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If .200 M solution of KNO3 and .100 M solution of NH4Cl were seperated by a semipermeable membrane what would be obserced when osmosis stops? A. Some KNO3 moved thr...

Question

If .200 M solution of KNO3 and .100 M solution of NH4Cl were seperated by a semipermeable membrane what would be obserced when osmosis stops? A. Some KNO3 moved through a semipermeable membrane. B. Some NH4Cl moved through a semipermeable membrane. C. The concentration of NH4Cl increased. D. Solutions on both sides of a semipermeable membrane contained some NH4Cl and some KNO3. E. The concentration of KNO3 increased.

if .200 M solution of KNO3 and .100 M solution of NH4Cl were seperated by a semipermeable membrane what would be obserced when osmosis stops? A. Some KNO3 moved through a semipermeable membrane. B. Some NH4Cl moved through a semipermeable membrane. C. The concentration of NH4Cl increased. D. Solutions on both sides of a semipermeable membrane contained some NH4Cl and some KNO3. E. The concentration of KNO3 increased.



Answers

The following pairs of aqueous solutions are separated by a semipermeable membrane. In which direction will the solvent flow? a. $A=0.48 M N_{a} C l ; B=55.85 \mathrm{g}$ of $N a C 1$ dissolved in $1.00 \mathrm{L}$ of solution b. $A=100 \mathrm{mL}$ of $0.982 M \mathrm{CaCl}_{2} ; \mathrm{B}=16 \mathrm{g}$ of $\mathrm{NaCl}$ in $100 \mathrm{mL}$ of solution c. $A=100 \mathrm{mL}$ of $6.56 \mathrm{mM} \mathrm{MgSO}_{4} ; \mathrm{B}=5.24 \mathrm{g}$ of $\mathrm{MgCl}_{2}$ in $250 \mathrm{mL}$ of solution

Question 82 in the chapter 11 of science chemistry, the signs in context. Okay, so question 82 minutes asking for these three scenarios. Which direction will the solvent flow? Okay. And so when we're talking about solvent movement, we're talking about Austin Moses. Okay, so you usually go, the solvent will flow from a low to high station. Or, in the case of osmotic pressure, it'll go from low to high. Osmotic. Okay. And just remember that the equation for osmotic pressure. So osmotic pressure equals our i r event hough factor. Okay, um, times our molar ity not morality. More polarity. Times are the are constant times temperature. Okay, but they're not giving us temperature will. So we can kind of ignore temperature and are so the our is a constant, and that's the same for every solution. So we can ignore our and we can also ignore t because they don't tell us, temperature. And we're gonna make the assumption that all of these solutions are in the same terrible. Okay, So, really, the only things we need to compare our event off factor for these solutions or the polarity okay. Or both. In the first Ah, In the first scenario, we both have an a c l O k. So we don't even have to worry about our, um our I Okay, our, um our event hough factor and the reason why is because event ha factor will be the same for both and ate meals. Is it growth? So, really, all we have to do is compare mole Aridjis in these. Okay, so we have one leader. That is that 0.48 Moller. I mean, we don't know the polarity here, so let's first off calculate similarity. It's in one leader. So all we need to know is how many moles of any scale we have. So we need to go to the periodic table and find the mole molecular weight of N A C l. Hey. And so when you do that, you should get mm. It should be 22.99 grams per mole, which is for a sodium plus 35.45 grams per mole for the, um, for the chloride. Okay. And that adds up to 58.44 grounds. Permal. Okay, so that's a molecular weight of sodium chloride. Um, and So to convert this into moles, we just do 55.85 grams times 58 0.48 which is our molecular weight grams over Mel's. I'm under rules, right, Because we want moles on top and we want grams to castle out. So this ends up being zero Ah, 0.955 moles. Okay. And since this is in one leader Ah, if we were to divide this by one leader is the same thing. So we end up with 0.955 as arm polarity, okay? And that is higher than the polarity that we have here in chains. That we know that for the first scenario, we're going to go from here to here. OK, so that's going from a to B on. Because in the problem, they say that this solution is ah, labeled a solution, and this solution be so it's gonna move from a to B is the solvents are going to go from low to high. Okay, well, let's move onto the next one. So in this case we have we're comparing any seal and calcium chloride. Okay, Um and so we first have to compare their event. Both. Okay, So calcium chloride will dissociate into calcium ions and to chloride ions. So it's I theoretical eyes three. Okay. And in the case of n a c l the theoretical I is to Okay, um, and we have molinari of calcium chloride already. So the only thing we need to do is find our mill aren t of our, um, sodium chloride and we The volume doesn't matter as long as we have malaria. So we have the molecular way of sort of accord from a previous problem so we can go ahead and just use that to calculate out our polarity. And so that's 16 grams times moles over 58.48 grams. And this ends up asst 0.2737 bulls. Okay. On minutes in 100 mills, 100 mills is the same as 0.1 leader. Okay, The way I did that was I just divided it by 1000 because there are 1000 mils in a leader. Okay? And so if I divided this by 0.1 leader, we would end up with the polarity of zero. So actually would be 2.7, 37 Moeller. Okay, so we see that the polarity is very high here. I mean, if we were to multiply that by two. The I factor. So if we're gonna try to calculate out the first part of this equation, um, this would end up being about So we multiply this by I, which was two. This ends up being 5.4 approximately. Whereas here, if we multiply on polarity and I, this ends up being 2.94 Okay, so the, uh, the product of our I and r M r polarity ends up being too when I four on this side and 5.4 on this side. Okay, these are the numbers were comparing. And again, we're going to go from high to low. Remember, we're ignoring our anti because we're making assumption it's the same for all solutions. I'm so osmosis will happen from low to high. Sorry. I mean says going to go this way again. So it's going to go from a and I go from a to B yet again. And so let's do the last problem. We're gonna have Teoh compare that caught Hoff factors and, um, polarity again. And so Davis Millet balls here. We can convert this two moles if you want. I'm And so there are 1000 mill immobile's in a mole. So if you just divide this by one 1000 you can get this and moles, um polarity. And so this would be 0.656 animals on our vat cloth here is just gonna be magnesium and the sulphate. Ah, dissociating. And so I ends up being to here. OK, now, for a magnesium chloride, you're gonna have one magnesium and to chloride ions dissociating. So I is three here, Okay, But we also don't know our polarities, so we need to calculate that out. Um, and the molecular weight for magnesium chloride. So molecular weight for magnesium chloride? I did that for you guys. I just added the molecular weight for magnesium and then to chlorides. I'm just ended up being a 95.205 grams per mole. Okay, well, let's move this. Let's zoom out a little bit. There we go. Okay. So that is our molecular weight. Okay, Now we just have to convert grams to moles and then find out our polarity, So that's gonna be 5.24 grams times our moules over 95.205 grams. Right. So Grams cancels out, and we end up with 0.550 point 055 moles of the magnesium. Cool ride. Okay. And this is in 250 milliliters. 250 million litres. This convert to 0.25 uh, leaders, right? We want it in. Leaders can again. I just did this by dividing 1000. Because there's 1000 mills in a leader to 0.25 leaders. Okay, His arm polarity for this is 0.22 Okay, which is much bigger than our magnesium sulfate. Okay, the magnesium sulphate, remember, was in mill. Uh, yeah, Mila molars. Um, and so that would be 1000 of a mole or 61 thousands of a Moeller. Okay. I mean, we end up with 610.22 so I don't even have to multiply it by the I factor was the magnesium chloride is obviously much more concentrated than are magnesium sulfate. Okay, you know that the solvents is gonna move again in this direction from a to be okay. And so that is how you solve

So in each of these scenarios, the solvent, which we're just gonna assume here is water is going to flow to even out the concentration. Since that means it's going to flow from a place that has a higher concentration of total number of particles to a concentration of lower, we'll lower number of total particles. And I make the point of saying total particles because there's a difference between 0.0.1 Moeller glucose that doesn't associate all in 0.1. Moller. You know k threepio four, which just states the four parts. This really creates point for Moeller just particles in general. And that's what plays a role here in, uh, the awesome noses of water, the Grady in process of trying to even out the concentrate. So for a first question, we have region a Region B and region A. We have 1.25 moller on a C L. And we have 1.5 Moeller, um, taste CEO. And so we're being told that doesn't look like the islands are going to move registering with water. And so, since both these associate two molecules their cat eye on an ion, there's no difference in there And so really, we're just like UK. Which of these concentrations is higher? Well, that's 1.5. So it's going to go. Pardon me? This is I drew the era wrong. I realized, um, it goes from a lower concentration to a higher concentration in order to dilute this high concentration. So the water here is gonna go from region a Region B in order to dilute this 1.5 Moeller and get it to be closer and concentration to the 1.25 Okay, next one, we have region A, which is 3.45 bowler calcium chloride. And so big. No, here is that this has not been associate two parts to this ocean three parts and Region B has 3.45 Moeller sodium bromide. And so here our concentrations are the same. So that's not a difference if that here the calcium chloride, we have one calcium and to chloride, so this breaks up into three parts. The sodium bromide only breaks up into two parts for every single mold that we put in, and so functionally A has a higher concentration of particles and so water is going to flow from Region B two region A. In order Teoh, I lewd that higher concentration in the process. It also makes apart the region and b'more concentrated. Since we're reducing the amount of water that dissolves. And for the last part, we have a which is 4.68 Moeller glucose and Region B, which is three. Moeller and I see out this has a higher concentration. But be careful, glucose. When you put it in, it stays glucose. For every one molecule of glucose, you just get one molecule of glucose solution. But here, every molecule of an A c l we get two molecules in solutions. So the functional concentration of particles. Here we have 4.68 times one apartment point point 68 But here we have three times to so six Mohler. And so even though this has like a quote unquote higher concentration region A. When it comes down to the amount of particles, we have more particles and be and so the water is going to flow rs the solvent. But I think in a sense, these species, um then be is gonna flow from an API dilutes the concentration

Are we kind of, ah, report question here that we're gonna try to answer based off of an osmosis illustration. And so let's pretend that we've got a container. And in the middle of our container is a semi permeable membrane ever remembering what's going to make it seem impermeable is the size and so over one have on the left side of our solution. Just gonna pretend is we've got some soul into molecules will make them smaller, since they're going to be the ones passing through the membrane. Do you have some on the left as well in the area, And then we're also going to have some saw you two molecules. So let's do on the left side, that's before saw your molecules and on the right side, let's do you just to. So what happens in osmosis is that whichever side is more concentrated is going to be the recipient of some of the solvent from the less concentrated side. So what will happen because our left side is more concentrated that some of the solvent from the right side will cross over that membrane to try to help dilute that concentration until they reach equilibrium. And so what happens is when the solvent starts flowing into that left side that's going to make the water level of the left side seem to go up, and the water level of the right side seemed to go down. And so what happens in that scenario is it looks like we have, you know, a greater amount on the left side, and what the goal has been is to try to pull a break. The concentrations. So no, this side of the left does have more saw you, which also takes up more space. But the addition of that extra water has hope dilute its concentrations that the two concentrations on either side can be equal into the level of the solution will stop rising once the two concentrations on the left and on the right have reached equilibrium all right and support. See when the level stops rising, there will be no net movement of water molecules across the membrane. They'll be able to move in both directions freely. But

So here we are, given information about specific association reaction where K. C. L. And it's solid form associates and UK plus in its acquis form plus chloride and nine so the K. S. P in this case would be equivalent to the concentration of species that change significantly. So it's equal to the concentration of K plus times the concentration of C. L minus. So since these dissolve in equal proportions, the concentration of K plus would be equivalent to the concentration of chloride. And were given information that 3.7 molars dissolves. So this expression would be like this. So in the end this would just be the scalability squared and we find that the K. S. P. Is equivalent to 13.69 for this specific solution. So now we're, for the second part, we're essentially going to stress this system with varying concentrations of are different species. So we're given information that we initially have volume of initial of 100 mL or 0.100 liters of our K. C. L solution or potassium chloride solution. And in the first case we're going to stress the system by adding 6.0 molars of hcl. So hydrogen at the proton would be involved in this case. So it's just the significant concentration will just be the chloride in terms of the common ion effect. And this would be 0.100 leaders of this hydro hydrochloric acid solution. So now we have to adjust for this concentration in terms of basically, since volumes are additive in this case, so we can see since we're essentially doubling the volume by adding this chloride, the concentration of chloride would be equivalent to 3.0 molars. Direct check is also it's possible to also do a direct check in this case. So the concentration of chloride would be three molars. So in this case let's rewrite our K. C. L expression. But in this case we already have a significant concentration record and iron and we know that the soluble it is going to be much less in this case. So temporarily, let's assume that uh when we have a certain amount of K. C. L. Dissolving, we're going to make the assumption that the S. Is much less than 3.0 Mueller's. So we can ignore this in regards to our expression here. And we see we can try to see if the assumption is justified. So we also need to remember that the K. S. P is about 13.7 would be equivalent to the scalability times 3.0 Mueller's 3.0. And if we plug this in, we see that the scalability is actually greater than 3.0 molars. So therefore, assumption is not justified, since the solid ability is not much less than three Mueller's. So we actually have to use the quadratic formula in this case. So you should rewrite this and we should actually include this term. So 13.7 equals the solid ability times 3.0 plus solid ability. So typically the quadratic formula is not our favorite, our favorite thing to use here. So we can use essentially use numerical iteration process. So we can just keep iterating for different numbers to essentially approach the actual solid ability. So you can try that in that case. So we see that two molars is a bit less than the actual sustainability. So we keep increasing our by increments in terms of plugging into our calculators. Yeah. And we find that essentially a solid ability of 2.50 molars is relatively close to the actual solid ability. If you use numerical iteration by plugging in for different values of S. Are solid ability here and we essentially eventually find value that's very close. So in our second case we're going to add a 12 molar hcl solution with still this volume. So as a result of basically having additive volumes, the effective concentration of chloride will be reduced in this case to 6.0 molars initially. And now we're going to have to consider the dissolution equilibrium again. So K. C. L solid associates into K plus acquis plus the chloride Equus in iron. So 6.0 molars this time, I'm just not going to make the assumptions and it seems like the concentrations are relatively large and the assumption won't be justified. So a certain amount of potassium chloride will dissolve. Then the concentration of cables will increase as well as the concentration of C. L minus. And this is equivalent to the K. S. P. And once again we can try to evaluate this by numerical iteration. So we have to essentially find values that approached values of the scalability that approach 13 lead this expression to approach 13.7. Yeah, so one point let's try 1.8 in this case. And and we see that our values about between 1.8 and 1.7. So we'll just call this about 1.8 molars. And this is our final answer. And we can basically compare this to our initial answer here. We found that we initially had a solid ability around 3.7 Mueller's. When we're adding chloride, it makes sense that the concentration dissolved, going to decrease to the common. I in effect adding an even higher concentration of chloride also makes sense to decrease the concentration even further, basically that dissolves. And this is our final answer.


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