Molarity is one of the most important concentration units in chemistry. It appears throughout IB Chemistry—from stoichiometry to titration, equilibrium, acids and bases, and reaction kinetics. Mastering molarity calculations makes many other solution-based problems much easier. This article breaks down the concept clearly and shows you exactly how to calculate molarity in a way that aligns with IB exam expectations.
What Is Molarity?
Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter (dm³) of solution.
The formula is simple:
Molarity (c) = moles of solute (n) ÷ volume of solution (V in dm³)
Units:
- M (moles per liter)
- mol dm⁻³ (official IB unit)
- mol/L (equivalent)
Molarity tells you “how much solute is dissolved in a given volume of solution.”
Step-by-Step: How to Calculate Molarity
Step 1: Calculate moles of solute
Depending on the data provided:
If given mass:
n = mass ÷ molar mass
If given volume × concentration (for solutions):
n = c × V
If given number of particles:
n = particles ÷ Avogadro’s number
Step 2: Convert solution volume to dm³
IB requires volume in dm³, not cm³.
Conversions:
- 1000 cm³ = 1 dm³
- 1 L = 1 dm³
If you skip this step, your molarity calculation will be incorrect by a factor of 1000.
Step 3: Apply the molarity formula
c = n ÷ V
Insert your moles and volume, and the result gives the solution’s molarity.
Worked Example (IB-Style)
Question:
What is the molarity of a solution prepared by dissolving 4.00 g of NaOH in water to make 250 cm³ of solution?
Step 1: Calculate moles
Mr of NaOH = 40.0
n = 4.00 g ÷ 40.0 g/mol = 0.100 mol
Step 2: Convert volume
250 cm³ = 0.250 dm³
Step 3: Calculate molarity
c = 0.100 ÷ 0.250 = 0.400 mol dm⁻³
This type of problem appears frequently in Papers 1, 2, and 3.
Rearranging the Molarity Formula
IB questions often require solving for different variables.
Find moles:
n = c × V
Find volume:
V = n ÷ c
These rearrangements are essential when working with titration calculations.
Molarity in Dilution Problems
Dilutions follow the formula:
c₁V₁ = c₂V₂
Where:
- c₁ = initial concentration
- V₁ = initial volume
- c₂ = final concentration
- V₂ = final volume
This relationship allows you to dilute stock solutions accurately.
Example:
5.00 cm³ of 2.0 M solution diluted to 100 cm³
c₂ = (2.0 × 5.00) ÷ 100 = 0.10 M
Molarity in Stoichiometry and Titrations
Because molarity is tied to moles, it is a key part of:
- Determining limiting reagents in solution
- Calculating titration endpoints
- Acid–base neutralization calculations
- Redox titrations
- Equilibrium constant (Kc) calculations
- Reaction rate expressions involving solutions
IB assessment frequently combines molarity with balanced equations.
Common Mistakes to Avoid
- Forgetting to convert cm³ → dm³
- Dividing by mass instead of moles
- Using moles of wrong reactant (limiting reagent errors)
- Confusing molarity (mol dm⁻³) with molality (mol kg⁻¹)
- Rounding too early—IB prefers proper significant figures
Accurate unit handling is the biggest factor in correct answers.
FAQs
Is molarity temperature dependent?
Yes. Because volume changes with temperature, molarity can vary. This is why standard solutions are prepared at controlled conditions.
What if the solute dissociates into ions?
Molarity refers to the concentration of the compound, not the ions. Ion concentrations are calculated only after dissociation, using stoichiometry.
Can solids have molarity?
No. Molarity applies only to solutions. Solid concentrations use different units such as mol per gram or mol per kilogram.
Conclusion
Molarity is a simple yet powerful way to express solution concentration. By converting volumes correctly, calculating moles accurately, and applying c = n/V, you can handle titration problems, dilution questions, and stoichiometric calculations with confidence. Mastering molarity is essential for success in IB Chemistry and for understanding how solutions behave in real laboratory scenarios.
