KEY
Activity 151 – 14
Units of Concentration
Directions: This Guided Learning Activity (GLA) focuses on chemical calculations related to solution
concentration. Part A gives the definitions of the most common concentration units: molarity, % mass
(m/m), and ppm. Part B uses these concentration units in mass and mole conversion problems. Part C
discusses solution dilution, and the application of C1V1 = C2V2 equation. The worksheet is accompanied
by instructional videos. See http://www.canyons.edu/Departments/CHEM/GLA/ for additional materials.
Part A – Basic Concentration Units
A solution is a homogenous mixture of two or more pure substances. A solvent is defined as the
substance present in greatest amount, while the solutes are present in smaller amounts. The concentration
of a solute is the amount of solute dissolve in the solution, and can be defined both qualitatively and
quantitatively. Qualitatively, a solution can be dilute, containing a small amount of solute, or
concentrated, containing a large amount of solute.
Three quantitative measures of solution concentration will be discussed here. They are:
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 (𝑴) =
% 𝒎𝒎𝒎𝒎 =
𝒑𝒑𝒑 =
𝒎𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔
𝒍𝒍𝒍𝒍𝒍𝒍 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔
𝒙 𝟏𝟏𝟏%
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔
𝒙 𝟏𝟏𝟔
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
Mass Percent & Parts Per Million: Recall that the solution is made up of both solute and solvent. The
mass of the solution is therefore mass of the solute plus mass of the solvent. Both the mass percent (%
mass) and the parts per million (ppm) concentrations can be found directly from the masses of the solute
and solution. Generally, ppm is used only for very dilute solutions.
Chemistry Guided Learning Activities
Activity 151 – 14
College of the Canyons
Page 1 of 6
Example #1. Calculate the % mass and ppm concentrations for each of the following solutions.
a. 2.35 g sucrose (C12H22O11) in 90.0 g of water.
% 𝒔𝒔𝒔𝒔𝒔𝒔𝒔 =
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟐. 𝟑𝟑 𝒈 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒙 𝟏𝟏𝟏% =
𝒙 𝟏𝟏𝟏% = 𝟐. 𝟓𝟓%
(𝟐. 𝟑𝟑 𝒈 + 𝟗𝟗. 𝟎 𝒈)𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒑𝒑𝒑 𝒔𝒔𝒔𝒔𝒔𝒔𝒔 =
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟐. 𝟑𝟑 𝒈 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒙 𝟏𝟎𝟔 =
𝒙 𝟏𝟎𝟔 = 𝟐𝟐𝟐𝟐𝟐 𝒑𝒑𝒑
(𝟐. 𝟑𝟓 𝒈 + 𝟗𝟗. 𝟎 𝒈)𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
b. 2.35 mg sucrose in 2.00 kg of solution.
𝟏 𝐠 𝐬𝐬𝐬𝐬𝐬𝐬𝐬
�
𝟏𝟏𝟏𝟏 𝐦𝐦
(𝟐. 𝟑𝟑 𝐦𝐦 𝐬𝐬𝐬𝐬𝐬𝐬𝐬) �
𝟐. 𝟎𝟎 𝐱 𝟏𝟏
𝟑𝐠
𝐬𝐬𝐬′𝐧
% 𝐬𝐬𝐬𝐬𝐬𝐬𝐬 =
= 𝟐. 𝟑𝟑 𝐱 𝟏𝟏−𝟑 𝐠 𝐬𝐬𝐬𝐬𝐬𝐬𝐬;
(𝟐. 𝟎𝟎 𝐤𝐤 𝐬𝐬𝐬′𝐧) �
𝟏𝟏𝟏𝟏 𝐠 𝐬𝐬𝐬′𝐧
�
𝟏 𝐤𝐠 𝐬𝐬𝐬′𝐧
=
𝟐. 𝟑𝟑 𝐱 𝟏𝟏−𝟑 𝐠 𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝐱 𝟏𝟏𝟏% = 𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎 %
𝟐. 𝟎𝟎 𝐱 𝟏𝟏𝟑 𝐠 𝐬𝐬𝐬′𝐧
𝟐. 𝟑𝟑 𝐱 𝟏𝟏−𝟑 𝐠 𝐬𝐬𝐬𝐬𝐬𝐬𝐬
𝐱 𝟏𝟏𝟔 = 𝟏. 𝟏𝟏 𝐩𝐩𝐩
𝐩𝐩𝐩 𝐬𝐬𝐬𝐬𝐬𝐬𝐬 =
𝟐. 𝟎𝟎 𝐱 𝟏𝟏𝟑 𝐠 𝐬𝐬𝐬′𝐧
Although we limit this GLA to discussing % mass and ppm on a mass basis, keep in mind that both mass
% and ppm can be defined on a mass solute/mass solution (m/m), mass solute/volume solution (m/v) or
volume solute/volume solution (v/v) basis.
Molar Concentrations: The molar concentration, or molarity, of a solution is used extensively in
stoichiometric calculations. The molarity calculation requires that the mass of solute be converted into
moles.
Example #2. Calculate molar concentration (M) of each of the following solutions.
a. 45.0 g of sodium chloride (NaCl) in 2.00 L of solution.
𝑴𝑴𝑴𝑴𝑴 𝑵𝑵𝑵𝑵 = 𝟒𝟒. 𝟎 𝒈 𝑵𝑵𝑵𝑵 �
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 =
𝟏 𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵
� = 𝟎. 𝟕𝟕𝟕 𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵
𝟓𝟓. 𝟒𝟒 𝒈 𝑵𝑵𝑵𝑵
𝒎𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵
𝟎. 𝟕𝟕𝟕 𝒎𝒎𝒎𝒎 𝑵𝑵𝑵𝑵
= �
� = 𝟎. 𝟑𝟑𝟑 𝑴
𝒍𝒍𝒍𝒍𝒍𝒍 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟐. 𝟎𝟎 𝑳 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
b. 62.8 g of urea (CH4N2O) in 250.0 mL of solution.
𝟏 𝐦𝐦𝐦 𝐂𝐇𝟒 𝐍𝟐 𝐎
(𝟔𝟔. 𝟖 𝐠 𝐂𝐇𝟒 𝐍𝟐 𝐎) �
� = 𝟏. 𝟎𝟎𝟎𝟎 𝐦𝐦𝐦 𝐮𝐮𝐮𝐮
𝟔𝟔. 𝟎𝟎 𝐠 𝐂𝐇𝟒 𝐍𝟐 𝐎
𝐌𝐌𝐌𝐌𝐌𝐌𝐌𝐌 =
𝐦𝐦𝐦𝐦𝐦 𝐬𝐬𝐬𝐬𝐬𝐬
𝟏. 𝟎𝟎𝟎𝟎 𝐦𝐦𝐦 𝐂𝐇𝟒 𝐍𝟐 𝐎
=
= 𝟒. 𝟏𝟏 𝐌
𝐋𝐋𝐋𝐋𝐋𝐋 𝐬𝐬𝐬′𝐧
𝟎. 𝟐𝟐𝟐𝟐 𝐋
Chemistry Guided Learning Activities
Activity 151 – 14
College of the Canyons
Page 2 of 6
Part B – Calculations using Solution Concentration
Solution concentration directly relates to the amount of solute present, and many calculations in solution
chemistry relate the amount of solute to the volume of the solution. To solve these problems, it’s
important to be familiar with the concentration definitions.
Practice:
How many moles of magnesium bromide (MgBr2) are found in 6.48 mL of a 0.135 M aqueous solution?
𝒎𝒎𝒎𝒎𝒎 𝑴𝑴𝑴𝒓𝟐
We know:
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 =
𝒍𝒍𝒍𝒍𝒍𝒍 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
Plugging in for molarity and volume (𝟔. 𝟒𝟒 𝒎𝒎 �
𝟎. 𝟏𝟏𝟏 𝑴 =
Solving for x:
𝒙 = �𝟎. 𝟏𝟏𝟏
𝟏𝑳
�
𝟏𝟏𝟏𝟏 𝒎𝒎
𝒙
𝟎. 𝟎𝟎𝟎𝟎𝟎 𝑳
= 𝟎. 𝟎𝟎𝟎𝟎𝟎 𝑳):
𝒎𝒎𝒎
� (𝟎. 𝟎𝟎𝟎𝟎𝟎 𝑳) = 𝟖. 𝟕𝟕 𝒙 𝟏𝟏−𝟒 𝒎𝒎𝒎 𝑴𝑴𝑴𝒓𝟐
𝑳
Alternatively, the concentration can be used as a conversion factor.
We know:
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 =
𝒎𝒎𝒎𝒎𝒎 𝑴𝑴𝑴𝒓𝟐
𝒍𝒍𝒍𝒍𝒍𝒍 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
Solving by dimensional analysis:
𝟔. 𝟒𝟒 𝒎𝒎 𝒔𝒔𝒔′𝒏 �
=
𝟎.𝟏𝟏𝟏 𝒎𝒎𝒎 𝑴𝑴𝑴𝒓𝟐
𝟏𝑳
𝟏 𝑳 𝒔𝒔𝒔′𝒏
𝟎. 𝟏𝟏𝟏 𝒎𝒎𝒎 𝑴𝑴𝑴𝒓𝟐
��
� = 𝟖. 𝟕𝟕 𝒙 𝟏𝟏−𝟒 𝒎𝒎𝒎 𝑴𝑴𝑴𝒓𝟐
𝟏𝟏𝟏𝟏 𝒎𝒎 𝒔𝒔𝒔′𝒏
𝟏 𝑳 𝒔𝒔𝒔′𝒏
Example #3. How much ammonium acetate (NH4NO3) is needed to prepare a 400.0 g solution with a
concentration of 7.50% NH4NO3?
𝟕. 𝟓𝟓 𝐠 𝐍𝐇𝟒 𝐍𝐎𝟑
(𝟒𝟒𝟒. 𝟎 𝐠 𝐬𝐬𝐬′𝐧) �
� = 𝟑𝟑. 𝟎 𝐠 𝐍𝐇𝟒 𝐍𝐎𝟑
𝟏𝟏𝟏 𝐠 𝐬𝐬𝐬′𝐧
Chemistry Guided Learning Activities
Activity 151 – 14
College of the Canyons
Page 3 of 6
Part C – Solution Dilution Calculations
In laboratories, often new solutions are prepared from concentrated stock solutions by dilution. When
dilutions are performed, the following equation can be used to easily calculate the concentration or
volume of the stock or dilute solution.
𝑪𝟏 𝑽𝟏 = 𝑪𝟐 𝑽𝟐
Where: C1 = concentration of stock solution
V1 = volume of stock solution
C2 = concentration of diluted solution
V2 = volume of diluted solution
Keep in mind that this equation is only used when a solution is diluted, and never to perform
stoichiometric calculations. Most solution calculations will not use this equation. The concentration can
be given in molarity, % mass, or ppm. Likewise, volume can be given in liters, milliliters, or another
convenient unit.
Practice:
What volume in mL of 4.00 M NaOH is needed to prepare 250. mL of dilute 0.200 M NaOH?
𝑪 𝟏 𝑽𝟏 = 𝑪 𝟐 𝑽𝟐
Stock solution: C1 = 4.00 M NaOH, V1 = ??
Diluted solution: C2 = 0.200 M NaOH, V2 = 250. mL
𝑽𝟏 =
𝑪𝟐 𝑽𝟐 (𝟎. 𝟐𝟐𝟐 𝑴)(𝟐𝟐𝟐. 𝒎𝒎)
=
= 𝟏𝟏. 𝟓 𝒎𝒎
𝑪𝟏
(𝟒. 𝟎𝟎 𝑴)
Example #4. 5.00 mL of a 0.500 M solution of lithium chloride is diluted to 25.00 mL. What is the final
concentration of the solution?
𝐂𝟐 =
(𝟎. 𝟓𝟓𝟓 𝐌 𝐋𝐋𝐋𝐋)(𝟓. 𝟎𝟎 𝐦𝐦)
𝐂𝟏 𝐕𝟏
=
= 𝟎. 𝟏𝟏𝟏 𝐌 𝐋𝐋𝐋𝐋
(𝟐𝟐. 𝟎𝟎 𝐦𝐦)
𝐕𝟐
Chemistry Guided Learning Activities
Activity 151 – 14
College of the Canyons
Page 4 of 6
A serial dilution is a process in which each diluted solution is treated as the stock solution for the next
dilution. Serial dilutions are a simple way for scientists to prepare solutions with a wide range of
concentrations.
Example #5. Solution A is prepared by diluting 1.00 mL of 0.160 M CaCl2 to 50.0 mL. Solution B is
prepared by diluting 10.0 mL of Solution A to 500.0 mL. What is the concentration of CaCl2 in Solution
B?
The stock solution is solution 1 and solution A is solution 2.
𝐂𝐀 =
(𝟎. 𝟏𝟏𝟏 𝐌)(𝟏. 𝟎𝟎 𝐦𝐦)
𝐂𝐬𝐬𝐬𝐬𝐬 𝐕𝐬𝐬𝐬𝐬𝐬
=
= 𝟎. 𝟎𝟎𝟎𝟎𝟎 𝐌
(𝟓𝟓. 𝟎 𝐦𝐦)
𝐕𝐀
Solution B is prepared using solution A as the stock, so for this calculation, the stock concentration
is 0.0032 M.
𝐂𝐁 =
(𝟎. 𝟎𝟎𝟎𝟎𝟎 𝐌)(𝟏𝟏. 𝟎 𝐦𝐦)
𝐂𝐀 𝐕𝐀
=
= 𝟔. 𝟒 𝐱 𝟏𝟎−𝟓 𝑴
(𝟓𝟓𝟓. 𝟎 𝐦𝐦)
𝐕𝐁
Chemistry Guided Learning Activities
Activity 151 – 14
College of the Canyons
Page 5 of 6
Part D – Extra Practice
1. Calculate the grams of solute necessary to prepare 200.0 mL of the following solutions:
50.0% (m/m) sucrose (C12H22O11), d=1.230 g/mL
50.0 mM sodium hydrogen phosphate (Na2HPO4)
1.46 M calcium chloride (CaCl2)
0.36 M glucose (C6H12O6)
2. Find the mass % concentration of sugar in a 28.735 g solution containing 3.429 g of dissolved sugar.
3. How many kilograms of sucrose are in 4.8 kg of a 40.0 % (m/m) sucrose solution?
4. What is the molarity of a solution that contains 32.4 g of potassium bromide in 1.5 L of solution? What
is the mass % (dsolution = 1.02 g/mL)?
5. Which solution contains more moles of sodium ions: 100.0 mL of 0.620 M sodium chloride (NaCl), or
100.0 g of 6.3% (m/m) sodium carbonate (Na2CO3)? Hint: Recall that soluble ionic compounds dissociate
in water. Refer to GLA 151-7 Complete and Net Ionic Equations for more guidance.
6. Benzene is a known carcinogen that can leak into groundwater. The maximum allowable level of
benzene in drinking water is 5 μg/L. The average American consumes 3.9 cups, or 0.92 liters, of water
each day. If benzene is present in an individual’s drinking water at 5 μg/L, how many grams of benzene
will that person consume in one year?
7. A technician wants to prepare a 175 mL of 0.84 M sodium hydroxide solution. What volume of the 5.0
M stock solution of sodium hydroxide does the technician need?
8. A 20.0 mL of a stock solution was diluted to 1.00 L. The concentration of the final solution was 40.0
mM. What was the concentration of the stock solution?
9. What mass of a 22.8 ppm lead (II) nitrate solution will contain 0.00020 mole of lead (II) nitrate?
10. Calculate the number of moles of nitric acid (HNO3) in 500. mL of a solution which is 30.0% HNO3
by mass. (dsolution = 1.18 g/mL).
Chemistry Guided Learning Activities
Activity 151 – 14
College of the Canyons
Page 6 of 6
Key for extra practice GLA 151-14
1. For this problem, you need to be familiar with the definition of each concentration unit, and be
able to manipulate units to get the correct units in the numerator and denominator of the
expression.
𝟏.𝟐𝟐𝟐 𝒈 𝒔𝒔𝒔′𝒏
𝟓𝟓.𝟎 𝒈 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
��
�
𝟏 𝒎𝒎 𝒔𝒔𝒔′𝒏
𝟏𝟏𝟏 𝒈 𝒔𝒔𝒔′𝒏
50.0% (m/m) sucrose: (𝟐𝟐𝟐. 𝟎 𝒎𝒎 𝒔𝒐𝒍′𝒏) �
= 𝟏𝟏𝟏 𝒈 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
For this conversion, recall that the metric prefixes can be applied to any metric unit, including
molarity. Therefore, a concentration of 1 M is equivalent to 1000 mM. We will represent 50 mM
𝟓𝟓.𝟎 𝒎𝒎𝒎𝒎 𝑵𝒂𝟐 𝑯𝑯𝑶𝟒
Na2HPO4 as �
�.
𝟏 𝑳 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
50.0 mM Na2HPO4 :
(𝟐𝟐𝟐. 𝟎 𝒎𝒎 𝒔𝒔𝒔′𝒏) �
𝟏. 𝟒𝟒 𝒈 𝑵𝒂𝟐 𝑯𝑯𝑶𝟒
𝟏 𝑳 𝒔𝒔𝒔′𝒏
𝟓𝟓.𝟎 𝒎𝒎𝒎𝒎 𝑵𝒂𝟐 𝑯𝑷𝑶𝟒
𝟏 𝒎𝒎𝒎
𝟏𝟏𝟏.𝟗𝟗 𝒈 𝑵𝒂𝟐 𝑯𝑷𝑶𝟒
��
��
��
�
𝟏𝟏𝟏𝟏 𝒎𝒎 𝒔𝒔𝒔′𝒏
𝟏 𝑳 𝒔𝒔𝒔′𝒏
𝟏𝟏𝟏𝟏 𝒎𝒎𝒎𝒎
𝟏 𝒎𝒎𝒎 𝑵𝒂𝟐 𝑯𝑯𝑶𝟒
=
For molar concentrations, it’s important to remember to convert to moles using molar mass. The
𝑿𝑿 𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔
𝑿𝑿 𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔
� or as �
�, since 1 L and 1000 mL are
molarity can be represented as �
𝟏 𝑳 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟏𝟏𝟏𝟏 𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
equivalent volumes.
𝟏.𝟒𝟒 𝒎𝒎𝒎 𝑪𝑪𝑪𝒍𝟐
𝟏𝟏𝟎.𝟗𝟗 𝒈 𝑪𝑪𝑪𝒍𝟐
��
�
𝟏𝟏𝟏𝟏 𝒎𝒎 𝒔𝒔𝒔′𝒏
𝟏 𝒎𝒎𝒎 𝑪𝑪𝑪𝒍𝟐
1.46 M calcium chloride: (𝟐𝟐𝟐. 𝟎 𝒎𝒎 𝒔𝒔𝒔′𝒏) �
𝟎.𝟑𝟑 𝒎𝒎𝒎 𝑪𝟔 𝑯𝟏𝟏 𝑶𝟔
𝟏𝟏𝟏.𝟏𝟏 𝒈 𝑪𝟔 𝑯𝟏𝟏 𝑶𝟔
��
�
𝟏𝟏𝟏𝟏 𝒎𝒎 𝒔𝒔𝒔′𝒏
𝟏 𝒎𝒎𝒎 𝑪𝟔 𝑯𝟏𝟏 𝑶𝟔
0.36 M glucose: (𝟐𝟐𝟐. 𝟎 𝒎𝒎 𝒔𝒔𝒔′𝒏) �
𝒎𝒎𝒎𝒎 𝒔𝒔𝒔𝒔𝒔
= 𝟑𝟑. 𝟒 𝒈 𝑪𝑪𝑪𝒍𝟐
= 𝟏𝟏. 𝟎 𝒈 𝑪𝟔 𝑯𝟏𝟏 𝑶𝟔
2. For this problem, we need to know % 𝒔𝒔𝒔𝒔𝒔 =
𝒙 𝟏𝟏𝟏%. The problem gives both the
𝒎𝒎𝒎𝒎 𝒔𝒔𝒍′ 𝒏
mass of the sucrose and the mass of the solution, so we simply plug the two values into the
expression. Sometimes, the mass of the solvent is given, and in these cases you’ll need to add the
mass of the solute and the solvent together to find the mass of the solution.
% 𝒔𝒔𝒔𝒔𝒔𝒔𝒔 =
𝟑. 𝟒𝟒𝟒 𝒈 𝒔𝒔𝒔𝒔𝒔
𝒙 𝟏𝟏𝟏% = 𝟏𝟏. 𝟗𝟗%
𝟐𝟐. 𝟕𝟕𝟕 𝒈 𝒔𝒔𝒍′ 𝒏
3. One of the advantages of concentration units is that they can be used as conversion factors to
convert between quantities. In this problem, we recognize that mass percent can be written as a
conversion factor. 40.0% means 40.0% of any mass. Because it is a percent measurement, we can
use any mass unit we choose. Because the mass of solution is given in kg, we will choose to represent
the mass % in kg.
𝟒𝟒. 𝟎 𝒌𝒌 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
(𝟒. 𝟖 𝒌𝒌 𝒔𝒔𝒔′𝒏) �
� = 𝟏. 𝟗 𝒌𝒌 𝒔𝒔𝒔𝒔𝒔𝒔𝒔
𝟏𝟏𝟏 𝒌𝒌 𝒔𝒔𝒍′ 𝒏
4. To calculate the molarity, we need to divide the moles of solute by the liters of solution. It may be
convenient to divide the mass by the volume, and then use dimensional analysis to convert the mass
of solute to moles of solute.
𝑴𝑴𝑴𝑴𝑴𝑴𝑴𝑴 =
𝒎𝒎𝒎𝒎𝒎 𝑲𝑲𝑲
𝟑𝟑. 𝟒 𝒈 𝑲𝑲𝑲
𝟏 𝒎𝒎𝒎 𝑲𝑲𝑲
= �
��
� = 𝟎. 𝟏𝟏 𝑴 𝑲𝑲𝑲
′
′
𝑳𝑳𝑳𝑳𝑳𝑳 𝒔𝒔𝒍 𝒏
𝟏. 𝟓 𝑳 𝒔𝒔𝒍 𝒏 𝟏𝟏𝟏. 𝟎 𝒈 𝑲𝑲𝑲
For the mass percent, we need to remember that the % mass measurement is based on the masses
of the solute and solution. The volume of the solution can be converted to mass by using the density
𝒈𝒈𝒈𝒈𝒈 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
as a conversion factor. The density of a solution is most often measured in
. Take care
𝒎𝒎 𝒔𝒔𝒔𝒔𝒔𝒔𝒔𝒔
to distinguish between the mass of the solution and the mass of the solute.
% 𝑲𝑲𝑲 =
𝒎𝒎𝒎𝒎 𝑲𝑲𝑲
𝒙
𝒎𝒎𝒎𝒎 𝒔𝒔𝒍′ 𝒏
𝟏𝟏𝟏%; �
𝟑𝟑.𝟒 𝒈 𝑲𝑲𝑲
𝟏 𝑳 𝒔𝒔𝒍′ 𝒏
𝟏 𝒎𝒎 𝒔𝒔𝒍′ 𝒏
�
�
�
�
�𝒙
𝟏.𝟓 𝑳 𝒔𝒔𝒍′ 𝒏
𝟏𝟏𝟎𝟎 𝒎𝒎 𝒔𝒔𝒍′𝒏
𝟏.𝟎𝟎 𝒈 𝒔𝒔𝒍′ 𝒏
𝟏𝟏𝟏% = 𝟐. 𝟏% 𝑲𝑲𝑲
5. For this problem we must use the concentration as one of several conversion factors that will
allow us to determine the number of sodium ions in each solution. To account for the fact that
sodium chloride contains one Na+ ion per formula unit while sodium carbonate contains two. This is
done by incorporating a mole Na+ ions to mole compound conversion factor.
𝟎. 𝟔𝟔𝟔 𝒎𝒎𝒎 𝑵𝒂𝒂𝒂 𝟏 𝒎𝒎𝒎 𝑵𝒂+
(𝟏𝟏𝟏. 𝟎 𝒎𝒎 𝑵𝑵𝑵𝑵 𝒔𝒔𝒍′ 𝒏) �
��
� = 𝟎. 𝟎𝟎𝟎𝟎 𝒎𝒎𝒎 𝑵𝒂+ 𝒊𝒊𝒊𝒊
𝟏𝟏𝟏𝟏 𝒎𝒎 𝒔𝒔𝒍′ 𝒏
𝟏 𝒎𝒎𝒎 𝑵𝑵𝑵𝑵
𝟔. 𝟑 𝒈 𝑵𝒂𝟐 𝑪𝑶𝟑
𝟏 𝒎𝒎𝒎 𝑵𝒂𝟐 𝑪𝑶𝟑
𝟐 𝒎𝒎𝒎 𝑵𝒂+
(𝟏𝟏𝟏. 𝟎 𝒈 𝑵𝒂𝟐 𝑪𝑶𝟑 𝒔𝒔𝒔′𝒏) �
�
�
�
�
�
𝟏𝟏𝟏 𝒈 𝒔𝒔𝒍′ 𝒏
𝟏𝟏𝟏. 𝟗𝟗 𝒈 𝑵𝒂𝟐 𝑪𝑶𝟑 𝟏 𝒎𝒎𝒎 𝑵𝒂𝟐 𝑪𝑶𝟑
= 𝟎. 𝟏𝟏 𝒎𝒎𝒎 𝑵𝒂+ 𝒊𝒊𝒊𝒊
The Na2CO3 contains more sodium ions.
6. This problem requires a number of conversion factors. The question asks how much benzene will
be consumed in 1 year, so we will start our dimensional analysis with the quantity 1 year. Another
important note is that the benzene concentration is given as 5 μg/L, or 5x10-6 g benzene/L solution.
𝟑𝟑𝟑 𝒅𝒅𝒅𝒅 𝟎. 𝟗𝟗 𝑳 𝒘𝒘𝒘𝒘𝒘 𝟓 𝝁𝝁 𝒃𝒃𝒃𝒃𝒃𝒃𝒃
𝟏 𝒈 𝒃𝒃𝒃𝒃𝒃𝒃𝒃
(𝟏 𝒚𝒚) �
��
��
�� 𝟔
� = 𝟎. 𝟎𝟎𝟎 𝒈 𝒃𝒃𝒃𝒃𝒃𝒃𝒃
𝟏 𝒚𝒚
𝟏 𝒅𝒅𝒅
𝟏 𝑳 𝒘𝒘𝒘𝒘𝒘
𝟏𝟎 𝝁𝝁 𝒃𝒃𝒃𝒃𝒃𝒃𝒃
7. This problem utilizes the dilution equation, 𝑪𝟏 𝑽𝟏 = 𝑪𝟐 𝑽𝟐. We will rearrange the variables to
isolate the variable of interest. Also, notice that the volume and concentrations need not be
converted to liters or molarity, respectively. As long as the units cancel by dimensional anlaysis,
they do not need to be converted into other units.
𝑽𝒔𝒔𝒔𝒔𝒔 =
(𝟎. 𝟖𝟖 𝑴)(𝟏𝟏𝟏 𝒎𝒎)
𝑪𝒅𝒅𝒅𝒅𝒅𝒅𝒅 𝑽𝒅𝒅𝒅𝒅𝒅𝒅𝒅
=
= 𝟐𝟐 𝒎𝒎 𝒔𝒔𝒔𝒔𝒔
(𝟓. 𝟎 𝑴)
𝑪𝒔𝒔𝒔𝒔𝒔
8. Again, this problem employs the dilution equation. The most difficult part of this problem may
be associating each value with a variable in the dilution equation.
C1 = concentration of stock solution = ??
C2 = concentration of diluted solution = 40.0 mM
V1 = vol. of stock solution = 20.0 mL
V2 = vol. of diluted solution = 1.00 L=1000 mL
Before substituting the values for the variables, we need to convert one of the volume measurement
units so these units will cancel.
𝑪𝟏 =
(𝟒𝟒. 𝟎 𝒎𝒎)(𝟏𝟏𝟏𝟏 𝒎𝒎)
𝑪 𝟐 𝑽𝟐
=
= 𝟐𝟐𝟐𝟐 𝒎𝒎 = 𝟐. 𝟎𝟎 𝑴
(𝟐𝟐. 𝟎 𝒎𝒎)
𝑽𝟏
9. In this problem, we need to convert 0.00020 moles of lead(II) nitrate to mass before we can use
the ppm measurement to find the mass of the solution containing 0.00020 moles of the compound.
𝟐𝟐.𝟖 𝒈 𝑷𝑷(𝑵𝑶𝟑 )𝟐
Recall that ppm can be written as a conversion factor: �
�.
𝟔
′
𝟏𝟏𝟏𝟎 𝒈 𝒔𝒔𝒍 𝒏
𝟎. 𝟎𝟎𝟎𝟎𝟎 𝒎𝒎𝒎 𝑷𝑷(𝑵𝑶𝟑 )𝟐 �
𝟑𝟑𝟑. 𝟐𝟐 𝒈 𝑷𝑷(𝑵𝑶𝟑 )𝟐
𝟏 𝒙 𝟏𝟎𝟔 𝒈 𝒔𝒔𝒔′𝒏
��
� = 𝟐𝟐𝟐𝟐 𝒈 𝒔𝒔𝒍′ 𝒏
𝟐𝟐. 𝟖 𝒈 𝑷𝑷(𝑵𝑶𝟑 )𝟐
𝟏 𝒎𝒎𝒎 𝑷𝑷(𝑵𝑶𝟑 )𝟐
10. As a reminder, mass % is the mass of the solute divided by the mass of the solution, not the
volume. To convert to mass, we use the density, which is the mass of the solution over the volume of
the solution.
(𝟓𝟓𝟓. 𝒎𝒎 𝒔𝒔𝒍′ 𝒏) �
𝟏. 𝟏𝟏 𝒈 𝒔𝒔𝒍′ 𝒏 𝟑𝟑. 𝟎 𝒈 𝑯𝑯𝑶𝟑
𝟏 𝒎𝒎𝒎 𝑯𝑯𝑶𝟑
��
��
� = 𝟐. 𝟖𝟖 𝒎𝒎𝒎 𝑯𝑯𝑶𝟑
′
′
𝟏 𝒎𝒎 𝒔𝒔𝒍 𝒏
𝟏𝟏𝟏 𝒈 𝒔𝒔𝒍 𝒏
𝟔𝟔. 𝟎𝟎 𝒈 𝑯𝑯𝑶𝟑