6.5Concentration of Aqueous Solution
Concentration of a solution is a measurement that shows the quantity of solute dissolved in a unit volume of solution, typically in \(1\text{ dm}^3\) of solution. The greater the amount of solute dissolved, the higher the concentration of the solution.
Learning Objectives
- State the meaning of concentration of aqueous solution.
- Solve numerical problems involving concentration of solution.
Conversion Between Concentration in \(\text{g dm}^{-3}\) and Molarity in \(\text{mol dm}^{-3}\)
Use molar mass to interconvert between mass concentration and molarity.
Brief Explanation
Understand the key definitions and formulas before solving numerical problems.
Concentration in \(\text{g dm}^{-3}\)
The mass of solute found in \(1\text{ dm}^3\) of solution.
\[\text{Concentration (g dm}^{-3}\text{)} = \dfrac{\text{Mass of solute (g)}}{\text{Volume of solution (dm}^3\text{)}}\]
Molarity in \(\text{mol dm}^{-3}\)
The number of moles of solute found in \(1\text{ dm}^3\) of solution. Molarity is represented by the unit \(\text{mol dm}^{-3}\) or molar (\(\text{M}\)).
\[\text{Molarity (mol dm}^{-3}\text{)} = \dfrac{\text{Number of moles of solute (mol)}}{\text{Volume of solution (dm}^3\text{)}}\]
| Conversion | Formula |
|---|---|
| Concentration \(\text{g dm}^{-3}\) → Molarity \(\text{mol dm}^{-3}\) | \[\text{Molarity (mol dm}^{-3}\text{)} = \dfrac{\text{Concentration (g dm}^{-3}\text{)}}{\text{Molar mass (g mol}^{-1}\text{)}}\] |
| Molarity \(\text{mol dm}^{-3}\) → Concentration \(\text{g dm}^{-3}\) | \[\text{Concentration (g dm}^{-3}\text{)} = \text{Molarity (mol dm}^{-3}\text{)} \times \text{Molar mass (g mol}^{-1}\text{)}\] |
| Number of moles, \(V\) in \(\text{dm}^3\) | \[n = MV\] |
| Number of moles, \(V\) in \(\text{cm}^3\) | \[n = \dfrac{MV}{1000}\] |
Try These Questions First
Answer mentally, then check the explanation.
What is the molarity of a solution containing \(0.4\text{ mol}\) of zinc chloride, \(\text{ZnCl}_2\), dissolved to produce \(2\text{ dm}^3\) of solution?
What is the concentration in \(\text{g dm}^{-3}\) of a \(0.5\text{ mol dm}^{-3}\) sulphuric acid solution, \(\text{H}_2\text{SO}_4\)? [Relative atomic mass: \(\text{H} = 1, \text{O} = 16, \text{S} = 32\)]
How many moles of lead(II) nitrate, \(\text{Pb(NO}_3)_2\), are present in \(200\text{ cm}^3\) of \(0.2\text{ mol dm}^{-3}\) lead(II) nitrate solution?
Common Mistakes
Avoid these errors when solving concentration problems.
Common Misconceptions
Compare each misconception with the correct fact.
"Mole" and "molar" mean the same thing.
"Mole" is the unit for the amount of substance (matter), while "molar" (\(\text{M}\) or \(\text{mol dm}^{-3}\)) is a unit of concentration representing the number of moles per unit volume of solution.
Solute mass and solute molar mass can be used interchangeably in concentration formulas.
Solute mass is the actual mass of solute dissolved, while molar mass is the mass of one mole of the substance. They must be substituted into the correct positions in the formula.
Summary
Key facts to remember.
- Concentration is the quantity of solute dissolved in a unit volume of solution (normally \(1\text{ dm}^3\)).
- Concentration in \(\text{g dm}^{-3} = \dfrac{\text{Mass of solute (g)}}{\text{Volume of solution (dm}^3\text{)}}\).
- Molarity in \(\text{mol dm}^{-3} = \dfrac{\text{Number of moles of solute (mol)}}{\text{Volume of solution (dm}^3\text{)}}\).
- Conversion: \(\text{Concentration (g dm}^{-3}\text{)} = \text{Molarity (mol dm}^{-3}\text{)} \times \text{Molar mass (g mol}^{-1}\text{)}\).
- Moles calculation: \(n = MV\) (if \(V\) in \(\text{dm}^3\)) or \(n = \dfrac{MV}{1000}\) (if \(V\) in \(\text{cm}^3\)).
Quick Activity
Complete all parts before checking your score.
A. Objective Quiz
Which of the following is the correct unit for molarity?
To convert the concentration of a solution from \(\text{g dm}^{-3}\) to molarity in \(\text{mol dm}^{-3}\), the concentration must be:
B. Fill in the Blanks
The concentration of a solution in unit \(\text{mol dm}^{-3}\) is also known as ____.
When the volume of a solution, \(V\), is given in \(\text{cm}^3\), the number of moles, \(n\), of solute is calculated using the formula \(n = \dfrac{MV}{\text{____}}\).
C. Match the Quantity with Its Definition or Formula
Drag each choice on desktop. On mobile, tap a choice and then tap the matching box.
Answer Choices
Concentration in \(\text{g dm}^{-3}\) Concentration in \(\text{mol dm}^{-3}\) Molarity \(\times\) Molar mass \(n = MV\)Definition / Formula
Keywords
Tap or hover over a term to read its meaning.
