# Molality Calculator URL: https://speedycalculator.com/calculators/chemistry/molality/ Last updated: 2026-10-30 > The molality calculator finds the concentration of a solution in moles of solute per kilogram of solvent, from moles, from the solute mass and molar mass, or from molarity and solution density, and also gives the mass percent, mole fraction and freezing and boiling point shifts for water. ## Inputs - Start from - Amount of solute (mol, mmol, µmol, nmol, kmol) - Mass of solute (g, mg, µg, kg, oz, lb) - Molar mass of solute (g/mol, kg/mol, Da (u), kDa) — Default: glucose C₆H₁₂O₆ - Mass of solvent (kg, g, lb, mg) - Molarity of the solution (mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³) - Density of the solution (g/mL, g/cm³, kg/m³, kg/L, lb/ft³) - Molar mass of the solvent (g/mol, kg/mol, Da (u), kDa) — 18.015 g/mol for water; used for the mole fraction. ## Results - Moles of solute - Molality - Mass of solvent per litre of solution - Mass percent of solute (w/w) - Mole fraction of solute - Freezing point depression if the solvent is water (i = 1) - Boiling point elevation if the solvent is water (i = 1) ## Formula ``` b = n_solute ÷ m_solvent (kg) · n = m ÷ M From molarity: b = c ÷ (ρ − c × M ÷ 1000) (c in mol/L, ρ in kg/L, M in g/mol) χ_solute = b ÷ (b + 1000 ÷ M_solvent) ``` ## Explanation Molality (symbol b or m, unit mol/kg, sometimes written "molal") measures how many moles of solute are dissolved in each kilogram of solvent. It looks similar to molarity, but the denominator is the mass of the solvent alone, not the volume of the finished solution. That small difference matters: masses do not change when a solution warms up and expands, so molality is independent of temperature and pressure. This is why molality appears in the equations for colligative properties – freezing point depression (ΔT_f = K_f · b · i), boiling point elevation (ΔT_b = K_b · b · i) – and in precise thermodynamic work such as activity coefficients and standard states. For dilute aqueous solutions, molality and molarity are nearly equal, because one litre of solution contains about one kilogram of water; for concentrated solutions they can differ greatly. You can start from the number of moles, from the mass and molar mass of the solute, or from a molarity and the solution density (from a density table or a hydrometer). The calculator also returns the mole fraction of solute, using the molar mass of the solvent (water by default). ### How to use the Molality Calculator - **Start from** - **Amount of solute** - **Mass of solute** - **Molar mass of solute** — Default: glucose C₆H₁₂O₆ - **Mass of solvent** - **Molarity of the solution** - **Density of the solution** - **Molar mass of the solvent** — 18.015 g/mol for water; used for the mole fraction. Results update as you type, every field has its own unit menu, and everything is calculated in your browser. ### How it works For the first two methods, moles of solute (n = m ÷ M) are divided by the solvent mass in kilograms. Starting from molarity, one litre of solution is taken: it weighs 1000 × ρ grams, of which c × M grams are solute, so the solvent mass is the difference. The mole fraction follows from b and the solvent molar mass. ### Worked example 10 g of glucose (C₆H₁₂O₆, M = 180.156 g/mol) is dissolved in 250 g of water. n = 10 ÷ 180.156 = 0.05551 mol, so b = 0.05551 ÷ 0.250 = 0.2220 mol/kg. The solution is 10 ÷ 260 = 3.85% glucose by mass, the mole fraction of glucose is 0.2220 ÷ (0.2220 + 1000/18.015) = 0.00398, and the water would freeze about 1.86 × 0.2220 = 0.41 °C lower than pure water. Starting from molarity instead: a 1.00 M glucose solution with density 1.06 g/mL contains 180.16 g glucose in 1060 g of solution, so 879.8 g of water and b = 1 ÷ 0.8798 = 1.137 mol/kg. ### Molality or molarity? Use molality whenever temperature changes (freezing, boiling, calorimetry over a range) or when you need colligative properties. Use molarity for everyday volumetric work – titrations, dilutions and making up solutions in volumetric flasks – where measuring a volume is easier than weighing the solvent. ### Tips - Always start from a balanced equation – coefficients give the mole ratio. - Convert everything to moles first, then use the ratio, then convert back. - Every field has its own unit menu – g, mg, mol, mmol, L and mL all work. - Everything is calculated in your browser. ### Common mistakes to avoid - Using mass ratios instead of mole ratios. - Forgetting water of crystallisation in the molar mass of hydrates. - Confusing molarity (per litre of solution) with molality (per kilogram of solvent). ### Inputs explained FieldWhat to enterStart fromChoose one: Moles of solute and mass of solvent, Mass of solute, molar mass and mass of solvent, Molarity and solution densityAmount of soluteUnits: mol, mmol, µmol, nmol, kmolMass of soluteUnits: g, mg, µg, kg, oz, lbMolar mass of soluteUnits: g/mol, kg/mol, Da (u), kDa. Default: glucose C₆H₁₂O₆Mass of solventUnits: kg, g, lb, mgMolarity of the solutionUnits: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³Density of the solutionUnits: g/mL, g/cm³, kg/m³, kg/L, lb/ft³Molar mass of the solventUnits: g/mol, kg/mol, Da (u), kDa. 18.015 g/mol for water; used for the mole fraction. ### What each result means - **Mass of solvent per litre of solution** = [Density of the solution]×1000 - [Molarity of the solution]×[Molar mass of solute] - **Mole fraction of solute** = [Molality]÷([Molality] + 1000÷[Molar mass of the solvent]) - **Freezing point depression if the solvent is water (i = 1)** = 1.86×[Molality] - **Boiling point elevation if the solvent is water (i = 1)** = 0.512×[Molality] ### Example values How the main result changes when **Amount of solute** changes and everything else stays at the example values above: Amount of solute (mol)Molality (mol/kg)0.1250.2220.18750.2220.250.2220.31250.2220.3750.2220.50.222 ## Worked example Inputs — Start from: Mass of solute, molar mass and mass of solvent; Mass of solute: 10 g; Molar mass of solute: 180.156 g/mol; Mass of solvent: 0.25 kg; Molar mass of the solvent: 18.015 g/mol. Results — Moles of solute = 0.055507 mol; **Molality = 0.22203 mol/kg**; Mass percent of solute (w/w) = 3.8462 %; Mole fraction of solute = 0.003984; Freezing point depression if the solvent is water (i = 1) = 0.413 °C; Boiling point elevation if the solvent is water (i = 1) = 0.1137 °C. ## Key terms - **Molality (b)**: Amount of solute per kilogram of solvent, mol/kg. - **Solvent**: The component present in the largest amount, usually water. - **Colligative property**: A property that depends on the number of dissolved particles, not their identity. - **Cryoscopic constant (K_f)**: Freezing point depression per mol/kg of solute; 1.86 °C·kg/mol for water. ## FAQ **What is the difference between molality and molarity?** Molality is moles per kilogram of solvent; molarity is moles per litre of solution. Molality does not change with temperature, molarity does (slightly) because liquids expand. **How do I calculate molality?** Convert the solute mass to moles and divide by the solvent mass in kilograms. 3 mol of solute in 0.5 kg of water gives 3 ÷ 0.5 = 6 mol/kg. **What is a 1 molal solution?** One mole of solute dissolved in one kilogram of solvent – for NaCl, 58.44 g in 1000 g of water. The total mass is 1058.44 g, and the volume is not exactly one litre. **How do I convert molarity to molality?** Use the density: b = c ÷ (ρ − c·M/1000) with ρ in kg/L. A 1.00 M glucose solution with ρ = 1.06 kg/L has b = 1 ÷ (1.06 − 0.180) = 1.14 mol/kg. **Why is molality used for freezing point depression?** ΔT_f depends on the ratio of solute particles to solvent, which molality measures without any temperature dependence. **What are K_f and K_b for water?** 1.86 °C·kg/mol for freezing point depression and 0.512 °C·kg/mol for boiling point elevation. Multiply by the van 't Hoff factor i for salts (about 2 for NaCl). **Is molality the same as molarity for dilute solutions?** Almost: for dilute aqueous solutions, 1 L of solution contains close to 1 kg of water, so the two numbers agree to within a few percent. **What formula does the Molality Calculator use?** b = n_solute ÷ m_solvent (kg) · n = m ÷ M. ## Sources - [IUPAC Compendium of Chemical Terminology (Gold Book)](https://goldbook.iupac.org/) - [OpenStax – Chemistry 2e](https://openstax.org/details/books/chemistry-2e) - [LibreTexts Chemistry](https://chem.libretexts.org/)