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The Henry's law constant for $\mathrm{O}_{2}$ is $1.3 \times 10^{-3} \mathrm{Matm}$ at $25^{\circ} \mathrm{C}$ . What mass of oxygen would be dissolved in a 40 -L aquarium at $25^{\circ} \mathrm{C},$ assuming an atmospheric pressure of 1.00 $\mathrm{atm}$ , and that the partial pressure of $\mathrm{O}_{2}$ is 0.21 atm?

0.35 $\mathrm{g}$

Chemistry 102

Chapter 11

Solutions and Colloids

Solutions

Rice University

University of Kentucky

Brown University

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Hello, everyone, this is Ricky and today we're working on problem number 24 and we have to solve for the massive auction oxygen dissolved in a 40 leader aquarium. So we can do this. Using Henry's law. Where we have the concentration of a gas is equal to a proportionality constant, constant times wth e partial pressure of the given gas. All right, so our were given in the problem that kay, our constant for oxygen, is 1.3 times 10 to be negative. Third Moeller atmosphere. Okay. And so then we have oxygen. Concentration of oxygen is equal to 1.3 times 10 to the negative. Third Moeller times a. T M. Moeller divided by PM times the partial pressure of oxygen which is 0.2 one atmospheres. And so we get that key concentration of oxygen gases 2.7 times 10 with a four Moeller. And now what we can do with this is calculate the number of moles of oxygen. So moles of 02 is equal to our concentration of 02 for more or loans. Leader of oxygen, I'm done. This is just some dimensional analysis, so we know about their 40 meters in the pink 40 liters bang bank. Knock that out and we get that There are 1.0 age times 10 to the negative second moles oxygen total and the tank. And lastly, to get the mass of oxygen, we have to multiply our total moles times the Moler Mass. Which for oxygen gas is about 32 grams for one more. It's a letter Mills multiply through and we get that the massive oxygen and detained point through five grams. Hope this video was helpful. I'll see you in the next.

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