Problem 103 - Introductory Chemistry Atoms First

Question

Your assistant tells you she measured out \(2.50\) moles of \(\mathrm{NaCl}\) and then added enough water to get \(500.0 \mathrm{~mL}\) of solution to prepare a \(5.00 \mathrm{M}\) solution of \(\mathrm{NaCl}\). (a) What was the mass of the \(\mathrm{NaCl}\) in grams? (b) Did she successfully prepare a \(2.5 \mathrm{M}\) solution? Prove your answer.

Step-by-Step Solution

Answer

(a) The mass of the NaCl in grams is calculated as: Mass (g) = Moles × Molar mass = 2.50 moles × 58.44 g/mol = 146.1 g (b) No, the student did not successfully prepare a 2.5 M solution. The observed molarity of the solution is: Molarity (M) = 2.50 moles / 0.5 L = 5.00 M Since the observed molarity (5.00 M) is not equal to the target molarity (2.5 M), the student did not prepare a 2.5 M solution.

Step 1: Find the mass of NaCl

We are given the amount of NaCl in moles which is 2.50 moles. We need to convert it to mass (grams). To do this, we'll use the molar mass of NaCl which is approximately 58.44 g/mol. The formula to find the mass is: Mass (g) = Moles × Molar mass Now, plug the values into the formula: Mass (g) = 2.50 moles × 58.44 g/mol

Step 2: Calculate the mass of NaCl

Now, we'll calculate the mass of NaCl: Mass (g) = 2.50 moles × 58.44 g/mol = 146.1 g So, 146.1 grams of NaCl were used in the solution preparation.

Step 3: Find the molarity of the prepared solution

The student aimed to prepare a 5.00 M solution, but we will verify if she actually prepared a 2.5 M solution or not. Molarity is defined by the formula: Molarity (M) = Moles of solute / Volume of solution (L) We know the number of moles of NaCl (2.5 moles) and the final volume of the solution (500.0 mL). To find the molarity, we need to convert the volume from milliliters to liters: Volume (L) = 500.0 mL × (1 L / 1000 mL) = 0.5 L Now, we calculate the molarity: Molarity (M) = 2.50 moles / 0.5 L

Step 4: Calculate the molarity of the prepared solution

Now, we'll calculate the observed molarity of the prepared solution: Molarity (M) = 2.50 moles / 0.5 L = 5.00 M The observed molarity of the prepared solution is 5.00 M.

Step 5: Check if the 2.5 M solution was prepared successfully

Now, we will compare the observed molarity with the target molarity (2.5 M) to answer part (b) of the exercise: Observed molarity: 5.00 M (calculated in step 4) Target molarity: 2.5 M (mentioned in the question) Since the observed molarity (5.00 M) is not equal to the target molarity (2.5 M), the student did not prepare a 2.5 M solution. So the answer to part (a) is 146.1 grams, and the answer to part (b) is no, the student did not successfully prepare a 2.5 M solution.

Key Concepts

Moles

A mole is a fundamental concept in chemistry that represents a specific quantity of particles, most commonly atoms or molecules. Think of it as a chemist's dozen. One mole contains exactly Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles. This might seem like a huge and daunting number, but it helps scientists use a measurable and consistent way to express chemicals en masse.

When measuring substances in a laboratory, chemists often prefer using moles over grams because it directly corresponds to the number of molecules involved in reactions. For example, in our original exercise, \(2.50\) moles of \(\mathrm{NaCl}\) were used, meaning that the student had \(2.50 \times 6.022 \times 10^{23}\) molecules of \(\mathrm{NaCl}\). This concept aids in predicting how much of each substance will react or be produced in a chemical reaction.

Molar Mass

Molar mass is another cornerstone concept for understanding chemistry solutions and reactions. It refers to the mass of one mole of a given substance, measured in grams per mole (g/mol). To find the molar mass, you add up the atomic masses of all the atoms in a molecule. Atomic masses can be found on the periodic table.

Let's consider sodium chloride (\(\mathrm{NaCl}\)) as an example. Sodium (\(\mathrm{Na}\)) has an atomic mass of about 23.0 g/mol, and chlorine (\(\mathrm{Cl}\)) has an atomic mass of about 35.5 g/mol. Combining these gives us a molar mass for \(\mathrm{NaCl}\) of approximately \(58.44\) g/mol.

Understanding molar mass is crucial when you need to convert between moles and mass, as seen in our exercise where the \(2.50\) moles of \(\mathrm{NaCl}\) were converted to \(146.1\) grams using the formula: \[\text{Mass (g)} = \text{Moles} \times \text{Molar Mass (g/mol)}\]This conversion is a fundamental step in preparing chemical solutions accurately.

Solution Preparation

Preparing a solution involves dissolving a known amount of solute, often in moles, into a solvent to achieve a desired molarity or concentration. Correct solution preparation is crucial in a chemistry lab for reactions to occur as expected.

In our exercise, \(2.50\) moles of \(\mathrm{NaCl}\) were added to enough water to form \(500.0\) mL – or \(0.5\) L – of solution. Adjusting the amount of solvent, typically water, determines the final concentration of the solution. Getting these measurements right is essential for the success of experiments, as even a slight deviation in concentration can yield different results. It's why precision in measuring the solute and solvent is so important in solution preparation.

Concentration Calculation

Concentration tells us how much solute is present in a given volume of solution. It's typically expressed as molarity (M), defining how many moles of solute are in one liter of solution. The formula for molarity is:\[\text{Molarity (M)} = \frac{\text{Moles of solute}}{\text{Volume of solution (L)}}\]This calculation is critical for determining whether the concentration matches what is required for a particular experiment or application.

In our case, to calculate if the solution prepared was indeed \(2.5\) M, we would check the moles of \(\mathrm{NaCl}\) (2.50 moles) and the volume of the solution (\(0.5\) L). Using the molarity formula, we find that the solution actually had a concentration of \(5.00\) M, not \(2.5\) M. This shows how measuring and calculating accurately is imperative for meeting target solution concentrations in chemistry.