 How many mills of a 0.20 molar glucose solution are needed to provide a total of 25.0 grams of glucose? You guys think you can do this one on your own, hopefully? So what does it show us? It says the concentration of the solution, right? So the molarity, and I'm just going to put C6H12O6. What is that molarity? 0.20 molar, okay? So that's 2-0 molar glucose. So what does that mean? That means 0.20, good job guys. Mol's up. Very good guys. Glucose per one liter of solution, right? S-O-L-N solution. Okay, so how many mills are needed to provide a total of 25 grams of glucose? Well, the first thing we're going to have to do is figure out how many grams of glucose we've got and one liter. So let's just figure that out. The mass of glucose, I'm going to keep all these six things for now. How about this? What's the units of this? What else? Is that it? Per liter. If you guys are not doing your units correctly, then you're going to get them wrong. And that's per liter of solution. So that's not 25 grams, that's more than 25 grams. So what we're looking for is the volume. That'll give us 25.0 grams. So what do we know? We know one liter of solution gives us 36.032 or 312 grams of glucose. But we're looking for how many grams? 25.0 grams. How did I know to do it that way? Because I wanted to cancel my units. So 25 divided by 36.03 and I get 0.693884. But that's liters, right? Not mills. So one liter per every thousand mills. Cancel, cancel. So I just take that, multiply it by a thousand, and of course you get 690, 694, but 6.9 times 10 squared. And why did I only do it for 268? It's pretty logical how you do it, right? So again, you know, how can I do these types of problems without ever seeing them before in my life? It's because I know the process of doing them. It's all the matter of a process of canceling units, okay? And using those conversion factors that you know. And if you guys can do that, then you should be good. Are there any questions on this one before I kill it? Okay, good job, guys.