How many moles and number of ions of each type are present in the following aqueous solution?

113 mL of .490 M aluminum chloride.

moles AlCl3 = M x L = ?

moles Al = the same since I see one Al atom in one molecule AlCl3.
moles Cl = 3x? since there are 3 Cl atoms in one molecule of AlCl3.

To determine the number of moles and number of ions in the given solution, we need to use the given molarity (concentration) of aluminum chloride and the volume of the solution. Aluminum chloride (AlCl3) dissociates into aluminum ions (Al3+) and chloride ions (Cl-) in aqueous solutions.

First, let's calculate the number of moles in the solution:

Step 1: Convert the given volume from milliliters (mL) to liters (L).
113 mL = 113/1000 L = 0.113 L

Step 2: Multiply the molarity by the volume to get the number of moles.
Moles = Molarity x Volume
Moles = 0.490 mol/L x 0.113 L
Moles = 0.05537 mol

Therefore, the solution contains 0.05537 moles of aluminum chloride.

Next, let's calculate the number of ions present in the solution:

Each formula unit of aluminum chloride, AlCl3, dissociates into one aluminum ion (Al3+) and three chloride ions (Cl-) in solution.

Number of aluminum ions (Al3+):
From the previous calculation, we have 0.05537 moles of aluminum chloride.
Since each formula unit of aluminum chloride contains one aluminum ion, the number of aluminum ions is also 0.05537 moles.

Number of chloride ions (Cl-):
Since there are three chloride ions in each formula unit of aluminum chloride, we can multiply the number of moles of aluminum chloride by three to get the number of chloride ions.
Number of chloride ions = 0.05537 mol x 3
Number of chloride ions = 0.16611 mol

Therefore, in the given 113 mL of 0.490 M aluminum chloride solution, there are approximately:
- 0.05537 moles of aluminum chloride
- 0.05537 moles of aluminum ions (Al3+)
- 0.16611 moles of chloride ions (Cl-)