Here is another example calculation for you to try. Solutions of acids and bases, are often sold in mass percent, sometimes that's called weight percent. For example a sodium hydroxide solution, might be advertised as being 50 weight percent. What is the sodium hydroxide molarity, if the density of the solution is 1.57 grams per mL? What do we need to do this calculation? Well first we know the definition of of molarity, it's moles of solute per liter of solution, in order to do this calculation, the tricky step is that we need to assume a certain volume of the solution. I'm going to assume that there is 100 mL of solution, because that's a convenient quantity. So, if I have 100 mL of solution, and there's 1.57 grams per one mL, then in this solution, I have 157 grams of solution, because my millilitres cancelled out. So now I know the mass of solution don't I? Remember the mass of solution, is the mass of the solute and the solvent. I haven't answered the question yet, have I? What is the sodium hydroxide molarity? Well I know that half of the solution, is the solute because it's a 50 weight percent solution. So if the total solution mass is 157 grams, half of that solution or 0.5 of the solution is Sodium Hydroxide. So the mass of sodium hydroxide must be 78.5 grams. I can convert a mass of sodium hydroxide into moles using the molar mass, of sodium hydroxide, which is 40 grams per mole. So with that information I can have my grams cancel, and figure out that there's 1.96 moles of sodium hydroxide in the solution. Remember my solution was 100 mL, that was the arbitrary volume that I chose. So, to calculate the molarity of the solution, I need to take the number moles of solute which I just calculated, and divided that by the volume of the solution, which here is 100 millimeters, I can covert that to liters, it's 0.100 liters. And that gives us a molar concentration of the solution, of 19.6 more. With such a concentrated solution, I'd probably would want to do some dilutions, in order to use that solution in the lab. Dilutions are governed by this equation, the product of the starting concentration, times the starting volume. Equals the product of the ending concentration, times the ending volume. So what I'm doing here is changing the volume of the solution, and determining the molar concentration at those two different volumes. And the way I'm changing the volume is by adding water, so I'm diluting the solution by adding water. The best way to practice with this, is with an example. Here's a review example problem. Concentrated phosphoric acid is 75% by mass, and has a density of 1.57 grams per milliliter. First, what is the molarity of concentrated phosphoric acid? And secondly, how many millileters of the concentrated acid would be requires, to prepare 1.5 liters of a 0.2 molar solution, of phosphoric acid? So there's two questions here. There are some equations, we need to answer these questions. One of those equations relates the mass percent, to the mass of the solute, and the mass of the solution. Another equation gives us the definition of molarity. And the third equation, is the one that we just looked at for diluting solutions. Let's start with the first question. What is the molarity of the concentrated phosphoric acid? Again, to do these conversions between concentration units, the best thing to do is assume an amount and you're often not told to do that, in the problem. And sometimes that's a stumbling block for students. So I'm going to assume that I have 100 grams of the solution, because it's 75% solute, I therefore know this is representing the 75%, that I have 75 grams of H3PO4 and 25 grams of water in this solution. I can convert the 75 grams of H3PO4 into moles, using the molar math of phosphoric acid, which is 97.97 grams per mole. In order to get the units to cancel here, I need to do a division, and that gives me an answer of 0.765 moles of H3PO4. Finally, if I have 100 grams of solution, I need to know what the volume of that is. Because molarity is moles per unit volume isn't it? So since I know the density of the solution, I can take the mass of the solution and divide it by the density, which causes the grams to cancel out and gives me milliliters for my answer, and I see that this particular sample of solution has a volume of 63.69 mL. What's the definition of molarity again? Moles of solute per liter of solution. Well I know the number of moles of solute here, I know the volume of the solution in liters if I just do the conversion up here to liters, which I know you can do easily at this point. So I can calculate that this particular solution, has a molar concentration of 11.87, express the three significant figures is 11.9 Molar. Now, we're ready to answer the second question, that was asked on the previous slide. How many milliliters of the concentrated acid solution would be required to prepare, 1.5 liters of a 0.2 molar solution of phosphoric acid? Well the initial solution we had was quite concentrated, wasn't it? It was 11.9 molar. But we know that we have this equation, C1V1 equals C2V2. If we know that the initial concentration is very high, we can solve that for the initial volume that we need of the stock solution, the more concentrated solution. And it's going to be C2V2 divided by C1, so, let's plug numbers into that expression now. So the volume that we need of the stock solution here, is going to be 25 mL. And what I did to calculate that is, I know the final molarity that I would like, that was in the problem, and I know the final volume I would like, that is in the problem so, all the final information was in the problem. And then, I divided that by the initial molar concentration, which I had calculated on the previous slide. My units cancel out, molarities cancel out, and that leaves me with liters, which I can convert to milliliters quite easily. Here's one for you to try. To what volume should a 12 mL, solution that is 18 molar sulphuric acid. Be diluted to prepare a titration solution, that has a concentration of 0.5 molar, take a few minutes to do some calculation, and please try to answer that now. Thank you for submitting your answer, of course to solve this problem we need the equation for diluting solutions. In this particular case, the volume is 12 mL and the molar concentration initially is 18 molar. We don't know what volume we need, because that's the unknown, but we know that we want a final solution that is 0.5 molar. So we're just going to plug in numbers and do a little algebra to calculate that this answer here, the concentration of the final solution, needs to be 432 mL. How many mL of water do I need to have, so that I can add my 12 mL of acid stock solution to it? Well, the total volume of the solution needs to be 432 mL. But some of that solution volume is going, to come from the stock solution. The more concentrated solution that I started with. In fact I calculated that I'm going to use 12 mL of that solution. Therefore I needed to have 420 mL of water. This concludes our discussion, of solution calculations and dilutions. This is also the last lecture of this course. I want to thank you for watching all of these lectures, and if you have participated in the quizzes, the essays, or the forums, I would like to thank you for that as well. I've had a wonderful time making these videos, and interacting with you on the forums. Finally there's one very important person to this course, who I would like to thank and that's Phun Li. Not only has she edited all of these videos, but she has also coded all of the weekly quizzes, she has helped you troubleshoot technical issues on the forum, and generally she has been extremely efficient and always positive in her work. So thank you Kun Li.