NAME __________________________________ INDEPENDENT PACKET: MOLARITY
Due: ________________
GOAL(S): When you have completed this packet, you will be able to:
1) determine the molar concentration (MOLARITY) of an aqueous solution when given any
TWO PIECES of the following information: a) moles of solute, grams of solute and / or volume of solution
2) determine the molarity of an aqueous solution given its density and percent composition of solute.
3) define the terms: aqueous solution , solute and solvent
There are 2 general types of Molarity problems. So, here's what you wish the text would tell you:
Cutting To The Chase
M O L A R I T Y
There are: and There are :
1 s t e p p r o b l e m s 2 s t e p p r o b l e m s
give you 2 of the 3 variables of the equation
Molarity = moles of solute
Liters of solution
If you are given M, Liters &/or moles of solute ask for or give you the number of grams of solute (rather than the number of moles). Use of the terms grams & molarity indicates that it’s 2 steps
then
Begin with the equation
Molarity = moles of solute
Liters of solution all you need to do is substitute into the equation the two given values and solve for the third.
AND
then, use unit cancellation to convert between grams

moles
Much of what you experience deals with "aqueous solutions". Apple juice, milk, soda, lake water, ocean water, tap water are all examples (complex and simple) of aqueous solutions.
As in all mixtures, an aqueous solution is a physical combination of a SOLVENT and a SOLUTE . The solvent is the substance which does the "dissolving" or "breaks up the solid". The solute is that which is dissolved. The solvent must always (powerful word) be in the greater quantity. In the case of aqueous solutions the solvent is water.
Anything water can dissolve to a certain point may be considered to be a water-soluble solute.
Aqueous solutions are mixtures . They are NOT substances. Rather they are physical combinations of at least
2 substances, which blend so as to become a homogeneous mixture .
An aqueous solution is NOT a substance, because the ratio between the solvent and solute can be along a pretty large continuum (although the solvent must always be in the greater molar concentration). The ratio is not fixed. Also, while bonds may be broken, no new bonds are made.
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As a general rule, when made correctly you would be unable to differentiate (visually) between a glass of pure water and a glass of concentrated salt water (NaCl
(aq)
).
When the symbol of the aqueous solution is written, the solute is recorded first and the water solvent is recorded with the abbreviation (aq). Thus, the formula NaCl
(aq) is known as a solution (a homogeneous mixture) of sodium chloride dissolved COMPLETELY in water.
These aqueous solutions may be described in terms of their concentration such as their MOLARITY. If you were to update your Concept Map of Matter you would have the following:
MATTER may be divided into 2 large categories
H e t e r o g e n e o u s M a t t e r H o m o g e n e o u s M a t t e r
exemplified by
Mixtures like emulsions & solid mixtures or like those
that often settle
is divided further into
two different groups
S u b s t a n c e s Mixtures
there are 2 types that don't settle out are called
out into layers
suspensions elements compounds solutions
like
alloys &
gas
mixtures made of only 1 type of made of 2 or more or for our class
atom different species
eg) sand + water eg) Fe
(s)
eg) C
2
H
6(g)
eg) aqueous
solutions
There are many types of homogeneous solutions. The " aqueous solution " is just one type. It is however, the most important to the understanding of the material contained in this Chemistry syllabus.
SUMMARY: An aqueous solution is not a chemical substance ... it is made up of multiple substances. physical, water
Aqueous solutions are ______________ combinations of ______________ and of solute, solvent some __________________. In an aqueous solution, water acts as the ________________. homogeneous
While they are not substances, aqueous solutions are ___________________, in that they are molarity
uniform in their properties. Concentration can be expressed in term of ______________.
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A) (symbol: M, or M ): a term of concentration which expresses the number of moles of solute completely dissolved in 1 liter of solution .
B) unit: may be: M , or M but it really is: moles/Liter or mol/L (read as moles per Liter)
1) the unit expresses the definition: M , or M is the number of MOLES per LITER of solution a) very often when you see M or M, you may "say" it as the word MOLAR
so, 0.50 M HCl(aq) is read or spoken as: a 0.50 MOLAR solution of HCl
C) Equation: M = # of moles of solute
# of Liters of solution
See Table T of your Reference Charts. But!
... notice how the definition gives you the
equation, and the equation gives the definition.
1)  The volume of the solution must be in LITERS
2) If volume is given in milliliters (mL), you must convert from one to the other.
Remember, 1 L = 1,000. mL
3) eg) Calculate the molarity of a 2.000 x 10 3 milliliter solution containing 3.0 moles of HCl(g)
M = # of moles of solute
# of liters of sol'n
M = 3.0 moles HCl____
2.0 Liters of sol'n
M = 1.5 M
4) eg) Calculate the molarity of 2.0 moles of HCl(g) dissolved in a 500 .
mL solution .
M = # of moles of solute
# of liters of sol'n
M = 2.0 moles HCl did you remember to convert the 500. mL of solution
0.500 Liters of sol'n
 to 0.500 Liters of solution ???????
M = 4.0 M
The answer has 2 sig figs because 2.0 has the fewer number of sig figs.
268

5) Practice for 1 step Molarity Problems (always begin with the equation, and then THINK!!) a) Calculate the molar concentration of a 900 .
mL solution in which 1.30 moles of
CuSO
4
are dissolved completely.
ANS: 1.44 M
Equation: M = moles
Liters
Substitution:
Answer: _________ M
Ans: 0.10 moles of NaCl
(note only 2 sig figs)
b) How many moles of NaCl are dissolved in a 0.20 M solution of 500. mL?
Equation:
Substitution:
Ans: _________ moles c) Calculate the volume (Liters) of a 0.500 M solution made with 3.50 moles of NaOH?
Equation:
Substitution:
Ans: 7.00 L
(note 3 sig figs)
Ans: ___________ Liters
SUMMARY: dividing, solute
The dimension molarity is found by _______________ the number of moles of _______ solution
by the number of liters of _________________. The most important thing to remember
Liters
is that the VOLUME of the solution must be expressed in the unit, ___________.
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
READ THIS SECTION ABOUT 2 STEP MOLARITY PROBLEMS!!!

C) Many times you will not be given the number of moles DIRECTLY .
When the question is a molarity question, BUT gives (or asks for) the number of grams of solute then you will need to use UNIT CANCELLATION.
Read the following question carefully. Note how the key words that clue you in on the fact that
it is a 2 step problem, have been identified.
1) eg) A student made a 5.0 L solution by dissolving 80.0 grams of NaOH(s) into water.
Calculate the molar concentration (molarity) of the solution. key words for
2-step problems
Start: You can recognize this as a molarity problem.
M = moles of solute
Liters of solution
Begin with the molarity equation
& try to fill in what you know…. b)
You know the Liters of solution, but nothing else that fits the equation.
However, you do step 1 know the grams
: calculate moles from grams;
, and you can convert to moles from grams. moles of NaOH = 80.0 grams | 1 mole of NaOH|
40 grams
⇚
(Na + O + H)
23g + 16g + 1g (calculating a mole mass) therefore: moles of dissolved NaOH = 2.00 moles
(note 3 sig figs based on the "given") c) step 2 : substitute into the M formula:
M = # of moles of solute
# of liters of solution
M = 2.00 moles NaOH
5.0 Liters of solution
M = 0.40 mol/L or 0.40 M or 4.0 x 10 each is equally correct, with 2 sig figs
-1
molar solution ….
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Ans: 1.19 M
3) Practice:
N.B. The following problems have been calculated using mole masses rounded to the nearest whole number,
for convenience. In lab, we would use the decimalized values. If you use the decimalized values for theses
problems, your answers would be just slightly different.
a) A 1.50 x 10
3
cm
3
solution has 100 . grams of KOH completely dissolved. Calculate
Ans 0.552 M did you recall 1 mL = 1 cm
3
the molarity of the solution key words for 2-step problems
Ans: 0.85 M b) Calculate the molar concentration of a 750 mL solution when 54.0 grams of NaNO
3(s)
Why only 2 sig figs?
are dissolved completely in water. clues for a 2-step problem c) Calculate the molarity of a 5.00 x 10
completely dissolved in water.
-1
L solution with 94.5 grams of C
12
H
22
O
11(s) clues for a 2-step problem
Ans 2.0 M d) Calculate the molarity of a 250 mL solution with 53 grams of Na
2
CO
3(s) completely
dissolved
Ans: 1.0 M e) Calculate the molarity of a 2.0 L solution with 320 grams of CuSO
4(s)
completely
dissolved.
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1) THIS IS A 1 STEP
*
: How many moles of NaI are required to make a 750. mL solution , of 0.80 M? ans. c
(*You are given M, and the volume in Liters, simply plug into the equation and solve for moles…
Do you need to use the molar mass?) a) 0.29 mole b) 7.0 moles c) 0.60 mole d) 0.10 mole ans. a
2) THIS IS A 2 STEP**: What is the number of grams of AlCl
3
dissolved in 0.500 L of a 0.900 M solution?
**(Note, you're asked for grams, but you must first use the equation to get to moles and then use
unit cancellation to calculate the grams... Will you need the molar mass?) a) 59.4 g b) 126 g c) 229 g d) 74.2 g ans d
3) How many grams of KOH are dissolved in a 2.0 x 10
3 mL solution of 1.5 M KOH(aq)?
How many steps is this? 1 or 2

How do you know?? a) 150 g b) 140 g c) 90 g d) 170 g ans b
4) A student used 11.0 grams of NaOH to make a 3.0 L solution. Determine the molarity of the solution. a) 2.5 M b) 0.092 M c) 0.50 M d) 1.8 M
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5) How many grams of LiCl are required to prepare 80.0 mL of a 0.150 M solution? a) 0.504 grams b) 0.761 grams c) 1.675 grams d) 3.19 grams ans. a
6) Calculate the number of liters of solution made from making a 0.50 M solution by dissolving 2 moles
of KOH into water. (How many steps is this? 1 or 2

How do you know??) a) 1 b) 2 c) 3 d) 4 ans. d
7) At 25ºC, 0.0018 grams of NaCl dissolves in enough 1-petanol (CH
3
CH
2
CH
2
CH
2
CH
2
OH) to give 0.10 L of
solution. Calculate the molar concentration of the solution.
(What do you have? What do you need?) a) 0.0031 b) 0.00031 c) 0.00016 d) 0.018 ans. b
8) A sugar-free soft drink contains 7.9 mg of saccharin, C
7
H
5
SNO
3
in 1.0 ounce (31 mL). What is the molar
concentration of saccharin in an ounce of the drink? (Note: you are given mg of solute, not grams….) a) 4.5 x 10 b) 1.4 x 10 -1 c) 3.2 x 10 -2 d) 1.4 x 10
-3
-3
M
M
M
M ans. d
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This section deals with a slightly more advanced look at concentration . It uses the issue of density as well.
When given the molarity and density of a solution , with a request to determine a volume or mass of solute, try this approach.
Always begin with M = mol/L…
Then Think about what you know and what you need to find!
• Assume 1 Liter of solution and use the density to determine the mass in grams of the solution.
• Use the % to determine what mass of that solution is actually due to the solute (the rest is due to water)
•
Calculate the moles of solute using unit cancellation and
•
Use the molarity equation (with a volume of 1.00 Liter) to find molarity. You'll note that this method
equates the number of moles from the last step to the molarity
9) What is the molarity of NaBr in a solution that is 40.0% NaBr and has a density equal to 1.42 g/mL?
• Think!
: The density tells you that there are 1.42 grams for 1 mLof solution, and 40.0% of the mass is NaBr ....
You want molarity
• Attack: The density given to you is in milliliters . But you need LITERS to use the molarity equation. Thus you must get
the density value into liters. Assume 1 Liter of solution and multiply the density value by 1,000. This does 2 things…
it increases the number of grams (the numerator) by 1,000, but it converts the mL unit to 1,000 mL or 1 Liter. Now,
you are poised to use the molarity equation. In the above example when I multiply 1.42 g/mL by 1,000 I get
1,420 grams/L and 40% of those 1,420 grams are NaBr …Hey! This can be solved!
How did I figure out that value?

Recall: 1,000 mL = 1.00 Liter

1.42 g/mL means that 1.00 mL of solution has a mass of 1.42 grams.

When 1.00 mL=1.42 grams & you assume 1,000 mL, then the mass of solution is 1,420 grams.
• Use the 40.0% of solution mass that is due to NaBr ...& find the grams of NaBr (0.400 x 1420) = 568 grams of NaBr
• Calculate by converting 568 grams of NaBr to moles using unit cancellation moles NaBr = 568 grams NaBr | 1 mol |
103 grams NaBr
• Use that mole value to plug into the molartiy equation ...and keep assuming 1.00 Liter of solution.... you’ll find
that as long as you have assumed 1 L of solution, the Molarity of the solution = the calculated number of moles...! Nice ans. a a) 5.51 M b) 6.42 M c) 5.78 M d) 6.89 M
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10) The density of a 72.0 % by mass
(or by weight) aqueous solution of ethanol (CH
3
CH
2
OH) is 0.8644 g/cm 3
Calculate the molarity of the solution.
.
• Think!: The density tells you that 1mL of solution is 0.8644 grams .. and 72.0% of the mass is ethanol .... You want molarity
• Assume you have 1.00 Liter of solution. So, you have * 864.4 grams of solution. (mass of 1.00L sol’n = the density x 1000)
• Use the fact that 72.0% of every 1.00 mL is ethanol. So ... use the percentage to determine the mass due to ethanol in solution
• Calculate the moles of ethanol by converting grams to moles
• Use the molarity equation ... assume 1.00 Liter of solution... ans. c a) 13.0 b) 12.2 c) 13.5 d) 14.1
11) What is the molarity of a 21.9% by weight of a AgNO
3
solution with a density of 1.220 g/cm
3
?
T hink: The density per mL of solution is 1.220 grams ... and 21.9% of the mass is AgNO
3
.... You want molarity a) 1.66 b) 1.42 c) 0.158 d) 1.57 ans. d
12) A 23.9% solution by weight of citric acid (H
3
C
6
H
5
O
7
) in water has a density of 1.10 g/cm 3 . Calculate the
molarity of the solution. a) 1.55 b) 1.37 c) 2.01 ans. b d) 0.773
13) The density of a 44.0% by mass aqueous solution of ethylene glycol (C
2
H
6
O
2
) is 1.0586 g/cm
3
. What
is the molarity of the solution? ans. c a) 4.35 b) 6.78 c) 7.50 d) 8.19
275

QUIZ YOURSELF ! (SELF-QUIZ #1)
___1. Which of the following represents a homogeneous mixture ? a) Br2(l) b) Na2S(s) c) LiF(aq) d) Ne(g)
___2. Calculate the molarity of a solution of sodium oxalate (Na
2
C
2
O
4
), containing 33.5 g of solute
in 100.0 mL of solution ? a) 1.50 M b) 2.50 M c) 3.75 M d) 4.00 M
___3. How many moles of NaCl are needed to prepare 5.0 L of a 2.0 M solution of NaCl
(aq)
? a) 2.5 b) 5.0 c) 10. d) 20.
4. Define the term: Molarity ______________________________________________________
____________________________________________________________________________
5. Define the term: Solute ____________________________________________________________
___6 The density of a 48.0% by mass aqueous solution of H
2
SO
4
is 1.3783 g/cm
3
. What is the molarity of the
solution? a) 6.75 M b) 10.2 M c) 8.50 M d) 4.06 M
ANSWERS: 1 c 2. b 3. c 4. a term of concentration referring to the number of moles of solute dissolved completely in 1 liter of solution. 5. the solid, liquid, or gas which is dissolved into the solvent. The solute is always in the lesser molar quantity. 6. a
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IF AT FIRST YOU DON'T SUCCEED......(SELF - QUIZ #2)
___1. What is the molarity of a solution of KNO
3 that contains 404 grams of KNO
3
in 2.00 liters of solution? a) 1.00 b) 2.00 c) 0.500 d) 4.00
___2. Calculate the total number of grams of KCl dissolved in 1.00 L of 0.200 molar solution. a) 7.46 g b) 14.8 g c) 22.4 g d) 29.8 g
___3. Calculate the molarity of a solution that contains 112 grams of KOH in 2.0 L of solution. a) 1.0 M b) 2.0 M c) 3.0 M d) 4.0 M
4 Define the term: Homogeneous Solution: ______________________________________________
__________________________________________________________________________________
5 Which is has the greater mass of NaOH:
a) 0.504 L of 0.0025 M NaOH
(aq)
OR b) 0.00393 L of 0.400 M NaOH
(aq)
?
___6) The density of a 56.0% by weight aqueous solution of methanol (CH
3
OH) is 0.9046 g/cm
3
. What is the
molarity of the solution? a) 17.0 M b) 8.75 M c) 15.8 M d) 11.6 M
ANSWERS: 1 b 2. b 3. a 4) a uniform mixture (when evaluated by the naked eye, for our course) composed of at least one solute evenly distributed/dissolved in solvent
5) b 6) c
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