Final answer:
To prepare a 5.0 M solution of aluminum oxide in a 6.5 L bottle, 3313.8 grams of aluminum oxide are required, based on the molarity and volume provided.
Step-by-step explanation:
To prepare a 5.0 m solution of aluminum oxide (Al2O3), you need to know the molar mass of Al2O3 and then calculate the mass required for the desired molarity and volume. The molar mass of Al2O3 is approximately 101.96 g/mol. Since molarity (M) is defined as the number of moles of solute per liter of solution, a 5.0 M solution contains 5.0 moles of solute per liter.
First, calculate the total number of moles needed for the 6.5 L volume:
5.0 moles/L × 6.5 L = 32.5 moles of Al2O3
Next, convert the moles of solute into grams:
32.5 moles × 101.96 g/mol = 3313.8 g of Al2O3
Therefore, you will need to mass 3313.8 grams of aluminum oxide to prepare the solution.