 0.75 mol of solute in 950 mL of
solution molarity?
16) Sodium chloride 501 N is prepared by dissolving 30.05 g NaCl in
1200 mL of solution molarity?
17) Molarity of 501 N contains 100.1 g potassium chloride dissolved
in 2.5 L of 501 N.
Ch16 HW#3 15 – 20
15) A solution of Na3Po4 contains 0.75 mol of solute in 950 mL of
solution molarity?
0.75mol
 .79M
.950 L soln
16) Sodium chloride 501 N is prepared by dissolving 30.05 g NaCl in
1200 mL of solution molarity?
17) Molarity of 501 N contains 100.1 g potassium chloride dissolved
in 2.5 L of 501 N.
Ch16 HW#3 15 – 20
15) A solution of Na3Po4 contains 0.75 mol of solute in 950 mL of
solution molarity?
0.75mol
 .79M
.950 L soln
16) Sodium chloride 501 N is prepared by dissolving 30.05 g NaCl in
1200 mL of solution molarity?
30.05 g NaCl 1mol NaCl

 .43M
1.2 L soln
58.5 g NaCl
17) Molarity of 501 N contains 100.1 g potassium chloride dissolved
in 2.5 L of 501 N.
Ch16 HW#3 15 – 20
15) A solution of Na3Po4 contains 0.75 mol of solute in 950 mL of
solution molarity?
0.75mol
 .79M
.950 L soln
16) Sodium chloride 501 N is prepared by dissolving 30.05 g NaCl in
1200 mL of solution molarity?
30.05 g NaCl 1mol NaCl

 .43M
1.2 L soln
58.5 g NaCl
17) Molarity of 501 N contains 100.1 g potassium chloride dissolved
in 2.5 L of 501 N.
212.5 g NaNO3 1mol NaNO3

 .83M
3.0 L
85.1g NaNO3
18) Molarity of soln containing 100.1g potassium chloride dissolved in
2.5L of soln.
19) How many moles of lithium bromide must be dissolved in 500mL
of solution to prepare 1.0M soln.
20) What mass of sucrose C12H22O11 is needed to make 300mL of a
0.5M solution?
18) Molarity of soln containing 100.1g potassium chloride dissolved in
2.5L of soln.
100.1g KC 1 mol KCl

 .537 M
2.5L
74.6g KCl
19) How many moles of lithium bromide must be dissolved in 500mL
of solution to prepare 1.0M soln.
20) What mass of sucrose C12H22O11 is needed to make 300mL of a
0.5M solution?
18) Molarity of soln containing 100.1g potassium chloride dissolved in
2.5L of soln.
100.1g KC 1 mol KCl

 .537 M
2.5L
74.6g KCl
19) How many moles of lithium bromide must be dissolved in 500mL
of solution to prepare 1.0M soln.
1.0 mol LiBr .5 L soln .5 mol LiBr 86.8LiBr



 43.4 g
1 L soln
1 mol
20) What mass of sucrose C12H22O11 is needed to make 300mL of a
0.5M solution?
18) Molarity of soln containing 100.1g potassium chloride dissolved in
2.5L of soln.
100.1g KC 1 mol KCl

 .537 M
2.5L
74.6g KCl
19) How many moles of lithium bromide must be dissolved in 500mL
of solution to prepare 1.0M soln.
1.0 mol LiBr .5 L soln .5 mol LiBr 86.8LiBr



 43.4 g
1 L soln
1 mol
20) What mass of sucrose C12H22O11 is needed to make 300mL of a
0.5M solution?
.5 mol suc .300 L .15 mol suc 342g suc



 51.3g suc
1 L soln
1 mol
Ch16.3 – Dilutions
- sometimes it’s useful to dilute a stock solution
The number of moles of solute does not change when a solution is diluted.
Ex1) How would you prepare 100mL of 0.40M MgSO4 from a stock soln
of 2.0M MgSO4?
HW#22) You need 250mL of 0.20M NaCl but you only have a solution
of 1.00M NaCl. How do you prepare it?
Ch16.3 – Dilutions
- sometimes it’s useful to dilute a stock solution
The number of moles of solute does not change when a solution is diluted.
Ex1) How would you prepare 100mL of 0.40M MgSO4 from a stock soln
of 2.0M MgSO4?
V2=100mL
M2=.40M
M1=2.0M
V1=?
M1V1 = M2V2
(2.0)(V1) = (.40)(100)
V1 = 20mL
HW#22) You need 250mL of 0.20M NaCl but you only have a solution
of 1.00M NaCl. How do you prepare it?
Ch16.3 – Dilutions
- sometimes it’s useful to dilute a stock solution
The number of moles of solute does not change when a solution is diluted.
Ex1) How would you prepare 100mL of 0.40M MgSO4 from a stock soln
of 2.0M MgSO4?
V2=100mL
M2=.40M
M1=2.0M
V1=?
M1V1 = M2V2
(2.0)(V1) = (.40)(100)
V1 = 20mL
HW#22) You need 250mL of 0.20M NaCl but you only have a solution
of 1.00M NaCl. How do you prepare it?
V2=250mL  .250L
M2=.20M
M1=1.00M
V1=?
M1V1 = M2V2
(1.00)(V1) = (.2)(.250)
V1 = .05L
Ch16 Quiz on Gas Laws
Review:
1) The volume of a gas-filled balloon is 1.0L when at a temp of 20oC
and 200kPa. Correct this volume to STP.
Ch16 Quiz on Gas Laws
Review:
1) The volume of a gas-filled balloon is 1.0L when at a temp of 20oC
and 200kPa. Correct this volume to STP.
V1=1.0L
V2=?
1 1
2 2
T1=20oC293K
T2=0oC273K
P1=200kPa
P2=101.3kPa
1
2
VP V P

T
T
(1)(200) V2  (101.3)

(293)
(273)
.683 V2  .371

.371
.371
V2  1.84L
Ex2) How many moles of gas are contained in a 2.0L balloon at 15oC
and 175kPa?
Ex3) What is the volume of 3.0g of oxygen gas at STP?
Ch16 HW#4
Ch16 HW#4
21. Experiment requires: 5 mL of 1 M KOH
Stock room has: 1 L of .5 M KOH
Can diluting the stock solution prepare you for the lab?
23. Stock solution: 2.0 M NaCl
How would you prepare 50 mL of .5 M NaCl?
M1 = 2.0M
V1 = ?
M2 = .50 M
V2 = 500mL
Ch16 HW#4
21. Experiment requires: 5 mL of 1 M KOH
Stock room has: 1 L of .5 M KOH
Can diluting the stock solution prepare you for the lab?
23. Stock solution: 2.0 M NaCl
How would you prepare 50 mL of .5 M NaCl?
M1 = 2.0M
V1 = ?
M2 = .50 M
V2 = 500mL
M1V1= M2V2
(2M)(V1) = (.5M)(500mL)
V1 = 125 mL
Ch16.5 Colligative Properties of Solutions
- depend on the # of particles in solution, NOT the type!
1. Vapor Pressure
- adding nonvolatile ( Don’t evaporate easily) solute
lowers the vapor pressure.
Because fewer of the volatile (do evap easily) solvent
particles are at the surface of the solution to evaporate.
Before Adding:
After Adding:
- you may recall that a liquid boils when it vapor pressure equals
atmospheric pressure.
The nonvolatile solute lowered v.p. = higher temp needed
to cause boiling.
2) Boiling Point Elevation
- B.P. of water is raised 0.512oC for every mole of nonvolatile
solute added to 1000g of water.
 0.512  C 


 1molal 


3) Freezing Point Depression
- F.P. lowers (for water) by 1.86oC for every mole of nonvolatile
solute added to 1000g of water.
 1.86  C 

(disrupts orderly arrangement in solid)  

 1molal 
All 3 of these colligative porperties are affected by the number of particles
in solution, not the type of particles.
Water wouldn’t care if you added NaCl or KCl.
Both contribute the 2 moles of ions to the solution.
NaCl(s)  Na+(aq) + Cl–(aq)
KCl(s)  K+(aq) + Cl–(aq)
Water would care if you added CaCl2 instead of NaCl,
because CaCl2 adds 3 moles of ions instead of 2!
CaCl2(s)  Ca+2(aq) + 2Cl–(aq)
Lab16.2 Molarity
All Red #’s are unique to your group!
1. Mass of batman & Robin
50.00 g
2. Mass of b,r,& sodium chloride dry
55.00 g
3. Mass of NaCI
5.00 g
Cales
1. # moles NaCI
5.00 g NaCI
1mol NaCI = .085 mol NaCI
58.5 g NaCI
2. Molarity
Exp: .085 mol NaCI = 17.1 M
.005 L
3. Accepted Molarity:
36.2g NaCI dissolved in 100mL soln (from
36.2 g NaCI
1 mol NaCI
= 6.2M
solubility)
.100L
58.5g NaCI
4. % error
accepted - exp
x 100% = 6.2 – 17.1 x 100% = 176%!
accepted
6.2
Questions
1. 24.03g sodium sulfate in 200 mL soln.
2. 10.35 g potassium chloride in 125 mL soln.
3. 5.00 g NaCI in 20.0 mL. Saturated?
Questions
1. 24.03g sodium sulfate in 200 mL soln.
Na+ S04-2
24.03 g Na2S04 1 mol Na2S04
.200L
___ g Na2S04
=
2. 10.35 g potassium chloride in 125 mL soln.
K+
CI-
3. 5.00 g NaCI in 20.0 mL. Saturated?
(Find molarity) M < 6.2 unsaturated
M = 6.2 saturated
M > 6.2 supersaturated
Do Lab16.2 questions for HW
Lab16.2 – Molarity of a Saturated Solution
- due tomorrow
Ch16.4 – Molality and Mole Fraction
- These are more common units for colligate properties
 moles 
moles of solute

Molality 
 
kilograms of solvent  kg 
Ex1) What is the molality when 2.5 mol of glucose is dissolved
in 1000g of water? What is molarity?
Ch16.4 – Molality and Mole Fraction
- These are more common units for colligate properties
 moles 
moles of solute

Molality 
 
kilograms of solvent  kg 
Ex1) What is the molality when 2.5 mol of glucose is dissolved
in 1000g of water? What is molarity?
2.5 mol
Molality 
 2.5m
1.0kg
2.5 mol
Molarity 
 2.5M
1.0L
Ex2) What is the molality when 29.25g of NaCl is dissolved
in 1.00kg of water?
Ex3) How many grams of potassium iodide must be dissolved in 500g
of water to produce a 0.600 molal KI solution?
Ex2) What is the molality when 29.25g of NaCl is dissolved
in 1.00kg of water?
29.25g NaCl 1 mol NaCl
 .500m
1.00 kg H 2 O 58.5g NaCl
Ex3) How many grams of potassium iodide must be dissolved in 500g
of water to produce a 0.600 molal KI solution?
.600 mol KI .500kg solvent 166.0g KI
1.0 kg solvent
1 mol KI
 49 .8 g KI
Mole Fraction
- the ratio of the moles of solute to the total # of moles in solution
Mole Fraction
NA
XA 
NA  NB
moles of one
moles total
Ex4) Compute the mole fraction of each component in a solution
of 1.25 mol ethylene glycol and 4.00 mol water.
(coolant)
Mole Fraction
Ch16 HW#5 24-28
Mole Fraction
- the ratio of the moles of solute to the total # of moles in solution
Mole Fraction
NA
XA 
NA  NB
moles of one
moles total
Ex4) Compute the mole fraction of each component in a solution
of 1.25 mol ethylene glycol and 4.00 mol water.
(coolant)
Mole Fraction
1.25 mol
X e.c. 
 .24
5.25 mol
4.00 mol
X water 
 .76
5.25 mol
Ch16 HW#5 24-28
Ch16 HW#5 24-28
24) Distinguish between 1 M and 1m solutions.
25) Find molality of a solution of 10g NaCl in 600g water.
26) Molality of solution of 5.0g calcium chloride in 200.0g water.
Ch16 HW#5 24-28
24) Distinguish between 1 M and 1m solutions.
1 mol solute
1M 
1 L solution
1 mol solute
1m 
1 kg solvent
25) Find molality of a solution of 10g NaCl in 600g water.
26) Molality of solution of 5.0g calcium chloride in 200.0g water.
Ch16 HW#5 24-28
24) Distinguish between 1 M and 1m solutions.
1 mol solute
1M 
1 L solution
1 mol solute
1m 
1 kg solvent
25) Find molality of a solution of 10g NaCl in 600g water.
10g NaCl
.6 kg
1 mol NaC
 .28 m
58.5g NaC
26) Molality of solution of 5.0g calcium chloride in 200.0g water.
Ch16 HW#5 24-28
24) Distinguish between 1 M and 1m solutions.
1 mol solute
1M 
1 L solution
1 mol solute
1m 
1 kg solvent
25) Find molality of a solution of 10g NaCl in 600g water.
10g NaCl
.6 kg
1 mol NaC
 .28 m
58.5g NaC
26) Molality of solution of 5.0g calcium chloride in 200.0g water.
5.0g CaCl2 1 mol CaCl2
 .23m
.200 kg
111.1g CaCl2
27) How many kg of water must be added to 9.0g oxalic acid, H2C2O4,
to prepare a 0.025 molal solution.
28) Would a dilute or concentrated sodium fluoride solution
have a higher boiling point?
27) How many kg of water must be added to 9.0g oxalic acid, H2C2O4,
to prepare a 0.025 molal solution.
9.0g H 2 C 2 O 4 1 mol H 2 C 2 O 4 1kg solvent
90.0 kg H 2 C 2 O 4 .25 mol H 2 C 2 O 4
 4kg H 2 C 2 O 4
28) Would a dilute or concentrated sodium fluoride solution
have a higher boiling point?
27) How many kg of water must be added to 9.0g oxalic acid, H2C2O4,
to prepare a 0.025 molal solution.
9.0g H 2 C 2 O 4 1 mol H 2 C 2 O 4 1kg solvent
90.0 kg H 2 C 2 O 4 .25 mol H 2 C 2 O 4
 4kg H 2 C 2 O 4
28) Would a dilute or concentrated sodium fluoride solution
have a higher boiling point?
Concentrated
In H2O, NaF breaks up as Na+ ions and F- ions,
they get in the way of the H2O trying to vaporize (boil).
H2O requires higher temp to boil.
More of them = higher B.P. (& lower M.P.)
Ch15,16 Rev
1. Which dissolve in water?
a. CH4
b. KCl
c. He
d. MgSO4
e. NaHCO3
2. Why does molten sodium chloride conduct electricity?
3. Name or write the formula
a. Na2B4O7.10H2O
c. MgSO4.7H2O
b. Na2CO3.H2O
4. A gas has solubility in water at 10˚C of 4.5g/L at 932kPa.
find the solubility at 638 kPa.
S1 = S2
P1 P2
5. Calc the molarity of 0.600g of NaHCO3 in 1500mL of solution.
6. Stock solution: 4.0M KNO3 . What volume must you dilute to make
50mL of 0.20M KNO3 solution?