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3. 3.00 & of Kci dissolved In 500.0 & of water t0 make an aqueous solution at 303 KWhat is the concentration of KCl in ppm? What Is the freezing point of th...

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3. 3.00 & of Kci dissolved In 500.0 & of water t0 make an aqueous solution at 303 KWhat is the concentration of KCl in ppm? What Is the freezing point of this solution 2.00 of 0.100 M KCI (p = 1.05 g/mL) is added? The cryoscopic constant for water is ckglmol Assume ideal behavior of the ons.What Is the vapor pressure torr of the solution descrlbed in (b) at 303 K? You will need to look up thc vapor pressure of water at thls temperature

3. 3.00 & of Kci dissolved In 500.0 & of water t0 make an aqueous solution at 303 K What is the concentration of KCl in ppm? What Is the freezing point of this solution 2.00 of 0.100 M KCI (p = 1.05 g/mL) is added? The cryoscopic constant for water is ckglmol Assume ideal behavior of the ons. What Is the vapor pressure torr of the solution descrlbed in (b) at 303 K? You will need to look up thc vapor pressure of water at thls temperature



Answers

An aqueous solution made up of $32.47 \mathrm{~g}$ of iron(III) chloride in $100.0 \mathrm{~mL}$ of solution has a density of $1.249 \mathrm{~g} / \mathrm{mL}$ at $25^{\circ} \mathrm{C} .$ Calculate its (a) molarity. (b) molality. (c) osmotic pressure at $25^{\circ} \mathrm{C}$ (assume $\left.i=4\right)$. (d) freezing point.

Here we have problem. 88. This question. Ask us to calculate the freezing point and boiling point for the following three solutions and assuming that the complete association of the solid All right part A, we have the mass off the salute F E sale three equal to 10.5 grams. And if we look up the Moler mass of this molecule, it's equal to 162.204 grams Permal and we have the mass off the water equal to 1.5 times 10 to the second. Graham, to convert this graham into kilogram will have the massive water equal to zero point 15 kilograms. So for the first solution, Effie, sale three is the solid and water is the solvent. All right. So in order to calculate the change of phrasing and boiling point, we will need to tackle the mill ality and to find the morality of a solution, we need to know the most of the solid. So the number off most in the letter in for the solid f e c l three is equal to the mass of f e C l three, divided by the Moler Mass of Easy agree. This is equal to 10.5 grams, derided by 1 62.204 grams per mole, and this is equal to your point 065 more. So this is a number off moves up the solid. Now the mass off water is the massive solvent, and we'll have the moles off the solid. We can calculate the morality. Formal ality is equal to number off most of the solid in the unit off mole divided by the mass of the soul mint in the unit of Kilogram. And this is equal to 0.0 65 more divided by zero point 15 killing ground. So this will tell us the morality of the solution equal. Choose your point for 33 Moeller. So that is the first morality for the solution. Now we have the formula Delta t off equal to i times. Okay, uh, times And where Delta t f is the Depression in freezing point K f is the freezing point Depression constant and the small M is the mill ality of the solution. And here the eye is the vents ho over factor. There's whole factor, so we had to rewrite the solution. Don't a t f. It's also equal to keep your minus tee up the solution. So the solution f e c l three on ionization produces four Ryan's So the I for this equation is equal to four and we also know for water. The Kate off is equal to 1.86 Sosa's per Moeller. So now we substitute this values into the formula and we can calculate the freezing point. Us. Don't a t f equal to I was able to four k I a physical to 1.86. So since degree per Moeller times, the mill ality of the solution would get from here 0.433 and this is equal to three points 22 Celsius degree. So this is the Depression of freezing point. So for water, the freezing point is for water. The temperature of freezing point is equal to zero Celsius degree, so t solution will be a quitter. T pure minus. Don't a t of equal to zero sells his degree minus 3.22 Celsius degree, and this is equal to negative 3.22. So sis decree. So this is the freezing point of the solution. So this is a freezing point for solution A. And now we are looking for the boiling point of the solution. So for boiling point, we have the formula Delta T B. You called to I times could be times smarter for adult A TB represent the elevation in boiling point in eyes Vince Hall factor K B is the boiling point elevation constant and the small M is the morality of the solution. And we also have the equation. Don't a TB you going to tea of solution minus t have the pure solvent. So for water we have K B. The value of K B for water is equal to zero point 512 Sosa's her Moeller so substituting this values into the solution we have Delta t be equal to ay So we said I is equal to four because of e. C. L three. Upon ionization will generate four Ryan's so four times K be equal to 0.512 cells. This promoter time Cem mil ality. Because your 0.433 and we get Delta t be equal. Choose your 0.86 So since degree. So the boiling point of water, we know that it's equal to 100 Celsius degree, so t of the solution will be equal to Delta T B plus t pure water. So that is there a 0.886 plus 100 and give us 100 0.886 See, also secret. So that ends our part, eh? A boiling point. So the boiling point for for part eight is 100.886 and the freezing point is negative. Three points to two. All right now we moved to Part B party. We have 3.5% of K C L. That means 3.5 grams of K C. L is in a solution of 100 gram. So the mass of water is equal to three point 100 minus 3.5. This is going to 96.5 grams of water and we also know the Moler mask for Casey l is equal to 74.553 grounds promote. So first we want to find the number off most for the salute. Casey L This is equal to mass divided by Muller Mass. So 3.5 grams divided by 74.553 grams formal. And that will give us their points. You for seven moves of Casey l So now we find the mill ality of the solution. Number off. Most off solid case E l divided by the mass of solvent was just a water kyul Graham. So this is equal to 0.47 moles divided by 96.5 iss Graham, we needed times 10 to the minus seven tend to the minor three to get a telegram. So by doing this calculation, we get the morality of the solution equal to zero point for a seven Moeller s o or Casey elp on ionization. It will give us potassium and chloride so to Ryan's so that I, for this solution, is a quint. You too, and we have the chaos off water equal to 1.86 cells is degree promoter. Now we've substituted state A into the formula of freezing point. So don't a t f is equal to I can, uh, times in I, as equipped to K F is equal to 1.86 times. The morality is zero point for 87 and this will give us 1.812 cells. This decree and we know for water. The freezing point is zero degree. So T solution is equal to keep your minus delta T f. That is equal to zero minus 1.812 This is equal to negative 1.812 So So, Stickley. So this is the phrasing point for part B. Now, for water we have, we're going to calculate for the solution be We also need to catch coid the boiling point. So we have K b of water. You could choose your coin 512 sources Stickley promoter and the boiling point for water is equal to 100 cells in secret. So we have don't a TB Equant you I k b times m It wanted to time 0.512 Celsius degree per Moeller times their point 487 Moller. And this will give us the Delta T go to point for 99 So so t of the solution, it's equal to don't a t b plus t. I'm disillusioned off the pure water. So this is equal to zero point for 99 plus 100 that will give us 100.499 cell cystic. And this number is the boiling point for part B. Now we move to you part, See? So for part C, we have zero 0.15 zero Moeller off mg of to and on ionization. It will give us mg two plus and two fluoride I So I for dissolution even help Constant. Is it going to three? So now, since the question already gave us the morality, we don't need to calculate the number of most and the mass of the soldiers. So we just simply delta t of equal to i n times k f times in equal to I could to three times k f for water is 1.86. So 60 quid pro Moeller times The Malala t is there a 0.15 and we have Delta t have equal to 0.837 Celsius degree, so t solution is equal to keep your minus delta T f. This is equal to the freezing point off water. It's Cyril degree minus 0.837 Celsius degree. This will give us negative 0.837 cell system and this is thief freezing point for a solution seat. And now moving to the boiling point would have delta t be equal to ay kay bee puns M that's equal to three time zero point 512 So cystic re promoter times the moral ality 0.15 And this will give us numbers equal to zero point 23 Cell cystic quick now t of the boiling point is equal to delta T B plus t pure. So we know Don't the TB equity of your 0.23 and the boiling point for water iss 100 Celsius degree. So we get a number off 100 points 223 cells of stupid. So this is our boiling point for a solution seat. So the boiling point and freezing point for solution see, is here boiling point and freezing point for solution be. And here we have the boiling point for a solution A and the freezing point for solution A is right here. So that is our problem

We used the freezing point depression to calculates the morality. Freezing point depression is 2.79 degrees Celsius. Constant for water is 1.8 63 Celsius. Promotable times The morality solving would give 1.5 Molo. We'll use this morality to calculate the boiling point elevation, which will be KB Times M through a 0.512 degree Celsius. I m times 1.5 m, which will yield 0.768 degrees Celsius. The boiling point of the solution will be 100 degrees Celsius plus your 0.7 60 degrees Celsius, which will be 100.768 degrees Celsius. That would be the answer for part. Okay, for part B determined the vapour pressure solution at 25 degrees Celsius. Um, we'll use the morality which we've got previously. So was the morality was 1.5 more of this saw Ute Burton 1.5 moles of the saw utes over, uh, kilogram H 20 And, um, we know that we have 1000 grams of water, so we need to calculate the more reach Too old. So 1000 grams of each 20 that comes from the morality. One kilogram into 1000 ground is the 1000 grams, divided by 18.0 grams per mole, which is the molar mass. Reach 20 gives 56 moles of water. So we know that there is 1.5 moles of Salyut in 56 moles of H 20 to calculate the mole fraction of each. Too old it will be 56 moles that we just calculated in the numerator 56 molds plus the 1.5 moles with saw you two in the denominator which will give us a mole fraction of 0.97 and then to calculate the partial pressure of the solution is equal to the mole fraction of H 20 times the pressure of H 20 Moulter actually calculated his 0.97 pressure of H 20 ver given is 23.76 millimeters recovery which will give us the pressure of the solution that is equal to 23.5 millimeters of mercury. Part c explain any assumptions that we have made in order to complete parts A and B. Uh, we're assuming that the solution is a non electrolyte, because when we solved for the morality, we're assuming that the Vantaa factor is one. Um, therefore, no, I ends would form. So defend how factors equal the one. And we didn't take that due account earlier in the question. We're also assuming that we're dealing with an ideal solution.

All right, so we've got some potassium chloride. Let's find out how many moles of each eye on. We're going to get from that. So first thing we're gonna do is go ahead change this two moles Using the molar mass 74 55. So that's gonna give us .04 69 moles of K C L. So that gives us .0469 moles K plus when it breaks up and the same amount more chloride and then we've got 0.5 moller And 60 ml or .06 L. So it gives us .03 00 malls of our calcium chloride. So that's going to give us .0300 moles of our calcium ion. But since there's two chlorides it will give us .0600 malls of our chloride ion. So we're going to take Those two moles and add them up. Okay, so that gives us a total of .0769 moles of chloride. Our total volume is 70 mL or .07 L. So the concentration of our chloride is 1.09 moller. Okay, so let's go ahead and take a look then at the potassium ion, That's .04 69 moles. We found this up here And again. We'll just divide that by our total volume, which is .07 L. So that's going to give us .670 more K plus. And then for the calcium we've got that for moles .0300. Right by our total volume. So that's going to give us .4-9 more calcium ion.

Problem. 83 asks you to calculate the freezing and boiling points of each solution using the following equations. So the stilt TF Ah equation here is the change and freezing temperature or the freezing temperature depression when a an electrolyte or assault is dissolved in water. So this I hear is the constant basically tells you how many moles of Salyut, um ions are found in the solution. This is a constant for a given morality of a given salt. This lower case M here is the morality of your solution to remember that morality is the number of moles I saw you per kilogram of solvent. And this k f here is a constant for, um, Saul Eudes dissolved in water. So, uh, the KF ISS 1.86 degrees Celsius per mole owl and very similar is this change and boiling point or boiling point elevation when salts or electrolytes are dissolved in water and instead it has a boiling constant, which for water is 0.512 degrees Celsius. Permal at so part a here asks you, um, the freezing and boiling point changes for a point. One moral potassium sulfate solution. Um, so they can plug what we know into these equations here. Um, because we're supposed to assume complete association of each of these electrolytes. Um, we know that a potassium sulfate becomes to potassium ions and one sulfate. So we know that our I factor here is three islands dissolved and solution. You know that our morality is 0.1. Then we know that our freezing constant here is 1.86 degrees Celsius Permal out. So that gets its 0.558 degrees Celsius. Um, as thief freezing point depression. Which means that our freezing point for this potassium sulfate or sulfide Sorry. Cell fetus? Yes, that Sofia is this for this is sulfide. My apologies. Um, so the freezing point of potassium sulfide is negative 0.558 degrees Celsius. It's very much the same for the boiling point elevation. Um, we can plug in, you know, and that gives us a boiling point. Elevation 0.0.154 degrees Celsius over pure water. Which tells us that the boiling point of potassium sulfide it's 100.154 degrees Celsius. So moving on apart be here, Probie asks, uh, what be freezing temperatures and the boiling temperature of 21.5 grams of copper chloride in 450 grand's or 4.50 times 10 to the second grams of water would be so to figure out the morality of copper core ride here, we first need thio know the molecular weight. Um, that's 134.45 grams per mole. And so we can figure out how many moles of copper chloride, um, are 21.5 grams by using this conversion, and that gives us 0.160 moles. So to figure out the morality, then we can take this 0.16 serum ALS of copper chloride. We know that we have 450 grams of water, and we know that we eventually want this kilograms so you could do that conversion that gives this 0.356 moral. So then, just like before, um, we can assume because, uh, this problem tells us to assume complete association again. We have three ions resulting from an association of copper chloride. They can plug what we know into the equation to figure out our freezing point depression been, are boiling point elevation, which gives us a freezing temperature for the solution of negative 1.99 degrees Celsius and a boiling point off 100.547 degrees Celsius. Finally, um, parte si asks the same of a 5.5% um, sodium nitrate solution by mass and water. And so, uh, percentage by mass recall means that you have that number of grams of Saul you for every 100 grams of solution. So what we can do here, then, is we know that we have 55 grams, huh? One kilogram. And we can figure out the number of moles of sodium nitrate by using the molecular weight conversion. So one more is 84.9947 grams, which gives us 0.647 moles per one kilogram or 0.647 allow. So again, um, there, too ions here. If we assume complete association, you can plug all of this into our temperature. Um, change equations, uh, freezing point depression and boiling point elevation. To find that the freezing temperature of this particular solution is negative 2.41 degrees Celsius and the boiling point ISS 100.663 south seas


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