Molality Calculator

What molality measures

Molality is moles of solute per kilogram of solvent. NIST's guide to the SI defines it as b = n / m, the amount of solute divided by the mass of the solvent, with the unit mol/kg. The important word is solvent: the kilograms counted are the liquid doing the dissolving, not the finished mixture.

Dissolve 58.44 g of sodium chloride in 1.000 kg of water and you have 1.000 mol of salt, so the molality is 1.000 mol/kg. The mass of the solution is 1.05844 kg, and that larger number never appears in the molality calculation.

NIST also notes that the word molal and the symbol m are obsolete and should be replaced by molality of solute B with the unit mol/kg. Older lab sheets still say 1 molal, so both forms appear in the wild. The number is the same.

Molality exists because mass does not expand when a solution warms up. A liter of solution does, which is why colligative properties such as freezing point depression are written in molality. The property depends on how many particles sit in a given mass of solvent, and that ratio stays put across the working temperature range.

Molality and molarity: two different denominators

Molarity divides moles by liters of solution. Molality divides moles by kilograms of solvent. In dilute water solutions the two land close together because a liter of water weighs about 1 kg, and that coincidence hides the difference until the solution gets concentrated or the temperature moves.

Water at 20 C has a density of 0.9982 kg/L, so a liter holds 998.2 g. At 50 C the density falls to 0.9880 kg/L, so the same liter holds 988.0 g. A solution that was 1.000 mol/L at 20 C becomes about 1.010 mol/L at 50 C if the solute mass is unchanged, because the volume grew while the moles did not.

The molality of the same solution does not move at all, since neither the moles nor the kilograms of solvent changed. That stability is the whole reason the colligative formulas use it.

Converting between molarity and molality

The bridge between the two units is the density of the solution. One kilogram of solvent carrying b moles of solute makes a solution of mass 1 + b x M kilograms, where M is the molar mass in kg/mol. Divide that mass by the density in kg/L and you have the volume. Molarity is the moles over that volume.

Take a 0.200 M sodium chloride solution with a measured density of 1.0075 kg/L. One liter holds 0.200 mol, which weighs 0.200 x 58.44 = 11.688 g. The water in that liter weighs 1,007.5 - 11.688 = 995.8 g, so the molality is 0.200 / 0.9958 = 0.2008 mol/kg. The gap from 0.200 is 0.4 percent, small but real.

Run it the other way on a 1.000 mol/kg sodium chloride solution at 1.0381 kg/L. The solution mass is 1.000 + 0.05844 = 1.05844 kg, the volume is 1.05844 / 1.0381 = 1.0196 L, and the molarity is 1.000 / 1.0196 = 0.9808 mol/L. Half a percent below 1 M, which is exactly the size of error that shows up in a lab report.

Two compact formulas cover both directions. From molarity to molality, divide c by (density minus c times M), with density in kg/L and M in kg/mol. From molality to molarity, multiply b by density and divide by (1 + b x M). Both need a density the label may not print, and that is the usual place a conversion stalls.

Freezing point depression and boiling point elevation

Dissolved particles lower the freezing point and raise the boiling point of a solvent. The size of the shift follows a straight line in molality: delta T equals the van't Hoff factor times the constant for that solvent times the molality. OpenStax Chemistry lists the constants used below.

A 1.000 mol/kg sodium chloride solution has i = 2, because each formula unit splits into one sodium ion and one chloride ion. The freezing point drops by 2 x 1.86 x 1.000 = 3.72 C, to -3.72 C, and the boiling point climbs by 2 x 0.512 x 1.000 = 1.024 C, to 101.02 C.

Sucrose does not split, so i = 1. Dissolve 34.2 g in 500 g of water and the moles are 34.2 / 342.30 = 0.0999, the molality is 0.0999 / 0.500 = 0.1998 mol/kg, and the freezing point falls by 1.86 x 0.1998 = 0.372 C. The same number of moles of salt would drop it twice as far.

The van't Hoff factor and real solutions

The factor i counts the particles a formula unit produces. Glucose and sucrose give 1, sodium chloride gives 2, calcium chloride gives 3, and a weak acid such as acetic acid gives a number between 1 and 2 that depends on how much of it has ionized.

Real solutions do not follow the ideal line for long. Seawater at a salinity of 35 g/kg holds about 0.599 mol of sodium chloride equivalent per kilogram, which the ideal formula turns into 2 x 1.86 x 0.599 = 2.23 C of depression. The measured freezing point of seawater is close to -1.9 C, roughly 15 percent less, because ions pair up and stop behaving as independent particles.

The gap grows with concentration. A 1 mol/kg sodium chloride solution freezes near -3.5 C rather than the ideal -3.72 C, and saturated brine at about 6 mol/kg freezes near -21 C against -22.7 C from the straight line. The percentage error shrinks as the solution saturates, but the absolute miss is still a degree, and a measurement is the only honest answer past roughly 1 mol/kg.

Weak electrolytes are the other trap. Acetic acid at 0.1 mol/kg ionizes only about 1.3 percent, so i sits near 1.01 rather than 2. Treating it as a strong acid and using i = 2 doubles the predicted temperature shift.

Road salt, antifreeze and pasta water

Rock salt on a driveway works through freezing point depression. Spread 1 kg of sodium chloride on 10 kg of water and the molality is 17.11 / 10 = 1.711 mol/kg, since 1,000 / 58.44 = 17.11. The ideal shift is 2 x 1.86 x 1.711 = 6.36 C, so the film of water on the pavement freezes near -6.4 C. Real brine freezes near -6 C, which is why salt stops helping in a deep cold snap.

Antifreeze uses the same physics with a different solute. Ethylene glycol has a molar mass of 62.07 g/mol, so 500 g in 2 kg of water is 8.056 mol and 4.028 mol/kg. The ideal shift is 1.86 x 4.028 = 7.49 C, and coolant charts put a 20 percent glycol mix near -9 C. The measured value is deeper than the ideal one here because glycol mixtures depart from the dilute model in the other direction.

Pasta water is the useful negative result. Ten grams of salt in four liters of water gives 0.0428 mol/kg, so the boiling point rises by 2 x 0.512 x 0.0428 = 0.044 C and the water boils at 100.04 C. Salting the water seasons the pasta; it does not cook it faster in any way a kitchen can detect.

Where the numbers go wrong

The first mistake is weighing the solution instead of the solvent. A recipe that says 1 kg of water plus 58.44 g of salt gives 1.05844 kg of solution, and dividing the moles by that number returns 0.945 mol/kg instead of 1.000. The error is 5.5 percent, larger than most lab tolerances.

The second is the gram and kilogram mix-up. Molality wants kilograms of solvent, so 500 g of water is 0.500 kg. Using 500 in the denominator shrinks the answer by a factor of 1,000 and turns a 0.200 mol/kg solution into 0.0002.

The third is the hydrate. Copper sulfate pentahydrate has a molar mass of 249.69 g/mol against 159.61 for the anhydrous form, and the water of crystallization counts as solute mass but also adds to the solvent when it dissolves. Weighing the pentahydrate while using the anhydrous molar mass overstates the moles by 56 percent.

The fourth is the electrolyte factor. Sodium chloride needs i = 2, calcium chloride needs 3, sugar needs 1, and a weak acid needs a measured value between 1 and 2. The fifth is unit substitution: dropping a molarity into a colligative formula gives an answer that looks right and drifts with the thermostat. Check that the denominator is kilograms of solvent before trusting the result.