AP Chemistry Molarity

The formula, the units, and four worked examples of the problems molarity shows up in.

Molarity is the only concentration unit AP Chemistry asks you to calculate, which makes it unusually important: it reappears in stoichiometry, titrations, equilibrium, and pH problems for the rest of the course. The formula is simple. The skill is knowing when to use it as a conversion factor between moles and volume, and that is what the worked examples here are built around.

The Molarity Formula and Its Units

Molarity (M) = moles of solute ÷ liters of solution, with units of mol/L. Rearranged, the same relationship gives the moles in any volume of solution: n = M × V, with V in liters. Two rules cover most errors: convert milliliters to liters before you use the formula, and use the volume of the solution, not the volume of solvent you started with. To turn grams of solute into moles, divide by molar mass, the same step used in mole conversions.

A solution of a specific molarity is made by dissolving the solute and then adding solvent until the total volume reaches the mark on a volumetric flask. Adding the solute to a full measured volume of water would give a slightly larger total volume and a lower molarity than intended.

Worked Example 1: Calculating Molarity From Mass and Volume

11.7 g of NaCl (molar mass 58.44 g/mol) is dissolved in water and diluted to a final volume of 250. mL. What is the molarity?

Moles: 11.7 g ÷ 58.44 g/mol = 0.2002 mol (one extra digit kept until the end).
Volume in liters: 250. mL = 0.250 L.
Molarity: 0.2002 mol ÷ 0.250 L = 0.801 M, rounded to three significant figures.

Worked Example 2: Finding Ion Concentrations From a Dissolved Salt

What are [Ca²⁺] and [Cl⁻] in 0.20 M CaCl₂, and how many moles of Cl⁻ are in 150.0 mL of the solution?

CaCl₂ is a soluble ionic compound that dissociates completely: CaCl₂ → Ca²⁺ + 2 Cl⁻.
[Ca²⁺] = 0.20 M, one ion per formula unit.
[Cl⁻] = 2 × 0.20 = 0.40 M, two ions per formula unit.
Moles of Cl⁻: 0.40 mol/L × 0.1500 L = 0.0600 mol.

The molarity of the compound is not the molarity of every ion in it. The subscripts in the formula set the ratio, the same reasoning used in net ionic equations.

Worked Example 3: Using Molarity as a Conversion Factor in a Neutralization

What volume of 0.100 M NaOH is needed to neutralize 25.0 mL of 0.150 M HCl?

Moles of HCl: 0.150 mol/L × 0.0250 L = 3.75 × 10⁻³ mol.
Mole ratio: HCl + NaOH → NaCl + H₂O, which is 1:1, so 3.75 × 10⁻³ mol of NaOH is needed.
Volume of NaOH: V = n ÷ M = 3.75 × 10⁻³ mol ÷ 0.100 mol/L = 0.0375 L = 37.5 mL.

This is stoichiometry with molarity doing the job molar mass does for solids: moles from volume on one side, volume from moles on the other, with the balanced equation between them. The full method is in stoichiometry.

Worked Example 4: Finding the Molarity When Two Solutions Are Mixed

100.0 mL of 0.50 M NaCl is mixed with 150.0 mL of 0.20 M NaCl. What is the molarity of the mixture? Assume the volumes are additive.

Moles from each: 0.50 mol/L × 0.1000 L = 0.0500 mol; 0.20 mol/L × 0.1500 L = 0.0300 mol.
Total moles: 0.0500 + 0.0300 = 0.0800 mol.
Total volume: 100.0 + 150.0 = 250.0 mL = 0.2500 L.
Molarity: 0.0800 mol ÷ 0.2500 L = 0.32 M.

The answer is between the two starting values, as it must be, and nearer to 0.20 M because there is more of that solution. Adding the two molarities, or averaging them without weighting by volume, gives the wrong result. For a single solution diluted with solvent, the shortcut M₁V₁ = M₂V₂ in the dilution calculations guide is faster.

Molarity Versus Molality on the AP Exam

Molarity (M) is moles of solute per liter of solution. Molality (m) is moles of solute per kilogram of solvent. They look alike, and many general chemistry websites cover both, but the AP Chemistry course framework assesses only molarity: calculations with molality, percent by mass, and percent by volume, along with colligative properties, are not required. The Unit 3 review lists these exclusions with the rest of the unit. If a practice problem asks for molality, it is not preparing you for the exam.

Common Molarity Mistakes

• Leaving the volume in milliliters. Divide by 1000 to get liters before dividing moles by volume.
• Using the solvent volume instead of the solution volume. Molarity depends on the final volume of the solution.
• Forgetting the subscript when finding ion concentration. 0.20 M CaCl₂ gives 0.40 M Cl⁻.
• Averaging molarities when mixing. Add the moles, add the volumes, then divide.
• Mixing up M and m. Capital M is molarity, per liter of solution; lowercase m is molality, which the exam does not test.

Molarity carries forward into later units. It is the starting point for pH calculations of strong acids and bases, and the absorbance concentrations in the Beer-Lambert law use the same units. The AP Chem Score Calculator is the place to see how your overall preparation might translate to a score.

Related Resources

• AP Chem Score Calculator
• Unit 3 Review: Properties of Substances and Mixtures
• AP Chemistry Dilution Calculations
• AP Chemistry Mole Conversions
• AP Chemistry Stoichiometry
• AP Chemistry Strong Acids and Bases
• AP Chemistry Significant Figures Rules
• AP Chemistry Study Guide

Frequently Asked Questions

This page is not affiliated with or endorsed by College Board. AP® is a trademark registered by the College Board.