formats · 5 min read
Molarity vs molality: definitions, formulas and when to use each
Molarity and molality differ by one word and by one physical quantity: liters of solution versus kilograms of solvent. That single change decides whether a concentration drifts when the temperature changes, and which unit fits a titration, a stock bottle label or a thermodynamic calculation.
This comparison puts the two definitions side by side, shows worked numbers for the same amount of solute, and explains when chemists reach for each one. It also covers mole fraction briefly, because the three concentration measures often appear together in problems.
Definitions and formulas
Molarity, usually written M, is moles of solute per liter of solution: M = n / V, where V is the final solution volume in liters. Molality, often written b and sometimes m, is moles of solute per kilogram of solvent: b = n / mass of solvent in kilograms. The difference is the denominator: solution volume for molarity, solvent mass for molality.
For the same amount of solute the numbers can differ by much more than a unit conversion. Half a mole of solute made up to 0.500 L of solution is 1.00 mol/L. Half a mole of the same solute dissolved in 1.000 kg of water is 0.500 mol/kg. Neither answer is wrong; they answer different questions about the same mixture.
Why temperature affects molarity but not molality
Liquids expand when heated, so a solution's volume at 25 °C is slightly larger than at 20 °C while the amount of solute stays the same. Because molarity divides by that volume, the same solution has a slightly lower molarity when warm. The change is usually small near room temperature, but it matters for precise work.
Molality divides by the mass of solvent, which does not depend on temperature. That is why molality is preferred for colligative properties and thermodynamic equations: the measurement is stable when the experiment is run at different temperatures. Molarity, by contrast, is convenient because solution volume is easy to measure with a volumetric flask and directly useful in reaction calculations.
When to use which
Use molarity for stock solutions, titrations and dilution calculations. The equation C1V1 = C2V2 is written in molarity because both sides describe moles per liter and the volume terms can be handled directly. When you prepare a solution in a volumetric flask and make it up to the mark, you are defining its molarity.
Use molality when the property depends on the amount of solvent rather than the solution volume: freezing point depression, boiling point elevation, vapor pressure lowering, osmotic pressure relationships and many thermodynamic expressions. Molality also appears in problems involving concentrated solutions, where the volume of the solution is much less intuitive than the solvent mass. A practical rule is that a procedure measuring solution volume with a flask asks for molarity, while a property driven by how much solvent is present asks for molality.
Converting between them and mole fraction
Converting molarity to molality requires the solution's density and the solute's molar mass. One liter of solution has mass equal to its density in kg/L; subtract the mass of dissolved solute and what remains is the solvent mass. With density ρ in kg/L and molar mass M_r in g/mol, b = M / (ρ − M × M_r / 1000) mol/kg.
Mole fraction is a third measure: χ(solute) = moles of solute divided by total moles of all components. It is dimensionless, needs no volume or mass measurements once the amounts are known, and is the natural unit for gas mixtures and thermodynamic activity. For a binary solution, mole fraction, molality and molar mass of the solvent can be converted into one another without knowing the solution density.
A worked comparison
Suppose 0.500 mol of glucose (molar mass 180.156 g/mol) is dissolved and made up to 0.500 L of solution. The molarity is 0.500 / 0.500 = 1.00 mol/L. If instead the same 0.500 mol is dissolved in 1.000 kg of water, the molality is 0.500 / 1.000 = 0.500 mol/kg. The two numbers differ because one divides by a solution volume and the other by a solvent mass.
To convert the 1.00 mol/L solution to molality we need its density; assume 1.02 kg/L for this example. The solute mass per liter is 1.00 × 180.156 = 180.156 g = 0.180 kg, so the solvent mass is 1.02 − 0.180 = 0.840 kg and the molality is 1.00 / 0.840 = 1.19 mol/kg. The calculation shows why a density is essential: without it, molarity and molality cannot be interconverted.