Molality Calculator

Calculate the molality of a solution by entering the amount of solute and the mass of solvent below.

Molality:

 
mol/kg
Learn how we calculated this below

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How to Calculate Molality

Molality represents the concentration of a compound in a solution measured as the amount of the substance in moles divided by the mass of the solute in kilograms. Molality is often expressed in moles per kilogram, or mol/kg, and is denoted using the letter m.

First used in 1923, molality is different from molarity in that molality is the moles of the solute in 1 kg of the solvent, while molarity is the moles of the solute in 1 L of the solvent.[1]

If the solvent is water, molality and molarity will be the same since 1 L of water weighs 1 kg.

In general, both of these terms are used to prepare and label standard liquid solutions that are frequently used in a laboratory. Both molality and molarity describe the strength of a chemical in solution; for example, a 2M solution is twice as concentrated as a 1M solution.

Molality Formula

Since it’s a ratio, you can calculate molality using the following formula:

molality (m) = moles of solute / kilograms of solvent

Thus, the molality m is equal to the amount of substance of solute in moles divided by the mass of solvent in kilograms.

Graphic showing the molality formula where the molality is equal to the moles of solute divided by the kilograms of solvent

For example, let’s calculate the molality of a solution with 3 moles of solute and 2 kilograms of solvent.

molality (m) = 3 moles solute / 2 kg solvent
molality (m) = 3 / 2 = 1.5 mol/kg

In this example, the molality of the solution is equal to 1.5 mol/kg. Notice that it does not matter what the solute and solvent are; the molality would be the same as long as there are 3 moles of that substance in 2 kg of the solvent.

References

  1. Lewis, G. N., Randall, M., Thermodynamics and the Free Energy of Chemical Substances, McGraw-Hill Book Company, Inc., 1923, New York and London, https://books.google.com/books?id=m81AAAAAIAAJ&ots=K6ziAMlj7Y&lr&pg=PR7