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16.2: Solution Concentration

Created by: CK-12

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.

\text{Percent by mass}=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100\%

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:

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

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.

\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}

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:

\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}

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.

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

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 NH3 is a “1.5 molar solution of ammonia.”

Sample Problem 16.2: Calculating Molarity

A solution is prepared by dissolving 42.23 g of NH4Cl 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 NH4Cl = 42.23 g
  • molar mass of NH4Cl = 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.

\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}
\dfrac{0.7893 \ \text{mol NH}_4\text{Cl}}{0.5000 \ \text{L}}=1.579 \ \text{M}

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 NH4Cl. Having four significant figures is appropriate.

Practice Problem
  1. What is the molarity of a solution for which 250. mL of the solution contains 10.0 g of Pb(NO3)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 (KMnO4). What mass of KMnO4 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 KMnO4 = 158.04 g/mol

Unknown

  • mass of KMnO4 = ? g

Moles of solute is calculated by multiplying molarity by liters. Then, moles is converted to grams.

Step 2: Solve.

0.250 \ \text{M KMnO}_4 \times 3.00 \ \text{L solution}=0.750 \ \text{mol KMnO}_4
0.750 \ \text{mol KMnO}_4 \times \dfrac{158.04 \ \text{g KMnO}_4}{1 \ \text{mol KMnO}_4}=119 \ \text{g KMnO}_4

Step 3: Think about your result.

When 119 g of potassium permanganate is dissolved in enough water to make 3.00 L of solution, the molarity is 0.250 M.

Practice Problem
  1. What mass of CaCl2 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 1-liter 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 1-liter volumetric flask are outlined and shown below (Figure below).

  1. The appropriate mass of solute is weighed out and added to a volumetric flask that has been about half-filled with distilled water.
  2. The solution is swirled until all of the solute dissolves.
  3. More distilled water is carefully added up to the line etched on the neck of the flask.
  4. 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).

mol1 = mol2

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

M1 × L1 = M2 × L2

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:

M1 × V1 = M2 × V2

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 M2.

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

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 one-fifth 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 (HNO3) 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 HNO3 (M1) = 16 M
  • V2 = 8.00 L
  • M2 = 0.50 M

Unknown

  • volume of stock HNO3 (V1) = ? L

The unknown in the equation is V1, the necessary volume of the concentrated stock solution.

Step 2: Solve.

\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}

Step 3: Think about your result.

250 mL of the stock HNO3 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.

Practice Problems
  1. 125 mL of a 2.55 M solution of Zn(NO3)2 is diluted with water to make the final volume 197 mL. Calculate the new molarity.
  2. 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. The image below (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 lower-case m written in italics.

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

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 (C6H12O6) in 250. g of water.

Step 1: List the known quantities and plan the problem.

Known

  • mass of solute = 28.60 g C6H12O6
  • mass of solvent = 250. g = 0.250 kg
  • molar mass of C6H12O6 = 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.

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

Step 3: Think about your result.

The answer represents the moles of glucose per kilogram of water and has three significant figures.

Practice Problem
  1. Calculate the molality of a solution containing 2.25 g of LiNO3 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

  1. How is a concentrated solution different from a dilute solution?
  2. Answer the following:
    1. As the temperature of a solution increases, what happens to its molarity? Explain.
    2. Why does temperature not affect the molality of a solution?

Problems

  1. What is the mass percent of an aqueous solution prepared by dissolving 12.0 g of solute into 40.0 g of water?
  2. 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?
  3. Calculate the molarities of the following solutions.
    1. 87.2 g of Na2SO4 in enough water to make 500. mL of solution
    2. 61.8 g of NH3 in enough water to make 7.00 L of solution
    3. 100. mL of ethanol (C2H5OH) in 500. mL of solution (The density of ethanol is 0.789 g/mL.)
  4. How many moles of KF are contained in 180.0 mL of a 0.250 M solution?
  5. Calculate how many grams of each solute would be required in order to make the given solution.
    1. 3.40 L of a 0.780 M solution of iron(III) chloride, FeCl3
    2. 60.0 mL of a 4.10 M solution of calcium acetate, Ca(CH3COO)2
  6. What volume of a 0.500 M solution of NaI could be prepared with 113 g of solid NaI?
  7. Calculate the molarity of the solutions prepared from the following dilutions.
    1. 125 mL of 2.00 M HCl is diluted to a volume of 4.00 L.
    2. 1.85 mL of 6.30 M AgNO3 is diluted to a volume of 5.00 mL.
  8. What volume of 12 M HCl is required to prepare 6.00 L of a 0.300 M solution?
  9. What is the molality of the following solutions?
    1. 171.9 g of Sr(NO3)2 is dissolved in 1.44 kg of water.
    2. 0.883 g of K3PO4 is dissolved in 40.0 g of water.
  10. A copper wire is dipped into 250. mL of a 0.500 M solution of AgNO3. The following single-replacement reaction occurs. \text{Cu}(s)+2\text{AgNO}_{3}{(aq)} \rightarrow \text{Cu(NO}_{3}{)}_{2}{(aq)}+2\text{Ag}(s) Assuming all of the dissolved AgNO3 reacts, what mass of the copper wire is consumed? (Hint: solve for the moles of AgNO3 using the molarity equation, then apply stoichiometry to find the mass of Cu.)

Further Reading / Supplemental Links

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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Date Created:

Aug 02, 2012

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Aug 21, 2014
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