1. Module 4 Intro

1.37. Page 3

Lesson 5

Module 4—Properties of Solutions

Read

 

The formula to use to calculate molar concentration. Lower case c equals lower case n divided by upper case V.

In Lesson 4 you learned to express concentration as % V/V, % W/V, % W/W, and in parts per million. While labels like percentage weight per volume are great for consumer products, and parts per million is often used for environmental work, chemists in a lab typically express concentration using another set of units. In a previous science course you learned about the mole, the chemical quantity used to express matter. Do you remember what a “mole” is? Later, in Modules 5 and 6, you will see that expressing quantities of substances using moles is favoured because these values can be used to predict quantities of other substances involved in a chemical reaction.

 

To review your understanding of the mole, read page 51 in your textbook.

 

amount concentration: the proportion of solute (in moles) to volume of solution

In this lesson you will learn how to calculate amount concentration. You will also learn how amount concentration can be applied to a wide variety of situations, such as dilution and solution preparation in a lab.

 

Amount Concentration

 

Amount concentration expresses the solute in number of moles and the volume in litres. In many cases, you will have to convert a given mass into a number of moles before finding the amount concentration.

 

Two formulas are shown. The one on the left is the formula to calculate moles. Moles, lower case n, equals mass, lower case m, divided by molar mass, upper case m. The formula on the right is the concentration formula described earlier in the lesson. Lower case c equals lower case n divided by upper case V.

 

Use this formula to find the number of moles.

n = number of moles (mol)
m = mass (g)
M = molar mass (g/mol)

Use this formula to find the amount concentration.

c = amount concentration (mol/L)
n = number of moles (mol)
V = volume (L)



Read "Amount Concentration" on page 205 in your textbook and work through “COMMUNICATION example 4.” Carefully work through the additional example below.

 

Example 1: Determine the amount concentration of 0.42 mol of magnesium sulfate dissolved in a 200-mL solution.

 



Example 2: Determine the amount concentration of 3.0 g of copper(II) nitrate dissolved in a 200-mL solution.

 

 

First, determine the number of moles of solute.

 

 

Now, calculate the amount concentration.

 

 

The amount concentration of the copper(II) nitrate solution is 0.080 mol/L.

 

Self-Check

 

SC 1. Determine the amount concentration of 6.3 mol of magnesium sulfate dissolved in 5.4 L of solution.

 

SC 2. Determine the amount concentration of 2.0 g of iron(III) chloride dissolved in 50 mL of solution.

 

SC 3. Determine the amount concentration of 150 mg of sodium chloride dissolved in 47.0 mL of solution.

 

Check your work.
Self-Check Answers

 

SC 1.

 



The amount concentration of the magnesium sulfate solution is 1.2 mol/L.

 

SC 2.

 

 

Now, calculate the amount concentration.

 

 

The amount concentration of the iron(III) chloride solution is 0.25 mol/L.

 

SC 3.

 

 

First, determine the number of moles of solute.

 

 

Now, calculate the amount concentration.

 

 

The amount concentration of the sodium chloride solution is 0.0546 mol/L.