 Okay, so to continue talking about freezing point depression, I should probably acknowledge that you've heard of freezing point depression before, most likely, and this is not the equation you're familiar with for freezing point depression. So most likely if you remember freezing point depression from having learned about colligative properties in a general chemistry course, you've probably seen an equation like this one delta T of freezing point depression is equal to some freezing point depression constant times a molality, maybe used a lowercase m, we're going to use a lowercase b for molality. So that equation that you're familiar with looks nothing like this equation. This equation does indeed tell us how to solve for the depressed freezing point as a function of some properties of the solution, namely the activity of the solvent. But it looks nothing like this one. So the good news is by making some simplifying approximations, we will be able to reduce this equation down to this one. And in particular, we're going to assume in several different ways that the solution is fairly dilute, meaning we're not pure solvent, but we haven't shifted very far off of the pure solvent, pure A line in making a relatively dilute solution whose freezing point is depressed by just a little bit. So let's start with this expression. First of all, in order to talk about the, rather than the freezing point itself, let's talk about the change in the freezing point. This I'll label it delta T of fusion, the amount by which the freezing point has changed when I go from the pure substance freezing point down to the slightly lower freezing point. The drop in temperature is the upper value, pure liquid freezing point minus the depressed freezing point, the freezing point in the solution, both of which show up in this equation. And another way of writing that equation would be to say that the depressed freezing point would be the pure solvent value minus the freezing point depression. So notice in this definition, even though the temperature has, we would normally define delta T since the temperature dropped, we'd normally think of delta T as being a negative value, but this delta T of fusion is the amount of our freezing point depression. We know that the freezing point is going to be depressed, so we go ahead and define this as a positive quantity, the amount by which the freezing point has been depressed. So that's a relatively common source of confusion. This delta T should come out to be a positive number of the amount of the depression. So if I use those definitions in this equation, this numerator, T naught minus T, that looks just like delta T, the freezing point depression, so I can rewrite log of activity is equal to minus enthalpy of fusion over R, numerator looks like delta T, denominator is going to look like T fusion, that's this quantity in the second equation, T fusion is T naught minus the change, so T naught minus delta T times T naught in the denominator. So now I've got this rewritten so that I can solve for the amount of freezing point depression rather than the freezing point itself. I've removed all the T fusions and replaced them with delta T's. Now comes the step at which we'll assume the solution is dilute to see what happens to simplify this expression when we have a dilute solution. So a few things happen in a dilute solution. First of all, if it's dilute, dilute again meaning I don't have very much solute in the solution. I've got nearly 100% solvent as a mole fraction. So we know from talking about, I suppose I should say approximately equal to, in a nearly pure solvent the activity of that solvent is going to be very close to its mole fraction. So I can replace that activity of A with mole fraction of A and that mole fraction of A is equal to 1 minus the mole fraction of the solute. So that's what I'm going to do on the left-hand side. I'm going to write, instead of log of activity, I'm going to write log of 1 minus mole fraction of solute. On the right side, I'll also use the fact that I'm in a dilute solution and in a dilute solution the temperature hasn't dropped by very much. For typical substances that we care about as chemists, their melting points are in the hundreds of Kelvin. Room temperature is 298 Kelvin. Things that we consider solids, they're going to melt at higher temperatures, they're going to melt at 3, 4, 5, maybe 1,000 Kelvin. So the melting points themselves are many hundreds of Kelvin typically. The change in the melting point, the freezing point depression, that's a number that's only going to be a few Kelvin typically. So the size of this delta T, although we're interested in the magnitude of it, it's interesting to us whether the freezing point is depressed by 2 Kelvin or 5 Kelvin. The magnitude of the change to the actual freezing point, 300 minus of your Kelvin is still a number pretty close to 300. So what that means is this quantity, T naught minus T, that's at least to a first approximation. I'm going to ignore this delta T, the size of this delta T relative to T naught. I can't ignore it as a quantity in the numerator, I can't replace it with zero, it's 100% larger than zero, but I can't replace it, I can't ignore it when subtracting it from T naught because it's only going to be a percent or so of that T naught value. So in the denominator I have this quantity in parentheses which I'll just call T naught, multiplied by another T naught, so I've got T naught squared. In the numerator is the delta T that I haven't ignored and I've still got out front enthalpy of fusion over R with a negative sign. Alright, so log 1 minus x of the solute is negative enthalpy of fusion over R times freezing point depression divided by freezing point of the pure solvent squared. And here I've definitely made a few approximations, these are only going to be true in dilute solution. As our next manipulation, now that I've got this written as log of 1 minus x, we can use again the dilute solution approximation and a dilute solution, x is going to be a small number, the mole fraction of the solute is a small number. So 1 minus x is a number that's fairly close to 1, so I can use a Taylor series approximation and say log of 1 minus x is, there's a negative sign here and here, log of 1 minus x is minus x plus x squared over 2 minus some x cubed and so on. Since x is small, I don't actually care about the rest of these terms, those themselves are small compared to x and I'm just going to replace log of 1 minus x with that first term, minus x and that's going to be valid when x is small compared to 1, which is the same thing as saying when I'm in dilute solution. So when I make that replacement, log of x becomes minus x and on the right hand side I have again enthalpy of fusion over r, freezing point depression divided by freezing point of the pure solvent squared. The negative signs cancel and if what I want to solve for is the freezing point depression, I'll leave that on one side by itself and move everything else to the other side so I've got an r and a t0 squared in the numerator brought up next to this mole fraction and then the enthalpy of fusion goes to the denominator and what I see is the freezing point depression, the thing we're interested in is r times freezing point squared for the pure solvent over enthalpy of fusion multiplied by mole fraction of solute. So that's beginning to look pretty good, that's in the form we expected it to be a concentration times some constants, not quite the same constant, not quite the same concentration. So to continue manipulating that a little bit, if we want to use this equation in terms of a molality rather than a mole fraction, we just need to remember that mole fraction of b, moles of b over moles of everything, we've already made multiple times this dilute solution approximation essentially saying that the amount of solute is small compared to the amount of solvent in a dilute solution. So that's about the same as nb over na, so I just ignored the amount of solute compared to the amount of solvent in a dilute solution. But notice that I can rewrite the moles of a, so I've got moles of b divided by moles of a, that would be mass of a divided by molecular weight of a. So I'm using lowercase m for a mass, uppercase m with a bar on it for the molar mass of a. This is the reason I don't want to use m for molality is because at this point I'm already using m for the mass of a. So rearranging these things now, moles of b divided by mass of solvent, moles of solute divided by mass of solvent, that's exactly the definition of a molality. So moles of b divided by mass of solvent I can write as molality of the solute. And this whole thing is molality of solute multiplied by the molar mass of a. Okay, so that work is all things I can do to replace the mole fraction of b if I don't want to use mole fraction if I'd rather use molality. So now my equation becomes change in the temperature, the freezing point depression is equal to the same r t0 squared over enthalpy of fusion that I had before. But now instead of multiplying by mole fraction I'll multiply by molality and molar mass or molecular weight. So I'll stick that molecular weight out front. So now freezing point depression is equal to a bunch of constants multiplied by the molality. If I take this collection of constants and I define that freezing point depression constant as molecular weight or molar mass of a multiplied by the gas constant multiplied by the melting point of a divided by the enthalpy of fusion of a, if I define that constant then what I have left is freezing point depression is equal to that particular constant multiplied by molality. And that's now in the same form as the equation we might have been expecting all along, much simpler equation of course. After all these manipulations the equation has become simpler, be aware of a couple different things. Number one, be aware of the fact that we've made this dilute solution approximation in order to turn this expression which is accurate, this is a correct expression, into a more approximate solution. We've in a number of places assumed that the solution is dilute and so we've made a series of approximations in order to get to the simpler equation. So this equation is fine if we're in dilute solution, this equation or equivalently this equation would be more accurate if we're not under dilute solution conditions. The other thing to notice is that this freezing point depression is another example of what we've called colligative properties and what makes it a colligative property is the size of this freezing point depression is proportional to the molality of the solute, the amount of solute I've dissolved in a solution. But notice that I don't need to know what the solute is, all I need to know is how many moles of it I have. If I dissolve one mole of sucrose in a kilogram of water or one mole of ethanol or one mole of anything else in a kilogram of water that gives me the same molality. So the identity of the solute doesn't matter. The identity of the solvent certainly does matter. The freezing point depression constant, this value, certainly depends on the solvent. All of these properties, molar mass is the molar mass of the solvent. Boiling point, I'm sorry, the melting point, that's the melting point of the solvent. Enthalpy effusion, that's the enthalpy effusion of the solvent. So I need to know multiple different properties of the solvent in order to calculate this freezing point depression constant. So as always with the colligative property, the size of the change in some property for the solution relative to the pure solvent depends on what the solvent is, depends on the chemical identity of the solvent, but only the amount of the solute and not the chemical identity of the solute. So that's freezing point depression. We can do a very similar derivation for the boiling point elevation, not the change in the freezing point, but the change in the boiling point. And so that's what we'll do next.