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 MLS of solution. All right, even though they don't directly say it, they want you to calculate the molarity of ions, so they want you to find ionic molarity or osmolarity. 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 change the 500 milliliters into liters. So remember 1 milli is 10 to the -3 liters, so that comes out to 0.500 liters. So now we have 0.500 liters here on the bottom and we have to find moles of chloride ions. So we're going to take the 58.1 grams of aluminum chloride and we're going to convert those grams of aluminum chloride into moles of aluminum chloride, one mole of aluminum chloride.
What is the mass of it? Well, we look on the periodic table, it's composed of 1 aluminum and three chlorines. According to the periodic table, the atomic mass of aluminum is 26.98 grams and that of chlorine is 35.45 grams. Multiply these numbers and when we add them together, we get 133.33 grams. That represents the mass of aluminum chloride, so grams of aluminum chloride cancel out.
Then we're going to convert moles of aluminum chloride, so one mole of aluminum chloride as within the formula, three chlorines. So that would be 3 moles of chloride ions. So when we multiply everything on top, divide by what's on the bottom, we get 1.3073 moles of chloride ion. Take those moles, plug them up here, and then when we divide we get 2.61 molarity and that would be the molarity of our chloride ions.
Now within our choices here, this has three sigfix. Here this has only one sig figs. But here we're just going to say 2.61 molar because just saying 3 molar I feel is too much of A rounding decision. So we're going to say 2.61 is the molarity of our chloride.