Question

In a laboratory experiment, a 10.0-mL sample of \(\mathrm{NaCl}\) solution is poured into an evaporating dish with a mass of \(24.10 \mathrm{~g}\). The combined mass of the evaporating dish and \(\mathrm{NaCl}\) solution is \(36.15 \mathrm{~g}\). After heating, the evaporating dish and dry \(\mathrm{NaCl}\) have a combined mass of \(25.50 \mathrm{~g}\). a. What is the mass percent \((\mathrm{m} / \mathrm{m})\) of the \(\mathrm{NaCl}\) solution? b. What is the molarity (M) of the \(\mathrm{NaCl}\) solution? c. If water is added to \(10.0 \mathrm{~mL}\) of the initial \(\mathrm{NaCl}\) solution to give a final volume of \(60.0 \mathrm{~mL}\), what is the molarity of the diluted \(\mathrm{NaCl}\) solution?

Short answer

a. 11.62% m/m b. 2.396 M c. 0.399 M

Step-by-step solution

Calculate the Mass of \(\text{NaCl}\)

Determine the mass of \(\text{NaCl}\) by subtracting the mass of the evaporating dish from the combined mass of the dish and dry \(\text{NaCl}\): \[ \text{Mass of NaCl} = 25.50 \, \text{g} - 24.10 \, \text{g} = 1.40 \, \text{g} \]

Determine the Mass of the \(\text{NaCl}\) Solution

Find the mass of the \(\text{NaCl}\) solution by subtracting the mass of the evaporating dish from the combined mass of the dish and solution: \[ \text{Mass of NaCl solution} = 36.15 \, \text{g} - 24.10 \, \text{g} = 12.05 \, \text{g} \]

Calculate the Mass Percent of \(\text{NaCl}\) Solution

Use the mass percent formula: \[ \text{Mass percent} = \left( \frac{\text{Mass of NaCl}}{\text{Mass of NaCl solution}} \right) \times 100 = \left( \frac{1.40 \, \text{g}}{12.05 \, \text{g}} \right) \times 100 \approx 11.62\text{%} \]

Calculate the Molarity of \(\text{NaCl}\) Solution

First, calculate the moles of \(\text{NaCl}\) using its molar mass (58.44 g/mol): \[ n = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{1.40 \, \text{g}}{58.44 \, \text{g/mol}} \approx 0.02396 \, \text{mol} \] Next, find the molarity using the formula: \[ M = \frac{n}{V} = \frac{0.02396 \, \text{mol}}{0.010 \, \text{L}} = 2.396 \text{ M} \]

Calculate the Molarity of the Diluted \(\text{NaCl}\) Solution

Use the dilution formula \(M_1 V_1 = M_2 V_2\): \[ 2.396 \, \text{M} \times 0.010 \, \text{L} = M_2 \times 0.060 \, \text{L} \] Solving for \(M_2\): \[ M_2 = \frac{2.396 \, \text{M} \times 0.010 \, \text{L}}{0.060 \, \text{L}} = 0.399 \, \text{M} \]

Key concepts

Mass Percent

Mass percent is a way to express the concentration of a solution by describing the mass of the solutes compared to the total mass of the solution.
In this exercise, we found that the mass of dry \(\text{NaCl}\) is 1.40 g and the mass of the \(\text{NaCl}\) solution is 12.05 g.
The mass percent formula is: \[ \text{Mass percent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100 \] Using our values: \[ \text{Mass percent} = \frac{1.40 \text{ g}}{12.05 \text{ g}} \times 100 \approx 11.62˜\text{%} \] This tells us that 11.62% of the \(\text{NaCl}\) solution is actually sodium chloride.
Mass percent is particularly useful when comparing the strength of various solutions.
It's a straightforward way to understand how concentrated a solution is.

Molarity

Molarity is another way to express the concentration of a solution.
It is defined as the number of moles of solute per liter of solution.
To calculate molarity, we need two things: the number of moles of the solute and the volume of the solution in liters.
First, find the moles of \(\text{NaCl}\) using its molar mass (58.44 g/mol): \[ n = \frac{\text{Mass of NaCl}}{\text{Molar mass of NaCl}} = \frac{1.40 \text{ g}}{58.44 \text{ g/mol}} \approx 0.02396˜\text{ mol} \] Next, use the molarity formula: \[ M = \frac{n}{V} = \frac{0.02396 \text{ mol}}{0.010 \text{ L}} = 2.396˜\text{ M} \] So, the molarity of the \(\text{NaCl}\) solution is approximately 2.396 M.

Dilution Formula

The dilution formula helps us find the new concentration of a solution after it has been diluted with more solvent.
The dilution formula is \[ M_1 V_1 = M_2 V_2 \] where \( M_1 \) and \( V_1 \) are the initial molarity and volume, and \( M_2 \) and \( V_2 \) are the final molarity and volume after dilution.
In this problem, the initial volume \( V_1 \) is 10 mL (or 0.010 L) and the final volume \( V_2 \) is 60 mL (or 0.060 L).
We found that \( M_1 \) is 2.396 M.
Using the formula: \[ 2.396 \text{ M} \times 0.010 \text{ L} = M_2 \times 0.060 \text{ L} \] Solving for \( M_2 \):
\[ M_2 = \frac{2.396 \times 0.010}{0.060} \approx 0.399 \text{ M} \] This tells us that after dilution, the molarity of the \(\text{NaCl}\) solution is approximately 0.399 M.

Laboratory Experiment

A laboratory experiment is a hands-on way to understand and verify theoretical concepts.
In this exercise, you performed measurements using specific instruments to find the mass of solutions and solutes.
Careful and precise measurement is crucial in these experiments to ensure accurate results.
First, you measured the empty evaporating dish.
After adding the \(\text{NaCl}\) solution, you weighed it again to find the combined mass.
Heating the solution helped evaporate the water, and weighing it again gave the mass of the dry \(\text{NaCl}\).
Through these steps, you systematically gathered data to calculate concentration.
This shows how theoretical calculations are verified through practical laboratory procedures.

Moles Calculation

Calculating moles is fundamental in understanding chemical reactions and solutions.
A mole represents \(\text{6.022} \times \text{10}^{23}\) entities (atoms, molecules, etc.).
To calculate moles, you need the mass of the substance and its molar mass.
For \(\text{NaCl}\), the molar mass is 58.44 g/mol.
Using the formula: \[ n = \frac{\text{Mass}}{\text{Molar mass}} \] In our case: \[ n = \frac{1.40 \text{ g}}{58.44 \text{ g/mol}} \approx 0.02396˜\text{ mol} \] This calculation is essential for finding molarity and understanding how much solute is present in a solution.
Knowing the moles helps in various other calculations, like determining the products of a chemical reaction and understanding its stoichiometry.