Molality vs Molarity — Why Molality Does Not Change with Temperature
Two concentration units, one letter apart, and a whole chapter of Class 12 depends on telling them apart. The difference is not a technicality: it is the reason every colligative-property formula in your textbook uses molality and not molarity.
The two definitions
Molality, m = moles of solute ÷ mass of solvent in kilograms
Two words differ, and both matter:
| Molarity (M) | Molality (m) | |
|---|---|---|
| Denominator measures | Volume | Mass |
| Denominator refers to | The whole solution | Only the solvent |
| Unit | mol L⁻¹ | mol kg⁻¹ |
| Changes when temperature changes? | Yes | No |
| Easy to prepare in a lab? | Yes — volumetric flask | Yes, but you must weigh the solvent |
| Used in | Titrations, reaction stoichiometry, pH | Colligative properties, elevation of boiling point, depression of freezing point |
Why only molarity changes with temperature
Heat a solution and the liquid expands, so its volume goes up. The mass of solute and the mass of solvent do not change at all — warming does not create or destroy matter.
Molarity has volume in its denominator: bigger denominator, smaller molarity. Molality has mass in its denominator, and that denominator never moved. That is the whole argument, and it is worth being able to write it in two sentences in an exam.
Illustration. Suppose a solution is 0.474 M and 0.500 m at 25 °C. It is warmed until the volume expands by 1.0%.
The same moles now occupy 1.010 L for every 1.000 L before, so
new molarity = 0.474 ÷ 1.010 = 0.4693
Molarity falls from 0.474 M to 0.469 M. Molality stays exactly 0.500 m.
(The 1.0% expansion is an assumed figure used to show the effect clearly — the real expansion depends on the liquid and the temperature range. The direction of the change is always this way for a liquid that expands on heating.)
This is also why molality is preferred whenever the experiment itself changes the temperature. Measuring a freezing point or a boiling point is a temperature change, and a concentration unit that drifts while you take the reading would be self-defeating. Hence:
Both use molality; neither uses molarity. That is not an arbitrary convention — it follows from the paragraph above.
The conversion formula
Converting between them needs one extra piece of information neither definition contains: the density of the solution, d, in g mL⁻¹.
m = (1000 · M) ÷ (1000 · d − M · Msolute)
Where Msolute is the molar mass of the solute in g mol⁻¹. You do not have to memorise these — both come out of one basis calculation, and doing it from first principles is safer. Worked example 1 does exactly that, then checks against the formula.
Worked example 1 — molality to molarity, from first principles
Question: A glucose solution is 0.500 m. Its density is 1.034 g mL⁻¹. Find its molarity. (M(C₆H₁₂O₆) = 180.156 g mol⁻¹)
Step 1 — choose a basis. "0.500 m" means 0.500 mol of glucose per kilogram of water, so take exactly 1000 g of water.
Step 2 — mass of the solute in that basis.
0.500 mol × 180.156 g mol⁻¹ = 90.078 g
Step 3 — mass of the whole solution (solvent + solute):
1000 g + 90.078 g = 1090.078 g
Step 4 — volume of the solution, using the density.
V = 1090.078 ÷ 1.034 = 1054.23 mL = 1.05423 L
Step 5 — molarity.
M = 0.500 mol ÷ 1.05423 L = 0.47428
Molarity = 0.474 M
Cross-check with the formula:
M = (1000 × 0.500 × 1.034) ÷ (1000 + 0.500 × 180.156)
= 517.0 ÷ 1090.078 = 0.47428 ✓ Identical.
Step 3 is the one people skip. Molality gave the mass of the solvent; molarity needs the volume of the solution, so the solute's own mass must be added in.
Worked example 2 — molality straight from laboratory data
Question: 18.0 g of glucose is dissolved in 250 g of water. Find the molality.
Step 1 — moles of solute. n = 18.0 ÷ 180.156 = 0.099913 mol
Step 2 — mass of solvent in kilograms. 250 g = 0.250 kg
Step 3 — divide. m = 0.099913 ÷ 0.250 = 0.39965
Molality = 0.400 mol kg⁻¹ (0.400 m)
Note what was not needed: the density, the final volume, the total mass of the solution. A balance is all molality requires. The molarity of this same solution would also need its volume measured.
Worked example 3 — the colligative-property payoff
Question: By how much does the freezing point of water fall for the 0.400 m glucose solution from Example 2? Take Kf for water as 1.86 K kg mol⁻¹.
Glucose is a non-electrolyte — it does not dissociate — so the van't Hoff factor i = 1 and we can use the simple form.
ΔTf = Kf × m = 1.86 × 0.400 = 0.744
ΔTf = 0.744 K, so the solution freezes at about −0.744 °C.
The molality used here came straight from Example 2 — no density, no volume, no temperature correction. Had the formula been written in molarity, we would have needed the density of the solution at its freezing point, which is precisely the awkward quantity nobody measures.
Worked example 4 — when are the two nearly equal?
Question: A 0.0100 M aqueous glucose solution has a density of essentially 1.000 g mL⁻¹. How close is its molality?
m = (1000 × 0.0100) ÷ (1000 × 1.000 − 0.0100 × 180.156)
denominator = 1000 − 1.80156 = 998.198
m = 10.00 ÷ 998.198 = 0.010018
m = 0.01002 mol kg⁻¹ — only 0.18% away from the molarity.
So for dilute aqueous solutions m ≈ M is a fair approximation: one litre of dilute solution is very nearly one kilogram of water. It fails once the solution is concentrated (the solute's mass stops being negligible) or the solvent is not water. Check, do not assume.
Common mistakes that cost marks
- Using the mass of the solution as the mass of the solvent. Molality needs solvent only. "18.0 g of glucose in 250 g of solution" means the solvent is 250 − 18.0 = 232 g. One word changes the answer.
- Leaving the solvent mass in grams. Molality is per kilogram. 250 g is 0.250 kg.
- Forgetting to add the solute mass when converting to molarity. Solution mass = solvent mass + solute mass.
- Using density as though it applies to the solvent. The d in the conversion formula is the density of the solution, not of pure water.
- Putting molarity into ΔTf = Kf m. Those formulas are defined with molality.
- Assuming m = M always. True only for dilute aqueous solutions, and even then only approximately.
- Writing M for molality. Capital M is molarity, small m is molality, and M often means molar mass too. Label your symbols so the examiner can see which you mean.
Which one should I use?
| Situation | Use | Reason |
|---|---|---|
| Titration, reacting volumes | Molarity | You measure out volumes with a pipette and burette |
| pH and equilibrium calculations | Molarity | Equilibrium expressions are written in mol L⁻¹ |
| Colligative properties; any work over a temperature range | Molality | The value stays fixed as the solution is heated or cooled |
| Very dilute aqueous work | Either | They agree to well under 1% — but say which you used |
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 12 | Solutions chapter — definitions, interconversion, colligative properties |
| JEE / NEET | Molality ⇄ molarity conversions with a given density; mole fraction links |
| IIT-JAM / CUET-PG | Colligative-property numericals and van't Hoff factor problems |
| GATE / CSIR-NET | Activity, chemical potential and non-ideal solution treatments, all built on molality |
Convert without dropping the solute mass. The concentration calculator handles molarity, molality, mass and density together, so the step everybody skips — adding the solute's own mass in — is done for you.
Open the Concentration (Molarity / Molality) Calculator →The Class 12 solutions chapter is short, predictable and heavily tested — one of the best places to pick up marks. ABC Chemistry teaches it in the Class 11–12 batches at the Gurugram centre and in online classes across India: abcchemistry.in.