Here we have a container in which we put a

solution of protein dissolved in water. This container has a membrane. This membrane allows

water to permeate but not the protein and so the green represents the protein solution.

And then to get to equilibrium, water is – there’s pure water surrounding this container. Water

permeates through the membrane, increases the level, creates a hydrostatic pressure

until we reach equilibrium and this hydrostatic pressure represents the osmotic pressure.

And so the question is, can we use this information to calculate the molecular weight of the protein?

What we’re going to do is standard relation for equilibrium, and that’s the fugacity of

water, the pure water, is equal to the fugacity of water in solution. So use the hat in solution.

And then we can write that in terms of the mole fraction of water and activity coefficient

to allow for non-ideality. And the fugacity of pure water at the pressure, so this means

evaluated at pressure P2. P2 means the pressure of the green solution. Now, it’s a reasonable

assumption that this is equal to one because the mole fraction of water is very high. And

then we can write that fugacity of water at pressure two in terms of the fugacity of water

in pressure one. So pressure one on the left side is the pure water. That’s at pressure

one. Right side, mole fraction of water. Fugacity of water at pressure two would be fugacity

of water at pressure one times the Poynting factor correction which is molar volume, the

difference in the pressures divided by the gas constant and the absolute temperature.

And so this term cancels this term. They both correspond to fugacity of pure water at pressure

one. So we have now an expression for the mole fraction of water in terms of the osmotic

pressure which is that pressure difference, P2 minus P1. So I’ve written the osmotic pressure

corresponding to this pressure difference here and then I just rearranged the equation

by bringing the exponential to the other side and therefore have this negative sign. So

I’m going to take the log of both side. I’m going to take advantage of a relation that

says when the mole fraction of water is close to one, the mole fraction of protein is very

small, then I can write this log as the mole fraction of water minus one. Well this difference

is minus the mole fraction of protein because mole fraction of protein plus the mole fraction

of water is equal to one. And then I’ve just rewritten it by multiplying both sides by

minus one. We need the osmotic pressure, we have to calculate that for this equation.

I’m going to call this equation ‘equation 1’. And then we need the mole fraction of

protein. And in the mole fraction of protein, we’re going to have a molecular weight. So

the osmotic pressure, if I represent osmotic pressure it’s just rho g h, the hydrostatic

pressure. So let me substitute in the values. And so what I’ve done is make the substitutions

and then I basically just converting units, kilograms, meters, seconds squared, Newtons,

so Newtons per meter squared is Pascal and I can use the gas constant in units of Pascal.

We now need to calculate the mole fraction of the protein to substitute into equation

1. And just from definition, it’s just the number of moles of protein over the number

of moles in solution of water plus protein. But we’re going to simplify this by saying

this term is very close to zero. Number of moles of protein because of it’s higher molecular

weight is really very small compared to the number of moles of water. So this mole fraction

is just the number of moles of protein over number of moles of water. And what we’re going

to do to make this calculation is divide by the total volume. When I do that, number of

moles of protein is going to be the mass of the protein, which we know in the problem,

divided by the molecular weight which is what we’re trying to determine, and this total

volume. Now the number of moles of water divided by the total volume is essentially the water

density, or the other way of writing it is the specific volume. So one over the specific

volume. This volume is cubic meters per mole. Volume of the liquid, it’s the same as in

equation 1. It’s the same volume that I have here. Mass of protein we know, the total volume

of protein we know, that’s the concentration of proteins. So let’s rewrite this equation.

So I’ve rewritten the equation and I can set that equal to equation one. Osmotic pressure,

molar volume over RT. So I’ve set these two equations equal and notice the molar volume

cancels. So in the equation, I can solve for molecular weight, I just have to calculate

the concentration of protein. So here I’ve calculated the concentration of protein given

2.5 grams and 600 cubic centimeters. I basically just did a unit conversion and I have kilograms

per cubic meter for concentration. I can substitute that in here and solve for molecular weight.

So let me rewrite the equation in terms of molecular weight. So I have that, now I’m

going to substitute in the values and be careful about units. So I’ve used absolute temperature

here. I have units now, they’re going to end up with kilograms per mole. So calculate 114

kilograms per mole which is 114,000 grams per mole. This is the molecular weight of

the protein. Probably not three significant figures accurate but this is the way using

osmotic pressure to get the molecular weight of a protein.