Post on 18-Mar-2018
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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.
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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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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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Chemadventure Chapter 11: Solutions
= 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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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
Heating Solutions
A solubility surprise:
Global implications
Most solids
Heating makesMore Soluble
Most gases Less Soluble
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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
Solutions
HgAgFillingsSolid-solid
saltsH2OOceanSolid-liq
Acetic acidH2OVinegarLiq-Liq
CO2H2OSodaGas-liquid
O2N2AirGas-gas
soluteSolventEx.Type
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Chemadventure Chapter 11: Solutions
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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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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
• 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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Chemadventure Chapter 11: Solutions
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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Chemadventure Chapter 11: Solutions
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
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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)
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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.
Provide your calculation and recipe below.
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
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2. Deliver 20 mL of a 31% v/v fruit juice solution to your instructor.
Provide your calculation and recipe below.
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.
Provide your calculation and recipe below.
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.
Provide your calculation and recipe below.
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.
Provide your calculation and recipe below.
Calculation Recipe
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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?
Provide your calculation and recipe below.
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?
Provide your calculation and recipe below.
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%?
Provide your calculation and recipe below.
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
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5. How would you concentrate 1 gallon of a 10%v/v chocolate milk solution up to 24%?
Provide your calculation and recipe below.
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?
Provide your calculation and recipe below.
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
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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?
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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?
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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.
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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).