A gas mixture contains 75.2% nitrogen and 24.8% krypton by mass. What is the partial pressure of krypton in the mixture if the total pressure is 745 mmHg?

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Hi, everyone. Welcome back. Our next question says a gas mixture contains 75.2 per cent nitrogen and 24.8 per cent krypton by mass. What is the partial pressure of krypton in admixture? The total pressure is 745 millimeters of mercury. A 67 millimeters H GB 74 millimeters HGC 55 millimeters HG or D 47 millimeters HG. So let's think about what we need here. We have a gas mixture, we're given the total pressure and we know that the total pressure of a gas mixture. So P subscript total, which in this case, we know is 745 millimeters of mercury is equal to the sum of the gas pressures of each individual gas in the mixture. So it would be the sum of the pressure of the nitrogen plus the pressure of the krypton. So in this case, it will be P N plus PK. Now, we have this percent telling us what percent of our mixture is nitrogen and what percent is Krypton. However, note that these are percents by mass. And when we think about gas mixtures, the mass of gas isn't the key quantity here, it's the number of particles because each of those gas molecules is so far apart the space they take up, it really doesn't matter the different mass. It matters how many particles, that's what determines the volume of our gas mixture. So we need to go from mass to particles. And of course, we use moles to do that. So what we do, we don't have a specific amount of mass we just have. So we'll invent a theoretical sample of this particular mixture. That's a 100 g samples, 100 g of gas. And if we have 100 g of gas total, well, we have mass percent, it would have 75.2 g of nitrogen since we know it should be 75.2% by mass. Now important to recall here when we're talking about nitrogen gas, then it exists as a diatomic molecule N two. So that's going to be important since we're using molar mass, we need to remember we have two nitrogen atoms in each molecule. Then that sample will also have 24.8 g of krypton. Now krypton remembers a noble gas. So it does not exist as a diatomic mole molecule. It's just by itself. So now if we imagine this theoretical 100 g sample, we know how many grams of each we have, we can now convert them to moles. Again, it doesn't matter. This is a different size sample. Than the one we're being asked about when we look at mole ratios, that's going to be the same no matter how, how much we have to start with. So let's convert our nitrogens. We have 75.2 g of N two multiply it by our conversion factor. Again, note that N two will mean it's double the atomic mass of nitrogen. And when we look that up, we're going to get that, that at molecular mass is 28 0.02 g of N two on the bottom, said the grams cancel out. And then on the numerator, we have one mole of N two. So we do that calculation. That's going to give us two 0.6838 moles of N two. It's actually approximately equal. I've left off a few decimal points. We're not going to round off to our significant figures until the end. But we know looking at our numbers that we're going to have a smaller number of significant figures. So I'm not going to carry out all the long six decimal places or whatever, just carry it out a little further. So then we just need to do the same thing with krypton. So we have 24.8 g of krypton multiplied by now, we could just look up simply our atomic mass here, which would be 83 0.80 g of crypto per one mole of crypto. So that's going to give me 0.29 59 moles of krypton. So now to look at the partial pressure of krypton, which is what I'm looking for, scroll up just a bit will say the pressure of krypton, it's going to equal to. And then we would say the mall fraction of krypton multiplied by the total pressure. So that will equal. Well, how do we get that mole fraction when we say we know we have in our theoretical sample, 0.2959 moles of krypton and we divide that by. So I'm just gonna leave it as moles, the total number of moles in the entire sample. So that would be 0.2959 moles. That's for our krypton and 2.6838 moles, firm moles of nitrogen gas. So that's our mole fraction multiplied by 745 millimeters of mercury roll up a little bit. And when we do that math in the parentheses, we have 0.2959 moles. And that would be divided by 2.9792 multiplied by 745 millimeters of mercury. And when I do that math, I end up so do that division multiplied by 745. And then I round it off to three significant figures. When I look at the significant figures I have here, I think it's 74.0 millimeters of mercury. So there's my final answer. I look at my multiple choice questions and I see that choice B is 74.0 millimeters of mercury. So there we go, we have our final answer again based on com coming up with a theoretical mass of the gas using my percents by mass to convert to a mole fraction. And then using that to calculate the partial pressure of the Krypton gas in this mixture. See you in the next video.

A gas mixture contains 75.2% nitrogen and 24.8% krypton by mass. What is the partial pressure of krypton in the mixture if the total pressure is 745 mmHg?

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Key Concepts

Dalton's Law of Partial Pressures

Mole Fraction

Ideal Gas Law