 ♫ ‘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
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+
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… ♫

• 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