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How many grams of sucrose $\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)$ must be added to $552 \mathrm{g}$ of water to give a solution with a vapor pressure $2.0 \mathrm{mm} \mathrm{Hg}$ less than that of pure water at $20^{\circ} \mathrm{C} ?$ (The vapor pressure of water at $20^{\circ} \mathrm{C}$ is $17.5 \mathrm{mmHg} .)$

$$1.3 \times 10^{3} \mathrm{g}$$

Chemistry 102

Chapter 12

Physical Properties of Solutions

Solutions

University of Central Florida

Rice University

University of Kentucky

Lectures

03:58

In chemistry, a solution is a homogeneous mixture composed of two or more substances. The term "solution" is also used to refer to the resultant mixture. The solution is usually a fluid. The particles of a solute are dispersed or dissolved in the solvent. The resulting solution is also called the solvent. The solvent is the continuous phase.

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In physics and thermodynamics, the natural tendency of a system to change its state is its tendency to increase the entropy of the system. It is a measure of the disorder in a system.

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At $60^{\circ} \mathrm{C}$…

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A solution is prepared by …

Hello, everyone. And thanks for joining me, Miss Hallstrom, for this demonstration of a vapor pressure calculation. First of all, vapor pressure is a collective property to see if I got up in here. Which means vapor pressure depends on how many the number of salyut particles that Aaron solution. I'm just gonna write parts right there. Number of cell, you particles that are in solution. I'll give you a little hint here. We're going to be using sucrose as our Salyut. And sucrose is a non electrolyte. And that's important for us because one model of molecule of sucrose when it dissolves, dissolves into one particle. Each molecule is a self contained particle, and it does doesn't associate into anything else. We'll talk more about the problem on their next's page of this, but I'm going to give you a calculation. The pressure of all right, this little nicer. The pressure of the Salyut is equal to the mole fraction of the solvent in this case, water. And what times the pressure of the solvent again, in this case, water. Here's the equation we're going to use. If you want to jot that down for reference, it would be a good time to do it. Pause it. Do that. I'm gonna go to the next page. I'm gonna jot this down again. Up here, Salyut equals Here's a mole fraction for a solvent. Multiply that by our so event. Okay, Now let's talk about her. Problem were asked to solve. Four are unknown. Are junk is grams of sucrose. How many grams of sucrose that's are unknown? Our givens in this problem are we have 552 grams of water. Remember, this is our I'm going to switch colors here. This is our Salyut. This is our so event. Oops. That's our solvent. We are also given the following scenario. We are told that the solution the solution vapour pressure is to Mm H G. And I don't have the correct numbers. Sig figs. Right here. So I better get that, um, 2.0 mm H g less than waters vapour pressure at 20 degrees. The whole systems at 20 degrees, which is waters vapor pressure. 20 degrees is 17.5 millimeters of mercury. So let's go ahead and fill a couple things in here. Now that we know him. R p for Salyut is 17.5 mm. H G minus 2.0 mm H g for 15.5 M h g. I can't get my g in there. Okay, we know that p for my solvent equals 17.5 Mm H g. These are two of the values that we need. So I have this and I have this The mole fraction. I'm gonna run to the next page and transcribe this. So p of Salyut was 15.5 mm h g p for the sol vent is 17.5 millimeters mercury and then we need our mole fraction for solvent. And remember, the mole fraction there are solvent is water, so it's going to be bowls of water. I'm gonna write this up here so I don't run out of room malls of water divided by moles of solution Moles of waters. Easy to figure out the moles of water. We're going to simply convert grams of water, two bowls of water as follows. We were given 552 grams of water and we're going to use 18.2 grams per mole, The molar mass of water and let me check my math here. I'm not going around this till later. 30 point 633 Moles of water. Now the bulls of solution is a little bit more challenging. How are we going to get bowls of solution? We know the moles of solution. Make a little note down here. The moles of solution is going to be equal to the bowls of water which we know, plus the moles of sucrose which we don't know. So we're going to substitute in this expression for our mole fraction of solvent denominator. Okay, let's go ahead and set up our equation, remember? All right, this down one last time. Solvent and pressure of our solvent. Okay, so we have 15.5 Elizabeth. Many millimeters mercury equals 30.633 bowls of water divided by 30.633 bowls plus x times 17.5 M m h g. Oops. Sorry about that. Okay, I'm going. Teoh, take a couple shortcuts here, but I'll discuss them really quickly. The first thing I'm going to Dio is divide each site by 17.5 millimeters mercury and I get 0.8 eight by seven equals 30.633 moles over 30.633 moles plus X. Next I'm gonna multiply each side by 30.633 moles plus X get too close to the edge. They're gonna have to squeeze this in. So excuse me, I'm gonna take this times 30.633 plus X. And this should remind you that you are going Teoh. Um, distribute this when I distribute this. When I take Let me write it down here again. Just so it's a little easier to see. I get 27 0.132 plus 0.8 857 X equals 30 633 I left units off here, but you know I'm dealing with balls now. I'm going to subtract this number from each side. Subtract this number and I get. When I subtract that number, I get zero point 8857 X equals 3.5 014 Divide each side by this switch colors one more time and I end up with three point nine 532 malls of sucrose. Now, this is not grams yet. We have one more step. I think you're going to see what this one is going to be. We need to convert moles to mass for sucrose. And to do that, we're going to use a bowler mass for sucrose of 3 42.3 grams per mole. So from our previous page 3.9532 moles of sucrose multiply that by 3 42.3 grams per mole. When I'm done, I get 1353. I can have three sig figs, so I'm gonna have 1.3. Let me double check this and see. Yep, It looks like three sig figs, 1.35 times 10 to the third grams of sucrose. That looks like a lot there. But I tried toe really include each and every step. So it's probably is not as much as you originally are thinking. It's not that tough to dio. And go ahead and combine all of this into one equation and you don't need to do each and every one of those math steps like I did it one at a time. Okay, thanks for joining

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