16.2: Solution Concentration
Lesson Objectives
 Use the terms concentrated and dilute to describe the relative concentration of a solution.
 Calculate the concentration of a solution as either a mass percent or a volume percent.
 Calculate the molarity of a solution. Use molarity to calculate the mass of solute needed to prepare a particular solution.
 Calculate the molarity of a diluted solution. Use the dilution equation to calculate the volume of a concentrated stock solution required for a particular dilution.
 Calculate the molality of a solution and distinguish molality from molarity.
Lesson Vocabulary
 concentrated
 concentration
 dilute
 molality
 molarity
Check Your Understanding
Recalling Prior Knowledge
 How is percent by mass of a compound calculated?
 How are grams of a substance converted to moles?
So far, you have studied how solutions can form and the limits to the amount of solute that can possibly dissolve in a solvent at a given temperature and pressure. In this lesson, you will learn how to quantitatively express the amount of solute present in any solution.
Percent Solutions
There are several ways to express the amount of solute present in a solution. The concentration of a solution is a measure of the amount of solute that has been dissolved in a given amount of solvent or solution. A concentrated solution is one that has a relatively large amount of dissolved solute. A dilute solution is one that has a relatively small amount of dissolved solute. However, these terms are relative, and we need to be able to express concentration in a more exact, quantitative manner. Still, concentrated and dilute are useful as terms to compare one solution to another (Figure below). Also, be aware that the terms “concentrate” and “dilute” can be used as verbs. If you were to heat a solution, causing the solvent to evaporate, you would be concentrating it, because the ratio of solute to solvent would be increasing. If you were to add more water to an aqueous solution, you would be diluting it, because the ratio of solute to solvent would be decreasing.
Solutions of a red dye in water are displayed from the most dilute (on the left) to the most concentrated (on the right).
One way to describe the concentration of a solution is by the percent of the solution that is composed of the solute. This percentage can be determined in one of two ways: (1) the mass of the solute divided by the mass of the solution, or (2) the volume of the solute divided by the volume of the solution. Because these methods generally result in slightly different values, it is important to always indicate whether a given percentage was calculated "by mass" or "by volume."
Mass Percent
When the solute in a solution is a solid, a convenient way to express the concentration is a mass percent (mass/mass), which is the grams of solute per 100 g of solution.



\begin{align*}\text{Percent by mass}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100\%\end{align*}
Percent by mass=mass of solutemass of solution×100%

\begin{align*}\text{Percent by mass}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100\%\end{align*}

Suppose that a solution was prepared by dissolving 25.0 g of sugar into 100 g of water. The percent by mass would be calculated as follows:



\begin{align*}\text{Percent by mass}=\dfrac{25 \ \text{g sugar}}{125 \ \text{g solution}} \times 100\%=20\% \ \text{sugar}\end{align*}
Percent by mass=25 g sugar125 g solution×100%=20% sugar

\begin{align*}\text{Percent by mass}=\dfrac{25 \ \text{g sugar}}{125 \ \text{g solution}} \times 100\%=20\% \ \text{sugar}\end{align*}

Sometimes, you may want to make a particular amount of solution with a certain percent by mass and will need to calculate what mass of the solute is needed. For example, let's say you need to make 3000 g of a sodium chloride solution that is 5% by mass. You can rearrange and solve for the mass of solute.



\begin{align*}\text{mass of solute}&=\dfrac{\text{percent by mass}}{100\%} \times \text{mass of solution}\\
&= \dfrac{5\%}{100\%} \times 3000 \ \text{g}\\
&=150 \ \text{g NaCl}\end{align*}
mass of solute=percent by mass100%×mass of solution=5%100%×3000 g=150 g NaCl

\begin{align*}\text{mass of solute}&=\dfrac{\text{percent by mass}}{100\%} \times \text{mass of solution}\\
&= \dfrac{5\%}{100\%} \times 3000 \ \text{g}\\
&=150 \ \text{g NaCl}\end{align*}

You would need to weigh out 150 g of NaCl and add it to 2850 g of water. Notice that it was necessary to subtract the mass of the NaCl (150 g) from the mass of solution (3000 g) to calculate the mass of the water that would need to be added.
Volume Percent
The percentage of solute in a solution can more easily be determined by volume when the solute and solvent are both liquids. The volume of the solute divided by the volume of solution expressed as a percent yields the percent by volume (volume/volume) of the solution. If a solution is made by taking 40. mL of ethanol and adding enough water to make 240. mL of solution, the percent by volume is:



\begin{align*}\text{Percent by volume}&=\dfrac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \\
&=\dfrac{40 \ \text{mL ethanol}}{240 \ \text{mL solution}} \times 100\% \\
&=16.7\% \ \text{ethanol}\end{align*}
Percent by volume=volume of solutevolume of solution×100%=40 mL ethanol240 mL solution×100%=16.7% ethanol

\begin{align*}\text{Percent by volume}&=\dfrac{\text{volume of solute}}{\text{volume of solution}} \times 100\% \\
&=\dfrac{40 \ \text{mL ethanol}}{240 \ \text{mL solution}} \times 100\% \\
&=16.7\% \ \text{ethanol}\end{align*}

Frequently, ingredient labels on food products and medicines have amounts listed as percentages (Figure below).
Hydrogen peroxide is commonly sold as a 3% by volume solution for use as a disinfectant.
It should be noted that, unlike in the case of mass, you cannot simply add together the volumes of solute and solvent to get the final solution volume. When adding a solute and solvent together, mass is conserved, but volume is not. In the example above, a solution was made by starting with 40 mL of ethanol and adding enough water to make 240 mL of solution. Simply mixing 40 mL of ethanol and 200 mL of water would not give you the same result, as the final volume would probably not be exactly 240 mL.
Molarity
Chemists primarily need the concentration of solutions to be expressed in a way that accounts for the number of particles present that could react according to a particular chemical equation. Since percentage measurements are based on either mass or volume, they are generally not useful for chemical reactions. A concentration unit based on moles is preferable. The molarity (M) of a solution is the number of moles of solute dissolved in one liter of solution. To calculate the molarity of a solution, you divide the moles of solute by the volume of the solution expressed in liters.


 \begin{align*}\text{Molarity (M)}=\dfrac{\text{moles of solute}}{\text{liters of solution}}=\dfrac{\text{mol}}{\text{L}}\end{align*}

Note that the volume is in liters of solution and not liters of solvent. When a molarity is reported, the unit is the symbol M, which is read as “molar”. For example, a solution labeled as 1.5 M NH_{3} is a “1.5 molar solution of ammonia.”
Sample Problem 16.2: Calculating Molarity
A solution is prepared by dissolving 42.23 g of NH_{4}Cl into enough water to make 500.0 mL of solution. Calculate its molarity.
Step 1: List the known quantities and plan the problem.
Known
 mass of NH_{4}Cl = 42.23 g
 molar mass of NH_{4}Cl = 53.50 g/mol
 volume of solution = 500.0 mL = 0.5000 L
Unknown
 molarity = ? M
The mass of the ammonium chloride is first converted to moles. Then, the molarity is calculated by dividing by liters. Note that the given volume has been converted to liters.
Step 2: Solve.


 \begin{align*}\text{42.23 g NH}_4\text{Cl} \times \dfrac{1 \ \text{mol NH}_4\text{Cl}}{53.50 \ \text{g NH}_4\text{Cl}}=0.7893\ \text{mol NH}_4\text{Cl}\end{align*}



 \begin{align*}\dfrac{0.7893 \ \text{mol NH}_4\text{Cl}}{0.5000 \ \text{L}}=1.579 \ \text{M}\end{align*}

Step 3: Think about your result.
The molarity is 1.579 M, meaning that a liter of the solution would contain 1.579 moles of NH_{4}Cl. Four significant figures is appropriate.
 What is the molarity of a solution for which 250. mL of the solution contains 10.0 g of Pb(NO_{3})_{2}?
In a laboratory situation, a chemist must frequently prepare a given volume of a solution with a specific molarity. The first task is to calculate the mass of the solute that is necessary. The molarity equation can be rearranged to solve for moles, which can then be converted to grams, as shown in Sample Problem 16.3.
Sample Problem 16.3: Finding the Necessary Mass of Solute
A chemist needs to prepare 3.00 L of a 0.250 M solution of potassium permanganate (KMnO_{4}). What mass of KMnO_{4} does she need to make the solution?
Step 1: List the known quantities and plan the problem.
Known
 molarity = 0.250 M
 volume = 3.00 L
 molar mass of KMnO_{4} = 103.10 g/mol
Unknown
 mass of KMnO_{4} = ? g
Moles of solute is calculated by multiplying molarity by liters. Then, moles is converted to grams.
Step 2: Solve.


 \begin{align*}0.250 \ \text{M KMnO}_4 \times 3.00 \ \text{L solution}=0.750 \ \text{mol KMnO}_4\end{align*}



 \begin{align*}0.750 \ \text{mol KMnO}_4 \times \dfrac{103.10 \ \text{g KMnO}_4}{1 \ \text{mol KMnO}_4}=77.3 \ \text{g KMnO}_4\end{align*}

Step 3: Think about your result.
When 77.3 g of potassium permanganate is dissolved in enough water to make 3.00 L of solution, the molarity is 0.250 M.
 What mass of CaCl_{2} is needed to make 600. mL of a 0.380 M solution?
Preparing Solutions
If you are attempting to prepare 1.00 L of a 1.00 solution of NaCl, you would obtain 58.44 g of sodium chloride. However you cannot simply add the sodium chloride to 1.00 L of water. After the solute dissolves, the volume of the solution will be slightly greater than a liter, because the hydrated sodium and chloride ions take up space in the solution. Instead, a volumetric flask needs to be used. Volumetric flasks come in a variety of sizes (Figure below) and are designed so that a chemist can precisely and accurately prepare a solution of one specific volume.
Volumetric flasks come in many sizes, each designed to prepare a different volume of solution.
In other words, you cannot use a 1liter volumetric flask to make 500 mL of a solution. It can only be used to prepare 1 liter of a solution. The steps to follow when preparing a solution with a 1liter volumetric flask are outlined below and shown in Figure below.
 The appropriate mass of solute is weighed out and added to a volumetric flask that has been about halffilled with distilled water.
 The solution is swirled until all of the solute dissolves.
 More distilled water is carefully added up to the line etched on the neck of the flask.
 The flask is capped and inverted several times to completely mix the solution.
Steps to follow in preparing a solution of known molarity. (A) Weigh out the correct mass of solute. (B) Dissolve the solute into the desired solvent in a partially filled volumetric flask. (C) Add more solvent until the fill line on the flask is reached, and then mix.
Dilutions
When additional water is added to an aqueous solution, the concentration of that solution decreases. This is because the number of moles of the solute does not change, but the total volume of the solution increases. We can set up an equality between the moles of the solute before the dilution (1) and the moles of the solute after the dilution (2).


 mol_{1} = mol_{2}

Since the moles of solute in a solution is equal to the molarity multiplied by the volume in liters, we can set those equal.


 M_{1} × L_{1} = M_{2} × L_{2}

Finally, because the two sides of the equation are set equal to one another, the volume can be in any units we choose, as long as that unit is the same on both sides. Our equation for calculating the molarity of a diluted solution becomes:


 M_{1} × V_{1} = M_{2} × V_{2}

Suppose that you have 100. mL of a 2.0 M solution of HCl. You dilute the solution by adding enough water to make the solution volume 500. mL. The new molarity can easily be calculated by using the above equation and solving for M_{2}.


 \begin{align*}\mathrm{M_2=\dfrac{M_1 \times V_1}{V_2}=\dfrac{2.0 M \times 100. mL}{500. mL}=0.40 \ M \ HCl}\end{align*}

The solution has been diluted by a factor of five, since the new volume is five times as great as the original volume. Consequently, the molarity is onefifth of its original value.
Another common dilution problem involves deciding how much of a highly concentrated solution is required to make a desired quantity of solution with a lower concentration. The highly concentrated solution is typically referred to as the stock solution.
Sample Problem 16.4: Dilution of a Stock Solution
Nitric acid (HNO_{3}) is a powerful and corrosive acid. When ordered from a chemical supply company, its molarity is 16 M. How much of the stock solution of nitric acid needs to be used to make 8.00 L of a 0.50 M solution?
Step 1: List the known quantities and plan the problem.
Known
 stock HNO_{3} (M_{1}) = 16 M
 V_{2} = 8.00 L
 M_{2} = 0.50 M
Unknown
 volume of stock HNO_{3} (V_{1}) = ? L
The unknown in the equation is V_{1}, the necessary volume of the concentrated stock solution.
Step 2: Solve.


 \begin{align*}\mathrm{V_1=\dfrac{M_2 \times V_2}{M_1}=\dfrac{0.50 \ M \times 8.00 \ L}{16 \ M}=0.25 \ L=250 \ mL}\end{align*}

Step 3: Think about your result.
250 mL of the stock HNO_{3} solution needs to be diluted with water to a final volume of 8.00 L. The dilution from 16 M to 0.5 M is a factor of 32.
 125 mL of a 2.55 M solution of Zn(NO_{3})_{2} is diluted with water to make the final volume 197 mL. Calculate the new molarity.
 What volume of a 1.50 M solution of KCl needs to be diluted in order to prepare 2.40 L of a 0.0750 M solution?
Dilutions can be performed in the laboratory with various tools, depending on the volumes required and the desired accuracy. Figure below illustrates the use of two different types of pipettes. The use of a calibrated pipette instead of a graduated cylinder improves accuracy. In the figure on the right, the student is using a micropipette, which is designed to quickly and accurately dispense very small volumes. Micropipettes are adjustable and come in a variety of sizes.
(A) A graduated glass pipette is used to perform a dilution. (B) A micropipette is used to deliver small volumes of a liquid.
Molality
A final way to express the concentration of a solution is by its molality. The molality (m) of a solution is the moles of solute divided by the kilograms of solvent. A solution that contains 1.0 mol of NaCl dissolved in 1.0 kg of water is a “one molal” solution of sodium chloride. The symbol for molality is a lowercase m written in italics.


 \begin{align*}\mathrm{Molality} \ (m)=\mathrm{\dfrac{moles \ of \ solute}{kilograms \ of \ solvent}=\dfrac{mol}{kg}}\end{align*}

Molality differs from molarity only in the denominator. While molarity is based on the liters of solution, molality is based on the kilograms of solvent. Concentrations expressed in molality are used when studying properties of solutions related to vapor pressure and temperature changes, as you will see in the subsequent lesson. Molality is used because its value does not change with changes in temperature. The volume of a solution, on the other hand, is slightly dependent upon temperature.
Sample Problem 16.5: Calculating Molality
Determine the molality of a solution prepared by dissolving 28.60 g of glucose (C_{6}H_{12}O_{6}) in 250. g of water.
Step 1: List the known quantities and plan the problem.
Known
 mass of solute = 28.60 g C_{6}H_{12}O_{6}
 mass of solvent = 250. g = 0.250 kg
 molar mass of C_{6}H_{12}O_{6} = 180.18 g/mol
Unknown
 molality = ? m
Convert grams of glucose to moles and convert the mass of the solvent from grams to kilograms.
Step 2: Solve.


 \begin{align*}\mathrm{28.60 \ g \ C_6H_{12}O_6 \times \dfrac{1 \ mol \ C_6H_{12}O_6}{180.18 \ g \ C_6H_{12}O_6}=0.1587 \ mol \ C_6H_{12}O_6}\end{align*}



 \begin{align*}\mathrm{\dfrac{0.1587 \ mol \ C_6H_{12}O_6}{0.250 \ kg \ H_2O}}=0.635 \ m \ \mathrm{C_6H_{12}O_6}\end{align*}

Step 3: Think about your result.
The answer represents the moles of glucose per kilogram of water and has three significant figures.
 Calculate the molality of a solution containing 2.25 g of LiNO_{3} dissolved in 600. g of water.
Molality and molarity have very similar values for dilute aqueous solutions because the density of those solutions is relatively close to 1.0 g/mL. This means that 1.0 L of solution has a mass of just about 1.0 kg. As the solution becomes more concentrated, its density will not be as close to 1.0 g/ml, and the molality value will be different than the molarity. For solutions with solvents other than water, the molality will be very different than the molarity. Make sure that you are paying attention to which quantity is being used in a given problem.
Lesson Summary
 Concentrated and dilute are qualitative terms that reflect the relative concentration of a solution, which is the amount of solute dissolved in a given amount of solvent.
 The mass percent of a solution is the mass of the solute divided by the mass of the solution. The volume percent of a solution is the volume of the solute divided by the volume of the solution.
 The molarity of a solution is the moles of solute divided by the volume of the solution in liters. Molarity is the most commonly used way to express the concentration of a solution that is used in a chemical reaction.
 Dilution is the process of adding solvent to a solution in order to decrease its concentration. The dilution equation can be used to calculate the molarity of diluted solutions or to determine the volume of a stock solution needed for a dilution.
 Molality is the moles of solute divided by the kilograms of solvent.
Lesson Review Questions
Reviewing Concepts
 How is a concentrated solution different from a dilute solution?
 Answer the following:
 As the temperature of a solution increases, what happens to its molarity? Explain.
 Why does temperature not affect the molality of a solution?
Problems
 What is the mass percent of an aqueous solution prepared by dissolving 12.0 g of solute into 40.0 g of water?
 What is the volume percent of a solution prepared by adding enough water to 200. mL of acetone to make a total volume of 1.60 L?
 Calculate the molarities of the following solutions.
 87.2 g of Na_{2}SO_{4} in enough water to make 500. mL of solution
 61.8 g of NH_{3} in enough water to make 7.00 L of solution
 100. mL of ethanol (C_{2}H_{5}OH) in 500. mL of solution (The density of ethanol is 0.789 g/mL.)
 How many moles of KF are contained in 180.0 mL of a 0.250 M solution?
 Calculate how many grams of each solute would be required in order to make the given solution.
 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl_{3}
 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH_{3}COO)_{2}
 What volume of a 0.500 M solution of NaI could be prepared with 113 g of solid NaI?
 Calculate the molarity of the solutions prepared from the following dilutions.
 125 mL of 2.00 M HCl is diluted to a volume of 4.00 L.
 1.85 mL of 6.30 M AgNO_{3} is diluted to a volume of 5.00 mL.
 What volume of 12 M HCl is required to prepare 6.00 L of a 0.300 M solution?
 What is the molality of the following solutions?
 171.9 g of Sr(NO_{3})_{2} is dissolved in 1.44 kg of water.
 0.883 g of K_{3}PO_{4} is dissolved in 40.0 g of water.
 A copper wire is dipped into 250. mL of a 0.500 M solution of AgNO_{3}. The following singlereplacement reaction occurs. \begin{align*}\text{Cu}(s)+2\text{AgNO}_{3}{(aq)} \rightarrow \text{Cu(NO}_{3}{)}_{2}{(aq)}+2\text{Ag}(s)\end{align*} Assuming all of the dissolved AgNO_{3} reacts, what mass of the copper wire is consumed? (Hint: solve for the moles of AgNO_{3} using the molarity equation, then apply stoichiometry to find the mass of Cu.)
Further Reading / Supplemental Links
 Molarity, (http://www.wisconline.com/Objects/ViewObject.aspx?ID=GCH1204)
 Solution Molarity, (http://www.kentchemistry.com/links/Math/molarity.htm)
 Molality, (http://www.kentchemistry.com/links/Math/molality.htm)
Points to Consider
Some of the physical properties of a solvent are altered by the process of dissolving a solute into that solvent. The vapor pressure, freezing point, and boiling point of a solution are different than that of a pure solvent.
 How is the vapor pressure of a solvent affected when a solute is dissolved?
 How can the freezing and boiling points of a solution be calculated?
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