How to Calculate Molality

Why molality exists when you already have molarity

Molality (m) gets pulled out of the drawer the moment temperature enters the problem. A flask of 1.00 M NaCl warmed from 20 °C to 80 °C is no longer 1.00 M — the water expanded, the volume crept up, the moles per liter dropped. A 1.00 m NaCl solution at 20 °C is still 1.00 m at 80 °C, because mass does not care about temperature. That is the entire reason every colligative-property formula you will meet — freezing point depression, boiling point elevation, the van’t Hoff equation in its proper form — is written in molality, not molarity.

The definition is short:

m = moles of solute / kilograms of solvent

Two tripwires hide in that one line. The denominator is solvent mass, not solution mass. And the units are kilograms, not grams. Miss either one and your answer is off by a factor of 1000 or by whatever the solute weighs.

The four-step method, with NaCl in water

Dissolve 15.0 g of NaCl in 500.0 g of water. What is the molality?

1. Identify the players. NaCl is the solute (M = 58.44 g/mol). Water is the solvent.
2. Solute → moles. 15.0 g ÷ 58.44 g/mol = 0.2566 mol NaCl.
3. Solvent → kilograms. 500.0 g ÷ 1000 = 0.5000 kg water.
4. Divide. 0.2566 mol ÷ 0.5000 kg = 0.513 m.

You have 0.513 mol of NaCl per kilogram of water — and you would still have 0.513 m if you took the flask outside on a cold day or warmed it on a hot plate.

Worked example: freezing point depression of road salt

Use that same 0.513 m NaCl solution. NaCl dissociates fully into Na⁺ and Cl⁻, so the van’t Hoff factor i = 2. For water, K_f = 1.86 °C·kg/mol.

ΔT_f = i × K_f × m = 2 × 1.86 × 0.513 = 1.91 °C

The solution freezes at 0.00 − 1.91 = −1.91 °C. This is exactly why salt scattered on icy sidewalks works: it depresses the freezing point of the meltwater enough to keep it liquid below 0 °C.

Worked example: converting 2.00 M H₂SO₄ to molality

The trap here: density. You cannot convert M to m without knowing how heavy a liter of solution actually is.

Concentrated sulfuric acid at 2.00 M has a density of 1.123 g/mL. Start with exactly 1 L:

1. Moles of H₂SO₄ = 2.00 mol.
2. Mass of H₂SO₄ = 2.00 mol × 98.08 g/mol = 196.2 g.
3. Mass of 1 L of solution = 1000 mL × 1.123 g/mL = 1123 g.
4. Mass of solvent (water) = 1123 − 196.2 = 926.8 g = 0.9268 kg.
5. Molality = 2.00 mol ÷ 0.9268 kg = 2.16 m.

Notice molality came out higher than molarity. That is the rule for any solution where the solute contributes appreciable mass: the kilogram of solvent is always less than the kilogram of solution, so dividing by the smaller number gives a larger result.

Worked example: boiling point elevation with glucose

Dissolve 50.0 g of glucose (C₆H₁₂O₆, M = 180.16 g/mol) in 200.0 g of water. K_b for water is 0.512 °C·kg/mol. What is the boiling point?

1. Moles glucose = 50.0 ÷ 180.16 = 0.2776 mol.
2. kg water = 0.2000.
3. m = 0.2776 ÷ 0.2000 = 1.388 m.
4. Glucose is molecular and does not dissociate, so i = 1.
5. ΔT_b = 1 × 0.512 × 1.388 = 0.711 °C.
6. Boiling point = 100.00 + 0.711 = 100.71 °C.

A noticeable but small bump — and the reason the canonical demo for boiling-point elevation uses concentrated sucrose or salt rather than dilute sugar.

Molality vs. molarity, side by side

Feature	Molality (m)	Molarity (M)
Definition	mol solute / kg solvent	mol solute / L solution
Temperature dependence	Independent	Volume changes with T, so M drifts
Used for	Colligative properties	Bench prep, titrations, kinetics
Requires density?	Only when converting from M	Yes, for unit conversions
Denominator	Solvent mass alone	Total solution volume

Traps that cost points on exams

• Solvent mass, not solution mass. If a problem gives you “100 g of solution containing 5 g of solute,” the solvent is 95 g, not 100 g. Subtract first.
• Grams to kilograms. Forgetting the ÷1000 turns a 0.5 m solution into a 500 m monster on paper.
• The lowercase/uppercase trap. “0.5 m” and “0.5 M” are not the same except in the limit of infinitely dilute aqueous solution. Read the problem carefully.

Practice

Work these by hand, then verify with the Solution Concentration Calculator:

1. 25.0 g KBr (M = 119.00 g/mol) in 750.0 g water — molality?
2. How much glucose (M = 180.16 g/mol) do you dissolve in 1.00 kg water for a 0.500 m solution?
3. Convert 1.50 M CaCl₂ (density = 1.13 g/mL) to molality.
4. Boiling point of 2.00 m NaCl in water (K_b = 0.512 °C·kg/mol, i = 2)?
5. An unknown solute (10.0 g) in 100.0 g water freezes at −3.72 °C. With K_f = 1.86 and i = 1, find its molar mass.