Solutions Part II
DHS Chemistry
Chapter 15
I. Concentrations of Solutions
• The concentration of a solution is a
measure of the amount of solute
that is dissolved in a given
quantity of solvent.
Amount of solute vs. amount of
water
Dilute vs Concentrated
Little solute
a lot of solute
Very Concentrated
Less Concentrated
Concentrated solutions
• A concentrated solution is one that
contains a high concentration of
solute.
Dilute solutions
• A dilute solution contains a small
concentration of solute.
Pictorial Representation
Pictorial Representation
• There are several ways to express
concentration. These include:
percent solutions (by volume or
mass), molarity, or molality.
A. Percent
Solutions
Percent Solutions
% solute =
amount of solute _
100
TOTAL amount of solvent
solution
3 types:
(%m/m) same units
(%v/v)
same units
(%m/v) needs to be g/mL
How much vinegar is just
acetic acid?
5% of vinegar is acetic acid
Percent by Volume
% by volume (% (v/v)) = volume of solute
100
Volume of solution
(solute + solvent)
units must be the same
Tip: watch out for the wording. You may
need to add the volume of the solute and
solvent to get the volume of the total
solution
Percent by Mass
% by mass (% (m/m)) = mass of solute
100
mass of solution
(solute + solvent)
*units must be the same
Tip: watch out for the wording. You may
need to add the mass of the solute and
solvent to get the volume of the total
solution
Percent Mass by Volume
% mass by volume (% (m/v) = mass of solute (g)
100
volume of solution (mL)
*units must g/mL
Ex 1: 20 mL of alcohol is diluted with
water to a total volume of 65 mL. What
is the percentage of alcohol, by
volume?
%(v/v) = Volume of solute
100
Volume of solution
%(v/v)=
?%
30.8%
100
Alcohol by
volume
= 20 mL alcohol
65 mL H2O
65 mL alcohol + water
20 mL
alcohol
30.8% of this solution is
alcohol. The rest is water.
Ex 2 : A solution containing 7 g
of NaCl in 165 g of solution.
158 g of solvent (water)
What is the percent of NaCl by
mass?
4.24%
? % NaCl (m/m)= 7 g NaCl
100
165 g solution
Solution = Solute + solvent
Solution = 7 g + 158 g
Children’s Dose vs
Adult Dose
Diphenhydramine hydrochloride
(active ingredient in allergy medicine
like Benadryl)
How do you feed a child medicine
when one tablet is too strong?
Liquid dose for children has been
diluted to 12.5 mg for every 5 mL of
medicine
What percent by mass of
diphenhydramine hydrochloride is
in the solution?
Liquid dose for children has been
diluted to 12.5mg for every 5mL of
medicine
.0125 g = .250% (m/v)
5 mL
100
EX 3: A saline solution containing 3.5 g of
NaCl in 62.5 mL of solution. What is the
percent of NaCl, by mass.
5.60%
? % = 3.5 g NaCl
100
62.5 mL solution
EX 3: A saline solution containing 3.5 g of
NaCl in 62.5 mL of solution. What is the
percent of NaCl, by mass.
5.60%
? % = 3.5 g NaCl
100
62.5 mL solution
Ex 4: What volume of ethanol is needed to
produce 120 mL of a 22.3% (v/v) ethanol
solution?
%(v/v)=
22.3 % ethanol by = 26.8
? mL ethanol
volume (v/v)
100
120 mL solution
Ex 5: What volume of a 2.8% (m/v) glucose
solution would you need to deliver to a
patient who needs 750 mg of glucose?
2.8 %
100
glucose by =
volume (m/v)
0.750 g glucose
? mL
26.8
mL glucose
solution
Practice
1.
If 10 mL of pure acetone is diluted with
water to a total solution volume of 200 mL,
what is the percent by volume of acetone
in the solution?
2. A bottle of hydrogen peroxide is labeled
3.0% (v/v). How many mL of H2O2 are in a
400.0 mL bottle of this solution?
3. Calculate the grams of solute required to
make 250 g of 0.10% MgSO4 (m/m).
4. A solution contains 2.7 g CuSO4 in 75 mL of
solution. What is the percent (m/v) of the
solution?
1. If 10 mL of pure acetone is diluted with
water to a total solution volume of 200 mL,
what is the percent by volume of acetone
in the solution?
%(v/v)=
? %% acetone by = 10 mL acetone
5.00
volume (v/v)
100
200 mL solution
2. A bottle of hydrogen peroxide is labeled
3.0% (v/v). How many mL of H2O2 are in a
400.0 mL bottle of this solution?
%(v/v)=
mL H2O2
3.00 %H2O2 by
= 12.0
? mL
volume (v/v)
100
400. mL solution
3. Calculate the grams of solute required
to make 250 g of 0.10% MgSO4 (m/m).
%(m/m)=
0.10 % MgSO4 by = 0.250
? g MgSO4
volume (v/v)
250 g solution
100
4. A solution contains 2.7 g CuSO4 in
75 mL of solution. What is the percent
(m/v) of the solution?
%(m/v)=
3.60
? %
100
CuSO4 by
volume (m/v)
= 2.7 g CuSO4
75 mL solution
Practice
1.
If 10 mL of pure acetone is diluted with
water to a total solution volume of 200 mL,
what is the percent by volume of acetone
in the solution? 5.00% acetone (v/v)
2. A bottle of hydrogen peroxide is labeled
3.0% (v/v). How many mL of H2O2 are in a
400.0 mL bottle of this solution?
12.0mL H2O2
3. Calculate the grams of solute required to
make 250 g of 0.10% MgSO4 (m/m).
0.250g MgSO4
4. A solution contains 2.7 g CuSO4 in 75 mL of
solution. What is the percent (m/v) of the
solution? 3.60% (m/v)
B. Molarity
Molarity
• Molarity (M) is the number of moles
of a solute dissolved per liter of
solution.
Molarity
• Molarity is also known as molar
concentration and is read as “
__#__ molar” (Ex. a 2M HCl
solution is read as two molar HCl”
• Note that the volume involved is
the total volume of solution, not
just the solvent.
Molarity
Molarity (M) = moles of solute
Liters of solution
M = mol
1
L
*if given grams, convert if to moles using the molar mass of the substance
Why are grams important?
Moles
Grams
Molar mass
___g = 1 mole
• You can not directly measure
moles, you must calculate the
mass in grams first
How to Prepare a Solution
To make 1.00 liter of a 1.00 molar
(1.0 M) solution:
1) add 1.0 mol of solute to a
volumetric flask
2) add about ¼ flask of distilled
water. Swirl the flask till the solute
is dissolved.
3) slowly add water until the final
volume reads 1.00 L
Molarity
EX 1. What is the molarity of a
solution that contains 8 moles of
CaCl2 in 50 mL of solution?
160M CaCl
M 2=
1
8 mol
mol
L L
0.05
Molarity
EX 2. How many grams of NaCl are
needed to make 500mL of a 0.2 M
solution?
.2 M
? mol
mol
M = .1
L L
1
0.5
0.1 mol NaCl 58.443 g NaCl =5.84 mol NaCl
1 mol NaCl
Using Molarity
Ex 3: A saline solution contains 0.90 g
NaCl in exactly 100 mL of solution.
What is the molarity of the solution?
Step 1: Calculate # moles
0.90g NaCl x 1 mol NaCl
=0.0154 mol NaCl
58.443 g NaCl
Step 2: mL  L
100 mL x 1 L
1000 mL
= 0.100 L NaCl
Ex 3 continued
Step 3: Calculate Molarity
0.154
M = 0.0154 mol
? M
M
L L
1
0.1
Ex 2: How many grams of solute
are present in 562 mL of 0.24 M
Na2SO4?
M = mol 
mol = M  L
L
mol =
0.24M Na2SO4 x .562L
= 0.135mol
Convert from Moles to Grams
0.135mol Na2SO4 | 142g Na2SO4 =
| 1 mol Na2SO4
= 19.2g Na2SO4
Practice
1. A solution has a volume of 2.0 L and contains
36.0 g of glucose. If the molar mass of
glucose is 180 g/mol, what is the molarity of
the solution? 0.100M glucose
2. How many moles of ammonium nitrate are in
335 mL of 0.425 M NH4NO3? 0.142mol
NH4NO3
3. How many grams of solute are in 250 mL of 2.0
M CaCl2 solution? 55.5gCaCl2
4. Describe how you would prepare 250 mL of a
0.2 M NaOH solution. Need 2.00g NaOH in
250mL of solution
1. A solution has a volume of 2.0 L and contains 36.0 g of glucose. If the
molar mass of glucose is 180 g/mol, what is the molarity of the solution?
Molarity = mol
L
Glucose = C6H12O6
Molar mass = 6(12.01g) + 12(1.008g) +
6(15.999g) =
Calculate moles:
36.0g C6H12O6 | 1 mol C6H12O6 = mol C6H12O6
| XXX g C6H12O6
Calculate Molarity: XXXmol C6H12O6 = 0.100M glucose
2.0 L
C. Dilutions
C. Dilutions
• You can make a less concentrated
solution by diluting it with solvent.
• The dilution reduces the grams of
solute per unit volume, but the
total amount of solute in solution
does not change.
Diluted Solutions
Before
Dilutions
After
Dilutions
Moles of solute before dilution = Moles of solute after dilution
Moles of solute = Molarity x volume
Dilutions:
M1V1 = M2V2
Ex: How many mL of a stock solution of 2.00
M MgSO4 would you need to prepare 100.0
mL of 0.400 M MgSO4?
Stock soln
(2M)(V1) = (0.400M)(100mL)
m1 v 1
m2
v2
V1 = 20mL of stock solution
Ex. 2: Describe how to prepare 100 mL of
0.400M MgSO4 from 2M MgSO4.
(see previous example)
Add 20mL of 2M stock solution in a
container and add solvent up to
the 100mL mark
Practice
1. How many mL of a stock solution of
4.00 M KI would you need to prepare
250.0 mL of 0.760 M KI?
47.5mL of 4.00MKI
2. What volume must you dilute to make
50.0 mL of 0.20 M KNO3 from 4.0 M
KNO3? 2.5mL of 4M KNO3
3. What is the molarity of a solution
formed when you add 200 mL of water
to 50 mL of 5.0 M HCl? 1.00M
1. How many mL of a stock solution of 4.00
M KI would you need to prepare 250.0 mL of
0.760 M KI?
(4M) (V1) = (0.760M) (250mL)
m1 v1
m2
v2
V1 = 47.5mL of 4M stock solution
2. What volume must you dilute to make
50.0 mL of 0.20 M KNO3 from 4.0 M KNO3?
(4M) (V1) = (0.20M) (50mL)
m1 v1
m2
v2
V1 = 2.5mL of 4M solution
3. What is the molarity of a solution formed
when you add 200 mL of water to 50 mL of
5.0 M HCl?
(5.0M) (50mL) = (M2) (250mL)
m1
v1
m2
v2
M2 = 1M of solution
EX. A chemist starts with 50 mL of a 0.40M
NaCl solution and dilutes it to 1000 mL.
What is the concentration of the dilute
solution?
Stock soln
(0.4 M)(50 mL) = (?M)(1000 mL)
M1
V1
M2
V2
M2 = 0.0200 M is the
concentration of the diluted
solution
Practice
1) What volume of a 3.00M KI stock solution
would you use to make 0.300 L of a
1.25 M KI solution? 0.125 L
2) How many milliliters of a 5.0M H2SO4
stock solution would you need to
prepare 100.0 mL of a 0.25M H2SO4?
5.00 mL
3) If you dilute 20.0 mL of a 3.0M solution
to make 100.0 mL of solution, what is
the molarity of the dilute solution?
0.600M
Practice 1
(3.00M) (V1) = (1.25M) (.300)
m1
v1
m2
v2
V1 = .125 L of 3M stock solution
Practice 2
(5M) (V1 mL) = (0.25M) (100 mL)
m1 v1
m2
v2
V1 = 5.00mL of 5M solution
Practice 3
(3.0M) (20mL) = (M2) (100mL)
m1
v1
m2
v2
M1 = 0.600M is the new
concentration