Concentration

Concentration is how much solute is dissolved in a given amount of solution or solvent. Chemical reactions between two solutions require knowing not just what's dissolved, but how much — and chemists express that amount in several different units depending on the situation. This article covers molarity, the MCAT's default concentration unit; how dilution works and the M₁V₁ = M₂V₂ relationship; and three other units you're expected to know — molality, percent by weight, mole fraction, and normality (along with the concept of an equivalent that normality depends on).

Key Takeaways

• Molarity (M) = mol solute / L solution — the MCAT's default concentration unit.

• Dilution adds solvent without changing the moles of solute present, so M₁V₁ = M₂V₂ relates a solution's concentration and volume before and after dilution.

• Molality (m) = mol solute / kg solvent — unlike molarity, it doesn't change with temperature, making it useful for thermodynamic calculations.

• Percent by weight = (mass solute / mass solution) × 100; mole fraction (X) = mol component / total mol.

• Normality (N) = equivalents solute / L solution = M × n, where n is the number of equivalents each mole of solute supplies (e.g., n = 2 for the diprotic acid H₂SO₄).

• An equivalent counts a specific reactive unit a molecule supplies — H⁺ or OH⁻ in acid-base reactions, electrons in redox reactions.

Molarity — The MCAT's Default Concentration Unit

Molarity (M) is defined as the moles of solute per liter of solution:

M = mol solute / L solution

Molarity is the concentration unit used most often in chemical equations and reactions, because it directly connects the amount of a substance (moles) to a volume you can measure in the lab.

Stock Solutions and Dilution

Chemical laboratories often buy concentrated stock solutions rather than dilute ones, since a concentrated solution takes up far less shelf space in a stock room for the same amount of dissolved solute. Before use, a stock solution is usually diluted — solvent (usually water) is added to lower its concentration.

The key insight behind dilution: adding solvent increases the solution's volume, but it doesn't add or remove any solute. The moles of solute present stay exactly the same before and after — only the volume changes, so the molarity drops. That conservation of moles is captured by the dilution equation:

M₁V₁ = M₂V₂

where M₁ and V₁ are the molarity and volume of the concentrated solution before dilution, and M₂ and V₂ are the molarity and volume of the solution after dilution. Because moles of solute = molarity × volume, and that quantity doesn't change during dilution, the product M₁V₁ before dilution must equal the product M₂V₂ after.

Worked example — You need 500 mL of 1.5 M HCl, and you have a 6.0 M HCl stock solution. How much stock solution do you need?

Solve M₁V₁ = M₂V₂ for V₁:

V₁ = (M₂V₂) / M₁ = (1.5 M × 0.500 L) / 6.0 M = 0.125 L = 125 mL

Measure out 125 mL of the 6.0 M stock solution, then add water until the total volume reaches 500 mL. The result is 500 mL of 1.5 M HCl.

Molality, Percent by Weight, and Mole Fraction

Molarity isn't the only way to express concentration. Three other units come up regularly:

• Molality (m): moles of solute per kilogram of solvent (not solution). m = mol solute / kg solvent. Because it's based on the mass of solvent rather than the volume of solution, molality doesn't change with temperature — a liquid's mass is constant, but its volume expands or contracts as temperature changes. This makes molality the preferred unit for thermodynamic calculations involving temperature changes.

• Percent by weight (%): the ratio of the solute's mass to the total mass of the solution, times 100. % = (mass solute / mass solution) × 100. This unit is common in industrial and commercial contexts, where concentrations are often specified as, for example, "5% saline."

• Mole fraction (X): the ratio of the moles of one component to the total moles of all components in the solution. X = mol component / total mol (all components). Mole fraction is used throughout chemistry, especially for gas mixtures and solutions.

Worked example — Suppose you dissolve 58.5 g of NaCl (1.00 mol, using NaCl's molar mass of 58.5 g/mol) into 1.000 kg of water, and the resulting solution has a total volume of 1.02 L. Express this solution's concentration in all three units above, plus molarity:

• Molarity: 1.00 mol solute / 1.02 L solution = 0.980 M

• Molality: 1.00 mol solute / 1.000 kg solvent = 1.00 m

• Percent by weight: mass of solute is 58.5 g; mass of solution is 58.5 g + 1000 g = 1058.5 g. (58.5 / 1058.5) × 100 = 5.53%

• Mole fraction: moles of water = 1000 g / 18.0 g/mol ≈ 55.5 mol. X(NaCl) = 1.00 / (1.00 + 55.5) = 0.0177

Notice that molarity and molality give numerically similar values here (0.980 vs. 1.00) — that's only because this solution is dilute and close to water's density of 1 kg/L. As concentration increases, molarity and molality diverge, since molarity's denominator (solution volume) changes with both temperature and how much solute is packed in, while molality's denominator (solvent mass) doesn't.

Normality and Equivalents

Normality (N) is the number of equivalents of solute per liter of solution: N = equivalents solute / L solution. Normality is especially useful in titration calculations and in reactions involving acids and bases.

An equivalent is a measure of a molecule's reactive capacity — how many moles of a specific reactive unit that molecule can supply or react with. In acid-base chemistry, the reactive unit is the H⁺ (or OH⁻) ion: one equivalent of an acid is the amount that supplies one mole of H⁺, and one equivalent of a base is the amount that supplies one mole of OH⁻. A monoprotic acid like HCl supplies one H⁺ per molecule, so one mole of HCl is one equivalent. A diprotic acid like H₂SO₄ supplies two H⁺ per molecule, so one mole of H₂SO₄ is two equivalents. The same logic extends to redox reactions, where the reactive unit is the electron: one equivalent is the amount of a substance that gains or loses one mole of electrons.

Because normality just counts equivalents instead of moles, it relates directly back to molarity through the number of equivalents each mole of solute provides:

N = M × n

where n is the number of equivalents per mole (sometimes called the valence factor — 1 for a monoprotic acid, 2 for a diprotic acid, and so on).

Worked example — What is the normality of a 0.500 M H₂SO₄ solution?

H₂SO₄ is diprotic, so n = 2 equivalents per mole:

N = M × n = 0.500 mol/L × 2 = 1.00 N

Common MCAT Mistakes

• Mixing up molarity and molality. Molarity divides by liters of solution; molality divides by kilograms of solvent. They're numerically close only in dilute aqueous solutions.

• Forgetting that dilution conserves moles, not concentration. M₁V₁ = M₂V₂ works because the moles of solute don't change when solvent is added — plugging in a final volume that already accounts for the added solute (rather than total solution volume) gives a wrong answer.

• Dropping the valence factor n when converting molarity to normality. N = M × n, not N = M — skipping n for a diprotic or triprotic acid/base understates the normality.

• Mixing up mass and moles in percent by weight versus mole fraction. Percent by weight uses masses in its ratio; mole fraction uses moles. Using mass where moles are required (or vice versa) gives a wrong ratio.

MCAT-Style Concept Check

Question: What is the normality of a 0.25 M solution of H₃PO₄, a triprotic acid?

• A) 0.25 N

• B) 0.50 N

• C) 0.75 N

• D) 1.00 N

Answer: C

Explanation: Normality equals molarity times the valence factor n, where n is the number of equivalents each mole supplies. H₃PO₄ is triprotic, supplying three H⁺ per molecule, so n = 3: N = M × n = 0.25 M × 3 = 0.75 N.

FAQ

What's the difference between molarity and molality?

Molarity (M) is moles of solute per liter of solution; molality (m) is moles of solute per kilogram of solvent. Because molality's denominator is a mass rather than a volume, it doesn't change with temperature, which is why it's preferred for calculations involving temperature changes.

How do you use M₁V₁ = M₂V₂ in a dilution problem?

Set the molarity and volume of the concentrated (stock) solution as M₁ and V₁, and the molarity and volume of the diluted solution as M₂ and V₂, then solve for whichever value is unknown. The relationship holds because dilution doesn't add or remove moles of solute — only the volume (and therefore the molarity) changes.

What is an equivalent, in the context of normality?

An equivalent is the amount of a substance that supplies one mole of a specific reactive unit — one mole of H⁺ or OH⁻ in acid-base reactions, or one mole of electrons in redox reactions. Normality counts equivalents per liter of solution rather than moles per liter.

Why doesn't molality change with temperature, when molarity does?

Molality's denominator is the mass of solvent, which doesn't change with temperature. Molarity's denominator is the volume of the solution, which expands or contracts as temperature changes, so molarity shifts slightly with temperature even though the amount of solute and solvent stays the same.