Describe how you would prepare \(250 \mathrm{~mL}\) of a \(0.707 M \mathrm{NaNO}_{3}\) solution.

Molarity is a fundamental concept in chemistry referring to the concentration of a solution. To calculate the molarity, you need to know the amount of the substance in moles and the volume of the solution in liters. The formula is expressed as:
\[\begin{equation}M = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\end{equation}\]
In the provided exercise, the student is asked to prepare a 0.707 M sodium nitrate (NaNO3) solution. This requires calculating the number of moles needed for 250 mL (0.250 L) of solution, based on the given molarity. By rearranging the molarity equation, the calculation becomes:
\[\begin{equation}\text{moles} = M \times \text{volume(L)}\end{equation}\]
By plugging in the numbers, \[\begin{equation}\text{moles} = 0.707 \frac{\text{mol}}{\text{L}} \times 0.250 \text{ L} = 0.17675 \text{ mol}\end{equation}\]
Understanding molarity calculation helps students to grasp how to manipulate the concentration of solutions—a critical skill in laboratory practices and various scientific applications.

Converting moles to grams is a straightforward process once the molar mass (molecular weight) of the substance is known. Each element's molar mass is found on the periodic table, and the molar mass of a compound is the sum of the molar masses of its constituent elements.
For the exercise at hand, students are converting the molarity of a NaNO3 solution into a measurable quantity for the laboratory. This involves using the molar mass of NaNO3, which is approximately 85 g/mol:
\[\begin{equation}\text{mass (g)} = \text{moles} \times \text{molar mass (g/mol)}\end{equation}\]
Thus, the mass of NaNO3 needed is calculated as:\[\begin{equation}\text{mass} = 0.17675 \text{ mol} \times 85 \text{ g/mol} = 15.02375 \text{ g}\end{equation}\]

Practical Tip:

In the lab, the level of precision should be appropriate for the equipment available. Therefore, the weight of NaNO3 is rounded to a practical value, such as 15.0 grams, to prepare the solution.

Once the required amount of solute is calculated and converted into grams, the next step is preparing the solution. Preparing a molar solution involves dissolving the solute in a solvent—typically water—to achieve the desired final volume and concentration.
Following the weight measurement of the solute, in this case, approximately 15.0 grams of NaNO3, the substance is dissolved in a volume of water that is less than the desired final volume of the solution. This is crucial as it ensures that all of the solute fully dissolves.
After the solute has fully dissolved, additional water is added until the total volume reaches the precise measurement required, which is 250 mL for our exercise. The container used for preparation should be calibrated for the volume or a separate measuring device should be used to ensure accuracy.

Safety Measures:

Always use personal protective equipment, such as gloves and goggles, and follow laboratory safety protocols during solution preparation. Additionally, the solution should be mixed thoroughly to ensure homogeneity. The end result is a 0.707 M solution of sodium nitrate ready for experimental use.