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Chemistry: Molality

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  • Type: Video Tutorial
  • Length: 10:17
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 110 MB
  • Posted: 07/14/2009

This lesson is part of the following series:

Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Physical Properties of Solutions (14 lessons, $22.77)
Chemistry: Characterizing Solutions (4 lessons, $6.93)

This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidation-reduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.

Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electron-rich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.

Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies molecule-based magnetism.

Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.

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When we dissolve a solute into a solvent, like magnesium chloride into water, as we saw in the previous example, again, we're preparing a concentration of solution and there are a number of different units that chemists use to describe that concentration. So, this time, we're going to describe concentration - instead of molar, moles per liter solution - we're going to talk about concentration in terms of molal. So that would be moles of solute per kilogram of solvent. We mentioned that the reason for people wanting to use that unit sometimes, is because it's independent of volume and so it's not sensitive to changes in temperature, which the volume of solvents can be affected by those temperature changes. So, let's think again about the idea of molal, where we take a known mass of substance - again, let's say magnesium chloride. We know, through molar mass, that that's converted into moles of substance. So we put the moles of substance into a known mass now of solution. We don't need to know anything about what the volume of the final solution is. We're conserving mass though. Mass in plus the mass of the solvent is the total mass of our final solution. So, let's see what a calculation would look like using these units. And, in fact, I think the most instructive thing for us to do is look at exactly the same solution that we just looked at and, by the way, if for some reason, you did not just look at this solution, you may want to review the tutorial that covers molarity. Just that dealt with magnesium chloride. So, again, what we're going to do is take 22.4 grams of magnesium chloride and dissolve it in 0.2 liters of water. We'll prepare this solution. We know that the density of water is 1 gram per milliliter, or cubic centimeter, and we also know that the density of the salt solution, once we make up the solution, is 1.089 grams per milliliter. What's the molality, then, of our solution? Now again, remember for molality we need to know ultimately kilograms or at least a mass of our solvent that we use to make up this material. So in this case, our calculation is relatively straightforward because we're given almost all of the numbers that we need right up front. This is actually, of the three concentration calculations we did with this solution, the easiest one in that we already know by just a simple calculation, what our mass of water is going to be and we've got our mass of magnesium chloride, so converting that to moles is straightforward also. So, let's go ahead and do this then. First of all, the rather trivial procedure of converting from volume to mass for water. Trivial in the sense that we know that there is a one to one relationship there. And so, if we take our volume, 0.200 liters and we multiply that by 1 gram per milliliter, and of course, the conversion that there are 1,000 milliliters in every liter, and then finally that there is one kilogram for one thousand grams, then we end up with a total of 0.200 kilograms, and that again would be our mass in kilograms now, of our solvent. Next we need the moles of our solute, so that would be 22.4 grams divided by 95.22 and that's grams per mole of, in this case, our magnesium chloride. And that's going to be equal to 0.235 moles now of magnesium chloride, and so we're just about there. We've got everything that we need to define molality - moles of solute, mass of solvent - so we just have 0.235 moles divided by 0.200 kilograms, and that gives us 1.18 moles per kilogram. And remember, moles per kilogram is molal. And we write a small "m" as opposed to the large "M" that we wrote for molarity.
So, if you go back to the value that we got from molarity here, you'll see that there is just a small difference. Units that you calculate or values you calculate from molarity and molality differ only by about 10% or less usually. So, you can make that comparison.
Now, let's go to a more interesting example where - a less intuitive example - where we have to convert now from one concentration unit to another. So let's consider this problem. Suppose that we know that the density of a 2.45 molar solution of methanol is 0.976 grams per milliliter. So, we're given density and remember, we have to have density if we're going to convert from molar to molal, or vice-versa, so that's a crucial number for us. What's the molality of the solution? Given that we know the molarity.
If you need to find molality, again, we must know the mass of the solvent before it was added to make up this solution. So, we have to backtrack a bit from what happened to get us to this solution value. In other words, let's think of a good strategy for this - let's pick a convenient amount of this solution - say one liter, because we know that this is defined as moles per liter. Let's take that as a convenient amount of the solution, figure out from that moles - well, that's trivial, that's going to be 2.45 moles per liter of solution - and figure out from that what mass of our solvent would have been in order to give us that concentration. Once we know the mass, we know the moles, we'll be able to figure out molality. So, first step is to figure out the mass of the solution in this case. Again, this is where density is going to come in. Because we know that 1 liter - and again, I'm arbitrarily going to decide - imagine I have 1 liter exactly of this solution. Because it's going to be convenient for me. One liter of solution and I multiply that by the fact that there's 1,000 milliliters to every liter, and now we multiply by density. This is going to convert us to mass. We have 0.976 grams per milliliter. And then, finally, we cancel out our units here, and that is equal to 976 grams. Now, what is this again? This is grams of our solution. We still need grams of solvent in order to figure out molal. So what we need to do is take our total mass, 976 grams of, again, our solution and we're going to subtract from this, the grams due to the solute. So, in this case, we know that we have 2.45 moles of solute in that liter - that hypothetical liter of solution we started with. And if we multiply that by molar mass, that would tell us number of grams. So, in this case, we're going to take 2.45 moles per liter - remember that's our units for M here - moles per liter - and multiply that by 32.04 grams per 1 mole of methanol (and in case I misspoke somewhere along the way, I might have gotten magnesium chloride out again - remember this is a solution of methanol now, not magnesium chloride). So, I've got my molar mass of methanol and I'm multiplying that again by my concentration. That's going to give me, then, grams of methanol in that liter of solution. And so that gives me a total, then, of 898 grams of water. Now, we're just about home.
This is the hardest step. So, let's just say this one more time. Total mass of 1 liter of solution minus the mass coming from the solute gives us mass of water. So the last thing I need to do is just take my ration of moles solute divided by mass of water and we're home free. So, I have 2.45 moles and I'm going to divide that by 898 - and that's, again, grams of water, grams of solvent, in other words - or to be correct, we need kilograms, so we'll just have a conversion factor of 1,000 grams per 1 kilogram, and that gives me a final 2.73 moles per kilogram, or 2.73 molal. So, again, notice that the concentration in terms of molal and concentration of molar is not too far apart from each other. That's a good double check on yourself. Those values should come out fairly close to each other.
So, once again, molal differs from molar in that it's per mass of solvent, rather than volume of solution. And, so, we need to go back, calculate in this case the mass of the solvent first, before we can do the ratio and finally arrive at our final concentration.
Physical Properties of Solutions
Characterizing Solutions
Molality Page [1 of 2]

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