5.3 Concentration and Solutions

Learning Objectives

  • Define minor and major components within solutions: solutes and solvents.
  • Understand how to represent concentrations through a variety of units: percentage concentration, parts per million, parts per billion, and molarity.
  • Use molarity calculations to determine the mass of solute needed to create solutions of specific concentrations.

Chemistry often involves the use of solutions to facilitate reactions. In a liquid solution, a compound (of any state of matter) is dissolved into a liquid medium. The major component of a solution is called the solvent. The minor component of a solution is called the solute. By major and minor we mean whichever component has the greater presence by mass or by moles. Sometimes this becomes confusing, especially with substances with very different molar masses. However, here we will confine the discussion to solutions for which the major component and the minor component are obvious. As a general rule of thumb for liquid solutions, water, alcohol and other liquids in large quantities are the solvents – while dissolved salts, acids, or anything added to a liquid is the solute.

Solutions exist for every possible state of matter combination of the solute and the solvent. Salt water, for example, is a solution of solid NaCl in liquid water; soda water is a solution of gaseous CO2 in liquid water, while air is a solution of a gaseous solute (O2) in a gaseous solvent (N2). In all cases, however, the overall phase of the solution is the same phase as the solvent.

Example: A solution is made by dissolving 1.00 g of sucrose (C12H22O11) in 100.0 g of liquid water. Identify the solvent and solute in the resulting solution.

Either by mass or by moles, the obvious minor component is sucrose, so it is the solute. Water — the majority component — is the solvent. The fact that the resulting solution is the same phase as water also suggests that water is the solvent.

Water is a polar solvent that is commonly used throughout chemistry thanks to it’s easy-to-reach boiling and melting points, alongside availability. It is so common that for chemical equations, a special subscript is used to quickly denote that a solvent is dissolved in a water solute: [latex]_{(aq)}[/latex]. This is referred to as ‘aqueous’. Commonly, acids (such as hydrochloric or nitric) exist as solutes in water solvents. As such, they are commonly shown as [latex]HCl_{(aq)}[/latex] or [latex]HNO_{3(aq)}[/latex] to showcase this fact, removing the need to mention water unless it is directly involved in the reaction.

On a technical note, [latex]H_2O[/latex] can never be considered aqueous – by definition, water cannot be dissolved by water. It is always referred to as a liquid (denoted by [latex]_{(l)}[/latex]). We will however let you decide whether water is wet!

Concentration and Dilution of Solutions

One important concept of solutions is in defining how much solute is dissolved in a given amount of solvent. This concept is called concentration. Various words are used to describe the relative amounts of solute. Dilute describes a solution that has very little solute, while concentrated describes a solution that has a lot of solute. One problem is that these terms are qualitative; they describe more or less but not an exact amount. There are a variety of ways to show a quantitative description of the concentration of a solution. Different methods of expressing concentration include:

  1. Percent concentration (%)
  2. Parts per million (ppm)
  3. Parts per billion (ppb)
  4. Molarity (M)

Percent concentration

Percentage concentrations are useful to show the direct comparison between a solvent and a solute. There are three ways of expressing concentration as a percentage.

  1. Mass/mass percent concentration, (m/m)% \[
    \frac{\textrm{mass of solute}\left(g\right)}{\textrm{mass of solution (g) }}\times100

    Example: What is the percent by mass concentration of sodium chloride solution prepared by dissolving [latex]3.25\textrm{g}[/latex] of sodium chloride in [latex]42.50\textrm{g}[/latex] of water?

    Masses of the solute and the solvent are given, and we are asked to calculate the percent concentration of sodium chloride by mass:\[\begin{align*}
    = & \frac{\textrm{mass of solute}\left(g\right)}{\textrm{mass of solution}\left(g\right)}\times100\\
    = & \frac{3.25g}{42.50g+3.25g}\times100\\
    = & 7.1\%
  2. Volume/volume percent concentration, (v/v)% \[
    \frac{\textrm{volume of solute}\left(mL\right)}{\textrm{volume of solution}\left(mL\right)}\times100\]

    Example: If ethanol [latex]8.00mL[/latex] is dissolved in enough water to give [latex]80.00mL[/latex] of the solution, what is the percent by volume concentration of ethanol in the resulting solution?

    We are asked to calculate ethanol concentration in percent by volume. The volumes of the solute and the solution are given: \[\begin{align*}
    = & \frac{\textrm{volume of solute}\left(mL\right)}{\textrm{volume of solution}\left(mL\right)}\times100\\
    = & \frac{8.00mL}{80.00mL}\times100\\
    = & 10.0\%


  3. Mass/volume percent concentration, (m/v)% \[
    \frac{\textrm{mass of solute}\left(g\right)}{\textrm{volume of solution}\left(mL\right)}\times100

    Example: If [latex]5.0g[/latex] of [latex]\ce{MgCl}_{2}[/latex] is dissolved in enough water to give [latex]200\textrm{mL}[/latex] of solution, calculate the concentration of [latex]\ce{MgCl}_{2}[/latex] in (m/v)%.

    We are asked to calculate the concentration of  [latex]\ce{MgCl}_{2}[/latex] in (m/v)%. The mass of the [latex]\ce{MgCl}_{2}[/latex] and the volume of the resulting solution are given: \[\begin{align*}
    = & \frac{\textrm{mass of solute}\left(g\right)}{\textrm{volume of solution}\left(mL\right)}\times100\\
    = & \frac{5.0g}{200mL}\times100\\
    = & 2.5\%


Parts Per Million and Parts Per Billion

Parts per million (ppm) and parts per billion (ppb) are typically used to describe trace quantities of solutes found within a solution. They represent the parts of an analyte found within one million or one billion parts of a medium respectively.

Parts can be in either mass or volume units. Both solute and solvent must have the same units. For example, one gram of solute is present in a million grams of solution, and one millilitre of benzene is present in a million millilitres of air. To illustrate: \[

Commonly, however, analytes are found dissolved in water. As the density of water is extremely close to  [latex]1~gmL^{-1}[/latex], it is often appropriate to report [latex]1~ppm[/latex] as equal to [latex]1~mgL^{-1}[/latex] for aqueous solutions near room temperature.

Example: Drinking water contains [latex]0.002g[/latex] of  [latex]\ce{Mg}^{+}[/latex] ions in [latex]100g[/latex] of water. What is this concentration in parts per million?

We are asked to calculate the given concentration in ppm: \[\begin{align*}
= & \frac{0.002g}{100g}\times1000000\\
= & 20ppm

Molarity (M)

Molarity is the number of solute moles present in one litre of a solution. This is the most common way to show the concentration of a solution within chemical laboratories thanks to its direct relationship to moles (which, as discussed in 5.2 ‘Molar Mass’, can be easily determined by mass). We can determine molarity by rearranging the following formula:


Where, [latex]n[/latex] is the number of moles [latex]\left(\textrm{mol}\right)[/latex]
[latex]c[/latex] is the molarity of the solution [latex]\left(\textrm{molL}^{-1}\right)[/latex]
[latex]v[/latex] is the volume of the solution [latex]\left(\textrm{L}\right)[/latex]

This is one of the most important formulas in chemistry! It allows us to determine the amount of moles of a substance present within a given concentration and the volume of a substance. However, rearranging allows us to determine the moles required to create a desired concentration for a given volume. As such, a common version of the above equation utilised by chemists is as follows:


Example: If [latex]2.50\textrm{mol}[/latex] of [latex]\ce{CuSO}_{4}[/latex] are dissolved in enough water to give a [latex]500.0\mathscr{\textrm{mL}}[/latex] solution, what is the molarity of the resulting solution?

We are given the number of moles present in [latex]500\textrm{mL}[/latex] of a solution and asked to calculate the molarity. Molarity means the number of moles present in [latex]1000\textrm{mL}[/latex] or [latex]1\textrm{L}[/latex] of solution. Recall that [latex]c=\frac{n}{v}[/latex]. Therefore:

= & \frac{\textrm{moles of solute}\left(mol\right)}{\textrm{volume of solution}\left(L\right)}\\
= & \frac{2.50mol}{0.5L}\\
= & 5.0M\

Have a look at the interactive simulation (Figure 5.3.1) to observe how moles and volume interact with each other to determine molarity:

Figure 5.3.1: Simulation by PhET Interactive Simulations, University of Colorado Boulder, licensed under CC-BY-4.0 (https://phet.colorado.edu).

In combination with [latex]n=\frac{m}{M}[/latex], these equations allow chemists to determine the mass of a solute needed to achieve a specific concentration within a solution. These two equations represent the majority of undergraduate chemistry calculations – try to remember both off by hard! Let’s observe how we can find the mass of a compound needed to achieve a desired concentration.

Example: How many grams of glucose are needed to make a [latex]250\textrm{mL}[/latex] of a [latex]1.5\textrm{M}[/latex] glucose solution? The molar mass of glucose is [latex]180.15\textrm{g/mol}[/latex].

Here, we are given the molarity of the solution and asked to calculate the mass of the solute present in a given volume. Let’s utilise [latex]n=cv[/latex] to determine the moles needed for the solution. Then, we will use the [latex]n=\frac{m}{M}[/latex] equation to convert moles to grams.

Number of moles of glucose present in [latex]250\textrm{mL}[/latex] solution: \[\begin{align*}
= & 1.50\textrm{mol/L}\times0.250\textrm{L}\\
= & 0.375\textrm{mol}

Grams of glucose present in [latex]250\textrm{ml}[/latex] solution:

n & =\frac{m}{M}\\
n\times M & =m\\
0.375\textrm{mol}\times180.15\textrm{g/mol} & =m\\
m & =67.6\textrm{g}

Example: How many millilitres of a [latex]1.00\textrm{M}[/latex] [latex]\ce{HCl}[/latex] solution would be required to obtain [latex]7.3g[/latex] of [latex]\ce{HCl}[/latex]? The molar mass of [latex]\ce{HCl}[/latex] is [latex]36.458\textrm{g/mol}[/latex].

We are given the molarity of the solution and asked to calculate the volume that the given grams of the solute would present. First, convert the grams of the solute into moles using [latex]n=\frac{m}{M}[/latex]. Then, molarity is used as the conversion factor to find the volume that the calculated moles would present.

required to produce 1.0 mol/L using 0.20 mol of HCl.” 

Number of moles of [latex]\ce{HCl}[/latex] corresponds to [latex]n=\frac{m}{M}[/latex], where, [latex]n=[/latex] number of moles [latex]m=[/latex] mass of the substance [latex]M=[/latex] Molar mass of the substance:

n & =\frac{m}{M}\\
n & =\frac{7.3\textrm{g}}{36.458\textrm{g/mol}}\\
n & =0.20\textrm{mol}

The volume that required to produce [latex]{1.0 mol/L}[/latex] using  [latex]{0.20 mol}[/latex] of [latex]\ce{HCl}[/latex]: \[\begin{align*}
\frac{1000\textrm{mL}}{1.00\textrm{mol}}\times0.20\textrm{mol} & =200\textrm{mL}

Dilution of Solutions

Dilution is a process used to lower the concentration of the original solution by adding solvents. The concentrated solution is known as the stock solution. However, the amount of solute present in the solution remains constant before and after the dilution; only the volume changes. The following equation shows the relationship between concentration and volume before and after dilution. This equation is built upon the fact that the same amount of solute is present in the solution before (stock solution) and after (diluted solution) the dilution: \[


[latex]C_{1}[/latex]= Concentration of the stock solution

[latex]V_{1}[/latex]= Volume of the stock solution

[latex]C_{2}[/latex]= Concentration of the diluted solution

[latex]V_{2}[/latex]= Volume of the diluted solution

Example: A chemist prepared a [latex]10\textrm{mL}[/latex] of [latex]1.00\textrm{M}[/latex] [latex]\ce{HCl}[/latex] solution by adding water to the [latex]3.0\textrm{M}[/latex] stock solution of [latex]\ce{HCl}[/latex]. How much stock solution should be taken for the dilution?

We are given the concentration of the diluted solution, volume of the diluted solution, and concentration of the stock solution and asked to calculate the volume of the stock solution used for the dilution. As the number of solute moles remains constant before and after the dilution, we can use the [latex]C_{1}V_{1}=C_{2}V_{2}[/latex] equation to calculate the volume of the stock solution: \[\begin{align*}
C_{1}V_{1} & =C_{2}V_{2}\\
3.0\textrm{M}\times V_{1} & =1.0\textrm{M}\times10\textrm{mL}\\
V_{1} & =\frac{1.0\textrm{M}\times10\textrm{mL}}{3.0\textrm{M}}\\
V_{1} & =3.3\textrm{mL}

Chemistry Is Everywhere: Preparing IV Solutions

In a hospital emergency room, a physician orders an intravenous (IV) delivery of 100 mL of 0.5% KCl for a patient suffering from hypokalemia (low potassium levels). Does an aide run to a supply cabinet and take out an IV bag containing this concentration of KCl (Figure 5.3.2)?

Not likely. It is more probable that the aide must make the proper solution from an IV bag of sterile solution and a more concentrated, sterile solution, called a stock solution, of KCl. The aide is expected to use a syringe to draw up some stock solution and inject it into the waiting IV bag and dilute it to the proper concentration. Thus the aide must perform a dilution calculation.

An IV bag attached to a drip system.
Figure 5.3.2: Medical personnel commonly must perform dilutions for IV solutions. Image attribution: “Infuuszakjes” © 2008 by Harmid is licensed under a Public Domain license.

If the stock solution is 10.0% KCl and the final volume and concentration need to be 100 mL and 0.50%, respectively, then to determine how much stock solution to use:

[latex]\begin{array}{rcl} (10\%)V_1&=&(0.50\%)(100\text{ mL}) \\ V_1&=&5\text{ mL} \end{array}[/latex]

Of course, the addition of the stock solution affects the total volume of the diluted solution, but the final concentration is likely close enough, even for medical purposes.

Medical and pharmaceutical personnel are constantly dealing with dosages that require concentration measurements and dilutions. It is an important responsibility: calculating the wrong dose can be useless, harmful, or even fatal!


In most cases, only a certain maximum amount of solute can be dissolved in a given amount of solvent. This maximum amount is called the solubility of the solute. It is usually expressed in terms of the amount of solute that can dissolve in 100 g of the solvent at a given temperature. Table 5.3.1 “Solubilities of Some Ionic Compounds” lists the solubilities of some simple ionic compounds. These solubilities vary widely: NaCl can dissolve up to 31.6 g per 100 g of H2O, while AgCl can dissolve only 0.00019 g per 100 g of H2O.

Table 5.3.1 Solubilities of Some Ionic Compounds.
Solute Solubility (g per 100 g of H2O at 25°C)
AgCl 0.00019
CaCO3 0.0006
KBr 70.7
NaCl 36.1
NaNO3 94.6

When the maximum amount of solute has been dissolved in a given amount of solvent, we say that the solution is saturated with solute. When less than the maximum amount of solute is dissolved in a given amount of solute, the solution is unsaturated. These terms are also qualitative terms because each solute has its own solubility. A solution of 0.00019 g of AgCl per 100 g of H2O may be saturated, but with so little solute dissolved, it is also rather dilute. A solution of 36.1 g of NaCl in 100 g of H2O is also saturated but rather concentrated. Ideally, we need more precise ways of specifying the amount of solute in a solution.

In some circumstances, it is possible to dissolve more than the maximum amount of a solute in a solution. Usually, this happens by heating the solvent, dissolving more solute than would normally dissolve at regular temperatures, and letting the solution cool down slowly and carefully. Such solutions are called supersaturated solutions and are not stable; given an opportunity (such as dropping a crystal of solute in the solution), the excess solute will precipitate from the solution.

It should be obvious that some solutes dissolve in certain solvents but not others. NaCl, for example, dissolves in water but not in vegetable oil. Beeswax dissolves in liquid hexane but not water. What is it that makes a solute soluble in some solvents but not others?

The answer is intermolecular interactions. Intermolecular interactions include London dispersion forces, dipole-dipole interactions, and hydrogen bonding. From experimental studies, it has been determined that if the molecules of a solute experience the same intermolecular forces that the solvent does, the solute will likely dissolve in that solvent. So, NaCl — a very polar substance because it is composed of ions — dissolves in water, which is very polar, but not in oil, which is generally nonpolar. Nonpolar wax dissolves in nonpolar hexane but not in polar water. This concept leads to the general rule that “like dissolves like” for predicting whether a solute is soluble in a given solvent (see Figure 5.3.3). However, this is a general rule, not an absolute statement, so it must be applied with care.

<a href="https://rmit.pressbooks.pub/rmitchemistrybridgingcourse/chapter/5-3-concentration-and-solutions/">Transcript.</a>
Figure 5.3.3: As potassium chloride (KCl) dissolves in water, the ions are hydrated. The polar water molecules are attracted by the charges on the K+ and Cl ions. Water molecules in front of and behind the ions are not shown. Transcript. Image attribution:OpenStax: licensed under a Creative Commons Attribution 4.0 International License.

Example: Would I2 be more soluble in CCl4 or H2O? Explain your answer.


[latex]\ce{I_{2}}[/latex] is a nonpolar molecule due to the similar electronegativities of the iodine atoms.

[latex]\ce{CCl_{4}}[/latex] is a nonpolar molecule: In [latex]\ce{CCl_{4}}[/latex], electrons are more attracted towards the chlorine atoms due to the high electronegativity of chlorine compared to the carbon. However, as four bonds are arranged symmetrically, the dipole moment of one bond is cancelled out by the opposite bond, creating a zero net dipolar moment. Therefore, [latex]\ce{CCl_{4}}[/latex], is nonpolar. View this video to learn why [latex]\ce{CCl_{4}}[/latex], is a nonpolar molecule.

[latex]\ce{H_{2}O}[/latex] is a polar molecule: In the water molecule, due to the high electronegativity of the oxygen atom, electrons are strongly attracted towards the oxygen atom. Additionally, the water molecule has a bent, or angular, structure due to the two lone pairs of electrons on the oxygen atom. This arrangement results in an asymmetrical charge distribution within the molecule, with the oxygen end being more negative and the hydrogen end being more positive. Therefore, the combination of the polar covalent bonds and the molecular geometry leads to the water molecule having a net dipole moment, making it a polar molecule. View this video to learn why [latex]\ce{H_{2}O}[/latex], is a polar molecule.

I2 is nonpolar. Of the two solvents, CCl4 is nonpolar, and H2O is polar. Since “like dissolves like,” nonpolar solutes tend to dissolve better in nonpolar solvents. So I2 would expected to be more soluble in CCl4.

Key Takeaways

  • Solutions contain at least two components: a large amount of solvent and a smaller amount of solute(s)
  • Species dissolved in water are called “aqueous solutions”.
  • There are four common measures of concentration: percentage concentration, parts per million, parts per billion, and molarity.
  • In chemistry, “like-dissolves-like”. Polar solvents will generally dissolve polar solutes, while non-polar solvents dissolve non-polar solutes.


Practice Questions


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