 Okay, so at this point we have quite a variety of different equations we could use to calculate the freezing point depression. Already by making a dilute solution approximation, we arrived at this already somewhat simplified form that lets us calculate the freezing point depression using several properties of the solvent, like its activity, its enthalpy of fusion, its melting point, but then by making a series of more approximations. First of all, converting the activity of the solvent into a relationship that uses the mole fraction of the solute. We lost this negative sign, so this is mole fraction of solute rather than activity of solvent. Then in case mole fraction is not a convenient concentration unit, we've converted that equation to molality, again in the dilute solution limit. And then finally we collected all these terms into one constant, the freezing point depression constant. So we have multiple different equations we can use. Obviously this one is the simplest. To show an example of how we use these equations and how we calculate a freezing point depression coefficient, let's go ahead and work an example. Let's say we're dissolving a solute in water, so let's first calculate the freezing point depression constant for water. Calculate these constants for water. We know for water it has a molar mass of 18 grams per mole, its melting point under standard conditions is zero degrees Celsius, 273 Kelvin, and its enthalpy of fusion is about 6 kilojoules per mole of water that melts. So those are the constants that we need to calculate the freezing point depression for water. So the question is what is that freezing point depression constant? So if we just combine those together as given by this expression, 18 grams per mole, 18.02 grams per mole, multiply that by the gas constant. Units of joules are going to be fine here. They will cancel some joules in the denominator. The denominator has this enthalpy of fusion, 6 kilojoules per mole is the way it was when I gave it to you. We can write that as 6,010 joules per mole and then we can see that the joules in the numerator, the joules in the denominator will cancel. What else do we have? I need to include the normal freezing point, so 273 Kelvin is also up here in the numerator, and now Kelvin will cancel Kelvin, but we do have to think a little more about units because if we want this freezing point depression constant to cancel some units in a molality, remember molality is moles per kilogram. So we've got units of a mass of grams up here. So we can also convert grams to kilograms. That will leave us with our final result in units of, let's see we should have a, one of these Kelvin cancels one of these Kelvin, we still have one left over. We have a moles in the denominator and we have a kilogram in the numerator. So if we combine all these numbers divided by 1000, the result that we get is the freezing point depression constant for water, 1.86 Kelvin kilogram per mole, or if we prefer since a shift in the freezing point of one Kelvin is the same as a shift of one degree Celsius and a mole under kilogram in the denominator, that's like molal. So freezing point depression constant for water works out to be 1.86 degrees Celsius per molal, same thing as 1.86 Kelvin kilogram per mole. So that's the number you may have used before, you probably have used before if you've done freezing point depression calculations, but now we see where that freezing point depression coefficient comes from. It comes from the various properties of water as a solvent. If we want to see what that type of calculation is good for, let's do an example of actually calculating the freezing point depression for an aqueous solution. So let's say we have a, let's use a 7 percent sugar water solution. So by mass 7 percent water and 93 percent, I'm sorry, 7 percent sucrose sugar, 93 percent water. So I'll use a solution with 7 grams of sucrose, 93 grams of water, and sucrose has a molar mass, that's some information that we're going to need of 342.3 grams per mol. The question is what is the melting point or the freezing point of that solution? We can calculate that, I'll finish that calculation up here. Of course the first step will be to calculate the molality of that solution. If we want to use a molality in this expression, the molality of that's going to be moles of sucrose divided by kilograms or the mass of a solvent. Moles of sucrose, we know the mass of sucrose, we know the molar mass of sucrose, so we know how many moles of sucrose we have, 7 divided by its molar mass. If we divide that by the mass of the solvent, 93 grams, or since we want this as a molality, I'll write that in kilograms, 93 grams is .093 kilograms of solvent. So do that calculation, 7 divided by 342.3 and .093, that gives us a molality of .22 mol, .22 moles per kilogram. Now that we have that molality, we can use it to calculate the freezing point. First the freezing point depression, the change in the freezing point, freezing point depression constant times the molality of the solute. So that's going to be our freezing point depression constant for the solvent for water, 1.86 Kelvin kilogram per mole, multiply that by the molality of the solvent of the solute, which we've just computed as .22 moles per kilogram, and that multiplication, 1.86 times .22, that gives us a freezing point depression of .41 Celsius or Kelvin. Now we have to stop and think that's just the change in the freezing point, that's not the freezing point itself. The freezing point of the solution is the standard freezing point minus the change. Remember freezing point depression, the name reminds us that the freezing point gets depressed, the freezing point goes down. So 273 Kelvin minus this freezing point depression of .41, freezing point is dropped by .41 from the standard freezing point, so we could say it's negative .41 Celsius or in Kelvin that's going to be 272.74 Kelvin. So that is an illustration of how to use this simple version of the freezing point depression constant. We can calculate the freezing point depression constant for any solvent that we'd like. The next step will be everything we've done for freezing point depression, we can also do for boiling points as well. So we'll move on to boiling point modifications next.