Molar Concentration

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

Instead of measuring the solute in grams, molar concentration measures the amount of solute in moles.

The equation is:

c = n/V

where:

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

The most common unit is:

mol/L

This can also be written as:

mol dm⁻³

because:

1 L = 1 dm³

So a solution with a concentration of:

2.0 mol/L

contains 2.0 mol of solute in every litre of solution.

From Mass Concentration to Molar Concentration

Previously, concentration could be expressed using mass:

mass concentration = mass of solute / volume of solution

or:

c = m/V

with units such as:

g/L

Molar concentration uses the same general idea, but measures the amount of solute in moles:

molar concentration = moles of solute / volume of solution

or:

c = n/V

with units:

mol/L

This is especially useful in chemistry because chemical equations describe reactions using mole ratios.

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What Does 1 mol/L Mean?

Suppose a sodium chloride solution has a concentration of:

1.0 mol/L

This means:

1 L of solution contains 1.0 mol NaCl

Since the molar mass of NaCl is approximately:

58.5 g/mol

1 L of this solution contains:

58.5 g NaCl

Similarly:

0.5 L contains 0.5 mol NaCl

0.25 L contains 0.25 mol NaCl

2.0 L contains 2.0 mol NaCl

The concentration remains:

1.0 mol/L

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Understanding the Equation

The central equation is:

c = n/V

For example, suppose:

n = 3.0 mol

and:

V = 2.0 L

Then:

c = 3.0 / 2.0

c = 1.5 mol/L

This means there are 1.5 moles of solute per litre of solution.

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Volume Must Usually Be in Litres

When using:

c = n/V

with concentration in mol/L, volume must be measured in litres.

Remember:

1000 mL = 1 L

Therefore:

500 mL = 0.500 L

250 mL = 0.250 L

100 mL = 0.100 L

50 mL = 0.050 L

25 mL = 0.025 L

To convert:

mL → L

divide by 1000.

To convert:

L → mL

multiply by 1000.

This conversion is one of the most important steps in molar concentration calculations.

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Calculating Molar Concentration

A solution contains:

0.50 mol NaCl

in:

2.0 L solution

Calculate the molar concentration.

Use:

c = n/V

Substitute:

c = 0.50 / 2.0

c = 0.25 mol/L

Answer

Concentration = 0.25 mol/L NaCl

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Worked Example: Smaller Volume

A solution contains:

0.30 mol KCl

in:

500 mL solution

First convert the volume:

500 mL = 0.500 L

Then:

c = n/V

c = 0.30 / 0.500

c = 0.60 mol/L

Answer

Concentration = 0.60 mol/L KCl

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Worked Example: 250 mL Solution

A solution contains:

0.125 mol NaOH

in:

250 mL

Convert:

250 mL = 0.250 L

Then:

c = 0.125 / 0.250

c = 0.500 mol/L

Answer

Concentration = 0.500 mol/L NaOH

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Rearranging the Molar Concentration Equation

The main equation is:

c = n/V

We can rearrange it.

Finding Moles

n = cV

Finding Volume

V = n/c

Therefore, the three useful forms are:

c = n/V

n = cV

V = n/c

These relationships are used repeatedly in solution chemistry.

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Calculating the Number of Moles

Suppose a solution has:

c = 2.0 mol/L

and:

V = 3.0 L

Calculate the number of moles.

Use:

n = cV

n = 2.0 × 3.0

n = 6.0 mol

Answer

Amount of solute = 6.0 mol

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Worked Example: Moles in 500 mL

A sodium chloride solution has:

c = 0.40 mol/L

and:

V = 500 mL

Convert:

500 mL = 0.500 L

Then:

n = cV

n = 0.40 × 0.500

n = 0.200 mol

Answer

n = 0.200 mol NaCl

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Worked Example: Moles in 50 mL

A hydrochloric acid solution has:

c = 1.5 mol/L

and:

V = 50 mL

Convert:

50 mL = 0.050 L

Then:

n = cV

n = 1.5 × 0.050

n = 0.075 mol

Answer

n = 0.075 mol HCl

Notice that even though the concentration is relatively high, the small volume contains a relatively small number of moles.

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Concentration and Amount Are Different

This is an important distinction.

Suppose:

Solution A

1.0 L of 1.0 mol/L NaCl

Moles:

n = 1.0 × 1.0 = 1.0 mol

Solution B

0.10 L of 5.0 mol/L NaCl

Moles:

n = 5.0 × 0.10 = 0.50 mol

Solution B is more concentrated, but Solution A contains more total moles of NaCl.

Therefore:

higher concentration does not necessarily mean more total solute.

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Calculating Volume

If concentration and moles are known:

V = n/c

For example:

How much solution is needed to contain:

2.0 mol NaCl

at a concentration of:

0.50 mol/L?

Calculate:

V = 2.0 / 0.50

V = 4.0 L

Answer

Volume = 4.0 L

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Worked Example: Finding a Smaller Volume

How much:

2.0 mol/L HCl

contains:

0.50 mol HCl?

Use:

V = n/c

V = 0.50 / 2.0

V = 0.25 L

Convert:

0.25 L = 250 mL

Answer

Volume = 250 mL

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Worked Example: Laboratory Volume

A chemist needs:

0.025 mol NaOH

from a solution with concentration:

0.50 mol/L

Calculate the required volume.

V = n/c

V = 0.025 / 0.50

V = 0.050 L

Convert:

0.050 L = 50 mL

Answer

Volume required = 50 mL

Molar Concentration and Particle Number

A mole represents a very large number of particles.

1 mol = 6.022 × 10²³ particles

Therefore, a:

1 mol/L NaCl solution

contains 1 mole of NaCl formula units per litre of solution.

That corresponds to approximately:

6.022 × 10²³ NaCl formula units per litre

Molar concentration therefore connects the macroscopic solution we can measure in the laboratory with the enormous number of particles present at the microscopic level.

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Concentrated and Dilute Solutions

Consider:

Solution A

0.10 mol/L

Solution B

1.0 mol/L

Solution C

2.5 mol/L

Solution C is the most concentrated.

Solution A is the most dilute.

At the particle level, equal volumes of Solution C contain more solute particles than equal volumes of A or B.

Connecting Mass and Moles

Many molar concentration problems begin with the mass of a solute rather than the number of moles.

We therefore need another equation:

n = m/M

where:

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

We can combine this with:

c = n/V

The pathway becomes:

mass → moles → molar concentration

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Multi-Step Example: NaCl Solution

A student dissolves:

11.7 g NaCl

to make:

500 mL solution

Calculate the molar concentration.

Use:

M(NaCl) = 58.5 g/mol

Find the number of moles

n = m/M

n = 11.7 / 58.5

n = 0.200 mol

Convert volume

500 mL = 0.500 L

Calculate concentration

c = n/V

c = 0.200 / 0.500

c = 0.400 mol/L

Answer

Concentration = 0.400 mol/L NaCl

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Multi-Step Example: NaOH Solution

A solution is prepared by dissolving:

8.0 g NaOH

to make:

250 mL solution

Calculate the molar concentration.

Use:

M(NaOH) = 40.0 g/mol

Calculate moles

n = 8.0 / 40.0

n = 0.200 mol

Convert volume

250 mL = 0.250 L

Calculate concentration

c = 0.200 / 0.250

c = 0.800 mol/L

Answer

Concentration = 0.800 mol/L NaOH

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Multi-Step Example: Calcium Chloride

A student dissolves:

22.2 g CaCl₂

to make:

400 mL solution

Use:

M(CaCl₂) = 111 g/mol

Calculate the molar concentration.

Moles

n = 22.2 / 111

n = 0.200 mol

Volume

400 mL = 0.400 L

Concentration

c = 0.200 / 0.400

c = 0.500 mol/L

Answer

Concentration = 0.500 mol/L CaCl₂

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Finding Mass from Molar Concentration

Sometimes we know concentration and volume but need the mass of solute.

The pathway becomes:

concentration + volume → moles → mass

Use:

n = cV

then:

m = nM

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Worked Example: Finding Mass of NaCl

What mass of NaCl is required to prepare:

500 mL

of:

0.20 mol/L NaCl?

Use:

M(NaCl) = 58.5 g/mol

Convert volume

500 mL = 0.500 L

Find moles

n = cV

n = 0.20 × 0.500

n = 0.100 mol

Find mass

m = nM

m = 0.100 × 58.5

m = 5.85 g

Answer

5.85 g NaCl

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Worked Example: Preparing NaOH

What mass of NaOH is needed to prepare:

250 mL

of:

0.40 mol/L NaOH?

Use:

M(NaOH) = 40.0 g/mol

Convert volume

250 mL = 0.250 L

Find moles

n = 0.40 × 0.250

n = 0.100 mol

Find mass

m = 0.100 × 40.0

m = 4.00 g

Answer

4.00 g NaOH

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Preparing a Solution in the Laboratory

To prepare a solution with a known molar concentration, a chemist may:

1. Calculate the required mass of solute.
2. Measure the solute using a balance.
3. Dissolve the solute in some solvent.
4. Transfer the solution to a volumetric flask.
5. Add solvent until the final volume reaches the calibration mark.
6. Mix thoroughly.

The final volume is the volume of the entire solution, not simply the volume of solvent initially added.

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Why Use a Volumetric Flask?

A volumetric flask is designed to contain a very precise volume of solution.

For example, a:

250 mL volumetric flask

allows a chemist to prepare a solution with a final volume of approximately:

250 mL

The solute is dissolved first, and then solvent is added until the bottom of the meniscus reaches the calibration line.

This gives a more accurate concentration than simply measuring water in an ordinary beaker.

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Volume of Solution vs. Volume of Solvent

Suppose the instructions say:

Prepare 250 mL of solution.

This does not mean:

Add the solute to 250 mL of water.

Instead:

Dissolve the solute and then add enough water to make the total solution volume 250 mL.

This distinction is important in accurate laboratory preparation.

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Comparing Molar Concentrations

Consider:

Solution A

0.50 mol solute in 1.0 L

c = 0.50 mol/L

Solution B

0.25 mol solute in 0.25 L

c = 1.0 mol/L

Although B contains fewer total moles, it is more concentrated.

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Same Concentration, Different Volumes

Consider:

Solution A

0.20 mol in 0.50 L

c = 0.40 mol/L

Solution B

0.80 mol in 2.0 L

c = 0.40 mol/L

The solutions have the same molar concentration.

Solution B contains four times as many moles because it has four times the volume.

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Molar Concentration During Dilution

Suppose we have:

0.50 mol solute

in:

0.50 L solution

Initial concentration:

c = 0.50 / 0.50

= 1.0 mol/L

Water is added until the total volume becomes:

1.0 L

The number of moles remains:

0.50 mol

New concentration:

c = 0.50 / 1.0

= 0.50 mol/L

The concentration has been halved.

The solute has not disappeared. The same number of moles is now distributed through twice the volume.

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Moles Are Conserved During Simple Dilution

When only solvent is added:

moles of solute before dilution = moles of solute after dilution

This idea leads to an important dilution relationship:

c₁V₁ = c₂V₂

where:

• c₁ = initial concentration
• V₁ = initial volume
• c₂ = final concentration
• V₂ = final volume

This equation is especially useful when preparing dilute solutions from concentrated stock solutions.

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Worked Dilution Example

A:

2.0 mol/L

solution has a volume of:

100 mL

It is diluted to:

500 mL

Find the new concentration.

Use:

c₁V₁ = c₂V₂

2.0 × 100 = c₂ × 500

Therefore:

c₂ = 0.40 mol/L

Answer

Final concentration = 0.40 mol/L

Because both volumes were expressed in the same units, there was no need to convert them to litres in this particular ratio calculation.

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Molar Concentration and Chemical Reactions

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

Consider:

HCl + NaOH → NaCl + H₂O

The mole ratio is:

1 mol HCl : 1 mol NaOH

Suppose we have:

100 mL of 0.50 mol/L HCl

Convert:

100 mL = 0.100 L

Moles HCl:

n = cV

n = 0.50 × 0.100

= 0.050 mol HCl

Therefore, complete reaction requires:

0.050 mol NaOH

This connects solution concentration directly to stoichiometry.

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Multi-Step Reaction Example

Consider:

2HCl + Mg → MgCl₂ + H₂

A student reacts magnesium with:

200 mL of 0.50 mol/L HCl

How many moles of hydrogen could theoretically form if magnesium is in excess?

Calculate moles HCl

200 mL = 0.200 L

n = cV

n = 0.50 × 0.200

= 0.100 mol HCl

Use the mole ratio

From:

2HCl → 1H₂

Therefore:

0.100 mol HCl × (1 mol H₂ / 2 mol HCl)

= 0.050 mol H₂

Answer

0.050 mol H₂

This demonstrates why molar concentration is so useful in chemical calculations.

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Multi-Step Problem: Concentration from Mass

A student dissolves:

9.8 g H₂SO₄

to prepare:

500 mL solution

Use:

M(H₂SO₄) = 98 g/mol

Calculate the molar concentration.

Find moles

n = 9.8 / 98

= 0.100 mol

Convert volume

500 mL = 0.500 L

Find concentration

c = 0.100 / 0.500

= 0.200 mol/L

Answer

Concentration = 0.200 mol/L H₂SO₄

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Multi-Step Problem: Required Mass

How much KOH is required to prepare:

750 mL

of:

0.20 mol/L KOH?

Use:

M(KOH) = 56.1 g/mol

Convert volume

750 mL = 0.750 L

Calculate moles

n = cV

n = 0.20 × 0.750

= 0.150 mol

Calculate mass

m = nM

m = 0.150 × 56.1

= 8.415 g

Answer

Approximately:

8.42 g KOH

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Molar Concentration and Ions

When ionic compounds dissolve, they separate into ions.

For example:

NaCl → Na⁺ + Cl⁻

A:

1.0 mol/L NaCl

solution produces approximately:

1.0 mol/L Na⁺

and:

1.0 mol/L Cl⁻

But consider:

CaCl₂ → Ca²⁺ + 2Cl⁻

A:

1.0 mol/L CaCl₂

solution produces approximately:

1.0 mol/L Ca²⁺

and:

2.0 mol/L Cl⁻

because each formula unit of CaCl₂ contains two chloride ions.

This becomes important in more advanced solution chemistry.

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Molar Concentration in Everyday and Scientific Applications

Molar concentration is used extensively in:

• analytical chemistry
• titrations
• pharmaceutical manufacturing
• environmental testing
• biochemical research
• industrial chemistry
• water analysis
• reaction stoichiometry

Knowing the concentration allows scientists to determine how many moles of a substance are present without evaporating the solution and weighing the solute.

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Common Mistakes

Using Millilitres Directly in c = n/V

If concentration is in mol/L, volume should be in litres.

Incorrect:

c = 0.2 / 250

Correct:

250 mL = 0.250 L

then:

c = 0.2 / 0.250

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Confusing Moles and Mass

The equation:

c = n/V

uses moles, not grams.

If mass is given, first calculate:

n = m/M

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Confusing Molar Mass and Molar Concentration

Molar mass has units:

g/mol

Molar concentration has units:

mol/L

They describe completely different quantities.

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Using Solvent Volume Instead of Solution Volume

Concentration is based on the final solution volume.

If 5 g solute are dissolved and the final solution is made up to 250 mL, use:

V = 250 mL

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Assuming Higher Concentration Means More Total Moles

A small volume of concentrated solution can contain fewer total moles than a large volume of dilute solution.

Always use:

n = cV

when comparing total amounts.

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Forgetting to Convert the Final Volume

If:

V = 0.075 L

and the question asks for mL:

0.075 L = 75 mL

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Rounding Too Early

Keep several digits during calculations and round the final answer appropriately.

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Using the Wrong Molar Mass

Be careful to calculate the molar mass of the entire chemical formula.

For example:

CaCl₂

contains:

• 1 Ca
• 2 Cl

not just one chlorine atom.

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A Problem-Solving Strategy

For molar concentration problems:

Identify what you know.

Look for:

• concentration
• moles
• volume
• mass
• molar mass

Convert units.

If using mol/L:

mL → L

Choose the correct relationship.

c = n/V

n = cV

V = n/c

If mass is involved:

n = m/M

or:

m = nM

Solve.

Substitute the values with units.

Check your answer.

Ask:

• Are the units correct?
• Does the size of the answer make sense?
• Did I convert mL to L?
• Did I use moles rather than grams?

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Key Terms

Molar concentration — The number of moles of solute present per unit volume of solution.

Concentration — The amount of solute present in a given volume of solution.

Mole — A unit for amount of substance equal to approximately 6.022 × 10²³ particles.

Solute — The substance dissolved in a solution.

Solvent — The substance that dissolves the solute.

Solution — A homogeneous mixture of solute and solvent.

mol/L — Moles per litre, a common unit of molar concentration.

mol dm⁻³ — Moles per cubic decimetre; equivalent to mol/L.

Molar mass — The mass of one mole of a substance, usually expressed in g/mol.

Dilution — Decreasing the concentration of a solution by adding solvent.

Stock solution — A solution of known concentration that can be used to prepare other solutions.

Volumetric flask — Laboratory glassware designed to prepare a precise volume of solution.

Volumetric pipette — Laboratory equipment designed to accurately transfer a specific volume of solution.

Concentration gradient — A difference in concentration between two regions.

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Key Takeaways

• Molar concentration measures moles of solute per volume of solution.
• The main equation is:

c = n/V

• The common unit is:

mol/L

• Because 1 L = 1 dm³:

mol/L = mol/dm³

• To calculate moles:

n = cV

• To calculate volume:

V = n/c

• When concentration is in mol/L, volume should normally be in litres.
• Convert mL to L by dividing by 1000.
• Molar concentration uses moles rather than grams.
• If mass is given, use:

n = m/M

before calculating molar concentration.

• If concentration and volume are known, moles can be found directly.
• If moles and concentration are known, the required solution volume can be calculated.
• A more concentrated solution contains more moles of solute per unit volume.
• A more concentrated solution does not necessarily contain more total moles.
• During simple dilution, the number of moles of solute remains unchanged.
• Molar concentration connects solution chemistry directly to stoichiometry and balanced chemical equations.

The central relationships are:

c = n/V

n = cV

V = n/c

and, when mass is involved:

n = m/M

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Check Your Understanding

Basic Molar Concentration

1. Define molar concentration.

2. State the common unit of molar concentration.

3. What does a concentration of 2.0 mol/L mean?

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

5. Why are moles particularly useful when describing chemical solutions?

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Calculate Concentration

Use:

c = n/V

6. Calculate the concentration of 2.0 mol solute in 4.0 L solution.

7. Calculate the concentration of 0.50 mol solute in 2.0 L solution.

8. Calculate the concentration of 0.25 mol solute in 500 mL solution.

9. Calculate the concentration of 0.075 mol solute in 250 mL solution.

10. Calculate the concentration of 0.020 mol solute in 50 mL solution.

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Calculate Moles

Use:

n = cV

11. How many moles are present in 2.0 L of a 0.50 mol/L solution?

12. Calculate the number of moles in 500 mL of a 1.2 mol/L solution.

13. Calculate the number of moles in 250 mL of a 0.80 mol/L solution.

14. Calculate the number of moles in 25 mL of a 2.0 mol/L solution.

15. A 0.150 L sample has a concentration of 0.40 mol/L. Calculate the number of moles.

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Calculate Volume

Use:

V = n/c

16. What volume of 0.50 mol/L solution contains 1.0 mol solute?

17. What volume of 2.0 mol/L solution contains 0.50 mol solute?

18. What volume of 0.40 mol/L solution contains 0.10 mol solute?

19. What volume of 1.5 mol/L solution contains 0.075 mol solute? Give your answer in mL.

20. What volume of 0.25 mol/L solution contains 0.050 mol solute? Give your answer in mL.

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Multi-Step Problems

Use:

n = m/M

and:

c = n/V

21. Calculate the concentration of a solution made by dissolving 5.85 g NaCl to make 500 mL solution. Use M(NaCl) = 58.5 g/mol.

22. Calculate the concentration of 8.0 g NaOH dissolved to make 500 mL solution. Use M(NaOH) = 40.0 g/mol.

23. Calculate the concentration of 9.8 g H₂SO₄ in 250 mL solution. Use M(H₂SO₄) = 98 g/mol.

24. Calculate the concentration of 11.1 g CaCl₂ in 200 mL solution. Use M(CaCl₂) = 111 g/mol.

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Finding Mass

25. What mass of NaCl is needed to make 1.0 L of 0.50 mol/L NaCl? Use M(NaCl) = 58.5 g/mol.

26. What mass of NaOH is required to prepare 250 mL of 0.20 mol/L solution? Use M(NaOH) = 40.0 g/mol.

27. What mass of KOH is required to make 500 mL of 0.40 mol/L solution? Use M(KOH) = 56.1 g/mol.

28. What mass of CaCl₂ is needed to make 200 mL of 0.25 mol/L solution? Use M(CaCl₂) = 111 g/mol.

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Analysis and Application

29. Solution A contains 0.50 mol in 1.0 L. Solution B contains 0.25 mol in 250 mL. Which is more concentrated? Show your calculations.

30. A student says that a 2.0 mol/L solution must contain more solute than a 1.0 mol/L solution. Explain why this is not necessarily true.

31. A student calculates the concentration of 0.20 mol in 200 mL as 0.001 mol/L. Identify the error and calculate the correct concentration.

32. Explain what happens to molar concentration when solvent is added without changing the number of moles of solute.

33. A 1.0 mol/L solution has a volume of 200 mL. It is diluted to 500 mL. Calculate the new concentration.

34. Explain why the number of moles of solute remains unchanged during simple dilution.

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

36. A chemist needs 0.10 mol NaOH. What volume of 0.50 mol/L NaOH should be used?

37. A student dissolves 4.0 g NaOH to make 200 mL solution. Calculate the molar concentration. Use M(NaOH) = 40.0 g/mol.

38. A 250 mL sample of NaCl solution has a concentration of 0.80 mol/L. Calculate the number of moles and then the mass of NaCl present. Use M(NaCl) = 58.5 g/mol.

39. Describe how you would prepare 500 mL of 0.20 mol/L NaCl solution, including the calculation of the required mass. Use M(NaCl) = 58.5 g/mol.

40. A student has 100 mL of 2.0 mol/L NaCl and adds water until the total volume is 400 mL. Calculate the initial number of moles, the final number of moles, and the final molar concentration.