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Pentasydra effect %fincreasing lemperature on Ihe dissolution = pentahydrale in water of copper (Il) sulfateWhat effect does solute have on the boiling solute have ...

Question

Pentasydra effect %fincreasing lemperature on Ihe dissolution = pentahydrale in water of copper (Il) sulfateWhat effect does solute have on the boiling solute have on the freezing point of a solvent? point of a solvent? What effect does a19.05.9 %f sodium chloride are soluble in 25.0 ml of sodium chloride that will dissolve in 135 ml of water; calculate the amount of waterWouedryou expect the solubility of sodium chloride the lemperature IS changed irom 25 %C (0 80 009 in water t0 increase or to

pentasydra effect %fincreasing lemperature on Ihe dissolution = pentahydrale in water of copper (Il) sulfate What effect does solute have on the boiling solute have on the freezing point of a solvent? point of a solvent? What effect does a 19.05.9 %f sodium chloride are soluble in 25.0 ml of sodium chloride that will dissolve in 135 ml of water; calculate the amount of water Wouedryou expect the solubility of sodium chloride the lemperature IS changed irom 25 %C (0 80 009 in water t0 increase or to decrease if



Answers

The copper(I) ion forms a chloride salt that has $K_{\mathrm{sp}}= 1.2 \times 10^{-6} .$ Copper (I) also forms a complex ion with $\mathrm{Cl}^{-} :$
$$\mathrm{Cu}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{CuCl}_{2}^{-}(a q) \qquad K=8.7 \times 10^{4}$$
a. Calculate the solubility of copper(I) chloride in pure water. (Ignore $\mathrm{CuCl}_{2}^{-}$ formation for part a.)
b. Calculate the solubility of copper(I) chloride in 0.10$M$ $\mathrm{NaCl}$.

This question were given a situation where we have assault. Um, the name of the salt is a chromium nitrate salt that at 15 degrees had a solid ability of 208 Gramps for 100 years of water. So what it's saying is, if I have a solution, if I have this salt solution, okay, at 15 degrees Celsius, then I can on Lee put up to 208 grams into solution. Okay, so that's 208 grams. And then if I try to add any more salt in that it's going to stay as a solid okay and because it's saturated, then the Salyut in solution and the salute that's not dissolved, dissolving the on the social later equilibrium. So you, if it's saturated, you can't have more than 208 grams in that one of your water, Okay? And what they're saying is that we made a solution at 35 degrees and at 35 degrees we're able to dissolve more than this. 208 were able to dissolve up to 324 right, so they have a lot more solution. There's no precipitate, and this contains 324 grams. And that's possible because we're at this higher temperature of 35 degrees C, and what happens is from 35. We called on the solution to 15 degrees C, and the question is, if no precipitate forms is this unsaturated stature your super saturated, well saturated? We already said containing will contain 208 grams and solution because we've exceeded that. This solution, which has 324 grams at 15 degrees Celsius, is now super saturated, so part of the answer is super saturated. Now, here in part B, we're taking the super saturated solution. Remember, it contains 324 grams for every 100 grams of water. And what we do is we take a spatula and we scratch the sides of it. And what happens then is we start to see crystals. What we've done when we scratch the sides of the glass is we've created a site where we concede the crystals basically give the crystals of place to begin forming. And so then what? We'll see along size of the along that scratch his information of crystals because we have seated a starting site for them. Okay, Um, so that's what happens when you scratch the glass. And what you'll see is that, um, you'll lose some of the crystals in from the solution, as they you'll lose some of the solid in the solution as it starts to form a Chris solid crystals along that scratch line. Finally, in part C were asked at equilibrium what massive crystals were formed will recall that at equilibrium, we're gonna, um we're gonna go back to saturate a solution. So we know that saturated solution contains 208 grams, and we started with 324. Okay, what will remain in solution is the 208 and what we started was the 324. So if we subtract, will get the mass of crystals that will form. So this difference is 100 in 16. Gramps and

All right, so this one is a freezing point depression problem. So let's go ahead and get our freezing point depression equation on the screen for its now. What they've done is they've told us that each of these three solutions were going to be working with have a mill ality of one. Now also, what we know will be the same as the KF. In each case, that's because we're dealing with water, so that will be 1.86 about it, but are 1.86 degrees Celsius Perm allow. So what will notice is going to be changing is the Van Hoff I factor, which represents harmony particles this particular compound forms in solution. So we'll go ahead and just kind of worked all three of our problems at the same time. So you've got K I c a c l two and then finally barium nitrate. All right, so right quick, let's go and figure out the eye for each of these. Potassium iodide is going to break up into two. Ion says I will be too calcium chloride will break up into three ions is I will be three, and then barium nitrate will also break up into three ions one barium and to nitrates. So now we're ready to solve for age of, um so have two times 1.86 times one for the 1st 1 that will get us a change in temperature of 3.72 degrees Celsius. And so what? This means if water normally freezes zero degree Celsius and we lower the freezing point by 3.72 degrees, that means that the new freezing point will be negative 3.72 degrees. And what we should expect to see is the larger the event off factor for the same mill ality. We should see a greater decrease in the freezing point. So three times 1.86 times one gets us 5.58 degrees Celsius. And again, if you start at zero and increase it by that much, your new freezing temperatures negative 5.58. Now, finally, we're wouldn't test our barium nitrate here, and this actually is taking the same form as the second part of this problem and so we should expect the same. The result

Well, the problem. 1 21. You have to calculate the mass percent of the water in the compound. Couple soon, Fi. Panda. Hydrate on. Then I arrived 1 21 here. Now you have this compound couple. So fight. Bend the hydrates. Yeah, so did hydrate compound. And then you have to calculate the mass of the percent of the water. So, you see, masked percent off water. It's not be much water off a Muslim ball. Right? So you have to try find water here on massive one water. It was 18 for your own 16 grand. I got buyout over fast, but holding on sitting here. So you divide it for the mask. Um, power so equal. 24. My oil 68 g couple so far. Ben High, right. And this water. And then you tie 100%. This must be equal. Let me Koppel a Okay. Yeah. Mhm. What is it, boy? 078%. Yeah, And when you hit, when the Kambale he's hit it and you will see this you will see what back? All right, people, for this Let low compound after what evaporate you get. Why come down here So you see what evaporated the same water in a combo. So the mass percent So my percent is lost equal. My percent of water equals 30 Sit boy. 078% in your in your book and as a key. I think they make me steak. Okay, on depend the way they calculate the band I the way they round up number. So you are different. He's very valuable, completely in a similar way to calculate


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