Osmolarity - Online Tutor, Practice Problems & Exam Prep

Osmolarity

Osmolarity, sometimes referred to as ionic molarity, represents the number of moles of ions per liter of solutions. So, it sounds kind of familiar. We know that molarity represents moles of solute per liters of solution. Now we're looking at solutes that have charges because now we're looking at ions. And we're going to say when it comes to osmolarity, there are a few different methods that we can employ to answer particular questions. Now in method 1, we have direct calculation of osmolarity. In this first method, we use the moles of ions and the liters of solution with its formula to calculate osmolarity. Now here, this is just simply osmolarity equals moles of ions over liters of solution. Again, pretty similar to molarity, which is just moles of solute over liters of solution. Now that we've gotten a basic understanding of what osmolarity represents, let's click on to the next video and start doing some example questions dealing with osmolarity under method 1.

Osmolarity = moles of ions liters of solution

Osmolarity Example 1

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.

Osmolarity

The second method involves determining the osmolarity from the molarity. Now we're going to say if the molarity of a compound is known, then the osmolarity for each of its ions can be determined by osmolarity equals the number of ions that are within that compound times the molarity of the compound overall. So this is yet another way that we can isolate osmolarity. Now that we've seen the second method, click on to the next video and let's take a look at a question where we're asked to calculate the osmolarity of a particular ion.

Osmolarity Example 2

Now remember, method 2, we can use osmolarity equals number of ions times the molarity of the compound. Here in this example, it says, what is the concentration of hydroxide ions in a 0.350 molar solution of Gallium Hydroxide, which has a formula of Ga(OH)₃. Alright. So we need to find the osmolarity of hydroxide ions.

The molarity of OH^- ions equals the number of hydroxide ions. If we take a look here at the formula, we see that there's a 3 here, meaning that there are 3 hydroxides within this parenthesis. So that would be 3 times the molarity of the original compound. Gallium hydroxide as a compound has a molarity of 0.350 molar. So now just do 3 times that value and when we do that we get 1.05 molar as the osmolarity of hydroxide ions. So it's as simple as that. If you know the molarity of the compound overall, use that to help you find the odd molarity of any one of its ions.

Osmolarity Example 3

Method 3 involves the number of ions from molarity. Now we're going to say here problems involving number of ions and molarity can use a given amount and converting factors to isolate an end amount. Now for those of you who haven't watched my videos on dimensional analysis and conversion factors, these terms basically form the foundation for a lot of the calculations you'll see in chemistry. Your conversion factor is basically just when you have 2 different units connected together. That is seen in molarity because molarity itself represents moles per liter, two units connected together. Our given amount is just a value that's given to us that has only 1 unit; that would be represented by these liters here. And then our end amount is just simply the value or the amount that we're trying to find at the end.

Now if we look at this question, remember when we have the word 'of' in between numbers, it means that we have to multiply them together. So, we're going to say here our given amount where we're starting is 0.120 liters and we're gonna have to utilize conversion factors to get to our end amount. Our end amount we're looking for is moles of calcium ions. Now we need to cancel out liters here. We're able to do that by utilizing the molarity given to us. The molarity is 0.450 molar, which means 0.450 moles of calcium phosphate per 1 liter of solution. So that's our first conversion factor. Then liters cancel out.

Now we have to just change moles of calcium phosphate into moles of just calcium ion. So here for every one mole of our entire compound, we can see that from the formula there's a little 3 there, which means that I have 3 moles of calcium ions. So then if we multiply everything out that's on top, I'll be left with 0.162 moles of calcium ion as my final answer.

Which of the following solutions will have the highest concentration of bromide ions?

• A. 0.10 M NaBr
• B. 0.10 M CaBr₂
• C. 0.10 M AlBr₃
• D. 0.05 M MnBr₄

How many milligrams of nitride ions are required to prepare 820 mL of 0.330 M Ba₃N₂ solution?

• A. 120 mg
• B. 320 mg
• C. 560 mg
• D. 760 mg
• E. 7600 mg

How many bromide ions are present in 65.5 mL of 0.210 M GaBr₃ solution?

• A. 7.95x10²² ions
• B. 2.48x10²² ions
• C. 7.95x10²³ ions
• D. 6.50x10²³ ions
• E. 3.91x10²³ ions