Molality is a measure of the concentration of a solution. It is defined as the number of moles of a substance (solute) dissolved in 1 kilogram of a liquid (solvent). Represented by the symbol m, it is widely used in physical chemistry, life sciences, and various practical applications. ^[1-4]

Formula

The formula for molality is given by: ^[1-4]

Molality (m) = (Number of moles of solute)/(Mass of solvent in kilograms)​

The number of moles of the solute can be determined using the following formula:

Number of moles (n) = (Mass of solute in grams)/(Molar mass of solute in grams per mole)

Units

The SI unit for molality is moles per kilogram (mol/kg). Thus, molality specifies the number of moles of solute present in one kilogram of the solvent.

Example: If 1 mole of salt is dissolved in 2 kilograms of water, the molality of the solution is:

m = 1 mol/ 2 kg = 0.5 mol/kg

A solution containing one mole of a solute per kilogram of solvent is called 1 molal solution (1 m).

1 mol/kg = 1 m

Advantages and Disadvantages of Molality

Molality offers several advantages as a concentration unit. It is based on the mass of the solvent and remains unaffected by temperature and pressure changes. This property ensures that the concentration remains constant even when the solution undergoes thermal expansion or contraction. ^[1-4]

Additionally, molality is useful in calculating colligative properties, such as vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. It provides accurate measurements even when the solution’s temperature changes.

Molality has several disadvantages that limit its practicality. One major drawback is its dependence on precise measurements of solvent mass. It can be challenging in experimental setups, especially when dealing with volatile or hygroscopic substances. Determining the mass of solvents and then the molality of solutions requires extensive calculations. This method is also less intuitive for expressing concentrations in dilute solutions, where volumes are typically easier to measure than masses.

Molarity vs. Molality

Molarity and molality are both widely used to express solution concentration. The table below highlights their key differences: ^[1-4]

Property	Molarity (M)	Molality (m)
Definition	Moles of solute per liter of solution	Moles of solute per kilogram of solvent
Formula	M = (moles of solute)/(volume of solution)	m = (moles of solute)/(mass of solvent)
Units	Moles per liter (mol/L)	Moles per kilogram (mol/kg)
Basis of Measurement	Total volume of the solution	Mass of the solvent only
Dependence on Temperature	Dependent	Independent
Ease of Measurement	Easier to use volumetric tools	Requires precise weighing of solvent
Common Usage	Reactions and titrations in labs	Colligative property and thermodynamic studies
Applicability	Limited by thermal expansion or contraction	Ideal for varying temperature or pressure

Example Problems with Solutions

Problem 1: Calculate the molality of a solution prepared by dissolving 10 g of NaCl (Molar mass = 58.44 g/mol) in 500 g of water.

Solution

1. Calculate the moles of solute:

Moles of NaCl = (Mass of NaCl)/(Molar mass of NaCl) = 10 g/58.44 g = 0.171 mol

2. Convert the mass of solvent to kilograms:

Mass of water = 500 g = 0.500 kg

3. Use the molality equation: 

Molality (m) = (Moles of NaCl)/(Mass of water)​

=> m = 0.171 mol/0.500 kg = 0.342 mol/kg

Problem 2: What is the molality of a solution containing 20 g of glucose (molar mass = 180 g/mol) dissolved in 200 g of water?

Solution

1. Moles of glucose = (Mass of glucose)/(Molar mass of glucose) = 20 g/180 g = 0.111 mol

2. Mass of water = 200 g = 0.200 kg

3. Molality (m) = (Moles of glucose)/(Mass of water)​

=> m = 0.111 mol/0.200 kg = 0.555 mol/kg

Problem 3: A solution contains 15 g of KCl (molar mass = 74.55 g/mol) dissolved in 300 g of water. What is the molality of the solution?

Solution

1. Moles of KCl = (Mass of KCl)/(Molar mass of KCl) = 15 g/74.55 g/mol = 0.201 mol

2. Mass of water = 300 g = 0.300 kg

3. Molality (m) = (Moles of KCl)/(Mass of water) = 0.201 mol/0.300 kg = 0.670 mol/kg

Problem 4: Determine the mass of urea (molar mass = 60 g/mol) needed to prepare 2 molal solution in 250 g of water.

Solution

2 molal solution = 2 mol/kg

1. Mass of water = 250 g = 0.250 kg

2. Rearrange the molality equation to find moles of solute:

Molality (m) = (Moles of urea)/(Mass of water)​

=> Moles of urea = m x Mass of water = 2 mol/kg x 0.250 kg = 0.500 mol

3. Mass of urea = Moles of urea x Molar mass of urea = 0.500 mol x 60 g/mol = 30 g