Molality Calculator

Molality Calculator – Concentration per Kilogram of Solvent

Molality tells you how many moles of solute are dissolved in each kilogram of solvent. This calculator works it out from what you actually measured in the lab, tells you how much solute or solvent to weigh out for a target molality, and converts molality to and from molarity, mass percent and mole fraction. Every result comes with a step-by-step derivation and a picture of how the solution splits by mass and by amount.

The molality formula

b = n_solute / m_solvent        [mol/kg]
n_solute = m_solute / M_solute

IUPAC uses the symbol b; many textbooks write m and call the unit "molal". Both mean the same number, so 0.500 mol/kg is a 0.500 molal solution. As a worked example, 10.0 g of sodium chloride (M = 58.44 g/mol) is 0.1711 mol. Dissolved in 250 g of water, that gives b = 0.1711 / 0.2500 = 0.6845 mol/kg.

Divide by the solvent, not the solution

The most common molality mistake is dividing by the mass of the whole solution. The denominator is the solvent alone. If you weighed the finished solution, subtract the solute first: m_solvent = m_solution − m_solute. The calculator does this for you when you choose "Total solution mass", and the composition bar draws a bracket under the solvent segment only, so you can see what the formula divides by.

Molality vs molarity

Molarity (mol/L) divides by the volume of solution. Volumes expand and contract with temperature, so molarity drifts slightly when a solution warms up or cools down. Masses do not change, which makes molality temperature independent. That is why the colligative property equations, such as freezing point depression (ΔTf = i·Kf·b) and boiling point elevation (ΔTb = i·Kb·b), are written in terms of molality.

To convert between the two you need the solution's density ρ and the solute's molar mass M:

c = 1000·ρ·b / (1000 + b·M)
b = 1000·c / (1000·ρ − c·M)

For dilute aqueous solutions the two numbers are almost identical, because a litre of dilute solution is close to one kilogram of water. They separate for concentrated solutions. Concentrated hydrochloric acid at 37 % w/w and 1.18 g/mL is about 12.0 mol/L but 16.1 mol/kg.

Mass percent and mole fraction

Both conversions use a basis of one kilogram of solvent, which holds b mol of solute weighing b·M grams:

w = 100·b·M / (1000 + b·M)          mass percent
x = b / (b + 1000 / M_solvent)       solute mole fraction

The mole fraction assumes a binary solution, one solute in one solvent, and counts formula units rather than separate ions.

Preparing a solution of known molality

To make a solution in the lab, weigh the solute and the solvent separately. You don't need a volumetric flask, which is one practical advantage of molality. For a target b, the solute needed for a given solvent mass is n = b × m_solvent. If you want a fixed total mass T of solution instead, the solute mass is T·b·M / (1000 + b·M). For example, 500 g of 1.00 mol/kg NaCl needs 27.61 g of salt and 472.4 g of water.

Particle molality and the van 't Hoff factor

Colligative effects depend on the number of dissolved particles, not on the number of formula units. Multiplying by the van 't Hoff factor i gives the particle molality i·b. An ideal NaCl solution has i = 2 and CaCl₂ has i = 3, while non-electrolytes such as glucose and urea have i = 1.