Concentration (mol/dm^{3}) (Higher Tier)

Definition of Concentration (mol/dm^{3})

Concentration is a measure of how much solute is dissolved in a given volume of solution. For GCSE Higher Tier, concentration is expressed in moles per cubic decimetre (mol/dm^{3}).

Concentration (c) is defined as the amount of solute (in moles, n) divided by the volume of solution (in cubic decimetres, V):

$$c = \frac{n}{V}$$

• Amount of solute, $n$, is measured in moles (mol)
• Volume of solution, $V$, is measured in cubic decimetres ($\mathrm{dm^{3}}$)
• Concentration, $c$, is measured in moles per cubic decimetre ($\mathrm{mol\,dm^{-3}}$)

Note: Concentration in grams per cubic decimetre (g/dm^{3}) is covered in a different topic and is not included here.

Calculating Concentration

The key formula linking amount of substance, concentration, and volume is:

$$n = c \times V$$

Where:

• $n$ = amount of solute in moles (mol)
• $c$ = concentration in moles per cubic decimetre ($\mathrm{mol\,dm^{-3}}$)
• $V$ = volume of solution in cubic decimetres ($\mathrm{dm^{3}}$)

Always ensure volume is in cubic decimetres before using the formula. To convert from cm^{3} to dm^{3}, divide by 1000:

$$V(\mathrm{dm^{3}}) = \frac{V(\mathrm{cm^{3}})}{1000}$$

You can rearrange the formula to find any one of the three variables:

• Concentration: $c = \frac{n}{V}$
• Amount of substance: $n = c \times V$
• Volume: $V = \frac{n}{c}$

For example, if you have 0.5 mol of solute dissolved in 1 dm^{3} of solution, the concentration is:

$$c = \frac{0.5}{1} = 0.5\,\mathrm{mol\,dm^{-3}}$$

For instance, if 0.3 mol of solute is dissolved in 0.6 dm^{3} of solution, the concentration is $c = \frac{0.3}{0.6} = 0.5\,\mathrm{mol\,dm^{-3}}$.

Practical Applications

Concentration calculations are essential in preparing solutions of known concentration for experiments and industrial processes.

• Preparing solutions: To make a solution of a specific concentration, you calculate the amount of solute needed for a given volume of solvent.
• Dilution: When diluting a solution, the amount of solute remains constant, but volume changes. The relationship is:

$$c_1 V_1 = c_2 V_2$$

Where $c_1$ and $V_1$ are the concentration and volume before dilution, and $c_2$ and $V_2$ are after dilution.

This allows you to calculate the new concentration or volume after dilution.

Concentration is also important in titrations, where solutions of known concentration are used to find the concentration of an unknown solution. (See the "Titration Basics" topic for detailed methods.)

Relationship to Amount of Substance

Concentration directly links the amount of substance (in moles) to the volume of solution. This is crucial for quantitative chemical calculations, such as determining how much reactant is needed or how much product can be formed.

For example, if you know the concentration of an acid solution and the volume used, you can calculate the moles of acid reacting. This helps predict the amount of product formed or the amount of another reactant required.

Understanding concentration in mol/dm^{3} is essential for balancing chemical equations quantitatively and for working out yields and limiting reactants (covered in other topics).

Example: Calculating concentration from moles and volume

Calculate the concentration of a solution containing 0.25 mol of sodium hydroxide dissolved in 500 cm^{3} of solution.

First, convert volume to dm^{3}:

$$500\,\mathrm{cm^{3}} = \frac{500}{1000} = 0.5\,\mathrm{dm^{3}}$$

Then use $c = \frac{n}{V}$:

$$c = \frac{0.25}{0.5} = 0.5\,\mathrm{mol\,dm^{-3}}$$