In This Lesson:
Molarity
(Lesson 2 of 4)
Today is Friday (!),
January 8th, 2016
Stuff You Need:
Calculator
Periodic Table
Polyatomic Ion
List
Worksheet
Pre-Class:
Earlier in the year I used hydrogen peroxide
in a lab but mentioned it was different from
normal peroxide. How come?
Also: Please find a worksheet and have your
solubility curve packets out.
Today’s Agenda
•
•
•
•
Concentrations
Molarity
Percent solutions
Dilutions
– In other words, today we’re going to talk about the
strength of solutions.
• Colligative Properties
• Where is this in my book?
– P. 480 and following…
By the end of this lesson…
• You should be able to calculate the
concentration of a solution in several different
ways.
• You should be able to calculate freezing-point
depression and boiling-point elevation.
Practice
• Molarity Practice Chart and Problems
– First three rows.
– Use the first column as a sample!
Molarity Practice Chart and Problems
Rest of key online…
Name
Calcium
iodide
Sodium
chloride
Formula
CaI2
NaCl
Electrolyte
or nonelectrolyte
Ions in
solution
Potassium Aluminum
carbonate hydroxide
K2CO3
Al(OH)3
Electrolyte Electrolyte Electrolyte Electrolyte
Ca2+, 2 I-
Na+, Cl-
2 K+, CO32- Al3+, 3 OH-
Glucose
Sulfuric
acid
C6H12O6
H2SO4
Nonelectrolyte
Electrolyte
N/A
2 H+, SO42-
Let’s start with the easy, shall we?
• Recently we talked about solutions, and today
we started by discussing the difference
between 3% and 30% H2O2.
• How are those numbers calculated?
http://www.acne.org/images/products/img-442.jpg
Percent Solutions
• There are two types of “__% Solutions.”
– Percent by volume.
– Percent by mass.
• Decide which to use by looking at the
denominator’s units.
• Let’s start with volume, and here’s your
practice problem:
• What is the percent by volume of 5 g of CaCl2
salt in 250 mL solution?
Percent by Volume
• What is the percent by volume of 5 g of CaCl2
salt in 250 mL solution?
• This is a percent by volume problem, since
we’re looking at a 250 mL solution.
• 5 g out of 250 mL is 2%.
• (5 g / 250 mL) * 100 = 2%
• If the solute is a solid or gas:
mass or volume of solute
Percent by volume 
 100
volume of solvent
Percent by Volume
• If the solute is a liquid, you have to use this formula:
volume of solute
Percent by volume 
 100
volume of solution
• As an example, imagine putting 5 mL of methanol
into 100 mL of water.
– The volume of solution (denominator) is now 105 mL.
• Fun fact: Liquids (or other stuff) that can be
infinitely mixed with others are known as miscible.
Percent By Mass
• Percent by mass is very similar.
• Note, however, that the denominator is the
total mass of solution (and not solvent).
– This is because volume is not meaningfully
impacted by a solid or gas solute, but mass is.
mass of solute
Percent by mass 
 100
mass of solution
Okay. Why?
• Why the differences in calculations?
• Imagine a glass of 50 mL of water.
– It’s also 50 g.
• If I shake 10 g of sodium chloride into it, does
the water level change?
– Nope.
• But does the mass change?
– Yep.
• Hence, volume of solvent but mass of solution.
Practice
• Molarity and Solution Percent worksheet
– #1, 4, 5 (#1 is challenging)
Molarity
• At some point this year you may have seen me
label acids or other liquids with something like:
– “6 M HCl”
• The big capital M stands for Molar, as in not your
teeth.
– 6 M HCl is pronounced “Six molar hydrochloric acid.”
• Molarity [that’s the name of this concept] is
another way to measure the concentration or
dilution of a solution.
– The higher the molarity, the stronger the solution.
Calculating Molarity
• Calculating molarity is fairly simple.
• Here’s the equation:
– M is the symbol for Molarity
moles of solute
M
liters of solution
Need a Reminder?
Molarity
Equation
Mole
Conversion
mol
g
M
L
mol
molar
mass
Molarity Practice Problem
• Intravenous (IV) saline solutions are often
administered to patients in the hospital. One
saline solution contains 0.90 g NaCl in exactly
100 mL of solution. What is the molarity of
the solution?
• 0.90 g NaCl = 0.0154 mol NaCl
• 100 mL = 0.100 L
• (0.0154 / 0.100) = 0.154 M
Molarity Practice Problem
• Household laundry bleach is a dilute aqueous
solution of sodium hypochlorite (NaClO). How
many moles of solute are present in 1.5 L of
0.70 M NaClO?
• 0.7 M = moles of NaClO / 1.5 L
• Solve for “moles of solute:”
• 1.05 mol NaClO
Practice
• Molarity and Solution Percent worksheet
– #2, 3
Practice
• Molarity Practice Chart and Problems
– Shaded four rows.
– Use the first column as a sample!
• Molarity Practice Chart and Problems
– Bottom two rows.
– For cation/anion molarity, multiply the coefficient
in front of each ion in the “Ions in Solution” row by
the molarity in the “Molarity of Solution” row.
– Use the first column as a sample!
Molarity Practice Chart and Problems
Rest of key online…
Name
Calcium
iodide
Sodium
chloride
Potassium Aluminum
carbonate hydroxide
Glucose
Sulfuric
acid
Ions in
Solution
Ca2+, 2 I-
Na+, Cl-
2 K+, CO32- Al3+, 3 OH-
N/A
2 H+, SO42-
Molarity
of
Solution
0.045 M
1M
0.50 M
0.75 M
1.5 M
0.51 M
Molarity
of Cation
1 x .045 M
= 0.045 M
1 M Na+
2 x 0.5 M
= 1 M K+
0.75 M
Al3+
-
2 x 0.51 M
= 1.02 M
H+
Molarity
of Anion
2 x .045 M
= 0.090 M
1 M Cl-
0.5 M
CO32-
3 x 0.75 M
= 2.25 M
OH-
-
0.51 M
SO42-
Computer Practice
• Okay, now it’s time to do a little reinforcement using
the computers.
• Head to Quia and open the quiz called:
– Salts and Solubility
• From there, it’s a pretty self-explanatory quiz that will
have you visiting another page at the same time (PhET,
an awesome website).
• You will be looking at various salts dissolving into water
and making molarity calculations based on them.
– Username: [firstnamelastname]832
– Password: Gleicher[house/apt. #]
Dilutions of Molar Solutions
• Sometimes chemists need to create dilutions
using known molarities and volumes.
• For example, during our Baking Soda/HCl lab, I
used 6 M HCl. However, the school receives
hydrochloric acid in big jugs of 12 M HCl (nasty
stuff).
• I needed to dilute it down to 6 Molar, and
here’s how I did…
Molar Dilutions
• Because we’re not changing how much solute
is around during a dilution (we’re just adding
water, the solvent), we can use the following
formula:
• M1V1 = M2V2
• In other words, like with gas laws, when
molarity goes up, volume typically goes down.
– This is Boyle’s Law, except with molarity instead of
pressure.
Molar Dilution Practice Problem
• How many milliliters of aqueous 2.00 M
MgSO4 solution must be diluted with water to
prepare 100.0 mL of aqueous 0.500 M
MgSO4?
• M1V1 = M2V2
• 2.00 M * V1 = 0.5 M * 100 mL
• V1 = 25 mL
Molar Dilution Practice Problem
• You put 2 moles of HCl into 312 mL of water.
If you wanted to make a 1 M dilution, to what
volume would the original solution need to be
diluted?
• Molarity of the solution = 6.41 M
• 6.41 M * 312 mL = 1 M * V2
• V2 = 2000 mL
Practice
• Molarity Practice Chart and Problems
– Reverse side (all)
Finally…
• …our last topic of the lesson. Let’s introduce it
with a photo:
http://www.bceo.org/images/SaltBrineSystem_e.JPG
So why use salt on a road?
• Salt on a road lowers the freezing point of water, thus
reducing icing.
– All those dissolved salt ions interfere with the water’s ability to
make an orderly crystalline arrangement and thus freeze.
• You can also use this phenomenon to make ice cream.
– Put your ingredients in a plastic bag and put the bag in a bucket of
ice water.
• The ingredients (milk, sugar) have lots of dissolved solutes.
– Add lots of salt to the ice water.
– The ice water temperature will drop as the ice absorbs heat from
the water, but the water can’t freeze.
– In turn, the ice water absorbs heat from the bag of ingredients,
which needs to drop significantly below freezing to…freeze.
– Eventually you get that nice softish ice cream texture.
Colligative Properties
• In fact, this is one of three main colligative
properties we’re going to learn:
1. Freezing-Point Depression
2. Boiling-Point Elevation
3. Vapor-Pressure Lowering
• A colligative property is one that depends only
on how much solute you have, not what kind
of solute it is.
– Like osmosis. You remember biology, right?
1. Freezing-Point Depression
• Every mole of dissolved solute particles lowers
the freezing point of 1000 g of water by 1.86°C.
• Example Problem:
• What is the freezing point of these three
solutions?
A. 1000 g of 1 M CaCl2 solution.
B. 58.44 g NaCl in 1000 g water.
C. 180.15 g C6H12O6 in 1 kg water.
• Well A is easy: -1.86°C.
1. Freezing-Point Depression
• Wait a sec…maybe A isn’t so easy.
• When a unit of CaCl2 dissolves, it creates three
ions: Ca2+, Cl-, and Cl-.
– In other words, 1 mol CaCl2 dissolves to 3 mol of
particles.
• 3 mol * 1.86°C/1 mol = -5.58°C.
• Similarly, for Part B, 58.44 g NaCl = 1 mol NaCl.
– That’s 2 mol of particles, so -3.72°C.
1. Freezing-Point Depression
• Now for Part C: 180.15 g C6H12O6 in 1 kg water.
• That’s 1 mol of sugar and it’s in one kg of water,
so it should be a freezing point of -1.86°C.
• But what about dissociation?
– Sugar doesn’t dissociate since it’s not an electrolyte,
so -1.86°C is correct.
1. Freezing-Point Depression
• One last freezing-point depression question:
• Suppose you’re the person in charge of Doylestown’s
road department, and therefore are in charge of the
de-icing process that happens usually every winter.
• You have two salt choices to spread on the roads:
NaCl or CaCl2.
• Assuming they’re equal in cost, which do you choose?
– CaCl2 – every one mole of calcium chloride dissociates into
three moles of ions, whereas only two are formed from
sodium chloride.
2. Boiling-Point Elevation
• Similarly, for every mole of solute in 1000 g of
water, the boiling point increases by 0.512°C.
– This is NOT why some recipes call for “salting”
water before boiling.
– Think of how much salt you’d have to add. It’s just
for taste.
• Our last colligative property is closely related.
3. Vapor-Pressure Lowering
• Recall that vapor pressure is the pressure of a
gas above a liquid in a container.
• Dissolved solutes will lower this pressure,
since lots of solutes will cause solvent
molecules to bind to them, making it harder
for evaporation to occur.
– Don’t worry about numbers here.
So…
• After all of this, there’s one more (confusinglynamed) calculation we should know how to do:
molality.
– Molarity is mol/Lsolution
– Molality is mol/kgsolvent
• These are equal when water is the solvent, but
different when it’s not.
• Molality is particularly useful for calculations of
colligative properties, especially for non-H2O stuff.
– Molality, oddly, uses the symbol m (like meters) or b.
Colligative Property
Equation
• ΔT = K * m * i
• ΔT = change in temperature
• K = constant
• Kb for boiling-point elevation
• Kf for freezing-point depression
• m = molality
• i = number of dissociated ions
• K is different based on the solvent
and is usually in units of °C/m.
• For water:
• Kb = 0.512°C/m
• Kf = 1.86°C/m
Solvent
Kb (°C/m)
Water
0.512
Ethanol
1.19
Benzene
2.53
Cyclohexane
2.79
Acetic Acid
3.07
Phenol
3.56
Nitrobenzene
5.24
Camphor
5.95
Solvent
Kf (°C/m)
Water
1.86
Acetic Acid
3.90
Benzene
5.12
Nitrobenzene
7.00
Phenol
7.40
Cyclohexane
20.2
Camphor
37.7
Example
• What is the freezing-point depression of a
glucose solution containing 20.0 g of C6H12O6
in 50.0 g H2O? (Kf = 1.86°C/m)
• 20 g C6H12O6 = 0.111 mol C6H12O6
• 0.111 mol C6H12O6 / 0.05 kg H2O = 2.22 m
• ΔTf = Kf * m * i
• ΔTf = 1.86 * 2.22 * 1 (since it’s molecular)
• ΔTf = 4.13°C (realistically it’s -4.13°C)
Example Deux
• 500 kg of CaCl2 are spread across Rt. 611 in winter. If
we assume that it comes in contact with 1000 kg of
H2O, what will the freezing point of that water become?
• Wow. Where to begin?
•
•
•
•
•
•
•
•
We have 500 kg of CaCl2 (solute). Let’s make it moles.
500 kg CaCl2 = 500,000 g CaCl2 = 4481.09 mol
4481.09 mol in 1000 kg H2O is 4.48 m.
BUT! CaCl2 splits into three ions in solution, so the real
molality is 4.48 * 3.
ΔT = Kf * m * i
ΔT = 1.86°C/m * 4.48 m * 3
ΔT = 24.99°C
So the new freezing point is -24.99°C.
Practice
• Colligative Properties Worksheet (except #3)
• Note: Some of these are challenging. Skip #3
for now.
• About #3:
– “25.0 mass percent solution” means that 25% of
the total solution mass is, in this case, ethylene
glycol.
– You’ll have to make up a “kg” value for water, but
be careful with it.
Closure – TEST QUESTION
• Determine the molar mass of a molecular
compound that will raise acetic acid’s boiling point
to 121°C when 25.8 grams of solute are added to
135.0 g acetic acid. Kb = 2.9°C/m, acetic acid’s
boiling point is 118.0°C.
• ΔTb = Kb * m
• ΔTb = 3°C
• 3°C = 2.9°C/m * m
• m = 1.03
Closure – TEST QUESTION
•
•
•
•
•
•
•
•
•
Okay, so the molality of the compound is 1.03.
We know that there were 135.0 g acetic acid.
m = mol / kg
1.03 = mol / 0.135 kg
mol = 0.13905
g / molar mass = mol.
We know 25.8 g = 0.13905 mol.
0.13905 = 25.8 g / molar mass
Molar mass = 185.54 g/mol
Another way to think of this…
• Colligative property problems almost always come
down to the two main equations:
T  K b  m
molsolute
m
kg solution
• From only reading the problem, I can fill in this
much:
121  118  2.9C/m  m
molsolute
m
0.135 kg
• Start with the left equation and you’re off!
Closure
• One last little thing.
• Once you have that solution, now freezing at a
lower temperature than normal, you have
yourself a eutectic mixture.
• You don’t need that word for a test, but it’s
fun nonetheless.
Closure Deux
• An enormously wide-ranging review video in
under five minutes:
– TED: Cynthia Chubbuck – The Chemistry of Cars