In this video, we're going to talk about colligative properties. Now you might be wondering, what is a colligative property? A colligative property is a property that is dependent on the concentration of solute particles and not the identity of those particles.
Now there's four colligative properties we're going to talk about. Boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure. So let's start with...
The first one, boiling point elevation. So what does that phrase tell you? Well, when you add solute particles to a solution, the boiling point will elevate or it will go up. Let's use water as an example here. Pure water has a boiling point of 100 degrees Celsius at sea level.
That is at an atmospheric pressure of 1 atm. The boiling point does depend on pressure. As you go up a mountain, the boiling point of water decreases. But at 1 atm, let me do that again, the boiling point is 100 degrees Celsius. Now what happens if we add salt to water?
As you add salt to water, the boiling point of the solution goes up. Salt is the solute and water is the solvent but combined they make up the solution Now just to give you some numbers here this is going to be the molality of The solution and this is going to be the boiling point if you use a one molar NaCl solution if you have that the boiling point is going to be approximately 101 degrees Celsius If you have a 5M NaCl solution, the boiling point will be approximately 105 degrees Celsius. Technically it's like 105.5.
And if you use a 10M NaCl solution, the boiling point is going to be 110 degrees Celsius. To calculate the boiling point, you could use this formula. The change in temperature is equal to Kb times m times i.
Kb is the boiling point elevation constant, m is the molality, i is the Van't Hoff factor. By the way, I think I said 105.5 for a 5m solution. It should be like 105.1. The Kb for water is 0.51. And if the molality is 5, and the Wendt-Hoff factor is 2, the reason why the Wendt-Hoff factor is 2 is because when sodium chloride dissolves in water, it creates two solute particles, Na plus and Cl minus.
So when you multiply these three numbers out, you're going to get 5.1. That's the change in temperature. So the original boiling point was 100. Thus, the boiling point is going to increase by 5.1. So it's going to be 105.1. So thus you can see why boiling point elevation is a colligative property.
It depends on the concentration of the solute particles. As the concentration of the solute particles go up, the boiling point of the solution goes up. And also it depends on the Van Gogh factor as well.
The more ions that are dissolved in a solution, the boiling point will go up as well. Now the next type of colligative property that you need to be familiar with is something known as freezing point depression. So think about what this expression tells you.
The word depression means something low, something that's down. So as you add salt to water, the freezing point, it goes down, it decreases. The formula for the freezing point depression is equal to negative KF. times the molality times the van't Hoff factor. The negative sign is to help you to see that the freezing point goes down.
If you don't use the negative sign, then Kf is going to have a negative value. So if you have a positive sign, Kf will be negative. In the case of water, it's going to be negative 1.86 degrees Celsius per meter.
molality. By the way, molality is equal to the moles of the solute particles divided by the kilograms of solvent. So that's how you calculate the molality of the solution, for those of you who might be wondering.
So as more salt is added to water, the freezing point, let's put FP for freezing point, it goes down. And this property is very useful, particularly in the wintertime. Perhaps, let's say for those of you who live up north, you probably noticed that during the wintertime, salt is added to the roads in order to prevent freezing.
And when you add salt to water, it makes it difficult for the liquid water molecules to freeze. And so that's why adding salt to ice, it helps ice to melt because the freezing point goes down. And that's why salt is typically added to the roads in the wintertime. It makes it difficult for ice to form, but easy for ice to melt. Now let's add some numbers to this situation.
So let's see the relationship between molality and freezing point. If the concentration is zero, that is the concentration of the solute, we have pure water. Water freezes at zero degrees Celsius.
Now, if we have a 1M sodium chloride solution, if you multiply negative 1.86 times 1 times the two ions in the sodium chloride solution, you're going to get 3.72. But if you apply the negative sign, it will be negative 3.72. So that means the freezing point will decrease by 3.72 units from zero.
So the new freezing point of the solution will be negative 3.72 degrees Celsius. If we were to increase the solute concentration by a factor of 5, the freezing point will increase by a factor of 5. That is the change in the freezing point. So it's going to be negative 18.6 degrees Celsius. Now let's say if we were to use a 5M solution but of a different substance, AlCl3.
The identity of the substance doesn't matter, but what matters is the quantity of the solute in the substance. Aluminum chloride can break up into four ions. We can get the aluminum 3 plus cation. So that's one ion, and we can get three chloride ions. So one formula unit generates four ions, which means our I value is four.
So if you multiply negative 1.86 times five times four, that will give you a freezing point, or the change in freezing point, which will be negative 37.2 degrees Celsius. But starting from zero... the new freezing point will be that value. So now you can see why freezing point depression is a colligative property because this property depends on the concentration of solid particles. As the concentration goes up the freezing point depresses it goes down.
It doesn't depend on the identity. of the solute particles, but it depends on the concentration. For instance, if I were to use a 1m solution of Ki, both of these have two ions per formal unit.
The identity is different, but the concentration is the same. Thus, the freezing point will be the same. And that's what makes it a colligative property.
It depends on the concentration of the solute particles and not the identity of the solute particles. Now let's move on to the next colligative property. The next one is vapor pressure.
This property also depends on the concentration of the solute particles. Now let's briefly talk about why. The vapor pressure of the solution is equal to the mole fraction of the solvent multiplied by the vapor pressure of the solvent.
Now the mole fraction of the solvent, let's say if we're using a salt water solution, water is going to be the solvent. So this is going to be the moles of H2O divided by the total moles of the solution, which will be the moles of... The salt that we have, let's use sodium chloride, plus the moles of water. So looking at this equation, we could see that the vapor pressure of the solution depends on the moles of solute particles.
So if we were to increase the moles of the solute particles, what's going to happen to the vapor pressure? Will it go up or will it go down? Well as we increase the moles of solute particles, we are increasing the value of the denominator of the fraction. A fraction has a numerator and it has a denominator. Whenever you increase the value of the denominator of a fraction, the value of the whole fraction goes down.
There's an inverse relationship between the two. So as we increase the moles of solute particles in the solution, the mole fraction of the solvent, that is this entire fraction, goes down in value. As the mole fraction of the solvent goes down in value, the vapor pressure of the solution goes down in value. And so we see an inverse relationship between solute concentration and the vapor pressure of the solution.
As the moles of the solute goes up, the vapor pressure of the solution goes down, making it a collicative property. Now let's consider the fourth one. The fourth colligative property we're going to talk about today is known as osmotic pressure.
Here's the formula for osmotic pressure. It's pi is equal to MRT times the Van Gogh factor I. Capital M is the molarity. R is the gas constant, 0.08206. and it's liters times atm divided by moles times kelvin so because r has a unit's kelvin the temperature has to be in kelvin keep in mind the kelvin temperature is the celsius temperature plus 273.15 the molarity is different than molality keep in mind molarity is the moles of solute divided by the liters of solution.
And let's contrast that with molality. Molality is equal to the moles of solute divided by the kilograms of solvent. So both of these quantities are different forms of concentration. But the result is the same. As we increase the concentration of the solute particles, the osmotic pressure is affected.
In this case, there is a direct relationship between the two. The osmotic pressure goes up. So let's review the four colligative properties.
Number one, boiling point elevation. As you increase the concentration of the solid particles, boiling point goes up. Number four, osmotic pressure. As you increase the molarity, osmotic pressure goes up. So for those two, there's like a direct relationship.
Now for number two and three, freezing point depression and vapor pressure, there's like an inverse relationship. If you increase the solute concentration, the freezing point goes down. And if you increase the solute concentration, the vapor pressure goes down. So let's just write that so you can see a summary.
So as we increase the concentration of the solute or the salt, The boiling point goes up and the osmotic pressure goes up. The vapor pressure goes down and what was the other one? Freezing point. The freezing point goes down as well.
So that's the relationship between the salary concentration and the four colligative properties. Now let's work on a few problems associated with boiling point elevation and freezing point depression. So in this example problem, we have 20 grams of sodium hydroxide dissolved in 200 grams of water. Our goal is to calculate the boiling point of the solution.
And we're given the boiling point elevation constant, Kb. It's 0.51. So what can we do here?
This is the formula that we need. The change in the boiling point is going to be equal to Kb times the molality of the solution times the Van't Hoff factor. The delta T is the difference between the boiling point of the solution and the boiling point of pure water. So I'm going to move this to the other side of the equation.
So the boiling point of the solution is going to be the boiling point of pure water plus Kb times m times i. So we know the value of Kb. And for the Van't Hoff factor, sodium hydroxide consists of two ions, the Na plus ion and the hydroxide ion. So when sodium hydroxide is dissolved, That one formula unit becomes two ions.
Thus the Van't Hoff factor is two for this problem. Now, the only thing that we're missing is the molality of the solution. Molality is moles of solute divided by kilograms of solvent.
So we're given the grams of the solute, 20 grams of sodium hydroxide. What we need to do is convert it to moles. To do that, we need the molar mass of NaOH.
So we need the periodic table. The atomic weight of sodium is 22.99. Let me get my periodic table just to make sure I have it correct.
And yes, that's it, 22.99. For oxygen it's 16, and for hydrogen it's 1.008. So this is 39.998, which I'm going to round to 40 because 20 and 40, it's like just half of each other. So one mole of sodium hydroxide has a mass of approximately 40 grams of sodium hydroxide. So the unit grams cancel, and now we have moles of solute.
We need to divide this by the kilograms of the solvent. The solute is NaOH, the solvent is water. So we need to convert 200 grams to kilograms. One kilogram is 1,000 grams. So anytime you need to convert from grams to kilograms, simply divide by 1,000.
200 divided by 1,000 is 0.2. So we're going to divide the moles of solute by 0.2 kilograms of solvent. So this here is going to give us the molality. of the solution. Now let's get rid of that and let's plug in the numbers.
So 20 divided by 40 that's 0.5. 0.5 divided by 0.2 is 2.5. So that is the molality of the solution. Now let's plug in everything into this formula. So we have the boiling point of water.
or at least we know what it is, the boiling point of pure water at atmospheric pressure of, let's say at sea level at an atmospheric pressure of 1 atm, that is 100 degrees Celsius. Now Kb is 0.51 with the units Celsius per molality times a 2.5 molal solution times a Waha factor of 2. So 0.51 times 2.5 times 2. That's 2.55. And then once we add that to 100, we're going to get the boiling point of the solution, which is 102.55 degrees Celsius. So that's how you can calculate the boiling point of a solution when you dissolve a solute in water.
Now let's move on to the next problem. Number two, determine the freezing point of a solution if 400 grams of aluminum chloride was dissolved in 1600 grams of water. And we're given the KF for water, it's 1.86.
So the boiling point of the solution is going to be the boiling point of the pure solvent, which is water again. But this time it's going to be plus. The Kf value times the molality times the Van't Hoff factor. Now we're dealing with aluminum chloride. When you dissolve aluminum chloride in water, it's going to break up into the aluminum 3 plus cation and we're also going to get three chloride ions.
So notice that we get a total of four ions per formula unit. Thus, the Wendt-Hoff factor is going to be 4 in this example. Now, let's calculate the molality.
So let's start with the grams of solute, just like we did in the last problem. So we need to convert grams of aluminum chloride into moles, which means we need the molar mass of AlCl3. So we have one aluminum atom and three chlorine atoms.
Using the periodic table, the atomic weight of aluminum is 26.98. And the atomic weight of chlorine is 35.45 times 3. So let's perform the computation. So this gives us a molar mass of 133.33 grams per mole. So this tells us that 1 mole of AlCl3 equates to a mass of 133.33 grams.
Now our next step is to divide the moles of solute by the kilograms of solvent. So we have 1600 grams of water. If we divide 1600 by 1000, it's going to give us 1.6 kilograms of solvent. So remember, the molality is equal to the moles of solute divided by the kilograms of the solvent.
so this is going to be 400 divided by 133.33 which is essentially 3 and 3 divided by 1.6 gives us this answer 1.875 and so that's the morality of the solution now let's plug in everything to get the final answer So the boiling point, I mean rather the freezing point of the solution is going to be the freezing point of water, which is 0 degrees Celsius plus KF. Now, whenever you add a solute to water. the freezing point goes down. The topic is freezing point depression.
When you add a solid to water, the boiling point goes up. That's the expression boiling point elevation. So in this problem, the Kf for water is going to be negative because it depresses or decreases the freezing point.
So this is going to be negative 1.86 times a molality of 1.875 times a Van't Hoff factor of 4. And so we're going to get negative 13.95 degrees Celsius. So that is the freezing point of the solution. Now let's move on to the last problem in this video. Which of the following solutions has the highest boiling point?
Is it A, B, C, or D? Well, let's write the equation. We know that the change in temperature is going to be equal to Kb times m times i.
Now, Kb is going to be constant for all of the solutions. We're assuming that the solvent is the same. most likely water.
So the only part that's different is the MI product because for each of the four solutions, I mean the four answers, the molality is different and the Vemha factor is different. So we got to find out which of these has the highest MI product value. So let's consider answer choice A. The molality is 0.35 and aluminum bromide has four ions.
One aluminum ion and three bromide ions. So if we multiply 0.35 by 4, that's going to be 1.4. Now let's move on to answer choice B.
The molality is 0.75, but glucose doesn't ionize into ions. Glucose is just a solute particle that doesn't ionize. So it has a Van't Hoff factor of 1. Now moving on to part C, we have a molality of 0.50, and calcium chloride has three ions, Ca and the two Cl-ions.
So 0.5 times 3 is 1.5. Finally, answer choice D, the molality is 0.8. Sodium chloride breaks up into two ions, so 0.8 times 2 is 1.6.
So therefore, Answer choice D is the answer because we have the highest MI value for that solution