Molality Calculator

Calculate molality, required solute mass, and colligative property changes - boiling point elevation and freezing point depression - for any solvent.

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Molality Calculator

Calculate molality, required solute mass, and colligative property changes - boiling point elevation and freezing point depression - for any solvent.

Molality -
Moles of solute -
Solvent mass (kg) -
Mass of solute needed -
Moles of solute needed -
Boiling point elevation (ΔTb) -
Freezing point depression (ΔTf) -
New boiling point -
New freezing point -

📊 What Is Molality?

Molality (symbol m) is a measure of solute concentration expressed as moles of solute per kilogram of solvent. Unlike molarity, which is based on volume (mol/L), molality is based on mass (mol/kg), making it independent of temperature and pressure. This is why chemists use molality - not molarity - when studying colligative properties such as boiling point elevation, freezing point depression, and osmotic pressure, all of which change with temperature.

The distinction between solvent mass and solution mass is critical. A 1 molal solution of NaCl dissolves 58.44 g of NaCl in exactly 1,000 g of water. The total solution mass is 1,058.44 g - but the molality calculation uses only the solvent mass (1,000 g = 1 kg). This is the standard chemist's definition per IUPAC.

Molality is dimensionally expressed as mol/kg but is often written without units as simply "m" (lowercase, italic). A 0.5 m solution contains 0.5 moles of solute per kilogram of solvent. The term "molal" (as in "0.5 molal") describes such a solution colloquially.

This calculator covers three tasks: computing the molality of a solution from mass and molar mass, back-calculating how much solute to weigh out for a target molality, and predicting the colligative property changes - boiling point elevation and freezing point depression - for six built-in solvents (water, benzene, ethanol, acetone, cyclohexane, acetic acid) or any custom solvent.

📐 Formula

m = n_solute / m_solvent(kg)
m = molality (mol/kg)
n_solute = moles of solute = mass of solute (g) ÷ molar mass (g/mol)
m_solvent(kg) = mass of solvent in kilograms
mass_solute = m × M × m_solvent(kg)
M = molar mass of solute (g/mol)
ΔTb = i × Kb × m    ΔTf = i × Kf × m
i = van't Hoff factor (1 for non-electrolytes; 2 for NaCl, KCl; 3 for CaCl₂; 1.9 for NaCl accounting for partial dissociation)
Kb = molal ebullioscopic constant (°C·kg/mol) — boiling point elevation
Kf = molal cryoscopic constant (°C·kg/mol) — freezing point depression

📖 How to Use

Steps to Calculate

1
Find Molality: Enter the mass of your solute in grams, its molar mass (g/mol), and the mass of your solvent (in grams or kilograms). The calculator returns molality in mol/kg and the number of moles of solute.
2
Find Solute Mass: Enter the target molality, the molar mass of the solute, and the mass of solvent. The calculator tells you how many grams and moles of solute to weigh out.
3
Colligative Properties: Enter the molality, the van't Hoff factor (1 for non-electrolytes like glucose; 2 for fully dissociating 1:1 electrolytes like NaCl), and select a solvent from the list or enter custom Kb and Kf values. The calculator shows the boiling point elevation, freezing point depression, and the new boiling and freezing points.
4
All three modes share the same Calculate button.

💡 Example Calculations

Example 1 — NaCl in water (Find Molality)

Dissolve 5.85 g of NaCl (molar mass = 58.44 g/mol) in 100 g of water.

1
Moles of NaCl = 5.85 / 58.44 = 0.1001 mol
2
Solvent mass = 100 g = 0.1000 kg
3
Molality = 0.1001 / 0.1000 = 1.001 mol/kg
Molality = 1.001 mol/kg
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Example 2 — Preparing 250 g of 0.5 m glucose solution (Find Solute Mass)

Glucose molar mass = 180.16 g/mol; solvent = 250 g of water (0.250 kg).

1
Moles needed = 0.5 × 0.250 = 0.125 mol
2
Mass needed = 0.125 × 180.16 = 22.52 g of glucose
Weigh out 22.52 g of glucose.
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Example 3 — Colligative properties of 1 m NaCl in water

NaCl dissociates into Na⁺ and Cl⁻, so i = 2. Water: Kb = 0.512, Kf = 1.86 °C·kg/mol.

1
ΔTb = 2 × 0.512 × 1 = +1.024 °C → new boiling point = 101.024 °C
2
ΔTf = 2 × 1.86 × 1 = 3.720 °C → new freezing point = −3.720 °C
Roads salted with NaCl don't freeze until about −3.7 °C lower than pure water.
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❓ Frequently Asked Questions

What is the difference between molality and molarity?+
Molarity (M) = moles of solute / litres of solution. Molality (m) = moles of solute / kilograms of solvent. Molarity depends on solution volume, which changes with temperature and pressure. Molality depends on solvent mass, which is temperature-independent. Use molality for colligative property calculations; use molarity for stoichiometry, titrations, and reactions where volume is the convenient quantity.
Why does molality use solvent mass, not solution mass?+
The IUPAC definition specifies kilograms of solvent (not solution) to keep the denominator independent of the amount of solute. If solution mass were used, adding more solute would change both the numerator (moles) and the denominator (total mass), making concentration calculations less direct. Using solvent mass isolates the concentration effect cleanly.
What is the van't Hoff factor (i)?+
The van't Hoff factor accounts for the number of particles a solute dissociates into in solution. Non-electrolytes (glucose, sucrose, urea): i = 1. Fully dissociating 1:1 electrolytes (NaCl, KBr, LiCl): i = 2 (theoretical); ~1.9 in practice due to ion pairing. CaCl₂ → Ca²⁺ + 2Cl⁻: i = 3 (theoretical). Weak acids (acetic acid): i slightly greater than 1, depending on degree of ionisation.
What is boiling point elevation?+
Boiling point elevation is the increase in a solvent's boiling point when a non-volatile solute is dissolved in it. The formula is ΔTb = i × Kb × m. For water, Kb = 0.512 °C·kg/mol. A 1 molal solution of a non-electrolyte (i = 1) in water boils at 100.512 °C instead of 100 °C. Electrolytes elevate the boiling point more because they increase the effective particle count.
What is freezing point depression?+
Freezing point depression is the decrease in a solvent's freezing point when a solute is dissolved. The formula is ΔTf = i × Kf × m. For water, Kf = 1.86 °C·kg/mol. Antifreeze (ethylene glycol) in car radiators exploits this effect. Road salt (NaCl, i ≈ 2) in a 1 m solution depresses the freezing point of water by approximately 3.72 °C. The Kf for cyclohexane is 20.0 °C·kg/mol, making it useful for measuring molar masses by cryoscopy.
How do I convert between molality and molarity?+
For dilute aqueous solutions, molality ≈ molarity because 1 L of water weighs ~1 kg. For the exact conversion: M = (m × ρ) / (1 + m × M_solute / 1000), where ρ is the solution density (g/mL) and M_solute is the molar mass (g/mol). Alternatively, m = (1000 × M) / (1000 × ρ − M × M_solute). This calculator works with molality directly - use the Molarity Calculator for molarity-based problems.
How do I calculate molality from mole fraction?+
If the mole fraction of solute is X_solute and molar mass of solvent is M_solvent (g/mol): m = (X_solute × 1000) / ((1 − X_solute) × M_solvent). For example, a mole fraction of 0.05 in water (M_solvent = 18.015 g/mol): m = (0.05 × 1000) / (0.95 × 18.015) = 50 / 17.11 = 2.92 mol/kg.
What is the Kb and Kf for common solvents?+
Water: Kb = 0.512, Kf = 1.86 °C·kg/mol. Benzene: Kb = 2.53, Kf = 5.12. Ethanol: Kb = 1.22, Kf = 1.99. Acetone: Kb = 1.71, Kf = 2.40. Cyclohexane: Kb = 2.79, Kf = 20.0. Acetic acid: Kb = 3.07, Kf = 3.90. All values are at 1 atm standard pressure. Cyclohexane's large Kf makes it useful for molar mass determination by cryoscopy (Beckmann method).
Can I use molality to determine molar mass?+
Yes - this is the cryoscopic method. Dissolve a known mass of unknown substance in a known mass of solvent, measure the freezing point depression ΔTf, then calculate: m = ΔTf / (i × Kf). Since m = mass_solute / (M_solute × mass_solvent_kg), rearranging gives M_solute = mass_solute / (m × mass_solvent_kg). For non-electrolytes (i = 1), this is straightforward and was historically used to determine molar masses of organic compounds before mass spectrometry.
What is the relationship between molality and osmotic pressure?+
Osmotic pressure (π) is given by π = iMRT, where M is molarity (not molality). However, for dilute aqueous solutions at 25 °C, M ≈ m. For a 1 molal NaCl solution (i = 2) in water: π ≈ 2 × 1 × 0.0821 × 298 = 48.9 atm. Molality is more precisely related to solvent activity (a₁ = exp(−m × M_solvent / 1000)), which is used in precise thermodynamic treatments of osmotic pressure and vapour pressure lowering.
Tags: Molality Calculator Molality Formula Mol per Kg Boiling Point Elevation Freezing Point Depression Colligative Properties Chemistry Calculator Calculator Formula Free Online

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What is molality and how is it different from molarity?

Molality (m) is the number of moles of solute per kilogram of solvent: m = n(solute) / kg(solvent). Molarity (M) is moles of solute per litre of solution. The key difference is the denominator: molality uses mass of the pure solvent, while molarity uses volume of the total solution. Molality does not change with temperature because it is based on mass, not volume. Molarity changes slightly with temperature as liquids expand and contract. For colligative property calculations, molality is always preferred.

What is the formula for molality?

Molality (m) = moles of solute / kilograms of solvent = n / m(solvent in kg). The moles of solute = mass of solute in grams / molar mass of solute. So the full formula is: molality = (mass of solute in g / molar mass in g/mol) / mass of solvent in kg. Units are mol/kg, also written as molal (m).

How do I calculate boiling point elevation using molality?

Boiling point elevation: delta Tb = Kb times m, where Kb is the ebullioscopic constant (solvent-specific) and m is the molality of the solution. For water, Kb = 0.512 degrees C per mol/kg. A 1 mol/kg NaCl solution (which dissociates into 2 particles, i = 2) raises the boiling point by 0.512 times 2 times 1 = 1.024 degrees C, so water boils at 101.024 degrees C instead of 100 degrees C.

How do I calculate freezing point depression?

Freezing point depression: delta Tf = Kf times m, where Kf is the cryoscopic constant. For water, Kf = 1.86 degrees C per mol/kg. A 0.5 mol/kg glucose solution (non-electrolyte, i = 1) lowers the freezing point by 1.86 times 0.5 = 0.93 degrees C, so the solution freezes at -0.93 degrees C instead of 0 degrees C. This principle explains why salt lowers the freezing point of ice on roads.

What units does molality use?

Molality uses units of mol/kg, often abbreviated as m (lowercase, italic). A 1 m (1 molal) solution contains 1 mole of solute dissolved in exactly 1 kilogram of solvent. Do not confuse this with molarity (M, uppercase) which uses mol/L. The unit mol/kg is also written as mol kg-1 in IUPAC notation.

How do I find the mass of solute needed for a target molality?

Rearrange the molality formula: mass of solute (g) = target molality times mass of solvent (kg) times molar mass of solute (g/mol). For a 0.5 mol/kg NaCl solution using 500 g of water: moles needed = 0.5 times 0.5 = 0.25 mol. Mass of NaCl = 0.25 times 58.44 = 14.61 g. Use the Find Solute Mass mode in this calculator to compute this instantly.

Why does salt lower the freezing point of water?

Salt (NaCl) dissolves into Na+ and Cl- ions, creating more solute particles than a non-ionic compound. More dissolved particles disrupt the formation of the ice crystal lattice, requiring a lower temperature to freeze. This is the colligative property called freezing point depression: delta Tf = Kf times i times m, where i = 2 for NaCl. A 0.5 mol/kg NaCl solution depresses the freezing point by 1.86 times 2 times 0.5 = 1.86 degrees C.

What is the van't Hoff factor and how does it affect molality calculations?

The van't Hoff factor (i) accounts for the number of particles a solute produces when dissolved. For non-electrolytes like glucose: i = 1. For NaCl: i = 2 (Na+ and Cl-). For CaCl2: i = 3 (Ca2+ and 2 Cl-). Colligative property formulas become: delta Tf = i times Kf times m and delta Tb = i times Kb times m. Without accounting for i, calculations for ionic compounds will underestimate the effect by a factor of 2 or more.

📌 Quick Tips

💡Molality uses mass of solvent (not solution), making it temperature-independent. This is why chemists prefer molality over molarity when working at varying temperatures, such as in cryoscopy or ebullioscopy experiments.
💡To convert between molality (m) and molarity (M), you need the solution density. Molarity = molality times molar mass of solute times density divided by 1000 plus molality times molar mass. For dilute aqueous solutions, molality and molarity are approximately equal.
💡For water as solvent, Kf = 1.86 degrees C per mol/kg and Kb = 0.512 degrees C per mol/kg. These are constants specific to the solvent, not the solute. Different solvents have different values. Benzene has Kf = 5.12 and Kb = 2.53.
💡When calculating freezing point depression for road salt (NaCl), remember the van't Hoff factor i = 2 (NaCl dissociates into Na+ and Cl-). So delta Tf = i times Kf times m = 2 times 1.86 times m. This doubles the depression compared to a non-electrolyte.
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