Examples of Molarity and Molality

Introduction to Molarity and Molality

Molarity (M) and molality (m) are both units of concentration used in chemistry to describe the amount of solute in a solution. Molarity is defined as the number of moles of solute per liter of solution, while molality is defined as the number of moles of solute per kilogram of solvent. Understanding these concepts is essential for accurate calculations in various chemical applications. Below are three diverse examples that illustrate the differences between molarity and molality in practical scenarios.

Example 1: Preparing a Saltwater Solution for a Biology Experiment

In a biology lab, a researcher needs to prepare a saltwater solution for an experiment involving marine organisms. The researcher decides to use sodium chloride (NaCl) and needs a 0.5 M (molar) solution in 2 liters of water.

To determine how many grams of NaCl are needed, the researcher calculates:

• Molar mass of NaCl = 58.44 g/mol
• Moles of NaCl required = Molarity × Volume = 0.5 moles/L × 2 L = 1 mole
• Mass of NaCl = Moles × Molar Mass = 1 mole × 58.44 g/mol = 58.44 g

The researcher dissolves 58.44 g of NaCl in enough water to make a total volume of 2 liters. This example clearly illustrates how molarity is applied in preparing solutions where the volume of the entire solution is known.

Notes:

• If the researcher needed to know how to prepare a 0.5 m solution using molality instead, they would need to account for the mass of the solvent (water) instead of the total volume of the solution.

Example 2: Calculating the Concentration of a Sugar Solution

A chemist is working on a project involving sugar (sucrose, C12H22O11) and needs to create a solution with a specific concentration of 2 m (molal) using 1 kg of water as the solvent. The molar mass of sucrose is approximately 342.30 g/mol.

To find out how much sucrose is needed, the chemist calculates:

• Moles of sucrose required = Molality × Mass of solvent (kg) = 2 moles/kg × 1 kg = 2 moles
• Mass of sucrose = Moles × Molar Mass = 2 moles × 342.30 g/mol = 684.60 g

The chemist dissolves 684.60 g of sucrose in 1 kg of water to create a 2 m molal solution. This example showcases how molality is particularly useful when the mass of the solvent is the primary consideration.

Notes:

• In this scenario, temperature changes can affect the volume of the solution, but molality remains constant as it is based on the mass of the solvent.

Example 3: Understanding the Impact of Temperature on Solutions

An environmental scientist is studying the impact of temperature on the concentration of a chemical pollutant in water. The scientist has a pollutant solution with a molarity of 1.5 M and wants to understand its molality at different temperatures, as the density of the solution may change with temperature.

Assuming the solution’s density is 1.02 g/mL at 25°C, the scientist can convert molarity to molality as follows:

• Total volume of 1 L solution = 1000 mL
• Mass of solution = Volume × Density = 1000 mL × 1.02 g/mL = 1020 g
• Mass of solute (assuming 1.5 M means 1.5 moles) = 1.5 moles × 58.44 g/mol = 87.66 g
• Mass of solvent (water) = Mass of solution - Mass of solute = 1020 g - 87.66 g = 932.34 g = 0.93234 kg
• Molality = Moles of solute / Mass of solvent (kg) = 1.5 moles / 0.93234 kg ≈ 1.61 m

By understanding the relationship between molarity and molality, the scientist can better assess the pollutant’s behavior under varying environmental conditions. This example emphasizes how temperature and density can affect the concentration units in practice.

Notes:

• As density changes with temperature, the molarity can also change, while molality will remain a more stable measure under varying conditions.