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 \(0.5000 \mathrm{~mol}\) of aluminum nitrate to a reaction? (c) If \(2.500 \mathrm{~L}\) of the prepared solution are required, how much more aluminum nitrate would you need? (d) Fifty milliliters of a \(0.450 \mathrm{M}\) solution of aluminum nitrate are needed. How would you prepare the required solution from the solution prepared in (a)?

Short answer

Answer: 402.57 g of additional aluminum nitrate is needed to prepare a larger volume of the solution.

Step-by-step solution

(a) Calculate the number of moles of aluminum nitrate

To find the maximum volume of solution that we can prepare, first determine the number of moles of aluminum nitrate. Use the molar mass of aluminum nitrate to convert the given mass to moles: \(\text{moles} = \frac{\text{mass}}{\text{molar mass}}\) The molar mass of aluminum nitrate (\(\text{Al(NO}_{3})_{3}\)) is \(26.98 + (3 \times 14.01) + (9 \times 16.00) = 213.00~\text{g/mol}\). Given the available mass of \(50.00 \text{g}\), the number of moles is: \(\text{moles} = \frac{50.00 \text{g}}{213.00 \text{g/mol}} = 0.2347 \text{mol}\)

(a) Calculate the maximum volume of solution

Using the molarity formula, find the maximum volume of the solution we can prepare: \(\text{Molarity} = \frac{\text{moles}}{\text{volume}}\) We need to prepare a \(0.8500 \mathrm{M}\) aluminum nitrate solution, and we have \(0.2347 \text{mol}\) of aluminum nitrate. Solving the equation for the volume, we get: \(\text{volume} = \frac{0.2347 \text{mol}}{0.8500 \mathrm{M}} = 0.2761 \mathrm{L}\). So, the maximum volume of solution that we can prepare is \(0.2761 \mathrm{L}\).

(b) Calculate the volume of solution needed for 0.5 moles of aluminum nitrate

Using the molarity formula, find the required volume of the prepared solution: \(\text{Molarity} = \frac{\text{moles}}{\text{volume}}\) We want to find the volume of the solution that contains \(0.5000 \text{mol}\) of aluminum nitrate in a \(0.8500 \mathrm{M}\) solution. Solving for volume, we get: \(\text{volume} = \frac{0.5000 \text{mol}}{0.8500 \mathrm{M}} = 0.5882 \mathrm{L}\) To find this volume in milliliters, multiply by \(1000 \frac{\text{mL}}{\text{L}}\), resulting in \(588.2 \text{mL}\).

(c) Calculate the additional amount of aluminum nitrate needed

We are given that we need to prepare \(2.500 \mathrm{L}\) of the prepared solution, but we can make only \(0.2761 \mathrm{L}\). First, use the molarity formula to find the number of moles of aluminum nitrate needed to prepare \(2.500 \mathrm{L}\): \(\text{Molarity} = \frac{\text{moles}}{\text{volume}}\) \(\text{moles} = \text{Molarity} \times \text{volume} = 0.8500 \mathrm{M} \times 2.500 \mathrm{L} = 2.125 \text{mol}\) We initially have \(0.2347 \text{mol}\) of aluminum nitrate. Subtract this from the required number of moles to find the additional amount needed: \(\text{additional moles} = 2.125 \text{mol} - 0.2347 \text{mol} = 1.890 \text{mol}\) Now, convert the moles to grams using the molar mass of aluminum nitrate: \(\text{additional mass} = \text{additional moles} \times \text{molar mass} = 1.890 \text{mol} \times 213.00 \text{g/mol} = 402.57 \text{g}\)

(d) Dilution calculation

We need to prepare \(50 \mathrm{mL}\) of a \(0.450 \mathrm{M}\) aluminum nitrate solution using the prepared \(0.8500 \mathrm{M}\) solution. Apply the dilution formula: \(\text{M}_{1}\text{V}_{1} = \text{M}_{2}\text{V}_{2}\) Converting \(50 \text{mL}\) to liters, we get \(0.0500 \text{L}\). We want to know the volume (\(\text{V}_{1}\)) of the prepared solution (\(0.8500 \mathrm{M}\)) that we need to dilute to obtain \(0.4500 \mathrm{M}\). Rearrange the formula and solve for \(\text{V}_{1}\): \(\text{V}_{1} = \frac{\text{M}_{2}\text{V}_{2}}{\text{M}_{1}} = \frac{0.450 \mathrm{M} \times 0.0500 \text{L}}{0.8500 \mathrm{M}}\) \(\text{V}_{1} = 0.02647 \text{L}\) To prepare the required solution, dilute \(0.02647 \text{L}\) (or \(26.47 \text{mL})\) of the \(0.8500 \mathrm{M}\) aluminum nitrate solution to a final volume of \(50 \text{mL}\).