Examples of Calculating Molality of a Solution

Explore practical examples of calculating molality in solutions, enhancing your understanding of this important chemistry concept.
By Jamie

Understanding Molality

Molality (m) is a measure of the concentration of a solute in a solution, defined as the number of moles of solute per kilogram of solvent. It is particularly useful in situations where temperature changes may affect volume, as molality is independent of temperature. In this article, we will explore three diverse examples of calculating molality to illustrate its application in real-world scenarios.

Example 1: Preparing a Salt Solution

In a laboratory setting, a chemist needs to prepare a saline solution for an experiment. The chemist decides to create a 1.5 m Molal NaCl (sodium chloride) solution using 500 grams of water.

To find the molality, the chemist follows these steps:

  1. Determine the number of moles of NaCl needed. The molar mass of NaCl is approximately 58.44 g/mol.

  • For 1.5 moles:

    Moles of NaCl = 1.5 m × 0.5 kg (500 g of water) = 0.75 moles

  1. Calculate the mass of NaCl required:

    • Mass of NaCl = Moles × Molar Mass = 0.75 moles × 58.44 g/mol = 44.58 grams

With these calculations, the chemist prepares the saline solution by dissolving 44.58 grams of NaCl in 500 grams of water, achieving a molality of 1.5 m.

Notes: Molality is particularly useful in colligative property calculations, such as boiling point elevation and freezing point depression.

Example 2: Analyzing Alcohol Concentration

A food scientist is analyzing the concentration of ethanol in a beverage. They want to determine the molality of the ethanol in a 250-gram sample that contains 10 grams of ethanol.

Here’s how the scientist calculates the molality:

  1. Calculate the number of moles of ethanol (C2H5OH). The molar mass of ethanol is approximately 46.07 g/mol.
  • Moles of ethanol = 10 g / 46.07 g/mol ≈ 0.217 moles

  1. Convert the mass of the solvent (water) from grams to kilograms. In this case, it is assumed that the remaining mass (250 g - 10 g = 240 g) is water:

    • Mass of water = 240 g = 0.240 kg

  2. Calculate molality:

    • Molality = Moles of solute / Mass of solvent in kg = 0.217 moles / 0.240 kg ≈ 0.904 m

The molality of ethanol in the beverage is approximately 0.904 m.

Notes: This calculation is useful for understanding the beverage’s properties, such as its freezing point and how it will behave under different temperature conditions.

Example 3: Industrial Use of Acid Solutions

An industrial chemist needs to create a concentrated sulfuric acid solution for a manufacturing process. They want a 3.0 m Molal solution using 1,000 grams of water.

To find the required mass of sulfuric acid (H2SO4), the chemist performs the following calculations:

  1. Calculate the number of moles of H2SO4 needed:
  • Moles of H2SO4 = 3.0 m × 1.0 kg (1,000 g of water) = 3.0 moles

  1. Determine the mass of H2SO4 required (the molar mass is approximately 98.08 g/mol):

    • Mass of H2SO4 = Moles × Molar Mass = 3.0 moles × 98.08 g/mol = 294.24 grams

Thus, the chemist prepares a solution by dissolving 294.24 grams of sulfuric acid in 1,000 grams of water, resulting in a molality of 3.0 m.

Notes: In industrial applications, knowing the concentration in molality is critical for ensuring proper reaction rates and product yields.