Ch15.1 – Water and Aqueous Solutions Ex1) Draw H 2 O structure shape: bond angle: polar:

Ch15.1 – Water and Aqueous SolutionsEx1) Draw H2O structure

shape:bond angle:polar:

Ch15.1 – Water and Aqueous SolutionsO shape bent

H H bond 105˚ O O polar yesH 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 yesH 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.

CgJC 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



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)



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 1-61. 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?



kJJOgHJOgH

3.11or 300,110.12260

10.5

2

2



Floats in its own liquid

kJJOgHJOgH

3.11or 300,110.12260

10.5

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.


Solution – solute dissolved in a liquid (or gas) solventAqueous solution – solute dissolved in water

8) Tablespoon of NaCl in water. ID solute & solvent.








8) Tablespoon of NaCl in water. ID solute & solvent.Solute – NaClSolvent – water



















polar cov ionic polar cov ionic nonpolar yes yes yes yes nof) CaCO3 g) Gasoline ionic nonpolar (not very soluble) no11) Electrolyte and nonelectrolyte







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.






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



sodium borate decahydrateHW#16a) Name Na2CO8

.10H2Osodium carbonate decahydrate

Ex 2) Formula for calcium chloride dihydrate Ca+2 Cl-1 2 H2O

CaCl2.2H2O

HW#15a) Formula for sodium sulfate decahydrate







CaCl2.2H2O

HW#15a) Formula for sodium sulfate decahydrate Na+1 SO4

-2 10 H2ONa2SO4

.10 H2O







CaCl2.2H2O

HW#15a) Formula for sodium sulfate decahydrate Na+1 SO4

-2 10 H2ONa2SO4

.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

Lab15.2 – Water of Hydration

- due in 2 days


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 octahydrateMg+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


particle size, filtration, ability to stay suspended.

14. Why don’t solutions exhibit Tyndall effect?


-2 . 7H20 Ba+2 OH-1 . 8H20





14. Why don’t solutions exhibit Tyndall effect?Particles too small to scatter light.


-2 . 7H20 Ba+2 OH-1 . 8H20







-2 . 7H20 Ba+2 OH-1 . 8H20MgSO4

.7H20 Ba(OH)2.8H20







-2 . 7H20 Ba+2 OH-1 . 8H20MgSO4

.7H20 Ba(OH)2.8H20


Mg+2 C03-2 3H20 magnesium phosphate

MgCO3.3H20 tetrahydrate






-2 . 7H20 Ba+2 OH-1 . 8H20MgSO4

.7H20 Ba(OH)2.8H20


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 SolutionsSolution 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

PS

PS

Lab16.1 – Solubility

- due in 3 days


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


Solute - gets dissolved (lesser amount)Solvent - does the dissolving (greater amount)

2. Name 3 factors that influence rate at which solute dissolve





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





3. What do to: a) make a saturated soln unsaturated Add more solvent b) make an unsaturated soln saturated Add more solute


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 …



S1 = S2 3.6g/L = 9.5g/LP1 P2 100kPa P2

P2=


6. Based on solids in table 16.4 (& lab) general statement about solubility & temp.As temp goes up, solubility goes up



S1 = S2 3.6g/L = 9.5g/LP1 P2 100kPa P2

P2=



7. Can a solution with undissolved solute be supersaturated? No. The solid solute present will draw the excess dissolved solute

out of the solution.


S1 = S2 3.6g/L = 9.5g/LP1 P2 100kPa P2

P2=



7. Can a solution with undissolved solute be supersaturated? No. The solid solute present will draw the excess dissolved solute

out of the solution.


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 )(solutionofLiterssoluteofmolesMMolarity

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 )(solutionofLiterssoluteofmolesMMolarity

0.4M ln 5

glucose 2

soLmol

Lmol0.15

5.58NaCl 1

100.090.0

gNaCl

molL

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 1molNa 142.1g

ln 1Na 2.05.1 SOg

SOSO

soLSOmolL

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




1 Cu @ 63.5 = 1 S @ 32.1 =

4 O @ 16.0 = g/mol

MsoL

mol 3.1 ln 750.0

KCl 0.1




1 Cu @ 63.5 = 1 S @ 32.1 =

4 O @ 16.0 = g/mol

MsoL

mol 3.1 ln 750.0

KCl 0.1

Lmol

soLmol

33.0 ln 5.1

MgCl 5.0 2




1 Cu @ 63.5 = 63.5 1 S @ 32.1 = 32.1

4 O @ 16.0 = 64.0 159.6 g/mol

MsoL

mol 3.1 ln 750.0

KCl 0.1

Lmol

soLmol

33.0 ln 5.1

MgCl 5.0 2

0.63M 6.159

14

400

4

44 gCuSO

molCuSOL

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 1molNaCl 58.5g

ln 1NaCl 5.00.1 g

soLmolL





ln 1NaCl 5.00.1 g

soLmolL

33

33 KN 1.101 KN 1molKN 101.1g

ln 1KN 2500.0 Og

OO

soLOmolL





ln 1NaCl 5.00.1 g

soLmolL

33

33 KN 1.101 KN 1molKN 101.1g

ln 1KN 2500.0 Og

OO

soLOmolL

22

22 CaCl 8.2 CaCl 1molCaCl 111.1g

ln 1CaCl 1.0250.0 gsoL

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






MsoL

mol 8.2 ln 250.0

NaCl 70.0






MsoL

mol 8.2 ln 250.0

NaCl 70.0

Lmol0.10

glucose 0.180glucose 1

0.2glucose 0.36

g

molL

g






MsoL

mol 8.2 ln 250.0

NaCl 70.0

Lmol0.10

glucose 0.180glucose 1

0.2glucose 0.36

g

molL

g

gmolgmol

molmolL

mol

56CaCl 1CaCl 1.111CaCl 50.0

CaCl 0.50 250.0.1CaCl 0.2

2

22

22

HW#14) (IC?) Calc moles and gramsa) 1.0L of 0.50M NaCl







ln 1NaCl 5.00.1 g

soLmolL





ln 1NaCl 5.00.1 g

soLmolL

33

33 KN 1.101 KN 1molKN 101.1g

ln 1KN 2500.0 Og

OO

soLOmolL





ln 1NaCl 5.00.1 g

soLmolL

33

33 KN 1.101 KN 1molKN 101.1g

ln 1KN 2500.0 Og

OO

soLOmolL

22

22 CaCl 8.2 CaCl 1molCaCl 111.1g

ln 1CaCl 1.0250.0 gsoL

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)

Solu

bilit

y g/

100g

H2O

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.




MLmol 79.soln 950.

75.0




MLmol 79.soln 950.

75.0

Mg

molLg 43.

NaCl 5.58NaCl 1

soln 2.1NaCl 05.30




MLmol 79.soln 950.

75.0

Mg

molLg 43.

NaCl 5.58NaCl 1

soln 2.1NaCl 05.30

Mg

molL

g 83.NaNO3 1.85NaNO3 1

0.3NaNO3 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?


2.5L of soln.



M537.KCl 74.6gKCl mol 1

2.5LKC 100.1g


2.5L of soln.




2.5LKC 100.1g

g 43.4mol 1

86.8LiBrLiBr mol 5.soln L .5soln L 1

LiBr mol 1.0


2.5L of soln.




2.5LKC 100.1g

suc 3.51mol 1

suc 342gsuc mol 15.L .300soln L 1

suc mol .5 g

g 43.4mol 1

86.8LiBrLiBr mol 5.soln L .5soln 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?



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?



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

TV

TV 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)

molalC

1512.0

molal

C1

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

Lab16.2 – Molarity of a Saturated Solution

- due tomorrow

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?

kgmoles

solvent of kilogramssolute 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?

kgmoles

solvent of kilogramssolute 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 5.25mol 1.25 Xe.c.

76.mol 5.25mol 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.




solution L 1solute mol 1 1M

solvent kg 1solute mol 1 1m






m28. NaC 58.5g kg .6 NaC mol 1 NaCl 10g






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?


28) Would a dilute or concentrated sodium fluoride solution have a higher boiling point?

OCH g4OCH mol .25 OCH kg 90.0

solvent 1kg OCH mol 1 OCH 9.0g

422

422422

422422

k


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 g4OCH 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 formulaa. Na2B4O7

.10H2O b. Na2CO3.H2O

c. MgSO4.7H2O

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 make50mL of 0.20M KNO3 solution?

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Ch15.1 – Water and Aqueous Solutions Ex1) Draw H 2 O structure shape: bond angle: polar:

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