Problem 104 - Introductory Chemistry Atoms First

Question

Your assistant tells you he measured out \(116.886\) \(\mathrm{g}\) of \(\mathrm{NaCl}\) and then added exactly \(1.00 \mathrm{~L}\) of water to it to prepare a \(2.00 \mathrm{M}\) solution of \(\mathrm{NaCl}\). Do you fire him or give him a promotion? Explain.

Step-by-Step Solution

Answer

After calculating the molarity of the NaCl solution: Molarity = \( \frac{\frac{116.886 \mathrm{g}}{58.44 \mathrm{g/mol}}}{1.00 \mathrm{L}} \approx 2.00 \mathrm{M} \) The calculated molarity is approximately 2.00 M, which indicates that the assistant has successfully prepared the desired 2.00 M solution of NaCl. Therefore, you should consider giving him a promotion.

Step 1: Calculate the Moles of NaCl Present in the Solution

We will calculate the moles of NaCl using the given mass and its molar mass. Molar mass of NaCl = 58.44 g/mol Given mass of NaCl = 116.886 g To calculate the moles of NaCl, divide the given mass by its molar mass: Moles of NaCl = \( \frac{116.886 \mathrm{g}}{58.44 \mathrm{g/mol}} \)

Step 2: Calculate the Volume of the Solution in Liters

The volume is given as exactly 1.00 L, so we can use this value directly in our calculations.

Step 3: Calculate the Molarity of the Solution

To calculate the molarity of the solution, divide the moles of NaCl by the volume of the solution in liters: Molarity = \( \frac{\text{moles of NaCl}}{\text{volume in liters}} \) Molarity = \( \frac{\frac{116.886 \mathrm{g}}{58.44 \mathrm{g/mol}}}{1.00 \mathrm{L}} \)

Step 4: Compare the Calculated Molarity with 2.00 M

The next step is to compare the calculated molarity of the NaCl solution with 2.00 M. If the calculated molarity is equal to or very close to 2.00 M, the assistant has successfully prepared a 2.00 M solution, and you can consider a promotion. If the calculated molarity is significantly different from 2.00 M, the assistant failed to perform the task, and you may consider firing him.

Step 5: Make an Informed Decision Based on the Calculation Results

After calculating the molarity of the solution, it is important to make an informed decision about the assistant's performance. Based on the result, decide whether to give a promotion or fire the assistant. The fundamental concept behind this exercise is to evaluate the accuracy of the solution preparation and the individual's knowledge of molarity and concentration.

Key Concepts

Solution Preparation

Understanding how to properly prepare a solution is crucial in many scientific and laboratory tasks. In this context, solution preparation refers to the process of accurately measuring and combining substances to achieve a desired concentration. This starts with knowing the desired molarity of the solution, which helps in calculating the amounts of solute and solvent needed.

The first step involves measuring out the precise mass of the solute. In this exercise, the solute is sodium chloride (\(\mathrm{NaCl}\)). It is essential to use a balance to measure exactly \(116.886 \mathrm{g}\) of \(\mathrm{NaCl}\) as specified. Following the mass measurement, the solute is dissolved in a specific amount of solvent, which is typically water. The given exercise specifies adding exactly \(1.00 \mathrm{L}\) of water for this preparation.

• Accurate measurement of both solute and solvent is key to replicable results.
• The choice of solvent (usually water) also influences the solubility and final concentration of the solution.

Proper solution preparation ensures that the chemical and physical properties of the solution are consistent and reliable for further use.

Concentration Calculation

Calculating concentration is an essential part of checking whether the solution has been prepared correctly. Concentration is a measure of the amount of solute present in a given quantity of solvent, which in laboratory settings is often expressed in molarity (M). Molarity is defined as the number of moles of solute per liter of solution.

To calculate the concentration or molarity of \(\mathrm{NaCl}\), first determine the moles of the solute. This is achieved by dividing the mass of \(\mathrm{NaCl}\) by its molar mass. For sodium chloride, the molar mass is \(58.44 \mathrm{g/mol}\). The calculation follows:\[\text{Moles of } \mathrm{NaCl} = \frac{116.886 \mathrm{g}}{58.44 \mathrm{g/mol}}\]

Once the moles are determined, the next step is calculating molarity using the formula:\[\text{Molarity (M)} = \frac{\text{moles of NaCl}}{\text{volume in liters}}\]For this case, the solution volume is \(1.00 \mathrm{L}\), simplifying our calculations:\[\text{Molarity} = \frac{\frac{116.886 \mathrm{g}}{58.44 \mathrm{g/mol}}}{1.00 \mathrm{L}}\]

• The concentration provides insight into whether the laboratory process achieves the intended chemical outcome.
• Constant verification by concentration calculations ensures that experimental runs maintain their integrity and repeatability.

Accurate concentration calculation is fundamental for anyone working in chemistry to validate whether the preparation was successful.

Chemical Accuracy

Chemical accuracy refers to ensuring that all measurements and calculations in chemical procedures are precise and accurate. It is crucial for achieving valid experimental outcomes, reproducibility, and reliability in scientific research.

In the context of this exercise, chemical accuracy is evaluated through careful measurement and calculation of the materials used. The weight of the \(\mathrm{NaCl}\) must be precise, as even a minor error can lead to a significant change in concentration. Furthermore, when adding the \(1.00 \mathrm{L}\) of water, it is pivotal to use precise measuring tools to prevent any deviation from the intended molarity.

• Accuracy in chemical measurements impacts the final properties of the solution, crucial for both education and professional laboratory settings.
• Careful adherence to procedure ensures safety and efficiency in laboratory operations.

By upholding high standards of chemical accuracy, we ensure that the intent behind laboratory exercises and professional chemical processes is fully realized, leading to consistent and dependable results.