Understanding Mole Fraction and Molality: An Example with 13.0% NaOH Solution

In this article, we will explore the process of determining the mole fraction (x) of a solute and the molality (m) for an aqueous solution that is 13.0% NaOH by mass. This example will provide a clear understanding of these concepts, which are crucial in chemical solutions and chemical engineering.

The Concept of Mole Fraction and Molality

Mole fraction (x) is the ratio of the number of moles of a particular component in a solution to the total number of moles of all components in the solution. Molality (m) is the number of moles of solute per kilogram of solvent. These terms are fundamental in understanding the properties of solutions and their behavior under different conditions.

Step-by-Step Calculation for 13.0% NaOH Solution

Determining the Mass of the Solution

Let's assume we have 100 grams of the solution, which is 13.0% NaOH by mass. This means that the mass of NaOH in the solution is 13.0 grams, and the mass of the solvent (water) is 87.0 grams.

Calculation of Moles of NaOH

The molar mass of NaOH is approximately 40.00 g/mol. Using this, we can calculate the number of moles of NaOH present in the solution:

Moles of NaOH mass of NaOH / molar mass of NaOH 13.0 g / 40.00 g/mol 0.325 mol

Calculation of Moles of Water

The molar mass of water (H2O) is approximately 18.015 g/mol. We can then calculate the number of moles of water:

Moles of water mass of water / molar mass of water 87.0 g / 18.015 g/mol ≈ 4.83 mol

Calculation of Mole Fraction of NaOH

The mole fraction of NaOH (xNaOH) is given by the ratio of the moles of NaOH to the total moles of NaOH and water:

xNaOH moles of NaOH / (moles of NaOH moles of water) 0.325 mol / (0.325 mol 4.83 mol) ≈ 0.063

Calculation of Molality of the Solution

Molality (m) is defined as the number of moles of solute per kilogram of solvent. We already have the moles of NaOH and the mass of water, which is 87.0 grams. Therefore, the molality is calculated as follows:

m moles of NaOH / mass of water (in kg) 0.325 mol / 0.087 kg ≈ 3.74 mol/kg

Additional Calculations for Further Understanding

For comparison, let's perform additional calculations based on the given data:

Molality: m moles of NaOH / mass of water (in kg) 0.25 mol / 0.08 kg ≈ 2.78 mol/kg

Mole Fraction of Solute: x moles of solute / total moles 0.25 mol / (0.25 mol 5 mol) ≈ 0.0476

Conclusion

In this detailed example, we have calculated the mole fraction and molality for a 13.0% NaOH solution. These calculations can be applied to various chemical solutions, enhancing our understanding of their properties and behavior. Understanding these concepts is essential for chemical engineering, pharmaceuticals, and many other scientific fields.

Key Points: Mole fraction is the ratio of the number of moles of a solute to the total number of moles of all components in the solution. Molality is the number of moles of solute per kilogram of solvent. The calculations involve determining the mass of solute and solvent, calculating their moles, and using these values to find mole fraction and molality.

Keywords: Understanding Mole Fraction, Understanding Molality, NaOH Solution