S1.4.5 Molar concentration

Molar Concentration: Definition, Units, and Applications

You’re making lemonade.
You carefully measure sugar and water, adjusting the sweetness to your taste.
If it’s too sweet, you add water; if it’s not sweet enough, you add more sugar.

What you’re doing is intuitively changing the concentration of sugar in the solution.

Molar Concentration: The Core Definition

Definition

Molar concentration

Molar concentration, commonly called molarity, measures how many moles of solute are dissolved in one cubic decimeter (dm³) of solution.

Mathematically, it is expressed as:

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

Where:

$C$ = molar concentration (in $m o l d m^{- 3}$ )
$n$ = amount of solute (in moles)
$V$ = volume of the solution (in dm³)

Note

This formula is fundamental for preparing solutions with precise concentrations and predicting reaction outcomes in solution-phase chemistry.

Why Use Molar Concentration?

Molar concentration bridges the macroscopic (measurable quantities like volume) with the microscopic (the number of particles in a solution).
For instance, knowing the concentration of a reactant helps calculate how much product a reaction will yield.

Tip

Always convert volume to dm³ (1 dm³ = 1,000 cm³) when using the molar concentration formula to avoid calculation errors.

Units of Concentration: Moles and Grams

Concentration can be expressed in various units depending on the context. Let’s explore the two most common: $m o l d m^{- 3}$ and $g d m^{- 3}$ .

1.Molar Concentration $C$ in $m o l d m^{- 3}$ :

This unit specifies the number of moles of solute per cubic decimeter of solution.

Example

Dissolving 1 mole of sodium chloride $N a C l$ in 1 dm³ of water results in a concentration of $1.0 {mol dm}^{- 3}$ .

2.Mass Concentration $ρ$ in $g d m^{- 3}$ :

Definition

Mass concentration

Mass concentration specifies the mass of solute (in grams) dissolved in one cubic decimeter (dm³) of solution.

It focuses on the actual mass rather than the number of particles, making it useful in laboratory preparations when measuring mass directly is easier than calculating moles.

Definition:
$ρ = \frac{m}{V}$
where:
- $ρ$ = mass concentration (g/dm³)
- $m$ = mass of solute (g)
- $V$ = volume of solution (dm³)
Converting Molar to Mass Concentration:
Since the number of moles $n$ is related to mass by the molar mass $M$ ( $n = \frac{m}{M}$ ), the mass concentration can also be calculated from molar concentration: $ρ = C \times M$ where:
- $C$ = molar concentration (mol/dm³)
- $M$ = molar mass of the solute (g/mol)

Example

Converting Between Units

A solution of 0.5 $m o l d m^{- 3}$ sodium chloride $N a C l$ has a molar mass of 58.44 $m o l d m^{- 3}$ .

To find its mass concentration:

$ρ = C \times M = 0.5 {mol dm}^{- 3} \times 58.44 {g mol}^{- 1} = 29.22 {g dm}^{- 3}$

This solution contains $29.22 g$ of $N a C l$ per dm³.

Common Mistake

A common mistake is forgetting to convert volume to dm³ when using $C = \frac{n}{V}$ . For instance, if the volume is given in cm³, divide it by 1,000 to convert to dm³.

Applications of Molar Concentration

Molar concentration is a practical tool used in laboratory experiments, industrial processes, and even daily activities.
Below are examples of common problem types involving molar concentration.

Example question

Calculating molar concentration

You dissolve $2.00 mol$ of glucose $C_{6} H_{12} O_{6}$ in $5.00 {dm}^{3}$ of water. What is the molar concentration of glucose in the solution?

Solution

Using $C = \frac{n}{V}$ :

$C = \frac{2.00 mol}{5.00 {dm}^{3}} = 0.400 {mol dm}^{- 3}$

The concentration of glucose is $0.400 {mol dm}^{- 3}$ .

Example question

Finding the Amount of Solute

How many moles of sodium hydroxide $N a O H$ are present in $250 {cm}^{3}$ of a $0.200 {mol dm}^{- 3}$ solution?

Solution

Convert the volume to dm³:

$V = \frac{250 {cm}^{3}}{1, 000} = 0.250 {dm}^{3}$

Now use (n = C \times V):

$n = 0.200 {mol dm}^{- 3} \times 0.250 {dm}^{3} = 0.0500 mol$

The solution contains $0.0500 mol$ of $N a O H$ .

Example question

Determining Solution Volume

What volume of a $2.00 {mol dm}^{- 3}$ sulfuric acid $H_{2} S O_{4}$ solution is required to provide $1.00 mol$ of solute?

Solution

Rearrange $C = \frac{n}{V}$ to solve for $V$ :

$V = \frac{n}{C} = \frac{1.00 mol}{2.00 {mol dm}^{- 3}} = 0.500 {dm}^{3}$

You need $0.500 {dm}^{3}$ or $500 {cm}^{3}$ of the solution.

Self review

A solution has a concentration of $0.100 {mol dm}^{- 3}$ . If you have $50.0 {cm}^{3}$ of it, how many moles of solute does it contain?

Practical Implications

Dilutions

Molar concentration is essential when preparing solutions of specific concentrations.
To dilute a stock solution, use the formula:

$C_{1} V_{1} = C_{2} V_{2}$

Where:

$C_{1}$ and $V_{1}$ : concentration and volume of the stock solution
$C_{2}$ and $V_{2}$ : concentration and volume of the diluted solution

Example question

Dilution

You have $100, {cm}^{3}$ of a $2.00 {mol dm}^{- 3}$ hydrochloric acid $H C l$ solution. How much water must you add to dilute it to $0.500 {mol dm}^{- 3}$ ?

Solution

Using $C_{1} V_{1} = C_{2} V_{2}$ :

$2.00 {mol dm}^{- 3}) (100 {cm}^{3}) = (0.500 {mol dm}^{- 3}) (V_{2})$

$V_{2} = \frac{200.0 {mol cm}^{3}}{0.500 {mol dm}^{- 3}} = 400 {cm}^{3}$

The final solution volume is $400 {cm}^{3}$ . Add $400 - 100 = 300 {cm}^{3}$ of water.

Reflection

Theory of Knowledge

How do numerical expressions of concentration enhance or hinder our understanding of chemical systems? In what scenarios might qualitative descriptions be more effective?

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S1.4.5 Molar concentration

Molar Concentration: Definition, Units, and Applications

Molar Concentration: The Core Definition

Why Use Molar Concentration?

Units of Concentration: Moles and Grams

1.Molar Concentration C in moldm−3:

2.Mass Concentration ρ in gdm−3:

Applications of Molar Concentration

Calculating molar concentration

Finding the Amount of Solute

Determining Solution Volume

Practical Implications

Dilutions

Dilution

Reflection

You've read 2/2 free chapters this week.

Introduction to Molar Concentration

1.Molar Concentration $C$ in $m o l d m^{- 3}$ :

2.Mass Concentration $ρ$ in $g d m^{- 3}$ :