♫ ‘Blue Hawaii’ ♫

by Alicia Tibbitt [ukulele playing] On the north shore of O’ahu,

in the midst of this miracle mile sits the famous wave, Pipeline. It’s here where the world’s best surfers

come to prove their skill And while they sit out in the water

and wait for that next set of waves to come, it can get kind of crowded out there But when those waves come

and that one surfer decides to drop in, it’s look out below, and the other ones

better get out of the way. Now, this situation actually reminds me of

the colligative properties of solutions. And these are properties

that solutions have that pure solvents do not,

and the reason is simply this, the solute particles get in the way. So, while you prepare

for this next lecture, take a look at some of these

infamous waves at Pipeline. Aloha, and welcome back. This lecture is the last one

we’ll do on solutions. And today, we focus

on the colligative properties. These are properties of solutions

that depend on the number of dissolved particles

and not their type. So, solutions have special properties

that pure solvents do not. And further, these properties only depend

on how many dissolved particles there are. It doesn’t matter if those particles

are sugar particles or if they are some other particle

like salt or CO2, or whatever. It just depends on how many particles

are in the solution. So, the properties of solutions

are different than properties of pure solvents. Now that’s interesting.

Solutions have these properties, but pure solvents do not.

Why is that? Well, the reason is actually

fairly simple. The solute particles just get in the way. Now as we discuss these

four colligative properties, try and see if you can predict how

these solute particles get in the way. The first colligative property

is vapor pressure lowering. Solutions have a lower vapor pressure

than pure solvents. I’ll remind you that we discussed

vapor pressure in Lecture 4. And we’ll just recall that

very briefly right here. Suppose you have a container

with pure water. And you cork the top

of that container. The water will vaporize –

some of it. Those more energetic particles

will vaporize. And once the vapor is up here

those particles will collide with the walls of the container and maybe lose a little bit

of kinetic energy. And some of those less energetic particles

will get trapped again by the solvent. And so, you have water vaporizing

and you’ll have water condensing, and once the system settles down,

the rate at which those two processes occur are equal and pointing

in opposite directions. And so the chemical equation

can be represented like this. Now, the pressure of the gaseous

water particles up here is called the vapor pressure. And, at about 25 degrees celcius

it’s about 24 Torr. Now, the rate of vaporization

actually depends on a couple of factors. It depends on the temperature – so the vapor pressure depends

on temperature, but more specifically,

the rate of vaporization depends on this. And also it depends on the surface area. Now, the vaporization is only occuring

at the surface here. Think of what happens

if we introduce a solute, say sucrose, or table sugar. So if you open up the cork and

you put some sucrose in there, then that sucrose will end up

being dissolved in the liquid. And if you put quite a lot

of sucrose in there, then you’ll have a lot of it dissolved

up here in the bulk phase, and also on the surface. And since it’s distributed

along the surface, it’s actually blocking

some of the surface. And, whereas water would normally

have access to those areas, it no longer has access

to those areas on the surface. And so, the rate of vaporization

is now less than what it was before. And, when you let

this system settle down, the rate of vaporization will equal

the rate of condensation, and you will have a less amount

of vapor up here. So the vapor pressure would be

quite a bit lower. Depending on how much solute

you have dissolved, you might see the vapor pressure

decrease to only 12 Torr. That’s if 1/2 of the surface is exposed. So if the solute particles

are blocking 1/2 of the surface, then the other have half,

water can freely vaporize. And so you’ll only get

1/2 of the vapor pressure. And that’s how it works. Now this little relationship

can be described by Raoult’s Law. Raoult’s Law tells us

that the vapor pressure above a solution equals the mole fraction

of the pure solvent times the vapor pressure

of the pure solvent. So, if the mole fraction of water

is 0.5 (or 1/2), that represents 1/2 of the surface

being available for water to vaporize. So you’ll get 1/2 times

the normal vapor pressure of water. And if the mole fraction is 3/4,

so 75% of the particles are water, then you would have 3/4

times the normal vapor pressure. And so on. So for instance, suppose you wanted to find

the vapor pressur of water above a solution of 1 mole glucose

dissolved in 9 moles water at 25 degrees Celcius. So the vapor pressure of water

above the solution would be the mole fraction of water

times the pure vapor pressure of water. And the mole fraction of water

is nine moles of water divided by the total moles,

which is ten. So you have 9/10 for the mole fraction. And you multiply that

times the vapor pressure of water at that temperature,

which is 23.9 Torr. So 9/10 times 23.9 Torr,

you end up with 21.5 Torr. So it has decreased a little bit. You still have most

of the vapor pressure, because most of the particles are water. Now, we can also describe

the vapor pressure, depending on the concentration

by this graph right here. This is a graphical representation

of Raoult’s Law. And…on the y-axis is plotted

the vapor pressure of water at 25 degrees Celcius,

and the x-axis is the mole fraction of water. So, for instance, at…

a mole fraction of 1, meaning it’s pure water,

the vapor pressure would be 23.9 Torr. So that’s this blue line. But at a mole fraction of 0.5,

then the vapor pressure, you go up to the solid pink line,

it would only be about… oh 13…12 maybe. So you can use this graph to predict the vapor pressure of water

above a solution. And again, if the mole fraction

is maybe 0.75, then you would go here to 0.75,

and go up here, and it’s about maybe 19 or so. So it’s less accurate to use the graph,

probably better to use the equation. But it still gives you a nice,

visual representation. You see, the line is a straight line. And when the mole fraction

decreases down to zero, the vapor pressure also

decreases down to zero. Now what are these dashed lines

above and below the solid pink line? Well those are deviations

from Raoult’s Law, depending on the strength

of the interaction between solute and solvent. Suppose the solute particles

have very strong interactions with the solvent particles. They would prevent the solvent

from vaporizing. So if the interactions between solute

and solvent are strong, the solvent will not want to vaporize. And the vapor pressure would be less

than what Raoult’s Law predicts. And the opposite occurs

if there are weak interactions between solute and solvent. And the solvent would

more readily vaporize, because whereas water is normally

interacting with other water particles, if you substitute those water particles

with solute particles, where the interactions

between the solute and water are very weak, then water will vaporize

more than what’s predicted. And so, this is the case

for weak solute-solvent interactions. In our previous example

of sugar and water, the sucrose was nonvolatile,

meaning it did not vaporize to any significant extent. However, sometimes the solute

is volatile, and so we need to account

for those situations. If the solute has a significant

vapor pressure, then it must also be accounted for. For example, find the total vapor pressure

over a solution that contains one mole water

and one mole ethanol, which is alcohol. The pure vapor pressure

for water is 23.9 Torr and that of alcohol, 44.6 Torr. Now we can imagine what this solution

looks like by this diagram right here. The water and the alcohol

completely mix, they are totally miscible. So, all of the molecules

will be intermingled, and likewise at the surface,

you’ll have water and alcohol molecules. Now the water wants to vaporize

with its own vapor pressure, however the alcohol,

at least some of it is in the way. And likewise with the alcohol,

it wants to vaporize, but the water is in the way. Now to find the total vapor pressure

above the solution, we need to add the vapor pressure

of water, which will be less

than what it normally is, along with the vapor pressure

of alcohol, which will be less

than what it normally is. So the total vapor pressure

will be the pressure due to water, this is applying Raoult’s Law for water, plus the vapor pressure of the alcohol,

Raoult’s Law for alcohol. Now the mole fraction

for water is 1/2, so 1/2 times the pure vapor pressure

of water, 1/2 times the pure vapor pressure

of alcohol. And adding these two components,

we get 34.25 Torr. Now this is the total pressure

above the solution. This pressure right here is actually

half-way between the two individual pressures. And that’s because we have

a mole fraction of 1/2. We can better understand this, perhaps,

by looking at the following graph, which plots vapor pressure on the y-axis

versus mole fraction of water on the x. Now, I’ve drawn two y axes here,

for convenience. If the mole fraction of water is 1,

then the pressure over the liquid, which is pure water in this case,

will be 23.9 Torr. And if the mole fraction of water is 0,

then we have pure alcohol, and the vapor pressure

of the pure alcohol is 44.6 Torr. Now, by connecting these two points

with a straight line, that’s actually accounting

for any mole fraction that we can consider. If the mole fraction is 1/2,

like in our example, then we go up to the line,

and we go over, and we see that we get 34.3 Torr. And this is actually half-way between

the two vapor pressures. We can consider any mole fraction. Suppose the mole fraction

of water is 9/10, or 0.9, we go up to the line, and we go over,

and we’re pretty close to water. If the mole fraction of water is 1/10,

and we go up to the line, and we’re really close

to that of alcohol. So any mole fraction

can be obtained here. Kind of a cool plot. Now this is the case

for any two liquid-liquid solutions, where both the liquids vaporize. Suppose we have component A,

which has its vapor pressure, and component B with its vapor pressure. Plotting mole fraction of B here,

when the mole fraction is 1, that means we have pure B,

then this is our vapor pressure, and if it’s 0, that’s our vapor pressure

which is the same as A. And if we connect this line

then we get the vapor pressure of any mole fraction that we want. Now suppose the two components

have strong interactions, then that will hinder the vaporization

of the two components, and the vapor pressure will be less

than expected. So if the interactions

between A and B are strong, then the vapor pressure will be less

than what we expect. We get this deviation

from what Raoult’s Law tells us. Raoult’s Law tells us to look

along the straight line. But if there are interactions,

then we can deviate this way. And likewise,

in the opposite case, if there are weak interactions

between the two components, then it will be easier

for them to vaporize and we’ll have a positive deviation. Now we can see what the effect

of adding a solute is on the vapor pressure

by looking at the phase diagram. Let’s look at the phase diagram

for water here, and see how the solute affects it. I’ll remind you, we discussed

phase diagrams in Lecture 5. Now, pure water has a phase diagram

given by the green lines here. This is the triple point, and we have our solid,

liquid and gas phases. Now, the vapor pressure curve

is the line going up this way, and this represents the pressure

where the liquid and gas are in equilibrium with each other. So, at this temperature,

water has a vapor pressure right here. At another temperature,

water has a vapor pressure right here. Now adding a solute to the mixture

lowers the vapor pressure. And it does that

at every single temperature. So at every single temperature,

the vapor pressure becomes lower, and so the net effect is to lower

the entire vapor pressure curve. Now interestingly, that’s not the only

part of the phase diagram affected. The solid-liquid equilibrium line

is also moved to the left. And if you study this

and you think about it you might wonder if this

changes the melting point. Well you’re correct – it does change

the melting point – in fact, it lowers it. Because, the temperature at which

solid is in equilibrium with liquid, by adding a solute, that temperature

becomes lower. So, the melting point also decreases. The second colligative property

is boiling point elevation. If you understood why the vapor pressure

of a solution is lower than that of a pure solvent,

then it’s easy to see why its boiling point is also higher. Let’s first recall

what the boiling point is. And it’s the temperature at which the vapor pressure is equal

to the atmospheric pressure. So if you imagine a container

that has some liquid in there, that liquid has a certain vapor pressure. Some of those particles at the surface

are vaporizing with a certain pressure. Now meanwhile, the atmosphere is pressing

down on the surface of the liquid. And if we increased the vapor pressure

of the liquid by heating the liquid up, then eventually,

if we heat it up enough, its vapor pressure will match

the atmospheric pressure. And when that occurs, it’s not just the surface particles,

which vaporize, but particles all throughout the liquid

will vaporize as well. And that’s why you see bubbles

start to form throughout the bulk liquid, because that’s vaporization

occurring everywhere. And that’s boiling. Now, when a solute

is addedto the liquid, that’s going to affect the vapor pressure,

and also affect the boiling temperature. And we can see how by looking at this diagram

right here, that’s going to affect the vapor pressure,

and also affect the boiling temperature. And we can see how by looking

at this diagram right here. We have pure water on the left

and a solution on the right. The solution will always have a lower

vapor pressure than the pure solvent. Meanwhile, the atmosphere

is always pressing down with a pressure of about one atmosphere. If we wanted to boil these two liquids,

we’d have to heat them up in order for their vapor pressures

to reach one atmosphere. Well, we know that the pure water will have to be heated up

to 100 degrees Celsius. And when it reaches 100 degrees Celsius

it’s vapor pressure is 1 atmosphere. And that’s why it starts boiling

at that temperature, because the vapor pressure

is matching 1 atmosphere, which is the atmospheric pressure. But the solution will have

a lower vapor pressure than that. And it won’t be boiling yet. It’s vapor pressure will still be

smaller than one atmosphere. To make it reach one atmosphere we have

to heat this one a little bit higher. And that’s why the boiling point

is higher for a solution. Now we can see this effect

on the phase diagram of a substance as well. If we look at the phase diagram

for a substance, any substance, it shows us the different phases

of the substance. And over here you have

your vapor pressure curve. These blue lines right here

represent the pure substance. And so for a pure liquid, in order for its vapor pressure

to equal one atmosphere, which is represented

by this horizontal line. it will have to be brought

to this temperature right here. When there’s a solution, the solution

will always have a lower vapor pressure, which is represented

by the pink lines right here. And so, in order for the solution to be brought up to one atmosphere

of vapor pressure, it needs to be raised

to this temperature. So you see the boiling point

has increased a little bit, as indicated by the phase diagram. Now, how much does it increase?

It’s a pretty simple relation. The change in the boiling temperature

is equal to the molality concentration times a constant, which is called

the boiling point elevation constant. And these constants are different

for every liquid, every solvent. And I’ve listed several solvents here

along with their constants. Here is some freezing point data

that we’ll get to in just a moment. But over on the right is boiling data. And there are several boiling temperatures

of the pure solvents along with their boiling point

elevation constants. So water boils at exactly

100 degrees Celsius. And, if you look at some

of the other liquids, it’s interesting that the ether boils

at 35 degrees Celsius, close to it. In fact, that’s around

95 degrees Fahrenheit. In fact, if you hold

a container of ether, you might be able to boil it

just with your hands. So, kind of interesting. Water has a boiling point

elevation constant of 0.512 degrees Celsius per molality. So every single molality will raise

the boiling point by 0.512 degrees. If the solution has

1 molality of particles the boiling point increases

by 0.512 degrees, and if there are

2 molality of particles it’s boiling temperature

increases by twice that much, 3 molality of particles

it increases by 3 times this much. It’s this many degrees per molality. So you have to multiply

the number of molality of particles times the boiling point

elevation constant. So for instance,

suppose you wanted to find the boiling point of 1.7 molality

aqueous sucrose solution. So this aqueous means

that water is the solvent. So we find the boiling point

elevation constant for water, and we multiply that by the concentration,

and we get 0.8704 degrees Celsius, which rounds to 0.87, and so,

that’s how much it increases. So we have to add that

to the original boiling temperature, and the new boiling temperature

is 100.87 degrees Celsius. So the solution has a slightly higher

boiling temperature than pure water. Now, the freezing point is also affected

by the presence of a solute, and we can see how that works

by this diagram. This is the third colligative property,

the freezing point depression… it gets lower. So, here ia a solution of some solute

dissolved in water. And this solution is being cooled down,

cold enough to start to form ice. It’s interesting that this ice

that’s forming amidst the solution is pure water. So, there are no solute particles

trapped in the ice, or relatively few. So pure water is what freezes

out of the solution. Now the reason it does this is because of those strong hydrogen bond interactions

between the water molecules, which help pull them together

to form the ice. That’s not really important

for this discussion right here, but that’s why it does it. And, since the pure water

is what freezes out of the solution, these solute particles

that are surrounding that ice are getting in the way. As this ice cube begins to grow

more and more water molecules need to find their way to the ice cube

to make it get bigger. Well, if the solute particles

are getting in the way, it’s gonna make it hard

for a water molecule to find its way to the ice. In fact at hotter temperatures,

that tends to stir up everything, and make it more difficult for water

to find its way to the ice. It would make it more difficult

for the freezing to occur. And so the solution would have to be

cooled down that much more, to allow the water to form the ice. And that’s why the freezing point

is lower for a solution than pure water. And we can also see this effect

on the phase diagram. Here’s that same phase diagram,

and you can see the solution is represented by the pink curve, and over here, this is

your freezing curve. Now the blue line is the pure substance,

and in order to melt or to freeze it would have to be

at this temperature right here. But for the solution, that shifts

the freezing curve to the left, and the freezing temperature is

a little bit colder for the solution. How much does it change

the freezing point by? Well, that’s also given by a similar

equation as we saw before. The change in the freezing temperature

is equal to the molality concentration times the freezing point

depression constant. Now, rather than

jump right in to an example I wanted to first quickly discuss

the case for solutions of ionic compounds. Because when you have

an ionic compound, that adds a little bit

of complication to the mix. For instance, when you take

one mole of sodium chloride, and you dissolve it in water,

we know that ionic compounds dissociate into their cations and anions. So one mole of sodium chloride

will give you two moles of particles. In other words, you’ll have

twice as many particles as what you might expect,

and that means you’ll have double the effect

on the colligative property, because remember colligative properties,

it doesn’t matter what type of particle it is,

it just depends on how many there are. So, when you dissolve

one mole of sodium chloride, you’re gonna get

twice as many particles. Now, there’s one little catch,

and that catch is, it seems like you would get

twice as many particles, but in reality, you’d only have

1.9 times as many. Now the reason that it’s not

two times as many particles is because…

if you imagine a container where you dissolve one mole

of sodium chloride in water and all of those cations become

separated from the anions. So you have all of your cations

and anions floating around independently of one another. It seems like you have

two moles of particles, and that would be a correct assumption,

however, if you look carefully, sometimes, every once in a while, you’ll see a cation

right next to an anion. So, it just happens, you know,

every once in a while, but a small small fraction

of those ions will be kind of close together,

and that’s kind of like one particle. And so, instead of there being

2 moles of ions, you’ll have a little bit less than 2,

it’s closer to 1.9. And this number over here is called

the Vant Hoff factor for sodium chloride. And all ionic compounds have

their own Vant Hoff factors. For instance, magnesium chloride

should dissociate into 3 ions, your magnesium 2+

and your 2 chloride ions, however your Vant Hoff factor shows

that it’s actually closer to 2.7. Iron (III) chloride

should give you 4 particles, your iron 3+ and your 3 chlorides,

but the Vant Hoff factor is actually 3.4. Now these Vant Hoff factors

are determined experimentally, by measuring what the

colligative property is, and going back and comparing it

to what you’d expect, and you’re able to work out,

you know, how many particles are actually in the solution. For instance, if you measure

the change in the boiling temperature, and you know the constant

for the solvent, then you should be able to determine

the molality of the particles, and that’s how you find

these Vant Hoff factors. So whatever the molality

of the particles…is, or whatever the molality

of the solute is you’d have to multiply it

by its Vant Hoff factor. So this equation for the

freezing point depression is modified a little bit to account

for the dissociation of ionic compounds. So you multiply the molality

times the Vant Hoff factor. And then you multiply it

by the freezing point depression constant. And that’s how it’s accounted for

for this colligative property. And the other colligative properties, you have to account

for ionic compounds as well. But we’ll just do it for this one case,

for the freezing point. So for instance,

find the melting point of 0.9 molality iron (III) chloride,

aqueous solution. So, the molality is 0.9,

but it’s iron (III) chloride, so we have the ionic compound

dissociating, not into 4 particles, like we might expect,

but it’s actually closer to 3.4. So when we multiply the Vant Hoff factor

times the molality concentration, times the freezing point

depression constant, you can go back for water,

and now you can find its freezing point depression constant

is 1.86 degrees Celsius per molality. When you multiply these out,

you get 5.6916 degrees Celsius, which will round to 6 degrees. So the freezing point gets lower

by 6 degrees Celsius. So you subtract this many degrees from the original

freezing point for water, and so the new freezing point

is -6 degrees Celsius. The fourth and final colligative property

is osmotic pressure. To see what osmotic pressure is

and how it works, let’s look at the following diagram. Here we have a U-shaped device,

in which we pour pure water on the left and some type of solution on the right. Now separating these two liquids

is some membrane, and we call that a semipermeable membrane. And that’s because it only allows

the solvent to pass through the membrane. So, water is allowed to pass through,

but the solute particles are not. So it stops the solute particles

from passing through. Now if we set this device up,

like that, at the very beginning,

these liquid levels are even. But what we’ll see,

if we wait a little while, is that water will end up flowing

in this direction, through the membrane, and these liquid levels

will end up changing, and the liquid level

on the solution side will end up being higher

than the pure solvent side. Well that’s kind of interesting,

because if water is able to freely travel through this membrane,

why does the water end up doing that? Why wouldn’t it just flow

back the other direction? What’s holding it up there? Well, what’s holding it up there is

the osmotic pressure that’s exhibited by the solute particles in solution. You see, a solution tries

to expand its volume by incorporating more solvent. And what’s doing that

are the solute particles. So these solute particles

are sort of moving around in this solution –

they’re exerting a pressure. And that’s what’s making the solution

expand its volume. And that’s what’s holding up

that extra weight of water. But let’s first take a closer look

at what’s going on on the molecular level. And let’s take a closer look

at this semipermeable membrane. You see, this semipermeable membrane

only allows the solvent to pass through. And if you think about the situation

right at the beginning, when you pour the two liquid in there,

and the two liquid levels are the same, let’s look at this membrane,

and on the left hand side of the membrane you’ll see solvent particles only. And these solvent particles,

a certain fraction of them, will be traveling that way,

and then a certain fraction will be traveling this way

through the membrane. Now, on this side of the membrane,

you will have, not only solvent particles, but also some solute particles,

and those solute particles are blocking up some of the membrane, so they’re occupying some of those sites

next to the membrane, and so you’re not gonna have as much water

traveling the other direction. So, on the left hand side there will be

a lot of water traveling this way. But on the right hand side, only a little bit of water

traveling the opposite direction. And that’s just based on the random motion

of the particles next to the membrane. So, what happens

is that water ends up moving this direction,

through the membrane. And it’s going to keep on doing that until finally there’s so much

weight of water up there, that the weight, sort of balances that tendency for it to travel

in this direction. So, this weight of water up here

is held in place by the osmotic pressure

of the solution. So, it’s the solute particles

that are, sort of, moving around and creating a pressure inside there. So the way you calculate this pressure

is kind of simple. The pressure, which is represented

by the Greek letter, Pi, is equal to the molarity concentration

times the gas constant, R, times the temperature. Now, if you recall

from first semester general chemistry, this equation might look kind of familiar. And that’s because, if you recall

the Ideal Gas Law, PV=nRT, that’s the pressure of a gas

times its volume is equal to the number of moles times the gas constant

times the temperature. And if we rearrange this equation

and solve for the pressure, we get the pressure

is equal to n/V RT. But n/V is the same thing

as the molarity. And so the pressure of an ideal gas

is equal to the molarity times R times T. And that is the same as the pressure

exerted by the solute particles inside of a solution. So, it’s very interesting how the pressure

of a solution is calculated using the same equation

as the pressure of an ideal gas. And it’s not really a coincidence, either. What it’s telling us

is that solute particles behave very similarly

to gas particles. In fact, it has been shown

that solute particles move around with similar velocities

inside of a solution as gas particles do in the air. And so, the pressure exerted by solute

particles on the walls of their container, would be similar to the pressure exerted by the gas particles

on the walls of their container. And that’s why these equations

are very similar. In fact, they’re basically equivalent. Now, solute particles move around and

exert pressure just like gas particles do. So that’s the take home message here. So we can calculate osmotic pressure,

fairly easily. If you know the molarity and you know

the temperature of the solution, then you can just plug it in

and calculate it. But let’s do a rather,

more interesting, calculation, involving osmotic pressure.

Here’s our example. A solution containing 35.0 grams

of the protein, hemoglobin, dissolved in water

to form 1 liter of solution. Now this exerts an osmotic pressure

of 10 Torr, at 25 degrees Celsius. Find the molar mass of hemoglobin. So here, you’re given the temperature

and you’re given the pressure. So, you have the temperature,

you have the pressure, and we always know

what the gas constant is. So we can calculate

the molarity of the solution. Right?

It’s not asking for the molarity, though. It’s asking for the molar mass. But, what you should remember

is that the molarity, which is equal to the number of moles

over the volume, can be expressed

in a different way. You see, the number of moles is also equal

to the mass divided by the molar mass. And so if we replace the number of moles

by mass over molar mass, and then all of that over the volume,

that’s also equal to the molarity. And this simplifies to that,

the molar mass comes down. And so the molarity is equal to the mass

over the molar mass and the volume. And so if you take this expression

for the molarity and you plug it in

for the osmotic pressure, you see the osmotic pressure equals ‘m’

over the molar mass and the volume times ‘R’ times ‘T’. And here, if you look at this expression

right here for the osmotic pressure, we know what everything is,

except for the molar mass. And if we rearrange this equation

and solve for the molar mass, we can then calculate it pretty easily, because we know all of the variables

over here on the right. So doing that… now we have to be careful

of our units too. So let’s be careful of our units

and let’s see how it goes. The molar mass is equal

to the mass, 35 grams, times the gas constant,

0.08206 liter atmosphere per Kelvin mole, and then times the Kelvin temperature. So 25 degrees Celsius is 298 Kelvin. In the denominator,

we have pressure and volume. But we want the pressure in atmosphere. So we take our 10 Torr

and we convert it to atmosphere. And then, we also put

the volume down there, 1 liter. And a lot of units cancel off,

you can see how they cancel off. And what you’ll end up with

are grams per mole. And you end up with 65,047 grams per mole,

which rounds to 6.50 times 10 to the 4th, or just 65,000 grams per mole –

that’s a pretty large molar mass. But again, we’re talking

about a very large protein molecule. There are several applications

of osmotic pressure, and we don’t have time

to go through all of them, but I tried to choose

a few interesting ones. The first one is a common application,

which involves medical solutions. And nurses out there, who have to make up

IVs and inject them into the body, well they have to be careful

and inject the right type of solution. If you inject too salty of a solution,

or too much saline, or too little, that can adversely affect the patient. And, you can imagine cells

in the body in the bloodstream, they have solute particles

inside the cell. And there are also solute

particles outside the cell. And if the concentration of solute outside

is similar to what’s inside, then there is no osmotic pressure

involved here. and so the cell

will maintain its size. But suppose a nurse were to inject

a very salty, or saline, solution. And the concentration of solute

is more outside the cell. Well this cell membrane can act

as a semipermeable membrane. And what happens, again, solvent

will always try to travel to the solution. And so the cytosol fluid

inside the cell will end up trying

to escape out the membrane and go into the extracellular fluid,

and so the cell will end up shriveling up. So, you don’t want that to happen. That’s called a hypertonic solution. A hypotonic solution

is actually opposite, where the nurse might inject

a very dilute IV fluid. And in that case, that can cause

the extracellular fluid to be more dilute relative

to the inside of the cell. And so in that case, solvent will end up

incorporating into the cell, and the cell will end up getting bigger

because more solvent is traveling in through the membrane. And that can cause cells

to increase in size, even rupture. So, we don’t want to have

a hypotonic solution either. You’d probably rather have

an isotonic solution. Right?

That doesn’t disturb the cells. Application two. Osmotic pressure is also the mechanism

by which water moves up a tree trunk. And here in Hawaii, we have

our nice beautiful palm trees, some of them are pretty tall. And, what happens is that these leaves

on top of the tree, water vapor ends up escaping the leaves

and into the vapor phase, and that leaves concentrated

leaf fluids inside the leaf cells. And when there are concentrated fluids

in those leaf cells, the pure solvent, or water,

which is in the ground, will end up traveling up

through the trunk to try to get into

the concentrated region. Remember, pure water, again always travels

towards more concentrated regions because of osmotic pressure pulling it. And so the osmotic pressure

will pull the water up through the trunk. In fact, palm trees may be 50 feet hight,

but some big trees, like in the Redwood Forest

of northern California, which can get up to 1…200 feet,

it has been estimated that osmotic pressure

up to 10 to 15 atmospheres is responsible for pulling

that water up those big trees. So, quite something. Now the lalst application

is making a pickle. You can imagine taking a cucumber.

The cucumbber has a nice skin coating, which can act as a semipermeable membrane. And if you surround a cucumber

by some salty solution, then the water inside the cucumber

will want to travel through the membrane to the salty area outside. And so the cucumber

will end up shriveling up And that’s how you make a pickle. I hope you have enjoyed this lecture

on the colligative properties. This does close up our second portion

of the course on solutions. In the next portion, we move

toward the chemical reaction rate, which deals with how fast

a chemical reaction goes. So we’re getting ready

to switch gears here. Stay tuned.

And aloha! ♫ ‘Blue Hawaii’ ♫

by Alicia Tibbitt ♫ Night and you, ♫ ♫ in blue Hawaii, ♫ ♫ the night is heavenly… ♫

## 4 Comments

## MADHU SINGH

helo

## Rhenius Paul

Sir you are great……..!From India…..!

## pia g.

lowkey looks like so ji sub. idk why

## Nitin Sharma

Sir ur a great