Molality Calculator
Moles per kilogram from a formula and two masses, molarity ⇄ molality with density, and the freezing and boiling point shifts your solution causes.
By Manzoor Ahmad, pharmacist · Updated How do you calculate molality?
Molality = moles of solute ÷ kilograms of solvent. Dissolve 18.02 g of glucose (C₆H₁₂O₆, 180.16 g/mol, so 0.1000 mol) in 500 g of water and the molality is 0.1000 ÷ 0.500 = 0.200 mol/kg. That solution freezes at −0.372 °C (ΔTf = 1.86 × 0.200).
Molality vs molarity
| Property | Molality (b) | Molarity (c) |
|---|---|---|
| Definition | mol solute per kg solvent | mol solute per L solution |
| Unit | mol/kg (m) | mol/L (M) |
| Changes with temperature? | No | Yes, slightly |
| Needs density to convert? | Yes | Yes |
| Typical use | colligative properties, osmolality | titrations, everyday lab work (molarity calculator) |
Freezing and boiling point constants (Kf, Kb)
| Solvent | Freezing point | Kf (°C·kg/mol) | Boiling point | Kb (°C·kg/mol) |
|---|
Worked example: unknown molar mass
2.00 g of a non-electrolyte dissolved in 25.0 g of benzene lowers its freezing point by 1.02 °C. Molality = ΔTf ÷ Kf = 1.02 ÷ 5.12 = 0.199 mol/kg, so moles = 0.199 × 0.0250 kg = 0.00498 mol, and M = 2.00 ÷ 0.00498 = 402 g/mol.
Molality FAQs
Molality (b or m) is the number of moles of solute per kilogram of solvent: b = n(solute) ÷ m(solvent, kg). Its unit is mol/kg, often written m (molal).
Convert the solute mass to moles with its molar mass, then divide by the solvent mass in kilograms. 18.02 g of glucose (180.16 g/mol) is 0.1000 mol; in 500 g of water the molality is 0.1000 ÷ 0.500 = 0.200 mol/kg.
Molarity is moles per liter of solution; molality is moles per kilogram of solvent. Molality does not change with temperature because masses don't expand, so it is preferred for colligative properties and precise thermodynamic work.
Use b = c ÷ (ρ − cM/1000), where c is molarity (mol/L), ρ is solution density (g/mL) and M is the solute molar mass (g/mol). A 1.00 M NaCl solution with density 1.04 g/mL is 1.019 mol/kg.
Use c = bρ ÷ (1 + bM/1000). For dilute aqueous solutions with density near 1 g/mL, molarity and molality are almost equal.
ΔTf = i × Kf × b, where i is the van 't Hoff factor, Kf the solvent's cryoscopic constant (1.86 °C·kg/mol for water) and b the molality. A 0.200 mol/kg glucose solution freezes at −0.372 °C.
ΔTb = i × Kb × b, with Kb = 0.512 °C·kg/mol for water. A 1.00 mol/kg NaCl solution (i ≈ 2) boils about 1.02 °C higher, at roughly 101.0 °C.
The number of particles each formula unit produces in solution: 1 for non-electrolytes like glucose or urea, about 2 for NaCl, 3 for CaCl2. Real values are slightly lower in concentrated solutions because ions pair up.
Rearrange to M = i × Kf × m(solute) ÷ (ΔTf × kg solvent). If 2.00 g of an unknown non-electrolyte lowers the freezing point of 25.0 g of benzene (Kf 5.12) by 1.02 °C, M = 5.12 × 2.00 ÷ (1.02 × 0.0250) = 402 g/mol.
Dissolved salt lowers the freezing point of water. Each mole of NaCl per kg of water gives about 3.7 °C of depression in theory, which is why road salt stops working well below about −10 °C.
Osmolality, the clinical cousin of molality, is reported in mOsm/kg. Plasma is about 285–295 mOsm/kg, and intravenous fluids and eye drops are formulated close to that value to avoid damaging cells.
No. Because it uses masses rather than volumes, molality stays the same when a solution warms or cools, unlike molarity.
Sources
Colligative results assume ideal, dilute solutions and ideal van ‘t Hoff factors; real electrolytes show smaller effects at higher concentrations. Atomic weights follow IUPAC conventional values.