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Problem 2 Easy Difficulty

Figure $\mathrm{P} 11.2$ shows a particle-level vicw of a scaled container partially filled with a solution of two miscible liquids: X (blue spheres) and $Y$ (red spheres). Which of the following statements about substances $\mathrm{X}$ and $\mathrm{Y}$ are true?
a. Y is the solvent in this solution.
b. Pure $Y$ has a higher vapor pressure than pure X.
c. The presence of $Y$ in the solution lowers the vapor pressure of $\mathrm{X}$.
d. If $Y$ were not present, there would be fewer total particles in the gas above the liquid solution.

Answer

B and D are true. A and C are false.

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Video Transcript

this is a problem. 11.2 at the end of chapter 11 in the chemistry is the science and contacts textbook. Okay? And so it asks. It gives us, um, some statements, and it wants us to know whether they're true or false based on this image. Okay, Um and so the first question is, why is the solvent so molecule? Why, which is a red molecule is assault into can remember. The solvent always dissolves the soul ute. So in the case of water and salt, water will be the solvent dissolving the soul. Youth being salt. Okay, um and so a solvent is usually the one with much more molecules. And so in this case, why is there's less why in the liquid phase and there are X so X is actually the solvent. And why is this all you This is false. Okay? It says be pure. Why has greater vapor pressure than X? Okay. And so if we were tohave the equation for the vapor pressure of solutions, remember, when you mix solutions, it follows Raldes law where the paper pressure of you know, a solution. And in this case, volatile solution is always the additive infractions of both volatile solutions. Remember these air Both volatile solutions, since both molecules go into the ass face, okay. And so we have p of solution equals the mole fraction of one, um, molecule right on its vapor pressure. It's standard vapor pressure on plus the same thing for the other solvent. Okay. And this can keep going if you have multiple volatile solutions in a mixture. Okay, on dso pure. Why, in this case, we have less a smaller mole fraction of pure Why, Yet we have equal amounts of, um, molecules in the gas face. So that means that this is true. So he is true. So let me write that over a year. Okay? B is true because we have a smaller mole fraction of pure why? Yet we still we have the equal amounts of pure home. Why? Molecules in the gas space. Okay, so that means that ax requires a higher mole fraction to get to that certain vapor pressure. Ok, um, so the standard vapor pressure of why is greater than the standard vapor pressure of X, you know, And that's why they asked for pure right. Okay, first see, it says Why lowers the vapor pressure of acts? Okay, um and so a lot of students will think this is true, because when you think about, um dissolving a solvent solute into assault vent, usually the vapour pressure is lowered. Right? So when we dissolve salt into water, we're lowering. OK, But in the case of two volatile solutions, what actually happens is that the vapor pressure of both will meet in the middle. Okay, so it's kind of it's gonna kind of average the vapor pressure of pure accident pure. Why into until me in the middle of the solution. Okay. And so remember, we already just We just stated that your wife has a greater vapor pressure than acts. So if we were to average accent, why, it would be somewhere in the middle, which is actually higher than the vapour pressure, that is, it's false. Okay, So when there are too volatile solutions being mixed, the, um, basil this be molecule that has a lower vapor pressure will actually get enhanced. Its be professional get enhanced. It'll go up closer to the vapor pressure of why, okay, And then D says without why there would be less gas particles, Okay. And so this this is true. And it's the same reason. As for C. Wright, with two volatile solutions, you would have an average of vapor pressure. And so, in the case of acts, the vapour pressure is increased slightly due to the high vapor pressure of while. Okay, so if we removed, why, then we would have the original. It would be pure acts, and we would have the vapour pressure of pure acts. Okay. And so without why, we would basically eliminate all of these acts particles here. Okay. Our sorry. All of these y particles and so in the solution would also eliminate them. Okay, um and so we would be left with three molecules of acts and solution. OK, um, and that would be it. And so there would be less gas particles in the, um, solution if why was not present? Okay, assuming that we're we're left with the exact same molecules. Number of molecules of acts in the beginning. Obviously, if we were to compensate for the loss of lie by adding more Axton, that would increase the amount of gas particles, um, in this figure, but, um, we're making the assumption that were just eliminating all of the wise. And so that would just decrease the amount of gas particles. Okay? And so that's how you solve 11 point.

University of Maryland - University College
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