Problem 44 - Chemistry The Central Science In Si Units

Question

What is the molarity of each of the following solutions: (a) \(15.0 \mathrm{~g}\) of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) in \(0.250 \mathrm{~mL}\) solution, (b) \(5.25 \mathrm{~g}\) of \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2} \mathrm{O}\) in \(175 \mathrm{~mL}\) of solution, \((\mathbf{c}) 35.0 \mathrm{~mL}\) of \(9.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) diluted to \(0.500 \mathrm{~L} ?\)

Step-by-Step Solution

Answer

(a) 175.2 M, (b) 0.1194 M, (c) 0.630 M.

Step 1: Calculate Molar Mass

First, find the molar mass of each compound. For \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\), calculate as follows: \(2(\text{Al}) + 3(\text{S}) + 12(\text{O})\). For \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O} \), calculate as: \(\text{Mn} + 2(\text{N}) + 6(\text{O}) + 4(\text{H})\). After calculations, \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}\) = 342.15 g/mol and \(\mathrm{Mn}\left(\mathrm{NO}_{3}\right)_{2} \cdot 2 \mathrm{H}_{2}\mathrm{O} \) = 251.01 g/mol.

Step 2: Convert Mass to Moles

Convert the given mass of each compound to moles using the formula: \( \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}} \). For (a), \(\frac{15.0 \, \text{g}}{342.15 \, \text{g/mol}} = 0.0438 \, \text{moles}\). For (b), \(\frac{5.25 \, \text{g}}{251.01 \, \text{g/mol}} = 0.0209 \, \text{moles}\).

Step 3: Convert Volume to Liters

Convert the volume of solution from milliliters to liters by dividing by 1000. For (a), \(0.250 \, \text{mL} = 0.00025 \, \text{L}\). For (b), \(175 \, \text{mL} = 0.175 \, \text{L}\).

Step 4: Calculate Molarity

Use the formula: \(\text{Molarity (M)} = \frac{\text{Moles of Solute}}{\text{Liters of Solution}}\) to find the molarity. For (a), \(\frac{0.0438 \, \text{moles}}{0.00025 \, \text{L}} = 175.2 \, \text{M} \). For (b), \(\frac{0.0209 \, \text{moles}}{0.175 \, \text{L}} = 0.1194 \, \text{M} \).

Step 5: Dilution Problem

For part (c), use the dilution formula \(M_1V_1 = M_2V_2\), where \(M_1 = 9.00 \, \text{M}\), \(V_1 = 0.0350 \, \text{L}\), and \(V_2 = 0.500 \, \text{L}\). Solve for \(M_2\): \( M_2 = \frac{M_1V_1}{V_2} = \frac{9.00 \times 0.0350}{0.500} = 0.630 \, \text{M} \).

Key Concepts

Molar Mass

Molar mass is a crucial concept in chemistry that refers to the mass of one mole of a substance. It provides a bridge between the mass of a substance and the number of moles. To compute molar mass, you sum the atomic masses of all atoms in a molecule. The atomic masses can be found on the periodic table.

• For example, in the compound \(\mathrm{Al}_2\left(\mathrm{SO}_4\right)_3\), the molar mass calculation involves adding up the atomic masses of Aluminum (Al), Sulfur (S), and Oxygen (O).
• The equation looks like this: 2(Al) + 3(S) + 12(O), because there are 2 Aluminum atoms, 3 Sulfur atoms, and 12 Oxygen atoms in one formula unit of aluminum sulfate.

Once the atomic masses are added, you obtain the molar mass, expressed in g/mol. This number allows you to convert between grams and moles in chemical calculations.
Understanding how to calculate molar mass is essential for finding the molarity of a solution, as molarity is expressed in moles per liter.

Mole Concept

The mole is a fundamental unit in chemistry used to express amounts of a chemical substance. It relates the mass of a substance to the number of particles it contains (atoms, molecules, etc.). The concept of the mole simplifies the count of microscopic particles by using the large number known as Avogadro's number, \(6.022 \times 10^{23}\).

• A mole of any substance contains exactly \(6.022 \times 10^{23}\) particles.
• Using the molar mass, chemists can convert a given mass of a substance into the number of moles: \(\text{Moles} = \frac{\text{Mass (g)}}{\text{Molar Mass (g/mol)}}\).

For example, in the step-by-step solution to the exercise, you calculated the number of moles for the compounds given. This calculation is pivotal because it allows you to work with macroscopic amounts of a substance and predict the outcomes of chemical reactions on a practical scale.
By understanding the mole concept, you can relate the mass of materials to the amounts in a very intuitive and consistent way, enabling precise recipe-scaling in chemistry.

Dilution Calculations

Dilution involves adding more solvent to a solution, which decreases the concentration of the solute within it. When performing dilution calculations, you aim to find the new concentration after changing the volume of the solution. The key formula used is \( M_1V_1 = M_2V_2 \), where \(M_1\) and \(V_1\) are the initial molarity and volume, and \(M_2\) and \(V_2\) are the final molarity and volume.

• To calculate the new molarity (\(M_2\)), rearrange the formula: \( M_2 = \frac{M_1 \times V_1}{V_2} \).
• This equation ensures that the number of moles of solute remains constant before and after the dilution.

For instance, in the exercise, you are given an initial molarity and volume of \(9.00 \, \text{M} \) and \(35.0 \, \text{mL}\) of \( \mathrm{H}_2 \mathrm{SO}_4\), which is then diluted to a final volume of \(0.500 \, \text{L}\). Plug these values into the formula to find the new molarity.
This concept is vital in preparing solutions in laboratories, ensuring the desired concentration is achieved for experiments and reactions. Understanding dilution helps maintain consistency across different experiments and make predictions about reaction outcomes efficiently.