In this example question, it says, calculate the molarity of chloride ions when dissolving 58.1 grams of aluminum chloride in enough water to make 500 mL of solution. Although they don't directly state it, they want you to calculate the molarity of ions. Here, molarity will equal the moles of chloride ions divided by liters of solution. We already have the volume of our solution; we just have to convert the 500 milliliters into liters. Remember, 1 milli is 10-3 liters, so that comes out to 0.500 liters.

Now, we have 0.500 liters as the denominator, and we need to find the moles of chloride ions. We will take the 58.1 grams of aluminum chloride and convert those grams into moles of aluminum chloride. One mole of aluminum chloride weighs, according to the periodic table, the sum of the atomic masses of 1 aluminum (26.98 grams) and 3 chlorine (35.45 grams each). When we add them up, we get 133.33 grams. This represents the mass of one mole of aluminum chloride.

With the grams of aluminum chloride, they cancel each other out in our conversion. Converting moles of aluminum chloride, note that one mole of aluminum chloride contains within the formula 3 moles of chloride ions. Thus, when we do the math on top, and divide by what's on the bottom, we get 1.3073 moles of chloride ions.

Plugging these moles into the molarity formula, dividing them gives us a molarity of 2.61. This value is the molarity of the chloride ions in the solution. In our considerations here, while the original values have different significant figures, it's more precise to represent the molarity as 2.61 molar. Using a rounded value of 3 molar might be too imprecise given the specific calculations involved.