Molality and Mole Fraction Calculations

Molality (m) is the number of moles of solute per kilogram of solvent, and mole fraction (χ) is the number of moles of one component divided by the total moles of all components in the solution. In brief: if 25.0 g of glucose (180.16 g/mol) is dissolved in 200.0 g of water, the molality is 0.694 mol/kg and the mole fraction of glucose is 0.0123.

Unlike molarity, molality and mole fraction do not change with temperature, because they are defined in terms of mass and amount of substance, not volume. This is why they are used in the study of colligative properties, such as boiling-point elevation and freezing-point depression. This article defines both units, compares them with molarity, and gives worked examples including a conversion between units.

What is molality?

Molality is defined as:

m = moles of solute / kilograms of solvent

The unit is mol/kg, often written as m, read “molal”. Note that the denominator is the mass of the solvent only, not the mass of the solution. A 1.00 m solution of NaCl is made by dissolving 58.44 g (1.00 mol) of NaCl in 1.000 kg of water. Because mass does not depend on temperature, the molality of a solution is the same at 5 °C and at 80 °C. Take care not to confuse the unit symbol m (molal) with the symbol for mass or with the meter.

What is mole fraction?

The mole fraction of component A in a mixture is:

χA = nA / (nA + nB + ...)

It is a dimensionless number between 0 and 1. The mole fractions of all components of a solution sum to exactly 1. Multiplying by 100 gives the mole percent. Mole fraction is the natural unit for Raoult’s law, which says that the vapor pressure of a solvent over a solution equals the mole fraction of the solvent multiplied by its vapor pressure when pure.

Comparison of concentration units

Worked examples

Example 1: molality and mole fraction of glucose

25.0 g of glucose, C₆H₁₂O₆ (180.16 g/mol), is dissolved in 200.0 g of water (18.015 g/mol).

1. n(glucose) = 25.0 / 180.16 = 0.1388 mol.
2. Molality = 0.1388 mol / 0.2000 kg = 0.694 mol/kg.
3. n(water) = 200.0 / 18.015 = 11.10 mol.
4. χ(glucose) = 0.1388 / (0.1388 + 11.10) = 0.1388 / 11.24 = 0.0123.
5. χ(water) = 1 − 0.0123 = 0.9877.

Example 2: mass of solute needed

How many grams of ethylene glycol (62.07 g/mol) must be added to 4.00 kg of water to make a 1.50 m solution? n = m × kg = 1.50 × 4.00 = 6.00 mol, and mass = 6.00 × 62.07 = 372 g.

Example 3: an ethanol–water mixture

A solution contains 46.07 g of ethanol (1.000 mol) and 54.05 g of water (3.000 mol).

• χ(ethanol) = 1.000 / 4.000 = 0.250 and χ(water) = 0.750.
• Molality of ethanol = 1.000 mol / 0.05405 kg = 18.5 mol/kg.

This shows that molality is not limited to dilute solutions, but is most useful for solutions in which one component is clearly the solvent.

Example 4: freezing point depression

The freezing point of a solvent drops by ΔT = K_f × m × i. For the solution in Example 1, K_f(water) = 1.86 °C·kg/mol and i = 1 for glucose, so ΔT = 1.86 × 0.694 = 1.29 °C. The solution freezes at about −1.29 °C.

How do you convert molarity to molality?

To convert, the density of the solution is needed. Consider 1.00 M NaCl, which has a density of 1.038 g/mL.

1. Take 1.000 L of solution. The mass is 1000 mL × 1.038 g/mL = 1038 g.
2. The mass of NaCl is 1.00 mol × 58.44 g/mol = 58.44 g.
3. The mass of water = 1038 − 58.44 = 979.6 g = 0.9796 kg.
4. Molality = 1.00 mol / 0.9796 kg = 1.02 mol/kg.

For dilute aqueous solutions molarity and molality are almost equal, since 1 L of the solution contains about 1 kg of water. For concentrated solutions the difference can be large.

Where are these units used?

• Colligative properties: freezing-point depression, boiling-point elevation and osmotic pressure.
• Raoult’s law: vapor pressure calculations use mole fractions.
• Thermodynamics: mole fractions appear in the equations for activity and chemical potential.
• Gas mixtures: the mole fraction of a gas is its partial pressure divided by the total pressure.

Where do the terms come from?

The concept of colligative properties was developed in the 1880s by François-Marie Raoult, who found that the freezing-point depression is proportional to the amount of dissolved substance in a given mass of solvent. Expressing concentration per mass of solvent makes the results independent of temperature, which is one reason why the molal scale became standard for such work. Raoult published his law of vapor pressure lowering in 1887. The Greek letter chi (χ) for mole fraction is a convention of physical chemistry.

Which mistakes are common?

• Using the mass of the solution: molality uses kilograms of solvent only.
• Mixing grams and kilograms: 200.0 g is 0.2000 kg.
• Assuming that molarity equals molality: this holds only for dilute aqueous solutions.
• Forgetting that mole fractions add up to 1: use this as a check.
• Counting ions as separate components in χ: decide whether you count formula units or particles, and keep to it.