Ch. 15 & 16 – WATER & SOLUTIONSVocabulary: read pages 504-507 & 518-523. Then define the vocabulary terms below and answer questions 1-3 on page 2.

solution:a homogeneous mixture consisting of particles dissolved in a solvent

suspension:a mixture from which some of the particles settle out slowly upon standing

colloid:(define and give one example)a mixture whose particles are intermediate in size between those in suspension and a solution

example:glue, milk, mayonnaise

emulsion: (define and give one example)the colloidal dispersion of one liquid in anotherexample:soap, detergent

solute:dissolved particles

solvent:the dissolving medium (often a liquid, but can be a gas or solid)

dilute:a solution containing a small amount of solute

concentrated:a solution containing a large amount of solute

solubility:the amount of a substance that dissolves in a given quantity of solvent

unsaturated:a solution that contains less solute than a saturated solution at a given temperature and pressure. (Clear)

saturated:a solution containing the maximum amount of solute for a given quantity of solvent at a constant temperature and pressure.An equilibrium exists between undissolved solute and ions in solution. (Clear, might have solid particles on the bottom)

***supersaturated:a solution that contains more solute than it can theoretically hold at a given temperature. Excess solute precipitates if a seed crystal is added. (Clear, very unusual, easily disturbed)

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Questions:

1. What is the difference between a solution and a mixture? A solution is where particles dissolve in a mediumand has only one phase. A mixture blends two or more substances but there may or may not be dissolving occurring, and a mixture may be more than one phase.

2. a. State Henry’s Law At a given temperature the solubility of a gas in a liquid is directly proportional to the pressure of the gas above a liquid.

b. Sketch a graph for the relationship described by Henry’s Law on the axis below. Is this a direct or an inverse proportion?

direct

c. Gases are most soluble in water at ____low______ temperature (high or low) and at ______high_______ pressure (high or low).

3. What factors affect the solubility of a substance in water?a. temperature (solids: solubility increases with T; gases: solubility decreases with T)

b. pressure (little effect for solids and liquids. Gases are more soluble at high pressure)

c. agitation (exposes surface area of solid particles so solubility increases)

CHOOSING THE RIGHT SOLVENT/SOLUTE COMBINATION: “Like Dissolves Like”

Review: Intermolecular Forces

1. nonpolar molecules: weak London dispersion forces

example: CH4, fats

properties: dissolve in nonpolar solvents such as acetone or gasoline – BUT NOT IN ALCOHOL OR WATER

2. polar molecules: dipole forces (stronger)

example: CH2O, PH3

properties: dissolves in polar solvents such as water and alcohol

P

S

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3. NEW: extremely polar molecules (hydrogen bonding) The attraction between a lone pair on N, O or F and H in another molecule

example: NH3, HF, H2Oproperties: * high melting and boiling points

* strong surface tension* high specific heat* dissolve in water

solute solventionic water

polar covalent (dipole) water, other polar covalent (alcohol)

nonpolar covalent (dispersion) nonpolar covalent

a. Which substances below would you predict to be water soluble?Which would be soluble in non-polar solvents?

type of compound (ionic, polar covalent or nonpolar covalent)

soluble in water (very polar solvent)?

soluble in alcohol?(polar – dipole; but not as polar as water)

soluble in nonpolar sovents(ex. ccl4)

K2SO4 I Y N N

C5H12 NP N N Y

Cu METAL N N N

C2H5OH P Y Y N

b. How and why do solutes dissolve in water?The positive sides of the water molecule are attracted to the negative ions while the negative side of the water molecule is attracted to the positive ion.So…. The solute-solvent forces of attraction need to be stronger than the solute-solute forces of attraction in order to dissolve.c. Do all ionic solids dissolve in water?NO – many are not. If the ions are more attracted to each other than to water, they are insoluble (e.g. carbonates, sulfates)http://www.youtube.com/watch?v=CLHP4r0E7hg&feature=relatedUNDERSTANDING SOLUBILITY CURVESThe solubility curve below is just for sugar and sodium chloride.

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a. Find the solubility of sugar at 20°C. 200 g sugar100g water

b. If I add 125g of sugar to 100g water at 20°C and stir, the solution will be unsaturated.

The solution will appear:clear

c. If I add 200g of sugar to 100g of water at 20°C and stir, the solution will besaturated

The solution will appear:clear

The solution will NOT be supersaturated because: the conditions of temperature and pressure are not appropriate – you need a high temperature to make a supersaturated solution, then gradually cool it.

d. The solubility of sugar in 100 g of water at 90 °C is 410 g of sugar in 100g of water. 400g of NaClis added to 100g of water at 90 °C. The solution is then slowly cooled to 20 °C. Two things could happen:

1. If solid sugar begins to precipitate, the solution is __saturated___.A total of _400 – 200 = 200___ g of sugar will precipitate.

2. If no solid appears, the solution is __supersaturated___.

solubility

Units of solubility: g solute

100 g solvent

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3. Reading Solubility Curvesa. What is the solubility of potassium nitrate at 50°C?

80 g KNO3

100 g H2O

b. What mass of potassium nitrate could dissolve in 250 g of water at 50 ºC?250 g H2O x80 g KNO3 = 200 g KNO3

100 g H2O

c. What mass of potassium nitrate could dissolve in 250g of water at 10ºC?250 g H2O x22 g KNO3 = 55 g KNO3

100 g H2O

d. What happens to the solubility of potassium nitrate as the temperature decreases?Decreases (as it is cooled, less can dissolve)

e. A saturated solution of potassium nitrate in 250g of water at 50 °C. It is then cooled to 10 °C. What mass of potassium nitrate will precipitate out of the solution?As it cools, the solution will undissolve and 145 g of precipitate would come out of the solution and sit at the bottom (use info from your answer in b and c)

f. Does the solubility of EVERY salt decrease with temperature?No. Some increase, some decrease.

g. Ammonia (NH3) is a gas. What happens to its solubility as the temperature increases?Solubility decreases

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DEMONSTRATION: SUPERSATURATED SOLUTION

http://www.youtube.com/watch?v=1y3bKIOkcmk&feature=related

1. Place 5 g of sodium acetatetrihydratein a large, clean, dry test tube(use funnel!) and add 1 mL of distilled water.

2. Place the test tube in a boiling water bath and heat until the crystals are dissolved.

3. Remove the test tube from the water bath; place a rubber stopper loosely on the mouth of the test tube.

4. Place the test tube in a room temperaturewater bath and let it cool to room temperature undisturbed (at least 20 minutes).

5. Carefully add one crystal of sodium acetate trihydrate.

6. Record your observations

7. Dispose of the sample in the waste container in the hood.

record your observations below

SOLUBILITY OF GASES

temperatureT ↑ solubility ↓temp increases, solubility of gas decreases becauseincreased T causes increased k. e., which breaksintermolecular bonds causing more molecules to escape.That’s why warm soda tastes flat, because CO2 has escaped.

pressureP ↑ solubility ↑pressure increases, solubility of gas increasese.g. soda – when you open the bottle, the gas “undissolves”so you see the bubbles

Also, the solubility of gases is more affected by changes in T and P than liquids or solids

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WATER

1. Draw a diagram of water molecules. Label the positive and negative poles, and use dotted lines to show hydrogen bonding between the molecules.

2. What does electronegativity have to do with hydrogen bonding?Makes H δ+ and O δ-, so they attract

3. What elements, when joined with hydrogen, can be found in molecules that hydrogen bond?N, O, F

4. Explain how hydrogen bonding in water leads the following properties:

adhesive forces the attraction of water to other substances - water sticks to the inside of a glass

cohesive forces the attraction of water to itself - water forms drops

high surface tension water supports the weight of a paperclip

high specific heat capacity (absorbs a lot of heat as it warms up) the ocean absorbs heat in the summer and releases it in the winter, making coastal areas more temperate (less temperature change) than inland. Water takes a long time (needs much energy) to boil.

solid water (ice) floats on liquid water water is less dense as ice as ice has an open hexagon structure

oil and water cannot mix oil does not have charges like water, so oil and water do not attract and therefore do not mix.

5. List two biological examples of hydrogen bonding.

a. holds together DNA helix b. capillary action in trees

6. (not in videos – you will need to Google or use your textbook)

What is a surfactant? Give one example of a surfactant.Any substance that interferes with the H-bonding between water molecules and reduces surface tension, e.g. soap

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UNITS OF CONCENTRATION

MOLALITY (m)

molality = moles solutekg solvent

1. Determine the molality of a solution prepared by dissolving 20 g of CaCl2 in 500g of water.20 g CaCl2 x 1 mol CaCl 2 = 0.18 mol CaCl2 500 g H2O = 0.5 kg H2O

111.1 g CaCl2

0.18 molCaCl2=0.36 mol CaCl2 = 0.360 m0.5 kg H2O

2. A student dissolves 42 g of sugar (C12H22O11) in 650g of ethanol. Determine the molality of the resulting solution.42 g C12H22O11 x 1 mol C 12H22O11 = 0.123molC12H22O11 650 g = 0.65 kg ethanol

342 g C12H22O11

0.123molC12H22O11=0.189molC12H22O11= 0.189 m0.65 kg ethanol

3. How many moles of sugar must be dissolved in 450g of ethanol in order to prepare a 0.75m solution?

STEP 1: Rewrite 0.75m as:

STEP 2: Set up cancellation and solve!0.75 molsugar = x mol sugar = 0.3375 mol sugar1 kg ethanol 0.45 kg ethanol

% SOLUTION

% = mass solute(salt )totalmass solutionx 100-OR- volume solute

total volumesolution x 100

Examples: vinegar, hydrogen peroxide, rubbing alcohol

1. Find the % of NaCl in a saline solution prepared by dissolving 15 g of sodium chloride in 125g of water.

15 g NaCl x 100 = 10.7%(15 + 125) g solution

2. Rubbing alcohol is 70% isopropyl alcohol and 30% water.

a. Why are alcohol and water able to mix together so well?They are both POLAR

b. What volume of isopropyl alcohol would be needed to prepare 450mL of rubbing alcohol?70 mL isopropyl alcohol= x mL = 315 mL isopropyl alcohol100 mL rubbing alcohol 450 mL rubbing alcohol

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DISSOCIATION OF IONIC SOLIDS IN WATER

i = number of ions (particles) formed when a compound dissociates

a. NaCl Na+ + Cl- 1 NaCl2 ions i =2

Not: NaCl + H2O

b. K2SO42K+ + SO42- 1 K2SO43 ions i =3

c. Pb(NO3)2 Pb2+ + 2 NO3- 1 Pb(NO3)23 ions i = 3

d. Al(C2H3O2)3 Al3+ + 3 C2H3O2- 1 Al(C2H3O2)34 ions i = 4

SOLUBILITY OF MOLECULAR (POLAR) SOLUTES IN WATER

C6H12O6does not dissociate!!! (MOLECULAR) i =1

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COLLIGATIVE PROPERTIES

colligative property:properties that depend on the number of solute particles dissolved in solution.

freezing-point depression:freezing-point goes down when a solute is dissolved in a solvent. Presence of solute particles makes forming an orderly pattern in a solid harder and requires release of more kinetic energy for solidification to occur.

boiling-point elevation:boiling-point goes up when a solute is dissolved in a solvent. Requires additional kinetic energy for solvent particles to overcome the new IMFs formed between the solute and solvent.

vapor pressure lowering:The solvent now has to break additional IMFs which takes more energy. Fewer solvent molecules have kinetic energy to escape as a vapor. Also, solute molecules occupy spaces at the surface of the liquid which prevents solvent molecules leaving.

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Calculation of Freezing and Boiling-Point for Solutions

ΔTf= kf.mi

change in freezing point = freezing point constant x molality x # of particles

ΔTb= kb.mi

change in boiling point = boiling point constant x molality x # of particles

1. Determine the freezing and boiling-point of a solution prepared by dissolving 1.2 moles of glucose (C6H12O6) in 875g of water.

a. Is solute ionic?No (i = 1)

b. Calculate molality

1.2 moles =1.37m0.875 kg

c. Determine change in boiling/freezing-point

BP: ΔTb= kb.m.i FP: ΔTf= kf

. m.i= 0.512°C/m • 1.37m • 1 = 1.86°C/m • 1.37m •1= 0.7 °C = 2.55 °C

d. Determine final boiling/freezing-point

BP: Tb = 100 + 0.7 = 100.7 °C FP: Tf = 0 – 2.55 = -2.55 °C

2. Determine the freezing and boiling point of a solution prepared by dissolving 1.2 moles of sodium sulfate in 875g of water.

a. Is solute ionic?Yes. Na2SO42 Na++ SO42- i = 3

b. Calculate molality

1.2 moles=1.37m0.875 kg

c. Determine change in boiling/freezing point

BP: ΔTb= kb.m.i FP: ΔTf= kf

. m.i= 0.512°C/m • 1.37m • 3 = 1.86°C/m • 1.37m •3= 2.1 °C = 7.6 °C

d. Determine final boiling/freezing-point

BP: Tb= 100 + 2.1 = 102.1 °C FP: Tf= 0 – 7.6 = -7.6 °C

Kf for water = 1.86 °C/m Kb for water = 0.512 °C/m

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MOLARITY

molarity = mol soluteL solution

molarity = mol solute per liter solution [] = “the molarity of”

Sample Problems:SOLID:a. Find the molarity of a solution prepared by dissolving 1.5moles of sodium nitrate in 350mL of water.

1.5/0.35 = 4.29 M

b. Find the molarity of a solution prepared by dissolving 10 g of sodium nitrate in 425mL of water.

n=m/M = 10/(23+14+48) = 0.118 mol/0.425 = 0.277 M

c. How many moles of sodium nitrate would be needed to prepare 250 mL of a 0.8 M solution?

STEP 1: Rewrite 0.8 M as:STEP 2: Solve using dimensional analysis:

0.8 M = 0.8 mol/1 L = 0.2 mol/250 mL

d. What volume of 0.9 M sodium nitrate can be made with 0.5 moles of sodium nitrate?

STEP 1:Rewrite 0.9 M as:STEP 2: Solve

0.9 M = 0.9 mol/1 L = 0.5 mol/0.556 L

e. What mass of sodium nitrate would you need to prepare 250mL of a 0.707M sodium nitrate solution, starting with solid sodium nitrate and water.

STEP 1: Rewrite 0.707M as:STEP 2: Solve

0.707 M = 0.707 mol/1 L = 0.17675 mol/0.25 L.0.17675 mol x 85 g/mol = 15.02 g NaNO3

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MOLARITY BY DILUTION – Making a weak solution from a stronger one

M1V1 = M2V2

PROBLEMS1.a. What volume of 12M HCl is needed to prepare 500 mL of a 0.5 M solution?

V1 = M2V2/M1 = (500 x 0.5)/12 = 20.8 mL

b. What volume of water would also be needed?

500 – 20.8 = 479.2 mL

2. Enough water is added to 350mL of 0.75M KOH to bring the volume to 600mL.Find the molarity of the new solution.

M2 =M1V1/ V2 = (350 x 0.75)/600 = 0.4375 M

3. Find the molarity of a solution prepared by adding 125mL of water to 375 mL of 2 M KNO3 solution.

M2 = M1V1/ V2 = (375 x 2)/500 = 1.5 M

TRY:1. What volume of 5 M NaBrsolution is needed to prepare 250mL of 0.15 M solution? What volume of water

is needed?

V1 = M2V2/M1 = (250 x 0.15)/5 = 7.5 mL volume water needed = 250 – 7.5 = 242.5 mL

2. Enough water is added to 1.2 L of 6 M HNO3 to bring to volume to 2.5 L. Find the molarity of the new solution.

M2 = M1V1/ V2 = (6 x 1.2)/2.5 = 2.88 M

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PRACTICE1. How many moles of magnesium sulfate must be dissolved in 250mL of water in order to prepare a 1.75M

solution?

Rewrite 1.75 M as:

0.438 mol

2. What mass of magnesium sulfate would be needed to prepare 1.5L of a 0.75M solution?

Rewrite 0.75M as:

135.3 g

3. Find the molarity of a solution prepared by adding 5 g of sodium hydroxide to 650mL of water.

0.192 M

4. What volume of 0.35M sodium hydroxide solution can be made using 15 g of solid sodium hydroxide?

Rewrite 0.35M as:

1.07 L

DILUTION

5. What volume of 0.75M potassium hydroxide solution is needed to make 350mL of a 0.15M solution? What volume of water is also needed?

70 mL potassium hydroxide and 280 mL water

6. A student adds 250 mL of water to 4 L of 1 M barium nitrate solution. What is the molarity of the resulting solution?

0.941 M

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REACTIONS IN SOLUTION

Single Replacement: A + BX(aq) AX(aq)+ B

1. Mg(s) + 2HCl(aq) MgCl2 (aq) + H2 (g)

2. 2Al(s) + 3CuSO4 (aq) Al2(SO4)3(aq) + Cu (s)

3. F2(g) + 2KBr(aq)2 KF (aq) + Br2 (g)

4. 2Na(s) +2H2O(l)2 NaOH (aq) + H2 (g)

Double Replacement: AX(aq) +BY(aq) AY(aq) + BX(s)

precipitate: a solidformed in a reaction

1. 2KBr(aq) + Cu(NO3)2(aq) 2 KNO3 (aq) + CuBr2 (s)

2. Fe(NO2)3(aq) + Na3PO4(aq) FePO4 (s) + 3 NaNO2 (aq)

General Rule for deciding which product is the solid:

All nitrates, nitrites and acetates dissolve in water – not the solidSalts containing alkali metals and ammonium dissolve in water – not the solidIf there is a heavy metal, it usually is part of the solid (if not with nitrate, nitrite or acetate)

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SOLUTION STOICHIOMETRY

1. What volume of 0.25M HNO3 is needed to react completely with 1 g of magnesium ribbon?

REACTION:__Mg (s) + 2 HNO3 (aq) Mg(NO 3)2 (aq) + H2 (g)____________

1 g Mgx 1 mol Mg x 2molHNO 3 = 0.0823 mol HNO3

24.3 gMg 1 mol Mg

0.25 mol =0.0823 mol1 L x L

So x = 0.0329 L

2a. A student prepares lead (II) chloride and sodium nitrite by reacting 0.55M lead (II) nitrite solution with excess sodium chloride solution. What volume of 0.55M lead (II) nitrite solution would be needed in order to form 0.5 g of lead (II) chloride?

REACTION: ____Pb(NO2)2 (aq) + 2 NaCl (aq) PbCl 2 (s) + 2 NaNO2 (aq)_________________

0.5 g PbCl2 x 1 molPbCl 2 x 1mol Pb(NO 2)2 = 0.00180molPb(NO2)2

278.2gPbCl2 1 molPbCl2

0.55 mol =0.00180mol1 L x L

So x = 3.3mL

b. Using the reaction above, what mass of lead (II) chloride would precipitate if 120mL of 0.55M lead (II)

nitrite solution react?

0.55 mol =xmol1 L 0.12 L

So x = 0.066molPb(NO2)2

0.066 molPb(NO2)2x 1molPbCl 2 x 278.2 g PbCl 2 = 18.4 g PbCl2

1 mol Pb(NO2)2 1 molPbCl2

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ANSWERS:½ pt per question = 5 pts total

1. KClO3

2. 100 g KClO3

3. 126 g KNO3

4. NaCl5. unsaturated6. 20 g7. NH3

8. KI9. KClO3

10. NaCl

HOMEWORK1

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HOMEWORK 2: MORE SOLUBILITY CURVE PRACTICE

1 pt per question = 5 pts total1. What is the solubility of the following solutes in water?

a) NaCl at 60ºC = 38 g NaCl / 100 g H2Ob) KCl at 40ºC = 39 g KCl / 100 g H2Oc) KNO3 at 20ºC = 32g KNO3 / 100 g H2O

2. Are the following solutions saturated or unsaturated? Each solution contains 100 g of H2O.

a) 31.2g of KCl at 30 ºC = unsaturatedb) 106g KNO3 at 60ºC = saturatedc) 40 g NaCl at 10 ºC = supersaturated (saturated works!)d) 150g KNO3 at 90 ºC = unsaturated

3. For each of the following solutions, explain how much of the solute will dissolve and how much will remain undissolved at the bottom of the test tube?

a) 180 g of KNO3 in 100 g of water at 80 ºC 170 g dissolved, 10 g undissolved

b) 180 g of KNO3 in 100 g of water at 20 ºC 32 g dissolved, 148 g undissolved

c) 60 g of NaCl in 100 g of water at 60 ºC 38 g dissolved, 22 g undissolved

4. A saturated solution of KNO3 is formed from one hundred grams of water. If the saturated solution is cooled from 90 °C to 30 °C, how many grams of precipitate are formed?200 – 45 = 155 g

5. A saturated solution of KCl is formed from 100 g of water. If the saturated solution is cooled from 90 °C to 40 °C, how many grams of precipitate are formed?52 – 39 = 13 g

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HOMEWORK 3:MOLALITY AND % CONCENTRATION5 pts total1. Determine the molality and mass % of solute in of each solution below:

a. 10.0g of magnesium sulfate dissolved in 250g of water.10 g MgSO4 x1 molMgSO4 = 0.08 mol MgSO4= 0.33 m

120 gMgSO4 0.25 kg H2O

10g MgSO 4 x 100 = 3.8%260 g solution

b. 44g of glucose (C6H12O6) mixed with 450g of ethanol(C2H5OH)44 g C6H12O6 x1 molC6H12O6= 0.24 mol C6H12O6 = 0.54 m

180 gC6H12O6 0.45 kg C2H5OH

44 C 6H12O6 x 100 = 8.9%494 g solution

c. 350g of water mixed with 250 g of ethanol (C2H5OH)250 g C2H5OH x 1mol C2H5OH = 5.43 mol C2H5OH = 15.5 m

46 gC2H5OH 0.35 kg H2O

250 C2H5OH x 100 = 41.7%600 g solution

2. What volume of hydrogen peroxide is in a 250 mL of 3% hydrogen peroxide solution?1 pt

3 mL hydrogen peroxide = x mL hydrogen peroxide100 mL solution 250 mL solution

3. What mass of sodium hydroxide must be added to 500g of water in order to prepare a 0.75m solution?

1 pt 0.75 molNaOH =x molNaOH = 0.375 mol1 kg H2O 0.5 kg H2O

0.375 molNaOH x40 g NaOH = 15 g NaOH1 molNaOH

15 g NaOH

Molality = 15.5 m ½ pt

Mass % = 41.7% ½ pt

Molality = 0.33m ½ pt

Mass % = 3.8% ½ pt

Molality = 0.54m ½ pt

Mass % = 8.9% ½ pt

7.5 mL hydrogen peroxide

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HOMEWORK 4: COLLIGATIVE PROPERTY PRACTICE4 pts total1. What is a colligative property? ½ pt

Properties that depend on the number of solute particles dissolved in solution2. Why is salt added to water when making pasta? ½ pt

Pasta cooks best at a higher temperature. By adding salt to the water, the boiling point of the water raises because the solute gets in the way of the water molecules making their way up to the surface and boil.

3. A student is making popsicles from Kool-Aid. He finds that the popsicles made with a strong Kool-Aid solution take longer to freeze than the popsicles made with a weak solution. Use colligative properties to explain why. 1 ptThe more solute, the lower the freezing point – the concentrated solution will take longer to freeze because there is more solute in those popsicles than in the weak solution pops.

4. Use the information in the box below to solve problems a & b.

a. Find the boiling point of a solution prepared by dissolving 25 g of potassium chloride in 500g of water.1 pt(Hint: first find the molality of the solution).

KClK+ + Cl- i = 2 BP: ΔTb= kb. m.i

= 0.512°C/m • 0.67m • 2= 0.68 °C

25 g KCl x1molKCl = 0.34 mol BP = 100 + 0.68 = 100.68 °C74.5 gKCl

m = 0.34 mol = 0.67m0.5 kg

b. Find the freezing point of a solution prepared by dissolving 0.75 moles of sugar in 500g of water.1 pt

sugar does not dissociate so i = 1 FP: ΔTb= kb. m.i

= 1.86°C/m • 1.5 m •1= 2.79 °C

m = 0.75mol = 1.5 m FP = 0 – 2.79 = -2.79 °C0.5 kg

ΔTf = kf m i ΔTb= kb m i

Kf for water = 1.86 °C/m Kb for water = 0.512 °C/m

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HOMEWORK 5

58 g NaCl x 1 molNaCl = 0.991 molNaCl58.5 g NaCl

M =0.991 = 0.991 M1

10 g AgNO3 x 1 mol AgNO 3 = 0.059 mol AgNO3

169.9 g AgNO3

M =0.059 = 0.118 M0.5

mol = molarity x volume = 0.5 x 2 = 1 mol1 mol KNO3 = 101.1 g

5 g KCl x 1 molKCl = 0.067 mol KCl74.6 g KCl

volume = mol = 0.067 = 0.268 Lmolarity 0.25

mol = molarity x volume = 0.1 x 0.1 = 0.01 mol0.01 molCuSO4

. 5 H2O = 2.5 g

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HOMEWORK 6: MOLARITY PRACTICE PROBLEMS

SOLID

1. Determine the molarity of a solution prepared by dissolving 45 g of CaCl2 in 450mL of water.

45 g x 1mole111 g = 0.41 moles

0.41 moles0.45 L = 0.91 M

2. What mass of sodium nitrate must be added to 750mL of water in order to prepare a 0.55M solution?

Rewrite 0.55M as: ______________________

0.55M = x

0.75L x = 0.41 moles x 85 g

1mol e = 35.1 g

3. What volume of 3.2M saline solution can be made using 2 moles of salt?

Rewrite 3.2M as: _________________________

3.2M = 2.0 moles

x x = 0.625 L

DILUTION

4. What volume of 12 M HCl solution is needed to prepare 550 mL of a 0.75 M solution? What volume of water is also needed to prepare the solution?

(12 M)(X) = (0.75M)(550mL), so X=34 mL of 12M HCl

550 mL(solution) – 34 mL (acid) = 516 mL of water

5. Find the molarity of a solution prepared by adding 150 mL of water to 250 mL of 0.75M sodium hydroxide solution.

(0.75M)(250 mL) = (X)(400 mL), so X= 0.469 M

6. What volume of water was added to 500 mL of 2 M sulfuric acid solution if a 1.5 M solution was produced?(2 M)(500 mL) = (1.5 M)(X), so X= 667 mL of solution

667 mL (solution) – 500 mL (acid) = 167 mL of water

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HOMEWORK 7: REACTION WRITING PRACTICE

PART I: Identify the reaction type and complete each aqueous reaction given below:

single displacement a. 2 K(s) + ZnSO4(aq) K2SO4 + Zn

single displacement b. Cl2(g) + 2 NaBr(aq) 2 NaCl + Br2

single displacement c. 3 Zn(s) + 2 Fe(NO3)3(aq) 3 Zn(NO3)2 + 2Fe

single displacement d. sodium + hydrochloric acid (write out the complete reaction below)

2 Na + 2 HCl 2 NaCl + H2

double displacement e. solutions of magnesium nitrate and potassium fluoride are mixed

Mg(NO3)2 + 2KF MgF2 + 2KNO3

PART II: Review of other reaction types. Label each reaction below as SYNTHESIS, DECOMPOSITION OR COMBUSTION. Then write out the balanced equation for each reaction.

decomposition a. calcium carbonate calcium oxide + carbon dioxide

CaCO3 CaO + CO2

synthesis b. barium reacts with nitrogen to form barium nitride

3 Ba + N2 Ba3N2

combustion c. propane (C3H8) burns in oxygen to form carbon dioxide and water

C3H8 + 5 O2 3 CO2 + 4 H2O

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