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3 If the solubillity of (NHA)SOa Is 57.5 g NHASOa per 1OOmL of water at 25,0"C determine whether solution that is prepared by dissolving 75.0 g of (NHA) SOa in...

Question

3 If the solubillity of (NHA)SOa Is 57.5 g NHASOa per 1OOmL of water at 25,0"C determine whether solution that is prepared by dissolving 75.0 g of (NHA) SOa in 200ml of HzO is saturated or unsaturatedHow much (NHaJSO4 remains dissolved in the solution prepared by adding 75.0 g (NHa)JSOa in 20OmL of HzO at 25.0*C?How much (if any) of the (NHa)SO4 precipitates out of the solution prepared by adding 75.0 9 (NHa)SOs in 200 ml of HzO at25.08C?

3 If the solubillity of (NHA)SOa Is 57.5 g NHASOa per 1OOmL of water at 25,0"C determine whether solution that is prepared by dissolving 75.0 g of (NHA) SOa in 200ml of HzO is saturated or unsaturated How much (NHaJSO4 remains dissolved in the solution prepared by adding 75.0 g (NHa)JSOa in 20OmL of HzO at 25.0*C? How much (if any) of the (NHa)SO4 precipitates out of the solution prepared by adding 75.0 9 (NHa)SOs in 200 ml of HzO at25.08C?



Answers

What volume of 0.250 $\mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}$ solution is required to neutralize a solution that contains 5.00 $\mathrm{g}$ of $\mathrm{CaCO}_{3} ?$

Okay to start with malls, there's modularity times leaders because polarities moles per liter. So we'll take the polarity which is .488 moller And multiply it by the volume and leaders. So that's gonna be .0500 leaders. So this is going to give us point 0244 moles. We'll go ahead and change moles to grams Molar masses to 94.20. So we'll see that we need 16 g mm. For our second example, we're going to start with four g of our ammonium sulfate. So let's go ahead and change grams to moles R n H 42. So four is 1 32 0.17. So that's going to give us .033 moles. And we're looking for the polarity which is moles per liter. So we'll take those malls And we'll divide by the leaders which is .4 L. And that will give us .0757 moller. And then finally we're starting with one .75 g of RCUS 04. Let's go ahead and change grams to malls. That's gonna be 159 0.62. Which gives us .0110 moles of RCUS 04. So leaders is going to be molds over polarity. So we'll take our moles And we're told the polarity of the solution is .0-50 moller. So that's going to be .439 L. Well just multiply that by a 1000 ml per liter And we'll get that we need 400 and to clean that up a little bit That we need 439 mm.

Let's first to the delusion calculation to figure out what the final concentration of the solution is. If we have 45.8 mil leaders of a 5.8 Moeller solution and we dilute it to one leader, which is 1000 mill leaders will end up with a 10000.266 Mohler potassium nitrate solution. Then to figure out the volume of this solution that contains 15 g. Potassium nitrate will start with 15 g. Potassium nitrate convert that to moles, potassium nitrate. Then we can use the polarity that we just calculated to convert from moles. Two moles potassium nitrate to leaders of potassium nitrate solution and we get 20.56 leaders. Potassium nitrate solution that contains 15 g potassium nitrate.

In this question, we need to provide the mass of the solid every time where the mass of the solution is provided, as well as the mass mass percentage concentration of the specific solution that we're using. So for instance, number eight. We're working with 85 grand solution With a 2% um mass mass percentage concentration of iron bromide. Now in order to solve these problems, we need to identify the appropriate unit factors that we need to use here. So for this purpose I will, first of all, you need to have a look at the different components that I can use in my unit factors. So if I have a 2% solution I can use the two g of iron remote the salute. I can use the 98 g of solvent And I can use the 100 g solution. Right? I need to calculate the mass of the solid. So I need to calculate the mess of the iron bro. My idea for this purpose, I need to start off with the mass of the solution, Which is 85.0 g of solution. Then in my unit factor I need to get rid of the solution part. So I need to have the 100 crams of solution here at the bottom so that I can divide with that and I need to end up with iron remind to here at the top. I will have the I in bromide and I know of having 100 and solution. I have two g of iron bromine. So if I do this calculation, I end up with 1.7 g of iron bromide right in number B I have 105 Grand solution and I have a five Um sodium carbonate solutions. The 5% is the mass mass percentage concentration of the solution. So let's have a look at the unit factor components here. First, if I have a five the same solution, I can make use of the five grams of solid, which is the Um sodium carbonate. You know to see 03. I can make use of the 95 grams of solvent And I can use make use of the 100 g solution depending on what I need to calculate in the end. Right? So I need to start off well I need to calculate the mess of the salute, which is and a two CO three sodium carbonate. So that must be equal to The mass of the solution that I have, which is one of 5.0 g. Which must be multiplied you with a unit factor. Um let me just Right solution. Year after the 105, It's one of 5.0 g solution. Everything is clear. So must multiply this with a unit factor. Well I have the solution, the 100 grand solution at the bottom so that I can get rid of the solution pot. And I need to end up with sodium carbon idea. So I know that in a 100 grand solution, these five g of sodium carbonate. If I do this calculation, we'll see that the solution unit cancels out. And I end up With the value of 5.25 g off sodium carbonite. Right? So let's just quickly recap in this question, I had to calculate the mass of the solid every time the mass of a of a specific solution um is given as well as the mass mass percentage concentration of that specific solution. So I had to identify the unit factor with which I was then able to multiply the mass of the solution in order to calculate the mass of the solid and the um the unit fact the components of the unit factor that I've identified to use here was the mass of the salute and the mass of the solution.

Hello, everyone. Thanks for joining me, Miss Hallstrom. As we discuss Celje Bility calculations today, we're going to focus on a problem that does that asks us to performance scalability calculation for a solution at two different temperatures. Before we get started, I'd like to you to take a quick look at this image I've included. This particular graph shows us the relationship of Celje bility to temperature, and this says it's for solids. But if you look right here, you'll see that solids are the solid lines. This one also includes some gases, which are beyond the scope of what we're going to take a look at today. But you'll solve lines or what we're going to consider. You'll notice that temperature is on our X axis and we're going from 0 to 100 which makes sense for water. And the substance we're going to be looking at today is potassium nitrate, which is the red line that goes from here to here from here to here. And if you take a look at this image, you'll see the potassium nitrate. Candle three is a substance most affected by a change in temperature on the Y axis were given selling Celje bility in terms of grams of Salyut per 100 grams of solvent. In this case, since these air all a quickest solutions, it's per 100 grams of water now on the line the quantities that we see on the line. For instance, if I'm looking right, I'm gonna try to pick something that sort of close to being right on the line so I can read it easily. This one's pretty close. Um, I can see that at 20 degrees C, about 29 grams of salt Salyut for both of these two, both K and 03 and any an a c L. That's the maximum amount of Salyut that can dissolve in 100 grams of water. That would be the amount to prepare a saturated solution. And a saturated solution contains the maximum amount of Salyut that can be dissolved. The problem were given today was gonna take me a moment to write down. We're going to be working with a solution of K and all three potassium nitrate, and we're starting with exactly and noticed, since we're given exactly a 100 grams l 1/4 sample of a solution at 75 degrees C and this solution is saturated. This is a saturated solution in 75 degrees C and were given to sell you abilities. First, we're given the soluble idiot 75 degrees Celsius were given that 155 grams potassium nitrate will dissolve to create a saturated solution in 100 grams water. Now we're gonna tow. We're told we're going to cool this solution to 25 degrees C and 25 degrees C were given The 38.0 grams of potassium nitrate will be present in a saturated solution at containing 100 grams of water. What we need to find out what we're as to find out is in the process of cooling down 100 grams of sample from 75 degrees to 25 degrees. How much Kono three is going to precipitate out of solution. How much is going to crystallize out of solution? So we're gonna make a plan to do this, and our plan is gonna have three steps. Our first part of our plan is we're going to find mass of Salyut and massive are solvent, which is water in 100 grams of a saturated can. All three solution at 75 degrees Celsius at Step one, our second step, we're going to find the massive Salyut of a saturated solution at 25 degrees Celsius. Given the mass of water from step one and last but not leave Super easy step, we're going to take the massive Salyut that could be contained at 75 degrees C minus the massive Salyut at 25 degrees C. To find out the mass of Salyut that will crystallize. So that's what we're going to do. Let's be again. And if you need Teoh, I'll try to jot all of our numbers back down again. Okay, so let's do our Step one again. I'm just going abbreviate here mass, Uh, Salyut and water in 100 grands saturated K and R three solution. Okay, In order to do that, this is actually going to be pretty good. And I'm going to give you again what we were given. 155 grams of Cano three will dissolve in 100 exactly 100 grams of water at 75 degrees C. Let's figure out what percentage k on all three. We have percentage of Kono three, I have 155 grams of candle three and to figure out the percentage of K and all three in the solution, I'm going to add this number and this number 1 55 grams plus 100 grams times 100 times 100. And my answer for this is going to be 60 0.8 percent. I knew I was gonna run out of room for my percent Here. We try to squeeze this in little bit. Times 100 equals 60.8 percent que on all three since we were given a 100 gram starting solution 60.8% of 100 grams. I'm sure you see where I'm going with This is 60.8 grams of K N 03 are initially present. This is the mass of Cannell three. It's initially present, since I have a 100 grams sample, 100 grams minus the 60.8 grams. That's K and all three gives me 39.2 grams of water initially present. And this is how much water we're gonna have present at both temperatures. The amount of water will not change for step two That's not a very nice too, but I think you can figure it out if I have. If we have 39.2 grams of water are solvent when we cool it to 25 degrees Celsius. How much Salyut for K Anil three. Can it contain? Okay, so this is a super simple calculation. Were given the following information. We know that 38.0 grams of K and all three will dissolve in 100 grams of water to give a saturated solution. So how many grams of Kono three will dissolve in 39.2 grams of water? Simple solve for the X and X gives us 14.9 grams of K and all three. I'll just switch pins for the last step here for Step three if you'll recall, and I'll just run back to the previous page. Here. Here is the amount of Kono three that was present. We started with 60.8 grams of candle three in a saturated solution. We're going to subtract 14.9 grams of candle three in our saturated solution. This one was at 75 degrees Celsius. The sun's at 25 and 45.9 grams of Cannell three will crystallize out of solution and that is our final answer. So this problem really doesn't have any really tough math in it. Um I think that the biggest problem that I see in a problem like this is that I'm going to go back to this page when I'm given 100 grams of saturated Enel three I see that people often forget that this is not Ah 100 grams saturated anal three solution. People forget that this is not 100 grams of water. It's 100 grams of solution. So make sure to read your questions carefully. Thanks so much for joining me. See you soon.


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