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Question based on question 11.64 (part (b) only) from text: Complete assignment question written here: pair of aqueous solutions are separated by semipermeable memb...

Question

Question based on question 11.64 (part (b) only) from text: Complete assignment question written here: pair of aqueous solutions are separated by semipermeable membrane (L.e; permeable to The following water only) In which direction will there be osmotic flow of water? Support your answer by showing anv calculations_ In class we saw that the net flow of water (osmotic flow) is towards the solution with the greater total particle concentration and; and that this concentration of total particles i

Question based on question 11.64 (part (b) only) from text: Complete assignment question written here: pair of aqueous solutions are separated by semipermeable membrane (L.e; permeable to The following water only) In which direction will there be osmotic flow of water? Support your answer by showing anv calculations_ In class we saw that the net flow of water (osmotic flow) is towards the solution with the greater total particle concentration and; and that this concentration of total particles is the value of"iM" to be careful when reading and writing concentration, this chapter in particular. It is important molestolute per solution and has units of mol/L; lowercase m indicates moles Uppercase M indicates solute per kilogram of solvent in the solution and has units of molkg: Solution #1: 100.0 mL of 982 M CaClz mLof an aqueous solution containing only 16.0 of NaCl Solution #2: 100.



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

So here we have a series of scenarios based on the solution that we have produced. And so what was the first thing we can know is that we have 58.44 g in one mole of sodium chloride. And so what we also need to know is the definitions for our salute our solvent. So the salute is just the substance that we can dissolve. And in this case that is sodium chloride, N A C L. Where the solvent is the substance that the salute is dissolved within. So what we have here is water. We have NACL. and H 20. So the 58.44 g or one mole of sodium chloride. We add this to the flask and so then we fill our flask up to the one later mark Where we create a one molar sodium chloride solution. So looking at the picture, we have less than one liter added. So if the same amount of solute is added to a smaller volume, then we have a solution that is more concentrated. So this would mean that the solution is greater than one Mola. So therefore see would be the right answer Where we have greater than one molar solution created because less solvent was added. Okay

For part A. When you to first calculate the morality of the solution and to do so was starting with 0.9 point 92% master volume solution of any C l 0.92 will assume 100 uh, in 100 g. That means 1000.92 g of any seal. It's convert this to malls 50.5 grams of any CEO present in one more and the mass of the solution. He's 100 g some of the solution minus the 0.92 g of any CEO, and we need to convert this to kilograms in this. Would you of the morality that's equal 2.16 Molo, calculate the freezing point depression. Any CL Vantaa factor is too constants 1.86 degrees Celsius per mole dualities 0.16 and this would yield minus 0.60 degrees Celsius for a freezing point. Depression. Now the value were given is point five two degrees Celsius. So this tells us that the advance off factor, uh, for in a C L is most likely, um, less then to And since they're pretty close together, um, 0.60 point 52 that who definitions are in agreement, they're fairly close together, so we can say that they will agree with one another For part B. We've got E, um Mixture here of several components. So let's calculate, um, we're starting with 3.5 g of any CEO and lets rip this two moles and then in one more of any CEO, we have to, um oh, uh, the irons one n in one cli in. So we have 10.12 more, uh, of the ions present from 3.5 g of any CEO. Let's keep going. We have 1.5 g of K c. L. That's all for moles of ions here, Casey Ella's of molar Masses 74.55 g in one more and it once again one more K. C. L will have two moles of irons and our, um, mole of Casey Lyons. Here workout 2.40 A mole, uh, of lines. The third component here is we have 2.9 g of sodium citrate eats in a three C six h, 507 molar mass. Here's to 58.7 grams of sodium C. six h 57 in one more in a three C six each, 57 and one more sodium citrate. There are formals of irons presence, and this would work out to, uh, 0.45 more, uh, ions present in sodium citrate. And lastly, we have 20 grand's of glucose, which is a non electrolyte C six h 12 06 discovered this to you moles here 1 82 moles of C 60 each 12 6 And with 380 programs more mass in one more. And this would work out to, um, 0.111 Moles C six, h 12 06 Now, using all of this, um, we can calculate the Malala d. So the morality here would be the moles of everything. Um, which will be the moles of any plus the moules K CEO, plus the mobile sodium citrate. It's this will be moles uses in here Moles, moles and bulls in the mass Here would be No, we're starting with, um, that was in grams minus. Now the mass of everything. 3.5 g and a CEO 1.5 g of K c e. 2.9 g of sodium century and 20 g of glucose and solving here for the morality. Uh, what? We need to convert this tequila grams 1000 g per 1 kg. And my morality here would work out 2.3 to 5 Molo and then using that the delta T f is going to be We've already got the ions present, so minus K f m minus 1.86 degrees Celsius promotable 0.3 to 5. Malone and my delta TF here works out to minus 0.60 degrees Celsius. So this is again close to the 0.52 Um, So, um, this is close to the value of minus 0.5 to 2 keys, two degrees Celsius with the difference, uh, being associate ID to, uh, the, um creation and band huffed factors as they're not, um, exactly all whole numbers. Um, since the value is in close agreements, we can conclude that the solution is isotonic


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