# Molality and freezing point relationship test

### Freezing-Point Depression to Determine an Unknown Compound | Protocol

For a solution with a liquid as solvent, the temperature at which it freezes to a solid is Discovery and Similarity · Quiz: Discovery and Similarity · Atomic Masses where K b is the molal boiling point constant and m is the concentration of the One valuable use of these relationships is to determine the molecular mass of. A solution's freezing point is related to the molality (m) of the solution. The equation that describes the relationship between freezing point depression and molality is: Add about 25 mL of deionized water to a 25 x mm test tube. 2. Insert a. Use the mathematical relationship between freezing point depression and solution molality. . The magnitude of the freezing point change is proportional to the molality of the . It may be easier to put the test tube in a beaker to take the mass.

In this case, i must be known, and the procedure is primarily useful for organic compounds using a nonpolar solvent.

Cryoscopy is no longer as common a measurement method as it once was, but it was included in textbooks at the turn of the 20th century.

As an example, it was still taught as a useful analytic procedure in Cohen's Practical Organic Chemistry of[5] in which the molar mass of naphthalene is determined using a Beckmann freezing apparatus.

Freezing-point depression can also be used as a purity analysis tool when analysed by differential scanning calorimetry. This is also the same principle acting in the melting-point depression observed when the melting point of an impure solid mixture is measured with a melting-point apparatussince melting and freezing points both refer to the liquid—solid phase transition albeit in different directions. In principle, the boiling-point elevation and the freezing-point depression could be used interchangeably for this purpose.

However, the cryoscopic constant is larger than the ebullioscopic constantand the freezing point is often easier to measure with precision, which means measurements using the freezing-point depression are more precise.

Also this phenomenon is applicable in preparing a freezing mixture for use in an ice-cream machine. For this purpose, NaCl or another salt is used to lower the melting point of ice. Last but not the least, FPD measurements are used in the dairy industry to ensure that milk has not had extra water added.

Milk with a FPD of over 0. This typically occurs simply because the solute molecules do not fit well in the crystal, i. It has to get pretty organized, so let's say it has to look something like this. The water molecules are going to have a regular structure where the hydrogen bonds dominate any kind of kinetic movement they want to do, and all the kinetic movement, they're just vibrating in place. So you have to get a little bit orderly right there, right?

And then, obviously, this lattice structure goes on and on with a gazillion water molecules. But the interesting thing is that this somehow has to get organized. And what happens if we start introducing molecules into this water? Let's say the example of sodium-- actually, I won't do any example. Let's just say some arbitrary molecule, if I were to introduce it there, if I were to put something-- let me draw it again. So now I'll just use that same-- I'll introduce some molecules, and let's say they're pretty large, so they push all of these water molecules out of the way.

So the water molecules are now on the outside of that, and let's have another one that's over here, some relatively large molecules of solute relative to water, and this is because a water molecule really isn't that big.

Now, do you think it's going to be easier or harder to freeze this? Are you going to have to remove more or less energy to get to a frozen state? Well, because these molecules, they're not going to be part of this lattice structure because frankly, they wouldn't even fit into it.

They're actually going to make it harder for these water molecules to get organized because to get organized, they have to get at the right distance for the hydrogen bonds to form. But in this case, even as you start removing heat from the system, maybe the ones that aren't near the solute particles, they'll start to organize with each other.

But then when you introduce a solute particle, let's say a solute particle is sitting right here.

It's going to be very hard for someone to organize with this guy, to get near enough for the hydrogen bond to start taking hold. This distance would make it very difficult. And so the way I think about it is that these solute particles make the structure irregular, or they add more disorder, and we'll eventually talk about entropy and all of that.

But they make it more irregular, and it's making it harder to get into a regular form. And so the intuition is is that this should lower the boiling point or make it-- oh, sorry, lower the melting point. So solute particles make you have a lower boiling point. Let's say if we're talking about water at standard temperature and pressure or at one atmosphere then instead of going to 0 degrees, you might have to go to negative 1 or negative 2 degrees, and we're going to talk a little bit about what that is.

Now, what's the intuition of what this will do when you want to go into a gaseous state, when you want to boil it? So my initial gut was, hey, I'm already in a disordered state, which is closer to what a gas is, so wouldn't that make it easier to boil? But it turns out it also makes it harder to boil, and this is how I think about it. Remember, everything with boiling deals with what's happening at the surface, and we talked about that in our vapor pressure.

So at the surface, we said if I have a bunch of water molecules in the liquid state, we knew that although the average temperature might not be high enough for the water molecules to evaporate, that there's a distribution of kinetic energies. And some of these water molecules on the surface because the surface ones might be going fast enough to escape. And when they escape into vapor, then they create a vapor pressure above here.

And if that vapor pressure is high enough, you can almost view them as linemen blocking the way for more molecules to kind of run behind them as they block all of the other ambient air pressure above them. So if there's enough of them and they have enough energy, they can start to push back or to push outward is the way I think about it, so that more guys can come in behind them.

So I hope that lineman analogy doesn't completely lose you. Now, what happens if you were to introduce solute into it? Some of the solute particle might be down here. It probably doesn't have much of an effect down here, but some of it's going to be bouncing on the surface, so they're going to be taking up some of the surface area.

And because, and this is at least how I think of it, since they're going to be taking up some of the surface area, you're going to have less surface area exposed to the solvent particle or to the solution or the stuff that'll actually vaporize.

You're going to have a lower vapor pressure. And remember, your boiling point is when the vapor pressure, when you have enough particles with enough kinetic energy out here to start pushing against the atmospheric pressure, when the vapor pressure is equal to the atmospheric pressure, you start boiling. But because of these guys, I have a lower vapor pressure. It is less energetically favorable to form a mixed lattice of solvent and solute particles. The solute particles remain in the solution phase.

Only solvent-solvent interactions contribute to lattice formation, so solvent-solute interactions reduce the rate of freezing compared to that of the pure solvent. The temperature at which freezing begins is the freezing point of the solution.

## Freezing and Boiling Points

The solution continues cooling as it freezes, but this continued decrease in temperature reflects the increasing concentration of solute in the solution phase.

Eventually, the solution temperature is so low and so little solvent remains in the liquid phase that it becomes favorable for the solute particles to form a lattice.

Once this point is reached, the temperature remains approximately constant until the mixture has frozen solid. The molar mass of the solute, and therefore the identify of the solute, can be determined from the relationship between the freezing point of the pure solvent, the freezing point of the solution, and the molality of the solution.

Molality, or m, is a measure of concentration in moles of the solute per kilogram of the solvent. This relationship depends on the the freezing point depression constant of the solvent and the number of solute particles produced per formula unit that dissolves. Molality can be expressed in terms of molar mass, so the equation can be rearranged to solve for the molar mass of the solute.

Plugging this into the freezing point equation allows the elucidation of the molar mass, once the temperature difference is known. Now that you understand the phenomenon of freezing point depression, let's go through a procedure for determining the molar mass of an unknown solute from freezing point temperatures.

The solute is a non-ionic, non-volatile organic molecule that produces one particle per formula unit dissolved, and the solvent is cyclohexane. To begin this experiment, connect the temperature probe to the computer for data collection. Insert the temperature probe and a stirrer into the sample container. Set the length of data collection and the rate of sampling. Allow sufficient time in the data collection for the sample to freeze. Set upper and lower limits of the temperature range to sample.

Add 12 mL of cyclohexane to a clean, dry test tube. Wipe the temperature probe with a Kimwipe. Insert the stopper assembly into the test tube such that the tip of the temperature probe is centered in the liquid and does not touch the sides or bottom. In a beaker, prepare an ice water bath. Then, start the temperature data collection.

Place the test tube into the ice water bath, ensuring that the level of liquid in the test tube is below the surface. Continuously stir the liquid at a constant rate. Once freezing begins, allow data collection to continue until the plot has leveled off at a constant temperature. This is the freezing point of pure cyclohexane. Remove the test tube from the ice water bath and allow it to warm to room temperature.

### Chemistry Molecular Weight by Freezing Point Depression

Once the cyclohexane has melted, accurately weigh the solid unknown material on weighing paper. Remove the stopper from the test tube and add the solid. Avoid allowing compound to adhere to the test tube. Replace the stopper and stir the solution until the solid is completely dissolved. It is important that no solid crystals remain. Set the parameters for data collection and prepare a fresh ice water bath. Start collection, place the test tube into the bath, and stir continuously at a constant rate.

Once freezing begins, the freezing point continues to decrease due to the increasing solute concentration. Continue collecting data until the slope of this decrease is evident. When the experiment has finished, allow the solution of the unknown compound to warm to room temperature and then dispose of it according to the procedures for organic waste.