 So let's go ahead and do this problem. It's calculating the molar mass of an unknown gas. So it says, if 2.56 grams of an unknown gas occupies a volume of 111 milliliters at 753 torr and 22 degrees Celsius, calculate the molar mass of the unknown gas. So I've written down all the information the problem gives us. Remember, we're going to, since this is a gas law problem, we're thinking PV equals nRT. So in order to do that, we want to remember what R is. 0.0821 liter ATM mole Kelvin. Of course, that's going to be given to you. And we're going to have to convert these units into units that we work with in the ideal gas law problem. So in order to do that, we have to remember that 760 torr equals 1 ATM. So well, we can do the rest pretty easily. 0.991, a 273 to this one. And that's now in Kelvin. So we've got this in liters, this in ATM, this in Kelvin. We've got R. But that's going to give us moles. So we don't want moles. We want molar mass. So what do we know? We know that if we take the molar mass like that, that equals the number of moles. So the mass divided by the molar mass. Can you set with that? Do you remember that? From way back then. So when we have this equivalency, we could just take this and put it into substitute for that n there. So now let's write that equation substituting that for n. So we've got PV equals, instead of n, we've got mass divided by molar mass like that times RT. So let's just look at this really quick. So we've got PVM molar mass RT. Look what we've got over here. Mass volume, so V, P, T, and R. The only thing we're missing is molar mass, right? So now we can just rearrange this, isolate that molar mass variable, and then solve for it. Does that make sense? Yeah. OK, good. I like your excitement on that one. OK, so do you figure it out how to do it now? OK, cool. So MRT divided by the molar mass like that. So I'm just going to rearrange this in one step. So we've got the molar mass is going to equal MRT divided by PV like that. Does that make sense? Yeah. OK. So now it's just a polygon chosen problem. So I'm actually going to erase all of that stuff that we've got there. We need, well, the mass, 2.56 grams times R, 0.0821 liter ATM per one mole temperature. So we've got the whole thing by pressure in one ATM like that. And multiply that by volume 0.1114 like that. Does that make sense? Yeah. OK, so now let's cancel and see what we've got left. So leaders cancels with leaders, ATM cancels with ATM. Kelvin cancels with Kelvin. So what do we have left? Grams divided by moles. That's good molar mass, right? So now all we do is use the calculator. 1, 5, 6, 2, 1, and then multiply that by 295. And I get an answer of 2,366, 564 grams per mole. So that seems like a rather high molar mass to me for gas, but that's the way it works. And that's the answer to that problem. You have a question? Yeah. Well, I guess we could do it that way. I don't know why I was getting this one, but I was also getting that one, I'm sure. Oh, that's the way to do it, OK? I got something, you know, something exactly. If I got 5, 6, 6, then I know. You must have rounded differently, OK? But can we also find out the number of moles by the PBRT? You could figure that out first and then divide. The number of moles over a mass. Yeah. So I did it all at once here by rearranging the equation. OK, but you could do it that way. I don't know why my math was off, though. I guess it's always a way of putting it in. Yeah, so we can try doing it that way if you want to. You got the right answer. Yeah, sure. I mean, yeah. Any other questions? OK, cool.