Molality Calculator

Formulas

n = m / M_w              (moles from mass and molar mass)
b = n / m_solvent_kg     (molality from moles and solvent mass)
b = m / (M_w × m_solvent_kg)   (combined formula)

Formula variables

n — amount of substance in moles (mol).
m — mass of the solute in grams (g).
M_w — molar mass of the solute in g/mol.
m_solvent_kg — mass of the solvent in kilograms (kg).
b — molality in moles per kilogram (mol/kg = m).

Molarity vs Molality

These two units look similar and sound similar, but they measure different things:

Molarity vs molality
Feature	Molarity (M)	Molality (m)
Definition	Moles per liter of solution	Moles per kilogram of solvent
Units	mol/L	mol/kg
Symbol	M (capital)	m (lowercase)
Temperature dependence	Yes (volume changes with temperature)	No (mass does not change)
Common use	General chemistry, titrations	Colligative properties, freezing/boiling point calculations

For dilute aqueous solutions at room temperature, M and m are numerically similar because 1 L of water weighs about 1 kg. For concentrated solutions or non-aqueous solvents, they diverge significantly.

Worked Examples

Example 1: Simple molality

Dissolve 5.844 g of NaCl (M_w = 58.44 g/mol) in 1.000 kg of water.

Moles: n = 5.844 ÷ 58.44 = 0.1000 mol.
Mass of solvent: 1.000 kg.
Molality: b = 0.1000 ÷ 1.000 = 0.1000 m.

Example 2: Molality in 500 g of water

Dissolve 18.016 g of glucose (M_w = 180.16 g/mol) in 500 g of water.

Moles: n = 18.016 ÷ 180.16 = 0.1000 mol.
Mass of solvent: 0.500 kg.
Molality: b = 0.1000 ÷ 0.500 = 0.2000 m.

Example 3: Mass needed for a target molality

How much NaOH (M_w = 40.00 g/mol) is needed to make 1.500 m solution in 250 g of water?

Moles needed: n = 1.500 × 0.250 = 0.375 mol.
Mass needed: m = 0.375 × 40.00 = 15.00 g.
Result: Dissolve 15.00 g NaOH in 250 g water.

Example 4: Freezing point depression

For a dilute aqueous solution, the freezing point depression is ΔT = K_f × b, where K_f = 1.86 °C·kg/mol for water.

For the 0.2000 m glucose solution: ΔT = 1.86 × 0.2000 = 0.372 °C. Freezing point = 0 − 0.372 = −0.372 °C.

Colligative Properties

Molality is used in calculations of colligative properties because these depend on the number of solute particles relative to the solvent, not on the total solution volume:

Freezing point depression: ΔT_f = K_f × m × i
Boiling point elevation: ΔT_b = K_b × m × i
Vapor pressure lowering: Raoult's law
Osmotic pressure: π = M R T (uses molarity, not molality)

Where i is the van 't Hoff factor (the number of particles the solute dissociates into; for NaCl, i ≈ 2).

Constants for Common Solvents

Freezing and boiling point constants
Solvent	K_f (°C·kg/mol)	K_b (°C·kg/mol)	Freezing point (°C)	Boiling point (°C)
Water	1.86	0.512	0.0	100.0
Ethanol	1.99	1.22	−114.6	78.4
Benzene	5.12	2.53	5.5	80.1
Cyclohexane	20.2	2.79	6.5	80.7
Acetic acid	3.90	3.07	16.6	118.1

Related Converters

Frequently Asked Questions

What is molality?

Molality (m) is the number of moles of solute per kilogram of solvent. It is used in calculations involving colligative properties.

How do you calculate molality?

Divide the moles of solute by the mass of solvent in kilograms: b = n / kg solvent. Or use b = m / (M_w × kg solvent) where m is mass in grams and M_w is molar mass.

Why use molality instead of molarity?

Molality is temperature-independent because it uses mass (which does not change with temperature). Molarity uses volume, which changes with temperature. For precise work involving temperature changes, molality is preferred.

Is molality the same as molarity in water?

Close but not identical. For dilute aqueous solutions at 20 °C, 1 L of water weighs approximately 1 kg, so molarity and molality are numerically close. For concentrated solutions, they diverge.

What are colligative properties?

Colligative properties are properties of solutions that depend on the number of solute particles, not on their identity. Examples: freezing point depression, boiling point elevation, vapor pressure lowering, osmotic pressure.

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