 Okay, so we're ready for another little presentation of sample problems. This one being working with molarity. It follows up on our mole video. Molarity is used to express the concentration of a solution in terms of moles of solute per liter of solvent. And for us, that solvent is usually going to be water and will refer to them as an aqueous solution, meaning made up in water. Now, there are other solvents that can be used for solutions, alcohols, a whole host of things. But for the sample problems that we're working, we're going to assume that these are aqueous solutions made up in water. So molarity is expressed in terms of moles per liter. So here we have one molar is equal to one mole per liter. So if we have one mole of an atom or molecule dissolved in one liter of solvent, then that is a one molar solution. And we can reviate this as one capital M is equal to one small case, mol over l per liter. So the big M stands for molar, meaning a concentration, and the small letter mol stands for moles, meaning based on how many moles the molecular weight that we've already worked on. So let's start working some samples to see how this works. So the question is, how would you prepare one liter of a one molar aqueous solution of potassium nitrate in the form of being KNO3? So here we go. So the first thing is to list the elements in the molecule and put out here how many of each we have. So here we're going to try to go to our periodic table and find potassium, which is 40.078 grams per mole. And oxygen nitrogen, which is 14.007 grams per mole. And oxygen, which is 15.999 grams per mole. So let's write these down. So the potassium was 40.078 grams. And the oxygen was 14.007 grams. And the oxygen was 15.999 grams. And the oxygen here is an example of 15.999. Some periodic tables using two places will round this off to 16.0. Some of the tutorials you'll be working with have oxygen as 16. And nitrogen as 14. Those numbers are fine. I'm going to be using, in this example, the three-digit numbers because that's how our little interactive periodic table comes up with them. But if you're using a periodic table rounded off, that's fine. So, and our third thing is to carry over how many of each we have because we're going to multiply by the number of atoms of each element. To come up in this case with 40.078, 14.007, and then 47.997. And for number four down here, we're going to sum those up to 102.082 grams per mole. So, that gives us the molecular weight of potassium nitrate in grams per mole. So, if one molar is equal to one mole per liter, then that's equal to 102.082 grams per liter. So, we will dissolve 102.082 grams of potassium nitrate in water to bring to volume liter. All right, so we'll dissolve it and then bring it to volume. So, for that question, how we prepare one liter, we weigh out 102.082 grams of potassium nitrate, dissolve it in water and bring it to volume of one liter. So, question two, how would you prepare one liter of three molar citric acid? The formula for citric acid is C6H8O7. So, first thing, carbon, hydrogen, oxygen. We have six carbons, eight hydrogens, and seven oxygens. So, periodic table. Well, oxygen, we're going to need that, so that's 15.999. Hydrogen is 1.0079. I'm going to round that up to 1.008. Our friend carbon is 12.011. So, let's put those values in. So, we just say carbon was 12.011. Hydrogen was 1.008. And oxygen is 15.999. All right, so then we're going to time some, how many we have? We've got six of those, eight of those, and seven of those. All right, so, I think we're in grams. Put my unit in. And so, we multiply those. So, for carbon, we get 72.066 grams. For hydrogen, we get 8.064 grams. And for oxygen, we get 111.993 grams. So, we sum those up, and we get 192.123 grams per mole of citric acid. All right, so, our question was, how do we make a one liter of three molars? So, if we have 192.123 grams per one liter equals one molar. So, we multiply that times three. 192.123 times three is equal to 576.369 grams per one liter is equal to three molar. All right, so, three molar. All right, so, three times one molar is equal to three molar. All right, so, if it takes 192.123 grams per liter for one molar, then it would take three times that amount in one liter for a three molar solution. All right, so, let's work on a variation on that, which asks us how do we prepare 500 milliliters of a three molar aqueous solution of sodium carbonate? Molecular formula NA2CO3. So, we list our elements, and we have two of those, one of those, and three oxygens. So, back to our periodic table. Oh, there's carbon when you use that, so it's 12.011, and we need sodium, which is 22.990, and then oxygen, which we've used before, which is 15.999. So, let's get those values in, and 15.999 grams. All right, we'll get our multipliers. There's two of these, one of those, and three of those. So, if we do the multiplication, 22.990 times two is 45.980, 12.011 times one stays 12.011, and then 15.999 grams times three is 47.97 grams. So, if we add up the individual elements, we get a molecular mass of 105.998 grams per mole. All right, so, if one molar equals one mole per liter, three molar is equal to three moles per liter. So, we're wanting a three molar solution. Now, if there's three moles per liter, we're only wanting to make 500 mills. So, how many moles are there in 500 mills? So, let's change our liter to mill. So, we've got three moles for 1000 milliliters is equal to how many in 500 milliliters? So, we've got three moles times 500 mill is equal to question mark moles times 1000 ml. So, we can get rid of that 1000 ml by dividing both sides by 1000. So, now the mills will cancel out. We've got 500 over 1000, which is 0.5. So, we've got three moles times equal to our unknown moles. So, that comes out to be 1.5 moles. So, we're going to need 1.5 moles. So, if there's 105.998 grams per one mole, how many grams are there in 1.5? So, 105.998 times 1.5 moles is equal to our unknown grams times one mole, right? So, basically dividing both sides to get rid of that mole on the right side. Then the moles cancel out. We're dividing by one. And so, we basically can 105.998 times 1.5 equals the number of grams, our unknown number, right? Which that comes out to be 158.982 grams, right? So, we've dissolved 158.982 grams in water and bring to volume 500 ml. So, we've got 1.5 moles and 500 ml is 3 molar, right? Of 1.5 moles and 500 ml would be equal to 3 moles in one liter. So, we did that in our two proportions. First, we decided if we're only making 500 ml, right? A 3 molar solution is 3 moles per 1,000 ml, but we're only making 500. So, how many moles do we need in 500 ml? We decided that was 1.5 moles. And what is 1.5 moles of the sodium? Carbonate at 1.5 moles then is 158.982 grams. Okay, so this problem asks us how will we make 1.25 liters of a 600 ml aqueous solution of sodium bicarbonate? And we have then the formula for sodium bicarbonate is Na8CO3. So, you look at these problems and, you know, we're going to be working with moles, so we're going to be wanting to know the molecular weight of sodium bicarbonate. And then you look at the question what is your volume, in this case 1.25 liters, and what is your concentration? 600 ml. So, those are the things you're looking for is how to get 1.25 liters at a concentration of 600 ml. So, first let's work on our sodium bicarbonate. So, we've got sodium, hydrogen, carbon, and oxygen. So, we've got 1 and 1 and 1 and 3. So, hopefully you've seen the periodic table often enough. You can find these sodium is 22.990. Hydrogen we decided was 1.008. Carbon we found as 12.011 grams. And oxygen 15.999 grams. So, times 1, times 1, times 1, times 3. So, that was going to say 22.90 grams. 1.008 grams. 12.011 grams. And 15.99 times 3 is 47.997 grams. So, we add those and we get 84.006 grams per mole of sodium bicarbonate. So, if 1 molar equals 1 mole per liter, 600 ml molar equals 600 ml per liter. So, when we're expressing molarity, it's always going to be expressed in terms of liter. So, if it's 1 molar, it's 1 mole per liter. If it's 600 ml, it's 600 ml per liter. If it's 5 molar, it's 5 moles per liter. So, when we see this 600 ml, we can just say that 600 ml per liter. So, if we want 600 ml per liter, and we're wanting to make 1.25 liters, how many millimoles do we need? Here's our cross-multiplying divide, or cross-multiplying. So, we get 600 millimoles times 1.25 liter is equal to our unknown millimoles times 1 liter. So, we'll get rid of that 1 liter by dividing both sides. So, we get 600 millimoles 1.25 liter divided by 1 liter is equal to our unknown millimoles. You saw me add the e there. Either mol or mol e is okay for moles in a small case. So, when we get rid of the liters, we divide by 1, we're going to get 600 millimoles times 1.25 is equal to our unknown millimoles. So, 600 times 1.25 is equal to 750 millimoles. So, we're going to want our 750 moles in our 1.25 mil. So, how do we figure out what our weight, how much we have to weigh out to get 750 millimoles? So, if we know back from up here, we get 84.006 grams per mole. So, 84.006 grams per 1 mole is equal to how many grams per 750 millimoles? Okay, so we need to convert this mole again to 1 mole equals 1,000 millimoles. So, we got 84.006 grams over 1,000 millimoles is equal to how many grams over 750 millimoles? Okay, so here's our cross-multiplying on our proportions. So, 84.006 grams times 750 millimoles is equal to our unknown grams times 1,000 millimoles. Okay, so we're going to divide both sides. We'll get rid of the grams there. So, we'll get rid of that into about 750 by 1,000. Then, we're going to get 84.006 grams times 0.75. We're going to get our unknown grams. So, when you multiply that, you get 63.005 grams is how much we need to give us the 750 millimole. So, we're going to dissolve 63.005 grams in water and bring to volume 1,250 mL, which is 1.25 liters. So, that's going to give us 1.25 liters at 600 millimoles. So, if we wanted 600 millimoles per liter to give us 1.25 liters, it would take us 755 millimoles. So, if there's 84.006 grams per mole, if we multiply that times 0.75, that means there's going to be 63.005 grams per 750 millimoles. And putting that into 1.25 liters makes the concentration 600 millimolar. Well, I hope these little sample problems help you in seeing the process you go through to work these problems. I know it takes a little getting used to about moles and millimoles and grams, but as I said in the earlier proportions videos, it's just proportions. You just have to start thinking in terms of that. So, there's a pretty systematic way of going about solving these problems. I say we have some interactive tutorials for you to work through some problems. And then in the text, there are a number of sample problems. So just work some problems so you get comfortable with them. Like I say, I hope this helps. And I say you have on the blackboard in this module, you have the interactive periodic table. So you can keep that, just kind of minimize in your tray and keep calling it up to get your values. Or if you have a paper periodic table, you can use that. There's also in the back of the textbook, there's a list of the atomic weights of the elements. So work some of these problems and have fun solving them. Thanks.