There are many ways to express concentration. The following sections focus on calculating concentration in |\text{g/L}| and percent (|\text{%}|). Practice questions and examples are included too.
Concentration in |\text{g/L}| refers to the ratio of the mass of the dissolved solute to the given volume of the solution.
|C=\dfrac{m_{solute}}{V_{solution}}|
where
|C:| concentration in grams per litre |(\text{g/L})|
|m_{solute:}| mass of the solute in grams |(\text{g})|
|V_{solution:}| volume of the solution in litres |(\text{L})|
A mass of |100\ \text{mg}| of potassium nitrate |(\text{KNO}_3)| is dissolved in |0.800\ \text{L}| of water. What is the concentration of the solution in |\text{g/L}|?
Potassium nitrate |(\text{KNO}_3)| is the solute and water is the solvent. Before calculating the concentration, identify the variables and convert the units.
||\begin{align}
m_{solute}&=\dfrac{100\ \cancel{\text{mg}} \times 1\ \text{g}}{1\ 000\ \cancel{\text{mg}}} = 0.100\ \text{g}\\
V_{solvent}&=0.800\ \text{L}\\
C&= \text{? g/L}
\end{align}||
Since the amount of solute is low compared to the amount of solvent, the volume of the solution does not change significantly after the solute is dissolved. Therefore, the given volume of solvent can be used as the volume of solution.
||V_{solution} \approx V_{solvent} = 0.800\ \text{L}||
Write the concentration formula and replace the variables with the given data values.
||\begin{align}
C&=\dfrac{m_{solute}}{V_{solution}}\\\\
C&=\dfrac{0.100\ \text{g}}{0.800\ \text{L}}\\\\
C&=0.125\ \text{g/L}
\end{align}||
The concentration of the solution is |0.125\ \text{g/L}.|
Sodium hydroxide |(\text{NaOH})| is used to prepare |250\ \text{mL}| of a |2.00\ \text{g/L}| aqueous solution. What is the mass, in grams |(\text{g}),| of sodium hydroxide required to prepare the solution?
Sodium hydroxide |(\text{NaOH})| is the solute and water is the solvent. Before calculating the concentration, identify the variables and convert the units.
||\begin{align}
C&=2.00\ \text{g/L}\\\\
V_{solution}&=\dfrac{250\ \cancel{\text{mL}}\times {1\ \text{L}}}{1\ 000\ \cancel{\text{mL}}} = 0.250\ \text{L}\\
m_{solute}&=\text{? g}
\end{align}||
Write the concentration formula and isolate the variable |m_{solute}.|
||\begin{align}
C&=\dfrac{m_{solute}}{V_{solution}}\\\\
m_{solute}&=C\times V_{solution}
\end{align}||
Finally, replace the variables with the given data values.
||\begin{align}
m_{solute}&=C\times V_{solution}\\
m_{solute}&=(2.00\ \text{g/}\cancel{\text{L}})\times (0.250\ \cancel{\text{L}})\\
m_{solute}&=0.500\ \text{g}\\
\end{align}||
The mass of sodium hydroxide |(\text{NaOH})| required to prepare the solution is |0.500\ \text{g}.|
Technically, chocolate milk is not a solution. It's not a homogeneous mixture, but rather a colloid. That said, the concept of concentration is easier to understand with delicious chocolate milk!
In this video, it is assumed that adding the solute (cacao powder) to the solvent (milk) does not affect the total volume of the solution (chocolate milk). Therefore, the final volume of chocolate milk |(V_{solution})| is estimated to be the same as the volume of the milk |(V_{solvent}).|
When the volume of the solution increases significantly after the solute is added to the solvent, the final volume of the solution used to calculate the concentration has to be determined.
Percent concentration is the percentage of the amount of dissolved solute compared to the given amount of solution. The amount can refer to mass or volume.
There are three types of percent concentration:
-
mass percent, also called mass/mass percentage |(\text{% m/m}),|
-
volume percent, also called volume/volume percentage |(\text{% V/V}),|
-
mass per volume percent, also called mass/volume percentage |(\text{% m/V}).|
The following table summarizes the formulas and the context for using each one.
|
|
Mass percent |(\text{% m/m})| |
Volume percent |(\text{% V/V})| |
Mass per volume percent |(\text{% m/V})| |
|---|---|---|---|
|
Formula |
||C=\left(\dfrac{m_{solute}}{m_{solution}}\right)\times 100|| |
||C=\left(\dfrac{V_{solute}}{V_{solution}}\right)\times 100|| |
||C=\left(\dfrac{m_{solute}}{V_{solution}}\right)\times 100|| |
|
Notes |
|
A |\text{m/V}| percentage corresponds to the mass of the solute per |100\ \text{mL}| of solution. |
|
|
Context |
Mass percent concentration is generally used when both the solute and the solvent are solid. |
Volume percent concentration is generally used when both the solute and the solvent are liquid. |
Mass per volume percent concentration is generally used when the solute is solid and the solvent is liquid. |
A metal block consists of |22.0\ \text{g}| of nickel |(\text{Ni})| and |84.0\ \text{g}| of zinc |(\text{Zn}).| What is the mass percent concentration of nickel in the metal block?
First, identify the variables.
||\begin{align}
m_{\text{Ni}}&=22.0\ \text{g}\\
m_{\text{Zn}}&=84.0\ \text{g}\\
C_{\text{Ni}}&=\text{?% m/m}
\end{align}||
For the concentration calculation, determine the total mass of the solid solution.
||\begin{align}
m_{solution}&=m_{\text{Ni}} + m_{\text{Zn}}\\
m_{solution}&=22.0\ \text{g} + 84.0\ \text{g}\\
m_{solution}&=106.0\ \text{g}
\end{align}||
Finally, write the formula for calculating the concentration of nickel |(\text{Ni})| in a solid solution and replace the variables with the given data values.
||\begin{align}
C_{\text{Ni}}&=\left(\dfrac{m_{\text{Ni}}}{m_{solution}}\right)\times 100\\\\
C_{\text{Ni}}&=\left(\dfrac{22.0\ \text{g}}{106.0\ \text{g}}\right)\times 100\\\\
C_{\text{Ni}}&\approx 20.8\text{% m/m}
\end{align}||
The metal block contains approximately |20.8\text{% m/m}| of nickel|(\text{Ni}).|
In a laboratory, sulphuric acid |(\text{H}_2 \text{SO}_4)| is used to prepare |125\ \text{mL}| of a |5.00\text{% V/V}| aqueous solution. How much sulphuric acid is required to prepare the solution?
Sulphuric acid |(\text{H}_2 \text{SO}_4)| is the solute and water is the solvent.
First, identify the variables.
||\begin{align}
C&=5.00\text{% V/V}\\
V_{solution}&=125\ \text{mL}\\
V_{solute}&=\text{? mL}
\end{align}||
Next, determine the correct formula and isolate the variable |V_{solute},| corresponding to the volume of the sulphuric acid dissolved in the solution.
||\begin{align}
C&=\left(\dfrac{V_{solute}}{V_{solution}}\right)\times 100\\\\
V_{solute}&=\dfrac{C\times V_{solution}}{100}
\end{align}||
Finally, replace the variables with the given data values.
||\begin{align}
V_{solute}&=\dfrac{C\times V_{solution}}{100}\\\\
V_{solute}&=\dfrac{(5.00\text{% V/V})\times (125\ \text{mL})}{100}\\\\
V_{solute}&= 6.25\ \text{mL}
\end{align}||
The volume of sulphuric acid |(\text{H}_2 \text{SO}_4)| required to prepare the solution is |6.25\ \text{mL}.|
The percent concentration of a solution is |2.5\text{% m/V}.| What is the concentration of the solution in grams per litre |(\text{g/L})|?
The percent concentration of |2.5\text{% m/V}| corresponds to a concentration of |2.5\ \text{g}| of solute per |100\ \text{mL}| of solution. First, write the ratio representing this situation.
||\begin{align}
\dfrac{2.5\ \text{g}}{100\ \text{mL}}
\end{align}||
The concentration in grams per litre |(\text{g/L})| corresponds to the mass of solute in grams |(\text{g})| per |1\ \text{L}| of solution. Since |1\ \text{L}| is equal to |1\ 000\ \text{mL},| write the ratio representing this relation.
|\dfrac{x}{1\ \text{L}}| or |\dfrac{x}{1\ 000\ \text{mL}}| where |x| is the unknown amount of solute in grams.
Establish a proportion and use cross multiplication to determine the mass of the solute dissolved in |1\ 000\ \text{mL}| of solution, since |2.5\ \text{g}| of solute is dissolved in |100\ \text{mL}| of solution.
||\begin{align}
\dfrac{x}{1\ 000\ \text{mL}}&=\dfrac{2.5\ \text{g}}{100\ \text{mL}}\\\\
x&=\dfrac{2.5\ \text{g}\times 1\ 000\ \cancel{\text{mL}}}{100\ \cancel{\text{mL}}}\\\\
x&=25\ \text{g}
\end{align}||
Therefore, |1\ \text{L}| of solution (i.e., |1\ 000\ \text{mL}|) contains |25\ \text{g}| of solute.
||\begin{align}
C=\dfrac{25\ \text{g}}{1\ \text{L}} = 25\ \text{g/L}
\end{align}||
The percent concentration of |2.5\ \text{m/V %}| corresponds to the concentration of |25\ \text{g/L}.|
Techniquement, le lait au chocolat n’est pas une solution. En effet, ce n’est pas un mélange homogène, mais plutôt un colloïde. Ceci dit, le concept de concentration se comprend mieux avec un bon lait chocolaté!
In this video, it is assumed that adding the solute (cacao powder) to the solvent (milk) does not affect the total volume of the solution (chocolate milk). Therefore, the final volume of chocolate milk |(V_{solution})| is estimated to be the same as the volume of the milk |(V_{solvent}).|
When the volume of the solution increases significantly after the solute is added to the solvent, the final volume of the solution used to calculate the concentration has to be determined.