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Ch15.1 – Water and Aqueous SolutionsO 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 Add soap (surfactants-surface active agents)it stick to its self, defies gravity interfere with hydrogen bonding, bead
spreads out.
coin coin
Ch15.1 – Water and Aqueous SolutionsO 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 Add soap (surfactants-surface active agents)to its self, defies gravity interfere with hydrogen bonding, bead
spreads out.
All liquids have some amount of surface tension.
coin coin
Heat CapacityWater has a high specific heat capacity(ability to absorb
energy without changing temp)
- Instead of energy going in and increasing average K.E. (molecular motion), causes bonds to bend, twist, wiggle, stretch.
Cg
JC p
18.4
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?
Ex) How much heat energy is required to vaporize 2.0g of water?2.0g 2260J = 4520J
1g
Ammonia – HV = 1370 J/gMethane – HV = 500 J/gThey 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/gMethane – HV = 500 J/gThey 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)
Evaporation - water has a large heat of vaporization - 2260J/g
Strong dipole forces, require lots of energy to break free.
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
Ch15 HW#1 1-61. 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?
6. What unique characteristic of ice distinguishes it from other solids?
kJJOgH
JOgH3.11or 300,11
0.1
2260
1
0.5
2
2
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?
Floats in its own liquid
kJJOgH
JOgH3.11or 300,11
0.1
2260
1
0.5
2
2
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 Review1) 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 = 12.36g2. Mass of glycerin & cylinder = 24.74g3. 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 Review1) 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 Mass of glycerin = 12.38g2. Mass of glycerin & cylinder = 24.74g3. Volume of glycerin = 9.8mL 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 = 9.04g 1cm3
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 – 127) 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 – 127) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solventAqueous 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 – 127) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solventAqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.Solute – NaClSolvent – 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 – 127) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solventAqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.Solute – NaClSolvent – 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 – 127) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solventAqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.Solute – NaClSolvent – 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 nof) CaCO3 g) Gasoline ionic nonpolar (not very soluble) no11) Electrolyte and nonelectrolyte
12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 127) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solventAqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.Solute – NaClSolvent – 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 nof) CaCO3 g) Gasoline ionic nonpolar (not very soluble) no11) Electrolyte and nonelectrolyte ions in solution no ions in solution conducts electricity does not conduct electricity12) Equation for how calcium chloride dissociates in water.
Ch15 HW#2 7 – 127) Distinguish between a solution and an aqueous solution
Solution – solute dissolved in a liquid (or gas) solventAqueous solution – solute dissolved in water
8) Tablespoon of NaCl in water. ID solute & solvent.Solute – NaClSolvent – 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 nof) CaCO3 g) Gasoline ionic nonpolar (not very soluble) no11) Electrolyte and nonelectrolyte ions in solution no ions in solution conducts electricity does not conduct electricity12) Equation for how calcium chloride dissociates in water.
CaCl2(s) Ca(aq)+2 + 2Cl(aq)
–
Ch15.3 – Suspensions, Colloids, MixturesSolutions - 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 mixtureLarge particle size (over 100nm)Scatters light (Tyndall effect)Particles separate (sediment forms)Particles can be filtered Ex: Muddy water
Colloids - Mixture is in-betweenMedium particle sizeScatters lightParticles don’t separate from liquidParticles cant be filtered
Exs: smoke, milk, marshmallows, gelatin, paint, aerosol sprays, whipped cream
Emulsions - colloidal dispersions of liquid in liquidsEx: 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 decahydrateEx 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 decahydrateEx 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 decahydrateEx 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 CaCl2.2H2O1 Ca @ 40.1 = 40.12 Cl @ 35.5 = 71.0 % water: 36.0 x 100% = 2 H20 @ 18.0 = 36.0 147.1
147.1
Hydroscopic – substance that pulls water out of the air(desiccant packs that come with your shoes and beef jerky)
Ch15 HW#3 13 – 17
Ch15 HW#3 13 – 1713. 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 Ba+2 OH-1 . 8H20
16b) Magnesium carbonate trihydrate c) Mg3(P04)2.4 H20
d) Calcium nitrate trihydrate e) CoCl2 .2H2O
Ch15 HW#3 13 – 1713. 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 Ba+2 OH-1 . 8H20
16b) Magnesium carbonate trihydrate c) Mg3(P04)2.4 H20
d) Calcium nitrate trihydrate e) CoCl2 .2H2O
Ch15 HW#3 13 – 1713. 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 Ba+2 OH-1 . 8H20
16b) Magnesium carbonate trihydrate c) Mg3(P04)2.4 H20
d) Calcium nitrate trihydrate e) CoCl2 .2H2O
Ch15 HW#3 13 – 1713. 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 Ba+2 OH-1 . 8H20
MgSO4.7H20 Ba(OH)2
.8H20
16b) Magnesium carbonate trihydrate c) Mg3(P04)2.4 H20
d) Calcium nitrate trihydrate e) CoCl2 .2H2O
Ch15 HW#3 13 – 1713. 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 Ba+2 OH-1 . 8H20
MgSO4.7H20 Ba(OH)2
.8H20
16b) Magnesium carbonate trihydrate c) Mg3(P04)2.4 H20
Mg+2 C03-2 3H20 magnesium phosphate
MgCO3.3H20 tetrahydrate
d) Calcium nitrate trihydrate e) CoCl2 .2H2O
Ch15 HW#3 13 – 1713. 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 Ba+2 OH-1 . 8H20
MgSO4.7H20 Ba(OH)2
.8H20
16b) Magnesium carbonate trihydrate c) Mg3(P04)2.4 H20
Mg+2 C03-2 3H20 magnesium phosphate
MgCO3.3H20 tetrahydrate
d) Calcium nitrate trihydrate e) CoCl2 .2H2O Ca+2 NO3
-1 3H2O Cobalt Chloride dihydrate
Ca(NO3)2.3H2O
Ch16.1 – Properties of Solutions
Solution FormationThe 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 soluteSaturated 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 miscibleIf they are insoluble immiscible
Factors Affecting SolubilitySolids 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
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
2
2
1
1
P
S
P
S
Ch16 HW#11. 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#11. 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#11. 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 dissolvesurface areaheat itstir 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#11. 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 dissolvesurface areaheat itstir 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#11. 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 dissolvesurface areaheat itstir 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 Don’t mix(likes dissolve likes) (polar and nonpolar)
5. How much NaCl can be dissolved in 750g H20 at 25˚c.36.2g NaCl = ___g NaCl100g 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 3.6g/L = 9.5g/LP1 P2 100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.36.2g NaCl = 271g NaCl100g 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 3.6g/L = 9.5g/LP1 P2 100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.36.2g NaCl = 271g NaCl100g 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 3.6g/L = 9.5g/LP1 P2 100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.36.2g NaCl = 271g NaCl100g 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 3.6g/L = 9.5g/LP1 P2 100kPa P2
P2=
5. How much NaCl can be dissolved in 750g H20 at 25˚c.36.2g NaCl = 271g NaCl100g 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 3.6g/L = 9.5g/LP1 P2 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 soluteConcentrated -contains a large amount of solute
The best unit to give exact concentration is molarity
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 = 23.01 Cl @ 35.5 = 35.5
58.5 g/mol
:nits U
)(
solutionofLiters
soluteofmolesMMolarity
Ch16.2 – Molarity (Give Lab results, graph tomorrow)A general measure of the concentration of a solution:
Dilute - contains only a small amount of soluteConcentrated -contains a large amount of solute
The best unit to give exact concentration is molarity
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 = 23.01 Cl @ 35.5 = 35.5
58.5 g/mol
:nits U
)(
solutionofLiters
soluteofmolesMMolarity
0.4M ln 5
glucose 2
soL
mol
Lmol0.15
5.58
NaCl 1
100.0
90.0
gNaCl
mol
L
gNaCl
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 ?
2 Na @ 23.0 = 46.0 1 S @ 32.1 = 32.1
4 O @ 16.0 = 64.0 142.1 g/mol
4242
4242 Na 34 Na 1mol
Na 142.1g
ln 1
Na 2.05.1SOg
SO
SO
soL
SOmolL
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.
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
MsoL
mol3.1
ln 750.0
KCl 0.1
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
MsoL
mol3.1
ln 750.0
KCl 0.1
Lmol
soL
mol33.0
ln 5.1
MgCl 5.0 2
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 = 63.5 1 S @ 32.1 = 32.1
4 O @ 16.0 = 64.0 159.6 g/mol
MsoL
mol3.1
ln 750.0
KCl 0.1
Lmol
soL
mol33.0
ln 5.1
MgCl 5.0 2
0.63M 6.159
1
4
400
4
44 gCuSO
molCuSO
L
gCuSO
HW#14) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
NaCl 29 NaCl 1mol
NaCl 58.5g
ln 1
NaCl 5.00.1g
soL
molL
HW#14) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
NaCl 29 NaCl 1mol
NaCl 58.5g
ln 1
NaCl 5.00.1g
soL
molL
33
33 KN 1.101 KN 1mol
KN 101.1g
ln 1
KN 2500.0Og
O
O
soL
OmolL
HW#14) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
NaCl 29 NaCl 1mol
NaCl 58.5g
ln 1
NaCl 5.00.1g
soL
molL
33
33 KN 1.101 KN 1mol
KN 101.1g
ln 1
KN 2500.0Og
O
O
soL
OmolL
22
22 CaCl 8.2 CaCl 1mol
CaCl 111.1g
ln 1
CaCl 1.0250.0g
soL
molL
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?
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
MsoL
mol8.2
ln 250.0
NaCl 70.0
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
MsoL
mol8.2
ln 250.0
NaCl 70.0
Lmol0.10
glucose 0.180
glucose 1
0.2
glucose 0.36
g
mol
L
g
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
MsoL
mol8.2
ln 250.0
NaCl 70.0
Lmol0.10
glucose 0.180
glucose 1
0.2
glucose 0.36
g
mol
L
g
gmol
gmol
molmol
L
mol
56CaCl 1
CaCl 1.111CaCl 50.0
CaCl 0.50 250.
0.1
CaCl 0.2
2
22
22
HW#14) (IC?) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
HW#14) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
NaCl 29 NaCl 1mol
NaCl 58.5g
ln 1
NaCl 5.00.1g
soL
molL
HW#14) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
NaCl 29 NaCl 1mol
NaCl 58.5g
ln 1
NaCl 5.00.1g
soL
molL
33
33 KN 1.101 KN 1mol
KN 101.1g
ln 1
KN 2500.0Og
O
O
soL
OmolL
HW#14) Calc moles and gramsa) 1.0L of 0.50M NaCl
b) 500mL of 2.0M KNO3
c) 250mL of 0.10M CaCl2
NaCl 29 NaCl 1mol
NaCl 58.5g
ln 1
NaCl 5.00.1g
soL
molL
33
33 KN 1.101 KN 1mol
KN 101.1g
ln 1
KN 2500.0Og
O
O
soL
OmolL
22
22 CaCl 8.2 CaCl 1mol
CaCl 111.1g
ln 1
CaCl 1.0250.0g
soL
molL
Ch16.2 – Molarity contEx1) 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
100 90 80 70 60 50 40 30 20 10 10 20 30 40 50 60 70 80 90 100
Temp (˚C)
So
lub
ilit
y g
/100
g H
2O
Temp 20 30 40 50 60 70 80 90 100
Actual Temp
Solubility
Previous Yrs
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?
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.
ML
mol79.
soln 950.
75.0
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.
ML
mol79.
soln 950.
75.0
Mg
mol
L
g43.
NaCl 5.58
NaCl 1
soln 2.1
NaCl 05.30
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.
ML
mol79.
soln 950.
75.0
Mg
mol
L
g43.
NaCl 5.58
NaCl 1
soln 2.1
NaCl 05.30
Mg
mol
L
g83.
NaNO3 1.85
NaNO3 1
0.3
NaNO3 5.212
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.
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?
M537.KCl 74.6g
KCl mol 1
2.5L
KC 100.1g
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?
M537.KCl 74.6g
KCl mol 1
2.5L
KC 100.1g
g 43.4mol 1
86.8LiBrLiBr mol 5.soln L .5
soln L 1
LiBr mol 1.0
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?
M537.KCl 74.6g
KCl mol 1
2.5L
KC 100.1g
suc 3.51mol 1
suc 342gsuc mol 15.L .300
soln L 1
suc mol .5g
g 43.4mol 1
86.8LiBrLiBr mol 5.soln L .5
soln L 1
LiBr mol 1.0
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 M1V1 = M2V2
M2=.40M (2.0)(V1) = (.40)(100)M1=2.0M V1 = 20mLV1=?
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 M1V1 = M2V2
M2=.40M (2.0)(V1) = (.40)(100)M1=2.0M V1 = 20mLV1=?
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 M1V1 = M2V2
M2=.20M (1.00)(V1) = (.2)(.250)M1=1.00M V1 = .05L V1=?
Ch16 Quiz on Gas LawsReview: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 LawsReview: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=?T1=20oC293K T2=0oC273KP1=200kPa P2=101.3kPa 2
22
1
11
T
V
T
V PP
(273)
)3.101(
(293)
(1)(200) 2 V
371.
371.
.371
.683 2 V
LV 84.12
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#421. 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 NaClHow would you prepare 50 mL of .5 M NaCl?
M1 = 2.0MV1 = ?M2 = .50 MV2 = 500mL
Ch16 HW#421. 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 NaClHow would you prepare 50 mL of .5 M NaCl?
M1 = 2.0MV1 = ?M2 = .50 MV2 = 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) solventparticles 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.
3) Freezing Point Depression- F.P. lowers (for water) by 1.86oC for every mole of nonvolatile
solute added to 1000g of water. (disrupts orderly arrangement in solid)
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)
molal
C
1
512.0
molal
C
1
86.1
Lab16.2 Molarity All Red #’s are unique to your group!1. Mass of batman & Robin 50.00 g2. Mass of b,r,& sodium chloride dry 55.00 g3. Mass of NaCI 5.00 gCales1. # moles NaCI
5.00 g NaCI 1mol NaCI = .085 mol NaCI58.5 g NaCI
2. Molarity Exp: .085 mol NaCI = 17.1 M
.005 L3. 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 6.2Questions1. 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?
Questions1. 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
Ch16.4 – Molality and Mole Fraction- These are more common units for colligate properties
Ex1) What is the molality when 2.5 mol of glucose is dissolved in 1000g of water? What is molarity?
kg
moles
solvent of kilograms
solute of molesMolality
Ch16.4 – Molality and Mole Fraction- These are more common units for colligate properties
Ex1) What is the molality when 2.5 mol of glucose is dissolved in 1000g of water? What is molarity?
kg
moles
solvent of kilograms
solute of molesMolality
m5.21.0kg
mol 2.5Molality
M5.21.0L
mol 2.5Molarity
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?
Ex3) How many grams of potassium iodide must be dissolved in 500g of water to produce a 0.600 molal KI solution?
m500.NaCl 58.5g OH kg 1.00
NaCl mol 1 NaCl g25.29
2
KI g 8.49
KI mol 1 solvent kg 1.0
KI 166.0gsolvent .500kg KI mol .600
Mole Fraction- the ratio of the moles of solute to the total # of moles in solution
Mole Fraction
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
BA
AA N N
N X
totalmoles
one of moles
Ch16 HW#5 24-28
Mole Fraction- the ratio of the moles of solute to the total # of moles in solution
Mole Fraction
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
BA
AA N N
N X
totalmoles
one of moles
24.mol 5.25
mol 1.25 Xe.c.
76.mol 5.25
mol 4.00 Xwater
Ch16 HW#5 24-28
Ch16 HW#5 24-2824) 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 24-2824) 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.
solution L 1
solute mol 1 1M
solvent kg 1
solute mol 1 1m
Ch16 HW#5 24-2824) 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.
solution L 1
solute mol 1 1M
solvent kg 1
solute mol 1 1m
m28. NaC 58.5g kg .6
NaC mol 1 NaCl 10g
Ch16 HW#5 24-2824) 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.
solution L 1
solute mol 1 1M
solvent kg 1
solute mol 1 1m
m28. NaC 58.5g kg .6
NaC mol 1 NaCl 10g
m23.CaCl 111.1g kg .200
CaCl mol 1 CaCl 5.0g
2
22
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.
28) Would a dilute or concentrated sodium fluoride solution have a higher boiling point?
OCH g4
OCH mol .25 OCH kg 90.0
solvent 1kg OCH mol 1 OCH 9.0g
422
422422
422422
k
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?ConcentratedIn 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.)
OCH g4
OCH mol .25 OCH kg 90.0
solvent 1kg OCH mol 1 OCH 9.0g
422
422422
422422
k
Ch15,16 Rev1. 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 b. Na2CO3
.H2O
c. MgSO4.7H2O