What Is Molar Concentration?

In chemistry, we often need to know how much of a substance is dissolved in a particular volume of solution.

Molar concentration tells us the number of moles of solute present in a given volume of solution.

It is often called molarity.

For example, a solution containing 1 mole of sodium chloride in a total solution volume of 1 dm³ has a molar concentration of:

1 mol/dm³

Molar concentration is especially useful because chemical equations describe reactions in terms of particles and moles rather than simply mass.

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Moles and Solutions

A mole is a measure of the amount of substance.

One mole contains approximately:

6.02 × 10²³ particles

These particles might be:

• Atoms.
• Molecules.
• Ions.
• Formula units.

For example:

1 mole of water contains approximately 6.02 × 10²³ water molecules.

1 mole of sodium chloride contains approximately 6.02 × 10²³ formula units of sodium chloride.

Molar concentration therefore tells us about the number of chemical particles present in a volume of solution.

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Calculating Molar Concentration

Molar concentration is calculated using:

concentration = moles ÷ volume

In symbols:

c = n ÷ V

where:

• c = molar concentration
• n = amount of solute in moles
• V = volume of solution

The volume must normally be expressed in dm³ when concentration is measured in mol/dm³.

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Units of Molar Concentration

A common unit is:

mol/dm³

This means:

moles per cubic decimeter

Since:

1 dm³ = 1 L

mol/dm³ and mol/L describe the same numerical concentration.

You may therefore see:

0.5 mol/dm³

or:

0.5 mol/L

depending on the convention being used.

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Understanding Molar Concentration

Suppose a sodium chloride solution has a concentration of:

2 mol/dm³

This means:

Every 1 dm³ of solution contains 2 moles of sodium chloride.

Similarly:

0.25 mol/dm³

means:

Every 1 dm³ of solution contains 0.25 mol of solute.

A larger molar concentration means more moles of solute are present per unit volume.

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Worked Example: Calculating Molar Concentration

A solution contains 0.5 mol of sodium chloride in 2 dm³ of solution.

Use:

c = n ÷ V

Substitute:

c = 0.5 ÷ 2

Therefore:

c = 0.25 mol/dm³

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Volume Conversion

Volumes are often given in cm³ rather than dm³.

Remember:

1000 cm³ = 1 dm³

Therefore:

cm³ → dm³: divide by 1000

For example:

250 cm³ ÷ 1000 = 0.250 dm³

So:

250 cm³ = 0.250 dm³

This conversion is extremely important in molar concentration calculations.

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Worked Example: Concentration from cm³

A solution contains 0.20 mol of solute in 500 cm³.

First convert the volume:

500 cm³ = 0.500 dm³

Now use:

c = n ÷ V

c = 0.20 ÷ 0.500

c = 0.40 mol/dm³

Therefore, the molar concentration is:

0.40 mol/dm³

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A Reliable Calculation Method

When solving molar concentration problems:

Step 1: Identify the number of moles.

Step 2: Identify the volume.

Step 3: Check the volume unit.

Step 4: Convert cm³ to dm³ if necessary.

Step 5: Select the correct equation.

Step 6: Substitute the values.

Step 7: Calculate.

Step 8: Include the correct unit.

Following these steps helps prevent unit errors.

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Rearranging the Equation

Starting with:

c = n ÷ V

we can rearrange the equation.

To calculate moles:

n = c × V

To calculate volume:

V = n ÷ c

Therefore, the three useful forms are:

c = n ÷ V

n = c × V

V = n ÷ c

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Worked Example: Calculating Moles

A solution has:

concentration = 0.50 mol/dm³

volume = 2.0 dm³

Use:

n = c × V

Substitute:

n = 0.50 × 2.0

Therefore:

n = 1.0 mol

The solution contains 1.0 mol of solute.

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Worked Example: Calculating Moles in a Smaller Volume

A sodium hydroxide solution has a concentration of:

0.40 mol/dm³

What amount of sodium hydroxide is present in 250 cm³?

First convert:

250 cm³ = 0.250 dm³

Now:

n = c × V

n = 0.40 × 0.250

n = 0.100 mol

Therefore:

0.100 mol of NaOH is present.

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Worked Example: Calculating Volume

A solution contains 0.30 mol of solute and has a concentration of 0.60 mol/dm³.

Use:

V = n ÷ c

Substitute:

V = 0.30 ÷ 0.60

V = 0.50 dm³

Therefore:

V = 0.50 dm³

or:

500 cm³

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Choosing the Correct Equation

Ask what quantity the question wants.

To find concentration:

c = n ÷ V

To find moles:

n = c × V

To find volume:

V = n ÷ c

Writing the equation first is usually safer than trying to perform the calculation mentally.

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Molar Concentration and Mass Concentration

Mass concentration and molar concentration describe solutions in different ways.

Mass concentration tells us the mass of solute per volume.

Typical unit:

g/dm³

Molar concentration tells us the amount of solute in moles per volume.

Typical unit:

mol/dm³

We can convert between them using molar mass.

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From Mass to Moles

The number of moles can be calculated using:

moles = mass ÷ molar mass

In symbols:

n = m ÷ M

where:

• n = moles
• m = mass in grams
• M = molar mass in g/mol

This equation is often combined with the molar concentration equation.

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Worked Example: Mass to Molar Concentration

A solution contains 5.85 g of sodium chloride in 500 cm³.

The molar mass of NaCl is approximately:

58.5 g/mol

First calculate moles:

n = 5.85 ÷ 58.5

n = 0.100 mol

Now convert the volume:

500 cm³ = 0.500 dm³

Calculate concentration:

c = 0.100 ÷ 0.500

c = 0.200 mol/dm³

Therefore:

concentration = 0.200 mol/dm³

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Multi-Step Concentration Problems

Some questions require several calculations.

A useful sequence is:

Mass → Moles → Concentration

If the mass of a substance is given:

First:

n = mass ÷ molar mass

Then:

c = moles ÷ volume

This allows us to convert a measured mass into molar concentration.

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Worked Example: Sodium Hydroxide

A student dissolves 4.0 g of NaOH and makes the final solution volume 500 cm³.

Calculate the molar concentration.

Relative atomic masses:

Na = 23

O = 16

H = 1

First calculate molar mass:

M(NaOH) = 23 + 16 + 1

M(NaOH) = 40 g/mol

Calculate moles:

n = 4.0 ÷ 40

n = 0.10 mol

Convert volume:

500 cm³ = 0.500 dm³

Calculate concentration:

c = 0.10 ÷ 0.500

c = 0.20 mol/dm³

Therefore:

concentration = 0.20 mol/dm³

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Worked Example: Copper(II) Sulfate

A student prepares 250 cm³ of solution containing 0.050 mol of copper(II) sulfate.

Convert volume:

250 cm³ = 0.250 dm³

Use:

c = n ÷ V

c = 0.050 ÷ 0.250

c = 0.20 mol/dm³

Therefore:

concentration = 0.20 mol/dm³

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Comparing Molar Concentrations

Consider two solutions.

Solution A

0.20 mol in 200 cm³

Solution B

0.30 mol in 500 cm³

We cannot simply compare the number of moles because the volumes are different.

Solution A

200 cm³ = 0.200 dm³

c = 0.20 ÷ 0.200

c = 1.0 mol/dm³

Solution B

500 cm³ = 0.500 dm³

c = 0.30 ÷ 0.500

c = 0.60 mol/dm³

Therefore:

Solution A is more concentrated.

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Concentration Is a Ratio

Molar concentration compares:

amount of solute

with:

volume of solution

For example:

0.1 mol in 0.5 dm³

has the same concentration as:

0.2 mol in 1.0 dm³

because:

0.1 ÷ 0.5 = 0.2 mol/dm³

and:

0.2 ÷ 1.0 = 0.2 mol/dm³

Doubling both moles and volume does not change the concentration.

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Molarity in Chemical Reactions

Molar concentration is particularly useful because chemical equations use mole ratios.

Consider:

HCl + NaOH → NaCl + H₂O

The equation shows a:

1 : 1

mole ratio between HCl and NaOH.

Therefore:

1 mol HCl reacts with 1 mol NaOH.

If we know the concentration and volume of an HCl solution, we can calculate how many moles of HCl are present.

We can then use the chemical equation to calculate how much NaOH is required.

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Worked Example: Using Molarity in a Reaction

Consider:

HCl + NaOH → NaCl + H₂O

A student has 100 cm³ of 0.50 mol/dm³ HCl.

How many moles of HCl are present?

Convert:

100 cm³ = 0.100 dm³

Use:

n = c × V

n = 0.50 × 0.100

n = 0.050 mol

Therefore:

0.050 mol HCl is present.

Since the reaction ratio is 1 : 1:

0.050 mol NaOH would be required for complete reaction.

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Worked Example: Using a Different Mole Ratio

Consider:

2HCl + Mg → MgCl₂ + H₂

Suppose 0.20 mol of HCl reacts completely.

The equation tells us:

2 mol HCl : 1 mol Mg

Therefore:

0.20 mol HCl : 0.10 mol Mg

So 0.20 mol of HCl requires:

0.10 mol Mg

This is why molar concentration is so useful: it allows solution measurements to connect directly to balanced chemical equations.

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From Solution Volume to Reacting Moles

A common chemistry calculation follows this sequence:

Concentration + Volume

↓

Moles

↓

Mole ratio from equation

↓

Moles of another substance

↓

Mass or volume required

This is an important connection between solution chemistry and stoichiometry.

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Challenge Example

200 cm³ of a 0.30 mol/dm³ sulfuric acid solution reacts with sodium hydroxide.

The equation is:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

First calculate moles of H₂SO₄.

Convert volume:

200 cm³ = 0.200 dm³

Use:

n = c × V

n = 0.30 × 0.200

n = 0.060 mol H₂SO₄

The equation gives:

1 mol H₂SO₄ : 2 mol NaOH

Therefore:

0.060 mol H₂SO₄ : 0.120 mol NaOH

So:

0.120 mol NaOH is required.

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Preparing a Solution of Known Molar Concentration

Chemists often need to prepare solutions with precise molar concentrations.

The general procedure is:

• Calculate the required number of moles.
• Convert the moles into mass.
• Measure the required mass of solute.
• Dissolve the solute in some solvent.
• Transfer the solution to a volumetric flask.
• Add solvent until the required final volume is reached.
• Mix thoroughly.

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Worked Example: Preparing a Solution

A student needs to prepare:

500 cm³ of 0.20 mol/dm³ NaCl

First convert:

500 cm³ = 0.500 dm³

Calculate moles:

n = c × V

n = 0.20 × 0.500

n = 0.100 mol

Molar mass of NaCl:

58.5 g/mol

Calculate mass:

mass = moles × molar mass

mass = 0.100 × 58.5

mass = 5.85 g

Therefore, the student needs:

5.85 g NaCl

The salt is dissolved and the final solution volume is made up to exactly:

500 cm³

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Why the Final Volume Matters

Suppose a student wants to prepare 250 cm³ of solution.

The student should not necessarily measure 250 cm³ of water and then add the solute.

Instead:

• Dissolve the solute in some water.
• Transfer it to a suitable volumetric container.
• Add water until the final solution volume is 250 cm³.

Molar concentration refers to the volume of the solution, not simply the original volume of solvent.

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Molar Concentration and Dilution

When water is added to a solution:

• The number of moles of solute stays the same.
• The total volume increases.
• The molar concentration decreases.

For example:

0.20 mol in 0.50 dm³:

c = 0.20 ÷ 0.50 = 0.40 mol/dm³

If diluted to 1.0 dm³:

c = 0.20 ÷ 1.0 = 0.20 mol/dm³

The number of moles has not changed.

Only the concentration has changed.

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Why Molar Concentration Is Useful

Mass tells us how heavy a sample is.

But chemical reactions depend on the number of particles involved.

For example:

58.5 g of NaCl

and

18.0 g of H₂O

have very different masses.

However, each amount is approximately:

1 mole

Therefore, each contains approximately the same number of formula units or molecules.

Molar concentration connects the measurable volume of a solution with the number of moles and therefore with the number of particles available to react.

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Laboratory Uses

Molar concentration is important in:

• Titrations.
• Acid-base reactions.
• Reaction-rate experiments.
• Equilibrium experiments.
• Analytical chemistry.
• Biochemistry.
• Industrial chemistry.

Knowing concentration allows chemists to calculate exactly how much substance is present in a measured volume.

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Interpreting Concentration Values

Suppose:

Solution A = 0.10 mol/dm³

Solution B = 0.50 mol/dm³

For equal volumes, Solution B contains:

5 times as many moles of solute

as Solution A.

This does not necessarily mean Solution B contains five times the mass, because the relationship between mass and moles depends on the molar mass of the substance.

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Worked Example: Same Molarity, Different Substances

Consider:

1 dm³ of 1.0 mol/dm³ NaCl

and:

1 dm³ of 1.0 mol/dm³ glucose

Both contain:

1.0 mol of solute

But their masses are different because NaCl and glucose have different molar masses.

Therefore:

same molar concentration does not necessarily mean same mass concentration.

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Common Mistakes

Using cm³ Directly

For concentration in mol/dm³, convert cm³ to dm³ first.

250 cm³ = 0.250 dm³

not 250 dm³.

Confusing Moles and Mass

Molar concentration uses moles, not grams.

If mass is given, convert mass to moles first.

Using the Wrong Volume

Use the final volume of the solution, not simply the amount of solvent added.

Forgetting Molar Mass

When converting mass to moles:

n = mass ÷ molar mass

Using an Unbalanced Chemical Equation

Stoichiometric calculations require a correctly balanced equation.

Ignoring Mole Ratios

A reaction does not always occur in a 1 : 1 ratio.

For example:

H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O

The ratio is:

1 mol H₂SO₄ : 2 mol NaOH

Forgetting Units

An answer such as:

0.25

is incomplete.

Write:

0.25 mol/dm³

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Check Your Understanding

1. Define molar concentration.

2. What is the common unit for molar concentration?

3. Write the equation connecting concentration, moles, and volume.

4. Convert 350 cm³ to dm³.

5. Calculate the concentration of 0.40 mol of solute in 2.0 dm³.

6. Calculate the concentration of 0.15 mol of solute in 300 cm³.

7. A solution has a concentration of 0.50 mol/dm³ and a volume of 0.40 dm³. Calculate the number of moles.

8. How many moles are present in 250 cm³ of a 0.80 mol/dm³ solution?

9. A solution contains 0.25 mol at a concentration of 0.50 mol/dm³. Calculate its volume.

10. Calculate the number of moles in 100 cm³ of 2.0 mol/dm³ HCl.

11. Explain the difference between mass concentration and molar concentration.

12. Why must mass sometimes be converted into moles before calculating molar concentration?

13. A solution contains 4.0 g NaOH in 250 cm³. Calculate its molar concentration. The molar mass of NaOH is 40 g/mol.

14. Explain why molar concentration is particularly useful when working with balanced chemical equations.

15. A student adds water to a solution. Explain what happens to the number of moles of solute and the molar concentration.

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Key Terms

• Molar concentration – number of moles of solute per unit volume of solution.
• Molarity – another term commonly used for molar concentration.
• Mole – amount of substance containing approximately 6.02 × 10²³ specified particles.
• Solute – substance dissolved in a solvent.
• Solvent – substance in which the solute dissolves.
• Solution – mixture formed when a solute dissolves in a solvent.
• mol/dm³ – moles of solute per cubic decimeter of solution.
• Molar mass – mass of one mole of a substance, usually measured in g/mol.
• Stoichiometry – quantitative relationships between substances in chemical reactions.
• Dilution – reduction in concentration by adding solvent.
• Volumetric flask – laboratory glassware used to prepare an accurate fixed volume of solution.

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Key Takeaways

• Molar concentration tells us how many moles of solute are present per unit volume of solution.
• Molar concentration is also commonly called molarity.
• A common unit is mol/dm³.
• Concentration = moles ÷ volume.
• Moles = concentration × volume.
• Volume = moles ÷ concentration.
• 1000 cm³ = 1 dm³.
• Convert cm³ to dm³ before using volumes in calculations involving mol/dm³.
• Mass can be converted to moles using moles = mass ÷ molar mass.
• Many problems follow the sequence mass → moles → concentration.
• Molar concentration allows solution volumes to be connected directly to mole ratios in balanced chemical equations.
• Different substances can have the same molar concentration but different mass concentrations.
• Dilution increases volume without changing the number of moles of solute.
• Therefore, dilution decreases molar concentration.
• Accurate molar concentrations are important in titrations, analytical chemistry, laboratory experiments, and industrial chemistry.
• Molar concentration is useful because chemical reactions depend on numbers of particles, and moles provide a practical way to measure those particles.