Solutions - TeacherTubecdn-media1.teachertube.com/doc601/582.pdf · Solutions Introduction: A ......

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Unit 11

Solutions Introduction: A

solution is a

homogenous

mixture. Perhaps the best way to get an idea just how common

solutions are is to go grocery shopping. There you will see

virtually every type of solution, in all sorts of colors and

types.

This unit is primarily a roll-up-your sleeves and learn how to

do things unit. Learn how to make a 25% (v/v) fruit juice

solution. Dilute a chocolate solution from 20% to 8%. Find out

why they salt the roads in the winter and just how much of an

effect it has.

Schedule:

Day Learn activity homework

1 How to make rock candy; introduction

to solutions

Rock candy lab

Intro to solutions

powerpoint

Ws 11.1: introduction to

solutions

Rock candy questions

2 Measure density of tiny objects Flink lab Complete lab

3 Theory of making, diluting, and

concentrating solutions

Solution concentration

powerpoint

ws

4 How to make and dilute solutions Making solutions activity ws

5 Why they put salt on the roads Colligative properties

6 Put it all together Test review Study for test

7 “ “ Solutions test Read next unit

introduction.

How do we make and modify solutions?

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Name__________________________ Period______________ Lab11.1: Rock Candy Lab

Rock Candy Lab

Introduction: Have you ever noticed on television how clean the

scientist’s labs are? How carefully they work with their solutions, their

little pipettes, or even with their cadavers?

Usually Hollywood hopelessly botches any attempt to portray scientists,

but in this case they got it right. Good science requires pure chemicals.

And to this day one of the best ways to prepare a pure solid sample is

by recyrstallization- the process of isolating pure crystals from a

solid/liquid solution.

In this experiment you will make rock candy at home and bring it in to share with your classmates.

This is an excellent example of recrystallization in action.

Rock Candy

A video can be seen here at http://chemistry.learnhub.com/lesson/9991-rock-candy-video

Prep: 15 min., Cook: 20 min., Stand: 14 days.

Ingredients

1 Pie tin

A clear cup

6 cups of sugar

4 wood skewers

Cardboard to cover cup

Masking tape

Recipe

2. Bring 2 cups of water to a boil, then add 6 cups of sugar and stir until

dissolved. Measure carefully, as our experiments suggest that as little a

measurement off by as little as 5% can make a huge difference in

crystallization time. Add any food-grade additives you like: food coloring,

cinnamon oil, vanilla, etc.

4. Fill a clear plastic cup with the hot sugar solution, cover with the

cardboard, and tape shut. Stick skewers in, not touching the bottom. Let

stand 10 to 14 days.

Safety Notice: Since no food is allowed in the lab,

this experiment must be performed at home.

http://chemistry.learnhub.com/lesson/9991-rock-candy-video

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Name_______________________________ period_________________ lab 1.1rockcandy

Use any reliable sources necessary on the web to answer the following questions:

1. Draw the chemical structure of glucose, fructose, and sucrose, with their chemical formula below

each structure.

glucose: formula:___________

fructose: formula:________

sucrose: formula:___________

2. Which of the above sugars is table sugar? Glucose, fructose, of sucrose?

2. It is said that a crystal resembles the structure of the molecule it is formed from. The hexagonal

quartz crystals from SiO2 are an example. Since we started from sugar crystals, dissolved them in

water, then let them reform slowly, the process is known as recrystallization. Based on the chemical

structure of table sugar, draw a prediction of what your crystals will look like.

Predicted close-up view of a sugar crystal

4. Explain in you own words why it may a good idea to coat your string in sugar to aid in

recyrstallization. The principle is known as nucleation, but please explain it without sounding like a

science geek.

5. Why do we use so much sugar and so little water in this recipe? The principle is known as

supersaturation, but explain it in your own words without sounding like a science teacher.

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Name___________________________ Period____________ rockcandylabnotebook

Rock Candy Lab Notebook

Please make daily notes as you observe your rock candy experiment progressing. You should include

observations, comparisons to other experiments, and document any changes, such as heating, adding

ventilation, or any substances added to the solution.

Date Observations Changes made to

experiment

Comparison to others

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1

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here

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4. How does it NOT happen? 5. Will heating or pressure help? 6. Make Crystals

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Chemadventure Chapter 11: Solutions

Solvent: Solute(s):

dissolver dissolved

What are they?Where are they?

Homogeneous mixtures

Only one thing visible

Everywhere!

yesbronze

nogranite

nowater

yesmouthwash

Solutions

What is in a solution?

Is it a solution?

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= solvent surrounding the solute

A molecular view of dissolving:

salt

waterPartly dissolved

Solvation: Fully dissolved

Electrolyte: Salt.Non-Electrolyte:

Not a salt (ex: sugar)

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SolubilityWhy don’t oil and water mix?

No!CH3CH2CH2CH2OHButanol

YesCH3CH2CH2OHPropanol

yesCH3CH2OHEthanol

yesCH3OHMethanol

Soluble in water?

FormulaNameG

reasie

r

Greasy watery

Rule of thumb: “like dissolves like”

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Heating Solutions

A solubility surprise:

Global implications

Most solids

Heating makesMore Soluble

Most gases Less Soluble

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Yes: Henry’s Law

If sol. Is 1g/L at 1 atm, it will be ______g/L at 2 atm

Everyone: if solubility is 3.45 g/L at 5.6 atm, what is the solubility at 1 atm?

3.45/5.61 = S2/1 S2 = 0.614 g/L

Is there a way to increase the solubility of ANY solution?

Solubility is proportional to pressure S ~ P

S1/P1 = S2/P2

(Henrys Law Worksheet)

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(Henrys Law Worksheet)

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How to make crystals:The process: recrystallizationThe principle: supersaturation

1. Make a hot supersaturated solution

and cool Or evaporate Or Reduce pressure2.

saturated

Watch a video online here or play the flv file here or the avi file here

The process of forming the very first crystal during crystallization is called nucleation

Fun nucleation video here

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Solutions

HgAgFillingsSolid-solid

saltsH2OOceanSolid-liq

Acetic acidH2OVinegarLiq-Liq

CO2H2OSodaGas-liquid

O2N2AirGas-gas

soluteSolventEx.Type

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Concentration

Moles of soluteMoles solution

Mole Fraction (X)

Moles of solute

Liter of solution

Molarity (M)

Volume of solute x 100

Volume of solution

% by volume (% v/v)

Mass solute x 100

Mass of solution

% by mass (% m/m)10 g NaCl90 g H2O 10% NaCl by Mass

10 mL juice90 mL H2O

58.5 g NaCl1L solution 1M NaCl

10% NaCl by Volume

58.5 g NaCl162 g H2O XNaCl = 0.1

Moles of solute

Kg of solventMolality (M)58.5 g NaCl

1 kg water 1m NaCl L1 only

L1 only

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Molarity Examples

• You have a 100.5 mL solution containing 5.1 g glucose (molar mass = 180.16 g/mol). What is the molarity of that solution?

• Solution • Molarity = moles of solute/L of solution

• Moles solute = 5.1 g glucose x 1 mole glucose/180.16 g glucose = 0.0283 moles glucose

• L of solution = 100.5 mL x 1L/1000 mL = 0.1005 L solution

• Molarity = 0.0283 moles/0.1005 L solution = 0.282M

Molarity ws

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Another example

• Make 100 mL of a 1M NaOH solution

NaOH g 4 solution liter 0.1 x NaOH mole

NaOH g 40 x

solutionliter

NaOH mole 1

Take 4 g NaOH; add water til 100 mL.

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Dilution• As solvent increases, concentration

decreases• C1 V1 = C2 V2

• Concentration may be Molarity, % v/v, % mass

• How can I dilute 53.4 mL of a 1.50M soln of NaClto make it a 0.800M solution?

• Easy: C1V1 = C2V2

• (1.50mol/L)(53.4mL)= (0.800mol/L)(V2)

• V2 = 100. mL

• (dilute to 100 mL to get 100 mL of a 0.8M soln)

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• How do solutes affect boiling and freezing point?

Solutes elevate the boiling point

Colligative PropertiesL2: concepts only. (L1 all)

Solutes lower the freezing point

(road salt)

collective

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Solutes elevate boiling point

BP elevation constant molality

= moles solute/Kg solvent

x pm Sugar = 1NaCl = 2CaCl2 = 3

Fewer solvent molecules on surface

“Particle molality”

What is the boiling point of a 2.75m aqueous NaCl solution?

Tb =Kbm x pm

Particle molality = 2 (easy to forget)

• = (0.512)(2.75 x 2)= 10.22 oC

• BP = 102.82 oC

L1 only

# of ions

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Solutes lower freezing point

FP depression constant molality= moles solute/Kg solvent

x pm Sugar = 1NaCl = 2CaCl2 = 3

Interfere with crystal formation

“Particle molality”

What is the freezing point of a 2.75m aqueous NaCl solution?

Tf =Kfm x pm

Particle molality = 2 (easy to forget)

• = (1.86)(2.75 x 2)= 10.22 oC

• FP = -10.22 oC

Tf = KfmL1 only

Next: Energy

12

Name: _______________________________ Date: _______ Period: ______ Lab11.2

Plastic Bead Density Activity

In this demonstration we will “flink” a plastic bead using sugar and water to determine the density of

the solution. We will then use our data to determine the concentration of the solution by mass,

volume, and molarity.

Procedure: Using a test tube, water, and sugar, Make your bead hover. Keep track of exactly how many mL of water and how many grams of sugar are in your solution. 1. Data

Volume of water: _______mL

Mass of water:__________g (same number since density of water is 1 g/mL)

Volume of sugar: ________g

Mass of sugar = _____mL (sucrose density = 1.59 g/mL). Show your calculation below:

Total volume of solution: _________mL (measure using a graduated cylinder.

Total mass of solution:___________g (add up mass of water and mass of sugar)

2. Calculations. Use the formulas on the first page for your calculations. You must show your work

for credit.

1. Determine the density of the bead.

2. Determine the % sugar in the solution by volume.

3. Determine the % sugar in the solution by mass.

4. Determine the molarity of the sugar

(C12H22O11) solution. Note that 1 mole of

sugar is 338 grams

Useful Formulas:

Density = mass/volume

% by Mass = (mass solute/mass solution) x 100

% by volume = (volume solute/volume solution) x 100

Molarity = M = moles solute/liters solution

Summarize your results here for credit: Results: 1. Density of bead: ___________2. % sugar in solution by volume:_________3. % Sugar in solution by mass:___________4. Molarity of sugar solution:__________

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Name_________________________ Period________________ ws 11.1

Talking about Solutions

For some of us, when we think of a chemist we imagine some person in a lab coat mixing chemicals

together. That person is preparing a solution, which is what this unit is all about. Read the paragraphs

below and then use your newfound knowledge to answer the questions that follow.

A solution is a homogeneous mixture, which means that only one thing is visible in the solution. The

substance that is dissolved is the solute, and the substance that does the dissolving is the solvent. On

a molecular scale dissolving involves the solvent surrounding the solute, and if they are salts they are

divided into their ionic components- this process is known as solvation. Solutes that are salts are

known as electrolytes. If the solutes are not ionic, like sugar for example, they are non-electrolytes.

In general if the solute has a

chemical structure similar to the

solvent, it will dissolve. For this

concept (solubility) it is said that

“like dissolves like”. In the organic

world one can identify, for example,

greasy (hydrocarbon chains), watery

(OH groups), and brick-like

(alternating double-bonded rings like

benzene) groups.

It has often been observed that one can, at least temporarily, add more solute to a solvent than it can

handle. For example, a maximum of 35.9 grams of table salt (NaCl) is soluble in 100 mL of water at 25 OC: this is a saturated solution; it has all the salt it can handle. But if you add it slowly you can get 40

or even 45 grams of salt to dissolve in water. It is now supersaturated, and the extra salt will

precipitate or “crash out of solution” by doing almost anything- it is ultra-sensitive. It’s almost like

the first molecule of salt needs to crystallize, but none of the molecules are volunteering. That first

crystal forming from a supersaturated solution is known as nucleation, and there are two ways to get

a volunteer. The classic way is to add a tiny amount of solid salt (known as a seed crystal)…kind of like

adding a volunteer; this is called heterogeneous nucleation. The second way is to just bump the

solution, or keep waiting, and finally a molecule decides to volunteer on its own. This is hard to

replicate because it is so sensitive, and is known as homogeneous nucleation. Finally, when the solute

precipitates slowly this is known as recrystallization, and is a great way to make super-pure crystals.

One can improve the solubility of a substance by changing the pressure or temperature of the

solution. For solid solutes in a liquid solution (like salt in water), solubility usually, but doesn’t always

increase with temperature. For example, more sugar will dissolve in water if you heat it up. Now

consider soda, where heating soda makes it go flat. So for gases dissolved in liquids, solubility

decreases as temperature increases. Globally, since more oxygen will dissolve in cold water than hot,

this explains why the nutrient-rich seas are near the poles. Finally, solubility increases with pressure-

by selling soda in pressurized containers more carbon dioxide can be added to soda, giving it a better

taste. This is known as Henry’s Law, which is our next topic.

Greasy: will dissolve in

greasy solvents

watery region

OH

watery: will dissolve in

watery solvents (like

water)

A “brick”: hard to

dissolve in anything.

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Name______________________________ Period__________ WS 11.1 Intro to solutions

Please read the previous page completely before answering the questions below

Use the word back to define or answer the following questions

1. Solution

2. Homogeneous (use twice)

3. Solvent

4. Solute

5. Solvation

6. Electrolyte

7. non-electrolyte

8. supersaturation

9. nucleation

10. heterogeneous

11. recrystallization

12. seed crystal

13. saturated

Predict if the following solutions are soluble (S) or insoluble (I).

Solute Solvent Soluble (S) or insoluble (I)?

14. methanol CH2OH Water (HOH)

15. naphthalene

ethanol (CH3CH2OH)

16. methane (CH4) Gasoline

(CH3CH2CH2CH2CH2CH3)

17. hexanol

(CH3CH2CH2CH2CH2CH2OH)

water

18. Describe three ways to increase the solubility of a solute in a solution.

19. Why do whales do most of their eating in the nutrient-rich waters near the poles?

20. What is Henry’s Law?

Word Bank- These may not be used at all, or

may be used more than once

_____A. A homogeneous mixture is a _______

_____B. Excessive solute temporarily

dissolved in a solvent results in _________

_____C. Adding a seed crystal results in ___________ nucleation

_____D. An ionic solute in an _________

_____E. A solution only has one thing visible- it is __________

_____F. A non-ionic solute is a ____________

_____G. The dissolvER in a solution is the _______

_____H. Crystal formation with no additive is ___ nucleation.

_____I. The substance that gets dissolvED in a solution is the ___.

_____J. Solvent molecules surrounding the solute molecules is ____.

_____K. When the first solute molecule precipitates is _____.

_____L. A solution that has all the solute it can dissolve is _____.

_____M. Add _______a to promote heterogeneous nucleation.

_____N. dissolving a solute then crystallizing again is ______.

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Name: ______________________________ Period_________ WS11.2

Henry’s Law

A sure-fire way to increase the solubility of any solute is to pressurize the solution. You see this in

action every time you open a can or bottle of soda: as soon as you open it the pressure in the

container goes down, so the solubility of the carbon dioxide in water goes down, and it precipitates as

bubbles. Put another way, soda makers can dissolve more carbon dioxide in their soda by pressurizing

the solution. Henry’s Law states this fact -that solubility is proportional to temperature -

mathematically. The formula is:

1 2

1 2

S S

P P

Where S1 and S2 are the initial and final solubilities, and the P’s are for pressure. Since the units

cancel and are always positive, any consistent units can be used.

1. 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 298 K, how much will dissolve in

1.0 L of water at 1.0 atm of pressure at the same temperature?

2. 1.8 g of a gas at 2.5 atm of pressure dissolves in 2.0 L of Carbon Tetrachloride at 420 K. What

pressure would the solution have to be at if 4.7 g of the same gas is dissolved at the same volume and

temperature?

3. 4.28 g of a gas at 1.7 atm of pressure dissolves in 2.3 L of water at 527 K, how much will dissolve in

1.0 L of water at 1.0 atm of pressure at the same temperature?

Example: I’d like to make some super-carbonated soda. The solubility of CO2 in water at room

temperature is 3.3 g CO2 per liter of solution at STP (273K, 1 atm). I’d like to triple that. How

much pressure should I apply?

Solution: This one you might be able to do in your head. To triple the solubility we need to triple

the pressure- from 1 atmosphere to 3 atmospheres:

1 2

1 2

S S 3.3 g /L 9.9 g/L (1 atm)(9.9 g/L); ; x = 3 atm

P P 1 atm x (3.3 g/L)

16

Name_________________________ Period________________ ws 11.3

Solution Making Activity

People are making solutions all the time. A cup of tea contains about 1% caffeine by mass, for

example. In this activity you will prepare and dilute several solutions.

Part one: Preparing solutions

Example Deliver 250 mL of a 10% v/v fruit juice solution to your instructor.

Provide your calculation and recipe below.

Solution: ten percent of the solution is juice so…

Calculation Recipe

(.1)(250 mL) = 25 mL juice

Dilute 25 mL of fruit juice to 250 mL with

water.

Deliver each of the following solutions up to your instructor.

1. Deliver 25 mL of a 5% v/v fruit juice solution to your instructor.


Calculation Recipe

The following formulas will be helpful:

mass solutePercent solution by mass x 100

mass solution

volume solutePercent solution by volume x 100

volume solution

Molarity = M = moles of solute

Liters of solution

Dilution formula: C1V1 = C2V2

Where

C = concentration. Usually in moles/liter, it can also be percent by mass, or

percent by volume

V = volume in liters

17

2. Deliver 20 mL of a 31% v/v fruit juice solution to your instructor.


Calculation Recipe

Part two: Diluting solutions

Example: Take ten mL of your 5% fruit juice solution from #1, dilute it to 3%, and deliver it to

your instructor.


Solution: use the dilution formula to find the total volume of your diluted solution.

Calculation Recipe

C1V1 = C2V2

(5)(10) = (3)(x); x = 16.7 mL

Dilute ten mL of your solution from #1 to 16.7

mL

4. Take 5 mL of your 31% fruit juice solution from #2, dilute it to 23%, and deliver it to your

instructor.


Calculation Recipe

Part three: Preparing solutions based on Molarity (L1 only).

Example: Prepare 80 mL of an aqueous 0.5M NaCl solution. Provide your calculation and recipe

below.

Solution: use the Molarity formula, molar mass of NaCl, and volume of your solution to find out

how many grams of salt you need.

Calculation Recipe

40 g NaCl0.5 moles NaCl x x 0.08 Liters solution = 1.6 grams NaCl

Liter of solution mole NaCl

Dilute 1.6 g NaCl to

80 mL with water.

5. Deliver 74 mL of a 0.7M NaCl solution to your instructor.


Calculation Recipe

18

Name________________________ Period_________________ WS 11.4 making solutions

Making Solutions Calculations

An essential skill for any scientist is the ability to make and modify solutions. To be sure your can

calculate how to prepare solutions, Use the formulas below to answer each question.

If you have any questions refer to worksheet 11.3

Type 1: Percent by mass and volume

1. How would you prepare 2 liters of a 35% m/m apple juice solution?


Calculation Recipe

2. How would you prepare 5 mL of an aqueous 31%v/v NaCl solution? Note that the density of NaCl is

2.16 g/mL?


Calculation Recipe

3. How would you prepare a 10% fruit juice solution from concentrate for any volume?

Calculation Recipe

Type 2: Concentration and dilution

4. How would you dilute 2 liters of a 35% m/m apple juice solution down to 19%?


Calculation Recipe

The following formulas will be helpful:

mass solutePercent solution by mass x 100

mass solution

volume solutePercent solution by volume x 100

volume solution

Molarity = M = moles of solute

Liters of solution

Dilution formula: C1V1 = C2V2

Where

C = concentration. Usually in moles/liter, it can also be percent by mass, or

percent by volume

V = volume in liters

19

5. How would you concentrate 1 gallon of a 10%v/v chocolate milk solution up to 24%?


Calculation Recipe

Type 3: Molarity, molality, and mole fraction (L1 only)

6. How would you dilute prepare 50 liters of 2M NaOH solution?

Calculation Recipe

7. How would you prepare 3 liters of a 4m vinegar (C2H4O2 in water) solution?


Calculation Recipe

8. Bonus Question: Provide a recipe for preparing 200 mL of a 25% v/v NaCl solution, then

concentrating it to 12M.

Calculation Recipe

20

Name: ______________________________ Period: _____ WS11.4

Colligative Properties WS I

L1 only

Directions: For each of the following questions use the appropriate relationship or equation to solve

the problem.

1. What are the boiling point and freezing point of a 0.625m Aqueous solution of any nonvolatile, nonelectrolyte solute?

2. What are the boiling point and freezing point of a 0.40m solution of sucrose in ethanol?

3. A lab technician determines the boiling point elevation of an

aqueous solution of a nonvolatile, nonelectrolyte to be 1.12°C.

What is the solution’s molality?

4. A student dissolves 0.500 mol of a nonvolatile, nonelectrolyte solute in

one kilogram of benzene (C6H6). What is the boiling point elevation of the resulting solution?

21

Name: _____________________________________ Period: _____ WS11.5

Colligative Properties WS II

Directions: For each of the following questions use the appropriate relationship or equation to solve

the problem.

1. What is the boiling point elevation and freezing point

depression of a solution containing 50.0 g of glucose

(C6H12O6) dissolved in 500.0 g of water?

2. What are the freezing point and boiling point of each of the

following solutions?

a. 2.75m NaOH in water

b. 0.586m of water in ethanol

c. 1.26m of naphthalene (C10H8) in benzene

3. A rock salt (NaCl), ice, and water mixture is used to cool milk and cream to make homemade ice

cream. How many grams of rock salt must be added to water to lower the freezing point 10.0°C?

4. What is the freezing point and boiling point of a solution that contains 55.4 g NaCl and 42.3 g KBr

dissolved in 750.3 mL H2O?

22

How to ace the solutions unit.

1. Fill in the blanks to review the vocabulary used in this unit:

Today I decided to make rock candy. I mixed sugar with water, so my solute is ________ and the

solvent is _________, and since the resulting mixture was clear and colorless it was

______________. It took a while for the sugar to dissolve, probably because the big chunks of

sugar made the molecular process of ___________ slow. I was surprised to see that this solution did

not conduct electricity; apparently sugar is a _________________. I was also surprised to see how

much sugar dissolved in water, sugar is highly _________________ in water. In fact I put so much

sugar in that when I shook the solution it spontaneously crystallized; apparently the solution was

__________________. Since I didn’t add a seed crystal to the solution, this is specifically known as

_______________ _______________. It was cool watching the first crystal form, that moment

known as ___________________.

I took one of my recrystallized sugar crystals and placed it under an atomic force microscope. I could

see numerous O-H groups, which reminded me of water. I can see why the saying “__________

___________ __________” is used to predict solute-solvent solubilities. I would predict sugar to

be ___________ in hexane (CH3CH2CH2CH2CH2CH3), and ____________ in ethanol (CH3CH2OH).

To increase that solubility, I could _____________, _____________, or add more

_____________, although one of these doesn’t always work (__________________). I know that in

the cases of gases, dissolved in liquids, solubility increases when the solution is _______________.

And pressurizing a solution to dissolve more solute is an example of __________ __________ in

action.

2. Henry’s Law: Solubility is proportional to _____________.

If the solubility of a solute in water is 2.8 g/L at 1 atmosphere pressure, and the pressure is

increased to 3 atmospheres, the solubility will increase to ________ g/L.

Solutions are ground zero in the chemical world- that’s where most of the action is. Gases are

difficult to contain, or even see. Solids don’t react well because of surface area issues. Solutions,

on the other hand, are easy to see, react, store, and work with. It’s no surprise, then, that

solutions are all around us. In the grocery store, at the gas station, in our bodies- solutions abound.

This has been predominantly a hands-on unit. Once we familiarized ourselves with the vocabulary,

we learned how to prepare solutions of different concentrations, and how to change their

concentration. We also learned how solutes affect the melting point and boiling point of solutions.

To help you ace this unit, we begin with a story to sharpen your language skills in this unit. Then we

present some situations where solutions need to be prepared and their concentrations adjusted.

And we finish with a road salt example of colligative properties in action.

Don’t forget to review your worksheets, PowerPoint’s, and labs before you take the solutions test.

Read the story below and fill in the blanks and answer the questions as you go. The story is

designed to include all of the new vocabulary and techniques you have learned.

23

3. Concentration and dilution

For all of these questions you have 29 grams of table salt in 500 mL of solution.

a. Describe how to prepare this solution

b. Calculate the percent salt by mass (water has a density of 1 g/mL)

c. Calculate the percent mass by volume (table salt has a density of 2.16 g/mL)

d. Calculate the molarity of the solution (table salt has a molar mass of about 58 g/mol)

e. This solution will have a (higher/lower) boiling point than pure water, and a (higher/lower)

freezing point than pure water.

f. Calculate the molality of the solution (L1 only; assume 950 g water).

g. Calculate the freezing point of this solution (L1 only; Kf H2O = 1.86 OC/m; use data above

for molality).

h. Calculate the boiling point elevation of this solution (L1 only; Kb H2O = 0.512 OC/m; use data

above for molality).

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