which-is-more-concentrated-a-10-mathrml-solution-with-20-mathrmg-of-sucrose-mathrmc_12-mathrmh_22--2

Name: which is more concentrated a 10 mathrml solution with 20 mathrmg of sucrose mathrmc_12 mathrmh_22 2
Uploaded: 2020-10-16
Duration: 4 min 13 s
Channel: Numerade
Description: Which is more concentrated: a 1.0 $\mathrm{L}$ solution with 20 $\mathrm{g}$ of sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ , or a 1.0 $\mathrm{L}$ solution with 20 $\mathrm{g}$ of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?$

Question

Which is more concentrated: a 1.0 $\mathrm{L}$ solution with 20 $\mathrm{g}$ of sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ , or a 1.0 $\mathrm{L}$ solution with 20 $\mathrm{g}$ of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?$

Vishal Sharma · Accepted Answer

in problem seven, which is more concentrated. So Ah, one littered solution with pentagram. So cross with Conta Graham So cross a former locks who crosses C 12 age 22 or 11 or one later Solution or one later solution are one later solution with Conte Graham Glucose formal off glucose is C six age 12 or six So in this case we first understand what is the majority? So malaria iti is the number off malls number off moles off salute divided by the volume off solution in leaders. Okay, so the molecular mass off so crosses So the molecular moss off So cross So cross is 3 40 to cram Permal and volume is also given That is one later and moss is Conte Graham So we find out the polarity. So Mueller iti is simply the number off moles It&#x27;s a number off more physical toe mosque our molecular mosque Okay, they end divided by the volume in later That is one later So on solving we get country upon 3 42 to 1 later. So on solving we get the majority is zero point 058 Moeller and in this case, one little solution with 20 g glucose or the molecular mass off glucose is 1 30 g per mole and morality is the number off. Moles off salute and number off. Moles off salute number off one&#x27;s physical toe. Weight over molecular weight into volume is again that is given one later. So volume is again given in the ocean. That is one later. So on solving this week find out the morality. So polarity is in this case off glucose is 0.111 Moeller So we can say that the one little solution. So in this case, we can say that the off one litter solution with Don t gram glucose with 20 g glucose that means C six h 12 don&#x27;t take around glucose. That means C six age 12 or six is more concentrated is more concentrated is more concentrated because, uh, see that in case off this Demel Arat e is in case off this across the minority is 0.58 mola And in case off the glucose theme, elaborate E is zero point Triple one Mueller So 0.111 Moeller. So that means the majority off this glucose solution is, uh, more than that off the majority off this across solution. So that means we can say that the A solution one later solution with 20 g glucose is more concentrated.

Angelina Chavez · Answer

in this problem were asked if a 46% sucrose solution by Mass has a higher concentration than a 1.5 mil Aridjis solution. We can determine this by doing a hypothetical calculation with our 46% sucrose solution. Let us first start by saying that we&#x27;re going to take 100 milliliters of our solution. That is 46% sucrose. We can determine the mass of this solution by using the density we were provided so we can go ahead and multiply, and we find that we have 121 grams. We know from our problem statement that are massive sucrose in the solution will be equal to this number times 46%. So when we go ahead and do that, we find that our massive sucrose is 55.66 grands. We can take the mass of sucrose that we calculated and divide by the molar mass to translate this into a number of moles rather than grams. So we find that our moles of sucrose are equal to 0.163 moles. Now we confined our polarity by taking our amount of moles of sucrose that we calculated and dividing by the volume of solution that we initially took and that is equal to 1000 milliliters or 0.1 leaders. So when we do this calculation, we find that are malaria. T is equal to 1.63 Moeller, and this is higher than the 1.5 sucrose solution that we were told about in the problem statement. So again, we can say that the 46% solution has a higher concentration than the 1.5 Mueller solution.

Emily Harris · Answer

All right. So you&#x27;re going to be taking a solution with a concentration given to us in molar iti and converting that mole fraction mass percent and morality. So this case, our solution consists of glucose, which is C 6 12 06 And, of course, the usual solvent of water. Just h 20 So we&#x27;re gonna start off by actually working part B first, which is the mass percent? No. Again, our mass percent is defined as the mass of the Salyut, divided by the total mass of the solution and, of course, will multiply by 100 at the end to convert our decimal answer to a percent. So when we&#x27;re given this concentration in polarity, what that means is we have 0.944 moles of our Salyut glucose per one leader of the solution. And so, first of all, let&#x27;s go ahead and convert our moles of glucose two grams of glucose by using the molar mass. The more mass will be based on the chemical formula of C six, h 12 06 And so what we&#x27;ll do is we&#x27;ll go to the periodic table and add up six Corbyn&#x27;s. That&#x27;ll be six times 4.1 well at a 12 Hodgins 12 times 1.81 plus six oxygen six times 16. And that gives us a total solar mass of 180 18 grams for mold. By solving that 180.944 times 1 80.18 That gives us an overall mass of glucose as 170 grams. Okay, and so we know that that will be the mass of our soil you&#x27;d for a mass person formula. Now we need to find the total mass of the solution, which will do using the density. So we&#x27;re told that we have one later of solutions, and they provide a density for us in grams for mill. So step one, let&#x27;s convert leaders to milliliters by knowing that there are 1000 milliliters within one later. Then what we can do is use the density as a conversion factor for everyone. Leader. It should have a mass of 1.6 to 4 grams of Excuse me nuts for one mill. There we go. So now we&#x27;ll go ahead and solve that out. So one times 1000 times 1.0, 64 that tells us that our total solution masses one of 62.4 grams that will go on the bottom of our fraction for the total mass of the solution. So let&#x27;s go ahead and solved. 1 70 divided by 1062.4 times 100. We&#x27;ll get us an overall mass percent. Ah, 16% solution. All right, so now let&#x27;s go ahead and go back and work part C over at it. Then we&#x27;ll go back and work part A as the last. So part C is where we&#x27;re at morality, which again is defined as moles of saw Ute per kilograms of soul bent. So here&#x27;s what we need to be thinking about. The solvent is only one part of the solution. So what we just found was the total mass of the solution 1062.4, and that is going to be composed of the masses of the solvent, plus the mass of the soul. You&#x27;d That&#x27;s all you in this case we found had a mass of 170 grams. And so what that means is the remainder of the mass must be due to the salt it so if we take 1062.4 and subtract 1 70 that tells us that are solvent Mass of water is 89 or is 809 2.4 grams. So we are you know, the moles of saw You earlier from normal aren t 0.944 moles and then for our kilograms of solvent. If it&#x27;s 809 2.4 brands, that means in kilograms it&#x27;ll be point a 9 to 4 grams or excuse me kilograms so 0.94 Ford about abaya 0.89 to 4 is going to give us a morality of 1.6 Millau All right, so we had so far apart be with 16% import See was 1.6 Millau Now let&#x27;s circle back around Once we&#x27;ve done that work and finish off with a which is the mole fraction and again typically if they&#x27;re asking for the mole fraction they&#x27;re going to want it in terms of the saw Ute which in this case is our glucose. So the mole fraction Lucas will be equal to the number of moles of glucose divided by the total moles so that would be moles of glucose plus see moles of water again We already know the moles of clickers to be 0.944 No, we need to find the moles of water from the grandes. So we know that the grants based on our work for part C of water is 8 92.4 grams. So what we&#x27;ll do to convert that to moles is use the molar mass of water, which is 18.2 grams per mole. What that gives us when we do by IHS 49 0.5 Moles of our throw One more sick fake on there and that&#x27;s what will be added at the bottom. Okay, so let&#x27;s go ahead and divide 0.94 forward about it by the some of 0.944 plus 49.52. And that gives us a mole fraction of 0.19 for our mole fraction of glucose. And again, that was part A. We were just working backwards

Ronald Prasad · Answer

Let&#x27;s find the mole percent of C 12 h 2011 in the mixture from Solution one. We&#x27;re told that we have a mill aridity of 0.1487 Moeller c 12 h 22 0 11 That tells us that we have 0.1 for 87 moles of C 12 h 22 0 11 in one leader of solution. So we this directly gives us the moles. See, 12 age 22 11 is 0.1487 Moles. Let&#x27;s take that and find the mass. See 12 h 22 0 11 0.1487 Moles Bowler Mass. 3 42.0 grams per mole in. This would yield 50.86 Um, Grams. Let&#x27;s so for the mass of the solution, we&#x27;re told that we have one leader, which is a quote to 1000 milliliters times the density. The density of the solution is 1.18 grams. Permit a leader field 1018 grams. Therefore, the mass of the H 20 would be for the 1000 and 18 grams. Subtract 50.86 grams. This would yield 9 67 0.14 grams of water. And but so for the moles of water in this week with a 9 67.14 graham smolder math of waters, 18 grams per mole, this would work out to 53.73 Uh, moles of water will come back to these numbers later. Let&#x27;s move on to solution to We&#x27;re told that we have a 10% uh, C 12 h 22 0 11 solution, which means that we have, ah, 10 grams of C 12 h, 22 0 11 in 100 grams of solution. So let&#x27;s calculate the moles, uh, see crows 10 grams moller mass equal to 0.0 do nine, 24 moles And that would be the moles of the sucrose. And then we can find the mass of the water would be the mass of the solution. Minus the mass of the sucrose should be 90 0.0 grams of water present itself for the moles of water, 19.0 grams 18 grams per mole from this worker to five moles of water. Now it&#x27;s like at the mixture. So the moles of C 12 h 22 0 11 In total would be, uh, the 0.1487 moles from solution one plus 3.0 292 formals from solution to. So this would yield 0.17 79 moles of sucrose in total, and the moles in total would be put 1487 moles plus 0.292 formals, plus the moles of water solution one is 53.73 moles and solution to was five moles. And so let&#x27;s add up the total number of moles. This would work out to 58.91 And now let&#x27;s calculate the, um, the mole percent of C 12 h 22 0 11 which should be with the 110.1779 moles divided by 58.91 moles, sometimes 100% and we get an answer, and we&#x27;ll do this too. Four significant figures that is more percent of 40.3 0 to 0%. So there is our mole percent in the mixture

David Collins · Answer

this calculation is also pretty much just a conversion calculation. We&#x27;ll start with our middle leaders and converted into leaders by dividing by 1000. And when we have leaders, we can use our molar ity as a conversion factor. Multiply the leaders by the 1.20 bowls per leader to get moles. Then, when we have most of Saul, you we can multiply it by the molar mass in order to get the mass of the Salyut 19.3 g glucose.

Nadia Lara · Answer

were asked to find. Given these conditions, how many milliliters of solution are needed for that? Notice that we have mass polarity, which means we&#x27;re gonna need the mold, a mass that is equal to 180.16 grams. And every more recall that polarity really represents more everyone Leader. So working through these units, I can see that I can find my way two volume leaders to do that. I&#x27;m going to start with the 25 grams of solitude I&#x27;m going to divide by the molar mass with me 180.16 grams and everyone more helps me to boost the unit&#x27;s cramps. Now that I know my polarity notice that this is a conversion factor. It can be used either has written or as the reciprocal shocking right. Is there a point to zero over warm for one over the 10.0? What helps me is if I divide my point is your balls. You can look at what it does to me. Units canceled this looking the leaders of solutions looking, listen to the calculator. I find that it is equal to 0.6 line leaders or 690 No leaders. Another way to do this problem might be to kind of think of it in a couple of different pieces. I know that polarity. The people too more over leaders. Well, we&#x27;ll do you one better. All right. And over B, which are the appropriate variables. Okay, so we have more room volume. We&#x27;re being asked about the volume. So one thing that I could do is sell for volume in this equation and find out volume is equal to the number of more divided by the polarity. So in order to find the volume first, I need the number of molds. I&#x27;m going to do that by dividing my grams of glucose by the molar mass. 180.16 grams. Hopefully, this feels a little bit more comfortable. This is equal ching 0.134 balls. Well, now we have the molds going Numerator divide by the polarity. Volume people to your 0.134 moles invited by the polarity which we said Woz with two old walls in here. This will do is my malls will cancel. I will be left over with units of one over. They&#x27;re a typical of leaders, just just leader. And this is equal to once again near a 0.69 leaders, or 619 little leaders noticed that we did the exact same process. 25 divided by 1 80 divided my pointe shoes here.

Problem

The Environmental Protection Agency has set the u…

Problem 7 Easy Difficulty

Which is more concentrated: a 1.0 $\mathrm{L}$ solution with 20 $\mathrm{g}$ of sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$ , or a 1.0 $\mathrm{L}$ solution with 20 $\mathrm{g}$ of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} ?$

Answer

1 L solution with 20 $\mathrm{g}$ of glucose

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