Solutions and Molarity

By the end of this lesson and the next few, you should be able to:

  • Describe the fundamental properties of solutions
  • Calculate solution concentrations using molarity
  • Perform dilution calculations using the dilution equation

Molarity

In preceding lessons, we focused on the composition of substances: samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances.

Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness (see image below). In the next set of lessons, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.

espresso

Sugar is one of many components in the complex mixture known as coffee. The amount of sugar in a given amount of coffee is an important determinant of the beverage’s sweetness. Image credit: Jane Whitney

Solutions

We have previously defined solutions as homogeneous mixtures, meaning that the composition of the mixture (and therefore its properties) is uniform throughout its entire volume. Solutions occur frequently in nature and have also been implemented in many forms of manmade technology. We will explore a more thorough treatment of solution properties in the chapter on solutions and colloids, but here we will introduce some of the basic properties of solutions.

The relative amount of a given solution component is known as its concentration. Often, though not always, a solution contains one component with a concentration that is significantly greater than that of all other components. This component is called the solvent and may be viewed as the medium in which the other components are dispersed, or dissolved. Solutions in which water is the solvent are, of course, very common on our planet. A solution in which water is the solvent is called an aqueous solution.

solute is a component of a solution that is typically present at a much lower concentration than the solvent. Solute concentrations are often described with qualitative terms such as dilute (of relatively low concentration) and concentrated (of relatively high concentration).

Concentrations may be quantitatively assessed using a wide variety of measurement units, each convenient for particular applications. Molarity (M) is a useful concentration unit for many applications in chemistry. Molarity is defined as the number of moles of solute in exactly 1 liter (1 L) of the solution:

\(M = \cfrac{\text{mol solute}}{\text{L solution}}\)

Calculating Molar Concentrations

A 355-mL soft drink sample contains 0.133 mol of sucrose (table sugar). What is the molar concentration of sucrose in the beverage?

Solution

Since the molar amount of solute and the volume of solution are both given, the molarity can be calculated using the definition of molarity. Per this definition, the solution volume must be converted from mL to L:

\(M = \cfrac{\text{mol solute}}{\text{L solution}} \)\(= \cfrac{0.133\text{ mol}}{355\text{ mL} × \frac{1\text{ L}}{1000\text{ mL}}} = 0.375 M\)

Deriving Moles and Volumes from Molar Concentrations

How much sugar (mol) is contained in a modest sip (~10 mL) of the soft drink from the example above?

Solution

In this case, we can rearrange the definition of molarity to isolate the quantity sought, moles of sugar. We then substitute the value for molarity that we derived in the example above, 0.375 M:

\(M = \frac{\text{mol solute}}{\text{L solution}}\)

\(\text{mol solute} = M × \text{L solution}\)

\(\text{mol solute} = 0.375 \frac{\text{mol sugar}}{\text{L}} × (10\text{ mL} × \frac{1\text{ L}}{1000\text{ mL}}) \)\(= 0.004\text{ mol sugar}\)


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