Calculate Mass Of NaNO3 For 1.50 M Solution Step-by-Step Guide

Introduction

In this comprehensive guide, we will delve into the step-by-step process of calculating the precise mass of sodium nitrate ($NaNO_3$) required to prepare 4.50 liters of a 1.50 M solution. This calculation is a fundamental concept in chemistry, particularly in the field of solution preparation. Understanding molarity and its relationship to mass and volume is crucial for accurate laboratory work and various chemical applications. We will leverage the molarity formula, molarity = (moles of solute) / (liters of solution), as the cornerstone of our calculations. By meticulously breaking down each step, we aim to provide a clear and concise explanation that will empower you to confidently tackle similar calculations in your chemistry endeavors.

Solution preparation is a critical aspect of chemistry, underpinning a wide array of experiments and industrial processes. The accuracy of these processes hinges on the precise measurement of reactants, and this is where molarity calculations come into play. Molarity, defined as the number of moles of solute per liter of solution, provides a convenient way to express the concentration of a solution. In this article, we focus on calculating the mass of sodium nitrate ($NaNO_3$) needed to create a specific solution, a common task encountered in chemistry labs. Sodium nitrate is a versatile chemical compound, finding applications in fertilizers, pyrotechnics, and food preservation. Its precise measurement is paramount in various chemical reactions and experiments. Therefore, mastering the calculation of the mass of $NaNO_3$ needed for a solution of a particular molarity is an essential skill for anyone working in the field of chemistry.

We will start by understanding the relationship between molarity, moles, and volume. Then, we will use the molar mass of $NaNO_3$ to convert moles into grams, thus arriving at the required mass. This process will not only provide the answer to the specific question but also illustrate the general methodology for calculating the mass of any solute needed to make a solution of a desired molarity and volume. Throughout the article, we will emphasize the importance of using the correct units and significant figures to ensure the accuracy of the final result. By the end of this guide, you will have a solid understanding of how to perform this type of calculation and be able to apply it to various scenarios in chemistry.

Understanding Molarity

Molarity is a fundamental concept in chemistry that quantifies the concentration of a solution. Specifically, it is defined as the number of moles of solute dissolved in one liter of solution. This concentration unit is denoted by the symbol 'M' and is expressed in units of mol/L (moles per liter). Understanding molarity is crucial for preparing solutions with specific concentrations, which are essential for chemical reactions, experiments, and various applications in industries such as pharmaceuticals and manufacturing. A 1.50 M solution, as mentioned in our problem, signifies that there are 1.50 moles of the solute dissolved in every liter of the solution. This precise definition allows chemists to accurately measure and control the amount of solute in a solution, ensuring the reliability and reproducibility of chemical processes.

The concept of molarity bridges the gap between the macroscopic world of measurable volumes and masses and the microscopic world of atoms and molecules. By using molarity, we can relate the mass of a solute to the number of particles it contributes to a solution. This connection is particularly important when dealing with chemical reactions, as reactions occur at the molecular level, and the number of moles of reactants determines the extent of the reaction. The molar mass of a substance, which is the mass of one mole of that substance, serves as the conversion factor between mass and moles. For instance, the molar mass of $NaNO_3$ (85.00 g/mol) tells us that 85.00 grams of $NaNO_3$ contains one mole of $NaNO_3$ molecules. This information is vital for calculating the mass of $NaNO_3$ needed to achieve a specific molarity.

When preparing solutions, it is important to consider the total volume of the solution, not just the volume of the solvent added. The solute also contributes to the total volume, and this must be taken into account to achieve the desired molarity. In practice, solutions are often prepared by dissolving the solute in a smaller volume of solvent and then adding more solvent until the final desired volume is reached. This ensures that the concentration of the solution is accurate. Moreover, it's essential to use appropriate glassware, such as volumetric flasks, which are designed to accurately measure specific volumes. Understanding these nuances of solution preparation, along with a solid grasp of molarity, is fundamental for successful work in chemistry and related fields. Therefore, a thorough understanding of molarity is not just an academic exercise; it is a practical skill with widespread applications.

Step-by-Step Calculation

To calculate the mass of $NaNO_3$ needed, we will follow a step-by-step approach, utilizing the molarity formula and the molar mass of $NaNO_3$. The first step involves rearranging the molarity formula to solve for the moles of solute. We know that molarity (M) = moles of solute / liters of solution. Therefore, we can rearrange this equation to find the moles of solute: moles of solute = molarity × liters of solution. This simple rearrangement is the key to connecting the given molarity and volume to the amount of solute required.

Next, we substitute the given values into the rearranged formula. We are given a molarity of 1.50 M and a volume of 4.50 L. Plugging these values into the equation, we get: moles of $NaNO_3$ = 1.50 M × 4.50 L. Performing this multiplication will give us the number of moles of $NaNO_3$ needed to prepare the solution. It's crucial to keep track of the units during this calculation to ensure that they cancel out appropriately, leaving us with the desired unit of moles. In this case, the liters (L) in the volume term will cancel out with the liters in the denominator of the molarity (mol/L), resulting in the final answer being in moles.

Once we have calculated the moles of $NaNO_3$, the next step is to convert moles to grams using the molar mass of $NaNO_3$. As provided, the molar mass of $NaNO_3$ is 85.00 g/mol. This means that one mole of $NaNO_3$ weighs 85.00 grams. To convert moles to grams, we multiply the number of moles by the molar mass: mass of $NaNO_3$ = moles of $NaNO_3$ × molar mass of $NaNO_3$. This conversion is essential because we typically measure chemicals in the lab by mass, not by moles. By performing this calculation, we can determine the exact mass of $NaNO_3$ that needs to be weighed out to prepare the 1.50 M solution. The final result will be in grams, which is the unit we need for practical laboratory measurements.

Detailed Calculation Process

Now, let's walk through the detailed calculation process. As we established earlier, the first step is to calculate the moles of $NaNO_3$ using the formula: moles of solute = molarity × liters of solution. We are given a molarity of 1.50 M and a volume of 4.50 L. Substituting these values into the equation, we get:

Moles of $NaNO_3$ = 1.50 mol/L × 4.50 L

Performing this multiplication, we find:

Moles of $NaNO_3$ = 6.75 mol

This result tells us that we need 6.75 moles of $NaNO_3$ to make the desired solution. It's important to note the units here: the liters (L) have canceled out, leaving us with moles (mol), which is the unit we were aiming for. This step demonstrates the importance of unit analysis in ensuring the correctness of the calculation.

The next step is to convert the moles of $NaNO_3$ to grams using the molar mass of $NaNO_3$, which is given as 85.00 g/mol. The formula for this conversion is: mass of $NaNO_3$ = moles of $NaNO_3$ × molar mass of $NaNO_3$. Substituting the values we have:

Mass of $NaNO_3$ = 6.75 mol × 85.00 g/mol

Performing this multiplication, we get:

Mass of $NaNO_3$ = 573.75 g

This result indicates that we need 573.75 grams of $NaNO_3$ to prepare 4.50 L of a 1.50 M solution. However, it's important to consider significant figures. The given values (1.50 M and 4.50 L) each have three significant figures, and the molar mass (85.00 g/mol) has four significant figures. According to the rules of significant figures, the final answer should have the same number of significant figures as the value with the fewest significant figures, which in this case is three. Therefore, we need to round our answer to three significant figures. Rounding 573.75 g to three significant figures gives us 574 g.

Final Answer and Conclusion

Therefore, to prepare 4.50 L of a 1.50 M $NaNO_3$ solution, you need 574 grams of $NaNO_3$. This calculation demonstrates the practical application of molarity and molar mass in solution preparation. By following the step-by-step approach, we first calculated the number of moles of $NaNO_3$ required and then converted moles to grams using the molar mass. The final answer was rounded to the appropriate number of significant figures, ensuring accuracy in the laboratory setting.

In conclusion, mastering the calculation of solute mass for solution preparation is an essential skill in chemistry. The ability to accurately prepare solutions of specific concentrations is crucial for a wide range of applications, from laboratory experiments to industrial processes. This article has provided a detailed explanation of how to calculate the mass of $NaNO_3$ needed to prepare a 1.50 M solution, highlighting the importance of molarity, molar mass, and significant figures. By understanding these concepts and following the step-by-step process outlined, you can confidently tackle similar calculations and ensure the success of your chemical endeavors. The principles discussed here are applicable to a variety of solutes and solutions, making this a fundamental skill for anyone working in chemistry or related fields. Remember to always double-check your calculations and units to ensure accuracy, and with practice, you will become proficient in solution preparation.

Keywords

Solution preparation, molarity calculation, mass calculation, sodium nitrate ($NaNO_3$), moles, molar mass, significant figures, chemical solutions, chemistry calculations, laboratory techniques.