Question

You are asked to prepare a \(0.8500 \mathrm{M}\) solution of aluminum nitrate. You find that you have only \(50.00 \mathrm{~g}\) of the solid. (a) What is the maximum volume of solution that you can prepare? (b) How many milliliters of this prepared solution are required to furnish \(0.5000\) mol of aluminum nitrate to a reaction? (c) If \(2.500 \mathrm{~L}\) of the prepared solution is required, how much more aluminum nitrate would you need? (d) Fifty milliliters of a \(0.450 M\) solution of aluminum nitrate is needed. How would you prepare the required solution from the solution prepared in (a)?

Short answer

Answer: The maximum volume of the \(0.8500~\mathrm{M}\) aluminum nitrate solution that can be prepared is \(276.1~\mathrm{mL}\).

Step-by-step solution

Calculate the molar mass of aluminum nitrate

To begin, we need to find the molar mass of aluminum nitrate (\(\mathrm{Al(NO_3)_3}\)). To do this, we can use the periodic table to find the atomic masses of each element and then multiply by the number of atoms in one mole of aluminum nitrate: Molar mass of \(\mathrm{Al(NO_3)_3} = 1 \times m_{Al} + 3 \times (m_{N} + 3 \times m_{O}) = 26.98 + 3 \times (14.01 + 3 \times 16.00) \approx 212.99 \frac{g}{mol}\)

Calculate the moles of aluminum nitrate in the given mass

Now that we have the molar mass, we can use the given mass of aluminum nitrate (\(50.00~\mathrm{g}\)) to find the number of moles: Moles of \(\mathrm{Al(NO_3)_3} = \frac{50.00}{212.99} \approx 0.2347~\mathrm{mol}\)

Find the maximum volume of the solution that can be prepared

Now we can solve for the maximum volume of solution that can be prepared with the given concentration (\(0.8500~\mathrm{M}\)) and amount of moles: Volume = \(\frac{0.2347}{0.8500} \approx 0.2761~\mathrm{L}\) or \(276.1~\mathrm{mL}\)

Calculate the volume required to furnish \(0.5000\) mol of aluminum nitrate to a reaction

We can use the relationship between the molarity, volume, and moles of a solution to find the volume needed for the reaction: Volume = \(\frac{0.5000}{0.8500} \approx 0.5882~\mathrm{L}\) or \(588.2~\mathrm{mL}\)

Calculate the additional amount of aluminum nitrate needed for \(2.500~\mathrm{L}\) of solution

To find the additional amount of aluminum nitrate needed for a larger volume of solution, we can use the molarity and desired volume: Additional moles of \(\mathrm{Al(NO_3)_3} = 0.8500 \times 2.500 = 2.125~\mathrm{mol}\) Since we already have \(0.2347~\mathrm{mol}\), we need an additional \((2.125 - 0.2347) = 1.8903~\mathrm{mol}\) Additional mass of \(\mathrm{Al(NO_3)_3} = 1.8903 \times 212.99 \approx 402.75~\mathrm{g}\)

Prepare the required \(0.450~\mathrm{M}\) solution from the solution prepared in (a)

To prepare the required \(0.450~\mathrm{M}\) solution, we can use the dilution formula: \(C_1V_1 = C_2V_2\), where \(C_1\) and \(V_1\) are the initial concentration and volume, and \(C_2\) and \(V_2\) are the final concentration and volume. We have \(C_1 = 0.8500~\mathrm{M}\), \(C_2 = 0.450~\mathrm{M}\), and \(V_1 = 276.1~\mathrm{mL}\). We need \(V_2\) for a \(0.450~\mathrm{M}\) solution: \(V_2 = \frac{C_1V_1}{C_2} = \frac{(0.8500)(276.1)}{0.450} \approx 513.437~\mathrm{mL}\) However, we only need \(50~\mathrm{mL}\) of the \(0.450~\mathrm{M}\) solution. Take \(50~\mathrm{mL}\) of the prepared solution, and dilute it with distilled water until the final volume reaches \(50 \times \frac{513.437}{276.1} \approx 93.2~\mathrm{mL}\).

Key concepts

Molar Mass Calculation

The molar mass is a foundational concept in chemistry that refers to the mass of one mole of a substance. It's particularly important when preparing solutions, as we need to convert between grams and moles. Molar mass is calculated by determining the sum of the atomic masses (from the periodic table) of all the atoms in one mole of the compound. For example, in the case of aluminum nitrate, \( \mathrm{Al(NO_3)_3} \), we add up the atomic mass of aluminum \(1 \times m_{Al}\) and the atomic masses of nitrogen and oxygen in the nitrate ions \(3 \times m_{N} + 9 \times m_{O}\).For students to accurately prepare a solution of aluminum nitrate, understanding this calculation is key. The exercise demonstrates that to find out how many grams in 50 grams of aluminum nitrate, we must first know the molar mass, which in this example is approximately \( 212.99 \frac{g}{mol} \) as calculated from the atomic masses.

Molarity Concentration

Molarity, denoted by \( M \), is the measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This measurement is essential when preparing solutions of a specific concentration or when carrying out reactions that require a certain molarity.

In the exercise we reviewed, molarity is used to ensure the solution of aluminum nitrate has the desired concentration of \(0.8500 \mathrm{M}\). To find the maximum volume of solution one can prepare with \(50.00 \mathrm{g}\) of aluminum nitrate, the number of moles of the compound is divided by the required molarity. Understanding how to calculate molarity, as well as its importance in solution preparation, is fundamental for students working on this type of problem.

Solution Dilution

Solution dilution is a process used to decrease the concentration of a solution. The concept is described by the dilution formula \(C_1V_1 = C_2V_2\), where \(C_1\) is the initial concentration, \(V_1\) is the initial volume, \(C_2\) is the final concentration, and \(V_2\) is the final volume after dilution. This is an important technique in both chemistry and biology for creating solutions of desired concentrations.

In our aluminum nitrate example, the student uses the concept of dilution to prepare a weaker \(0.450 M\) solution from a stronger \(0.8500 M\) solution. By understanding the necessary calculations and the principle that the number of moles of solute remains constant during dilution, students can learn how to adjust concentrations efficiently and accurately.

Stoichiometry of Reactions

Stoichiometry involves the calculation of reactants and products in chemical reactions. It is based on the conservation of mass and the principle that elements are neither created nor destroyed in chemical reactions. This concept is crucial when determining the amount of reactants needed or predicting the quantity of products formed.

Relating back to the exercise, when students are asked how many milliliters of aluminum nitrate solution are required to furnish \(0.5000\) mol for a reaction, stoichiometry principles help them to use the known molarity to find the required volume. It is the bridge between chemical equations and practical applications like creating a reaction's reactants in the lab. Grasping stoichiometry allows a student to approach reactive scenarios methodically and understand the quantitative relationships within a reaction.