A solution prepared by dissolving 100.0 g of a mixture of sugar 1C12H22O112 and table salt (NaCl) in 500.0 g of water has a freezing point of - 2.25 °C. What is the mass of each individual solute? Assume that NaCl is completely dissociated.

Video transcript

Hi everyone for this problem. It reads when a 100 g mixture of glucose and potassium chloride was dissolved and 1.25 kg of water, the resulting solution has a freezing point of negative 2.79°C, calculate the mass of each cell. You consider potassium chloride to be fully dissociated in the solution. So what we want to answer here is the mass of each cell you. So what this means is our salutes our glucose and potassium chloride, which means our solvent is water. And in order for us to calculate the mass of each saw you. Let's go ahead. And first, because we're given the freezing point, let's go ahead and use the freezing point depression to find the total morality of the solids in solution. Because the total morality of the solution is actually the osmolarity of the solution. And the way that we find that is osmolarity is equal to I times M. Where I is the van Hoff constant. M and M is morality. So, we'll have a formula then where the change in temperature, the change in temperature for the freezing point is going to equal negative, I times K F, times M and K F. Is the mole al freezing point depression constant. Alright, so since the solvent is water, the value of K F can be looked up in the in a table. And when we look up this value for water, which is our solvent, we're going to get K F is equal to 1. degrees Celsius times kilograms per mole. Okay, And this is the value for water. So the free and one other thing we need to know is that the freezing point of pure water is 0°C. Okay, so freezing point. So let's write here pure water. So we have the moolah freezing point depression constant K. F. For water and the freezing point is zero degrees Celsius. So with the formula that we have, we can write out that the change in temperature for the freezing point is equal to the temperature of the solution minus the temperature of the solvent. Okay, so let's go ahead and solve for this first, the change in temperature for the freezing point. Okay, so we're going to get the temperature for the solution was given in the problem and that is negative two, 2.79°C. And the temperature of the solvent is 0°C. Alright, so that means then our change in temperature for the freezing point is equal to negative 2.79°C. Alright, so putting the values of the change in temperature of the freezing point and the K. F constant. We can find out the I times M. So we can find out the Van Hoff's constant, times the morality. Okay, and so we're gonna plug these values in that we just saw four into this equation. So what we get is the change in temperature for the freezing point, which we just calculated is 2.79°C is going to equal I times M Times are KF value which we set for pure water is 1.86°C times kg per mole. Alright, so let's isolate I times M. Okay, so I times M if we move this to one side is going to equal -2.79°C over -1.86°C times kilograms per mole. Alright, so that means the value for I times M is going to equal 1.50 mo per kilogram with that we can with that we know that the total morality of the ions slash particles and solution is 1.50 moles per kilogram. So this is important information that we need to know. So now what we can do Step two because this was step one to solve this problem. So step two now is we can use the mass of water to find the total moles of ions or particles in the solution. So what we're going to do is let's convert this to the total moles of ions slash particles. Okay, so we'll say here that the total moles of ions slash particles Is going to equal. So let's start off with the solvent. Okay, so we have 1.25 kg of water. That's what we're told in the problem. So this is going to be our starting point. So 1-1.25 kg. Okay, and we just said that there is 1.50 most per most of solute per one kg of solvent. Alright, this right here. So we're going to use that right here so that we can set up a dimensional analysis. Okay, so there's 1.50 moles of Saw you, her one kg of solvent. So here we see that. So we can go ahead and write 1.25 kg of solvent, which is water. And we'll see here that the kg of solvent cancel. And we're left with Moles of saw you. So what we're going to get here as an answer for total moles is equal to 1.875 moles. So with that now that we know the total moles, we can find the individual moles of the solute. We're going to use the atomic masses from the periodic table to find the molar masses. Alright, so let's go ahead and write out the molar masses for both of our salutes. So the molar mass of glucose. Using the periodic table, Our molar mass for glucose is 180.18 g per mole. And the molar mass of potassium chloride. Using the periodic table is equal to 74. 83 g per mole. Alright, so let's say that there are XG of glucose. So we'll say there's x grams of glucose and why grams of potassium chloride. Okay. So if we assume that then we can write out an equation which will tell us that x plus Y equals 100 g. And we're going to isolate for one of these variables. So let's go ahead and isolate for X. If we want to isolate for X, that means X is going to equal 100 minus y. Okay, so the molds of each substance now we have everything set up so that we can calculate the molds of each substance. So we're going to need to use some substitution here. So what this means is the moles of glucose. So the moles of glucose Is going to equal x over 180. moles. And the moles of potassium chloride is going to equal to y Over 74.5483 moles. So it's going to equal that over the molar masses that we wrote up above. Okay, And now we can set this equal to the total moles. So this equation that we wrote for each we're gonna set it equal to the total moles. And so what that means is When we write this out, total moles is going to equal these two, set to the actual total moles that we calculated, which is 1.875. Okay, so that 1.875. We calculated it right here. So we're gonna set those equal to our total moles. So our total moles is going to equal x over 180. mole plus two Y over 74. mol And this is going to be equal to our total moles which is 1.875. Okay, so now we're just going to simplify so that we can solve for our one of our variables. So the variable that we're going to solve for here is let's simplify. Okay, so what we're going to get when we simplify this is 100 - over 180.18 plus two y Over 74. mol. Excuse ME, Yep. Yes, 75 74.5483. We can take out the moles. We don't need to write that now. So this is going to be equal To 1.875. So we're just simplifying. So if we simplify the left side some more, what we're going to get is 0. -0. Or 0. Y plus 0. eight. Why is equal to 1.875. Alright, so for the left side, when we simplify we'll get 0. 1278 Y is equal to one -0.555. Okay, so let's go ahead and simplify some more. So 0.0 to 1 to 78 Y is equal to 1.32. And now we want to solve for the variable Y. So we'll go ahead and divide both sides by 0.21278. And when we do that, what we'll get is why is equal to 62. g. And remember up above we said that we're going to let y equal our grams of potassium chloride. Okay, so here we said, why is our grams of potassium chloride? And now that we know why we can solve for X because X plus Y equals 100 g. And so that means to software X. It's going to be 100 minus Y. Okay, so we know what why is and why is equal to g, which we said is our grams of potassium chloride. So now let's go ahead and calculate X. We know that X is equal to let's do this in a different color. We know that X is equal to minus Y. So that means X is going to be equal to 100 minus g. So X is equal to 38 g And we said that X is equal to our grams of glucose. Okay, so our grams of glucose. Alright, so now let's go ahead and write out our final answers. Okay, and then we'll go ahead and highlight that and be done. So for glucose, which we said is represented by X. Glucose is equal to 38g, and potassium chloride Is equal to 62g. So that is going to be the mass of each saw. Ute. Okay. That is it for this problem. I hope this was helpful.