Answer:The mass of aluminum sulfate required to produce 455 mL of 0.310 mol/L of aluminum ions is approximately 94.7 grams.
Step-by-step explanation: To determine the mass of aluminum sulfate that is required to produce 455 mL of 0.310 mol/L of aluminum ions, you will need to first calculate the number of moles of aluminum ions that are required. You can use the following steps:
Convert the volume of the solution to liters. Since there are 1000 mL in 1 L, the volume in liters is:
volume (L) = 455 mL / 1000 mL/L = 0.455 L
Calculate the number of moles of aluminum ions. Using the concentration of the solution in mol/L, the number of moles of aluminum ions is:
moles (mol) = concentration (mol/L) x volume (L)
= 0.310 mol/L x 0.455 L = 0.139 mol
Calculate the mass of aluminum sulfate required to produce this many moles of aluminum ions. Aluminum sulfate contains two aluminum ions per molecule, so the number of moles of aluminum sulfate required is twice the number of moles of aluminum ions. The mass of aluminum sulfate required can then be calculated using the molar mass of aluminum sulfate:
mass (g) = moles (mol) x molar mass (g/mol)
= (0.139 mol x 342.1 g/mol) / 2
= 94.7 g
Therefore, the mass of aluminum sulfate required to produce 455 mL of 0.310 mol/L of aluminum ions is approximately 94.7 grams.
Keep in mind that this is the mass of aluminum sulfate required to produce the desired concentration of aluminum ions in the solution, and that the actual mass of the solution may be larger due to the presence of water or other dissolved substances.