In this example we're going to calculate the freezing point depression of water, we add
304.7 grams of glucose to a liter of water, and what we're going to use as an approximate
equation, this is the mole fraction of water in the mixture, the ideal gas constant, the
freezing point of pure water, and it's of course the same value that's here, this is
the freezing point of the mixture, this is what we're trying to determine, and this is
the enthalpy change of fusion, and let's make some assumptions, the big one is ideal solution,
in a separate screen cast we go through more details of where this equation came from.
In order to do this calculation we're going to need to calculate the mole fraction of
water, so what we'll do is calculate the mole fraction of the solute, so we need to determine
the number of moles of the solute, we have 304.7 grams of the glucose, determine its
molecular weight, 180.2 grams per mole, so this corresponds to 1.691 moles of glucose.
We have 1000 grams of water in a liter, and it's 18 grams per mole, so this is 55.55 moles
of water. So the mole fraction of our solute, 1.691 moles of glucose over the total number
of moles, which is the water plus the glucose and this gives us a mole fraction of 0.0295
which means that the mole fraction of water is just 1 minus this, so the mole fraction
of water that we'll substitute into the equation 0.9705. So I reproduced the equation in here,
and then I rewrote it in order to solve for what we're interested in, which is the change
in the freezing temperature, so now we're going to substitute in the values into this
equation, and so I've put in the gas constant in units that are consistent with the heat
of fusion value, the freezing point of pure water is zero degrees centigrade, 273 Kelvin,
has to be absolute temperature, and then the log of the mole fraction of water, the units
cancel, so the right side has units of Kelvin, and if we then calculate this number, minus
3.09 Kelvin. So the temperature change, it would mean the temperature where the mixture
freezes is 269.9 Kelvin, or minus 3.1 essentially centigrade, we've dropped the freezing point
by that small amount. There's alternate ways that you'll see this freezing point depression
equation written. So I reproduced the equation here, we can write this as one minus the mole
fraction of the solute, now it turns out that when X is small, this left side of the equation
is essentially minus X of the solute, just a mathematical relation, and so this is often
the format where of course delta T is just T for the solution, minus T of the pure material,
and if we use this equation we get almost the same temperature change, we get minus
3.04 decrease in temperature. The other thing to keep in mind is if we were to add a salt
that disassociates then we have to account for that in the mole fraction because if we
have, for example, potassium chloride, and we added 1.691 moles, then we have to account
for the fact that we're essentially adding twice as many moles, so we're adding really
3.382 of ions, potassium ions and chloride ions and we have to then use that number in
the calculation.