How does raoults law work

The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards. You could prove to yourself that that does not affect what we have been looking at by re-drawing all these diagrams with the slope of that particular line changed.

You will find it makes no difference whatsoever. We can calculate the vapor pressure of the solution in two ways, depending on the volatility of the solute. If the solute is volatile, it will exert its own vapor pressure and this amount is a significant contribution to the overall vapor pressure of the solution, and thus needs to be included in the calculations.

On the other hand, if it is nonvolatile, the solute will not produce vapor pressure in solution at that temperature. These calculations are fairly straightforward if you are comfortable with stoichiometric conversions. Because the solute is nonvolatile, you need only determine the change in vapor pressure for the solvent. Using the equation for Raoult's Law, you will need to find the mole fraction of the solvent and the vapor pressure of the pure solvent is typically given.

The vapor pressure of water alone is What is the new vapor pressure of Kool-Aid? To solve for the mole fraction, you must first convert the 2 L of water into moles:.

Calculate the vapor pressure of a solution made by dissolving The vapor pressure of pure water is A bigger change that the glucose example above. The only difference between volatile and nonvolatile solutes, is that the partial pressure exerted by the vapor pressure of the volatile solute and the vapor pressure of the solvent must be accounted for.

The sum of the two will give you the total vapor pressure of the solution. What are the partial pressures of benzene and toluene in a solution in which the mole fraction of benzene is 0. What is the total vapor pressure? The vapor pressure of pure benzene is Now that we know the mole fractions and vapor pressures, this problem is a cinch.

Solve for x H2O. If you rearrange the Raoult's Law equation, you can solve for P o. Jim Clark Chemguide. This observation depends on two variables: the mole fraction of the amount of dissolved solute present and the original vapor pressure pure solvent.

Ideal vs. Nonideal Solutions Raoult's Law only works for ideal solutions. Why Raoult's Law works If you look review the concepts of colligative properties, you will find that adding a solute lowers vapor pressure because the additional solute particles will fill the gaps between the solvent particles and take up space. Limitations on Raoult's Law In practice, there's no such thing as an ideal solution!

However, features of one include: Ideal solutions satisfy Raoult's Law. In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent.

The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event! The nature of the solute There is another thing that you have to be careful of if you are going to do any calculations on Raoult's Law. Raoult's Law and Colligative Properties The effect of Raoult's Law is that the saturated vapor pressure of a solution is going to be lower than that of the pure solvent at any particular temperature.

How to Calculate the Vapor Pressure of a Solution We can calculate the vapor pressure of the solution in two ways, depending on the volatility of the solute. Nonvolatile solutes These calculations are fairly straightforward if you are comfortable with stoichiometric conversions. The total number of moles is therefore There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy.

Because of the level I am aiming at, I'm just going to look at the simple way. Remember that saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again.

A certain fraction of the solvent molecules will have enough energy to escape from the surface say, 1 in or 1 in a million, or whatever. If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time. But it won't make any difference to the ability of molecules in the vapour to stick to the surface again. If a solvent molecule in the vapour hits a bit of surface occupied by the solute particles, it may well stick.

There are obviously attractions between solvent and solute otherwise you wouldn't have a solution in the first place.

The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapour phase - it is less likely that they are going to break away, but there isn't any problem about them returning. If there are fewer particles in the vapour at equilibrium, the saturated vapour pressure is lower.

Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. In practice, there's no such thing! However, very dilute solutions obey Raoult's Law to a reasonable approximation. The solution in the last diagram wouldn't actually obey Raoult's Law - it is far too concentrated. I had to draw it that concentrated to make the point more clearly. In an ideal solution, it takes exactly the same amount of energy for a solvent molecule to break away from the surface of the solution as it did in the pure solvent.

The forces of attraction between solvent and solute are exactly the same as between the original solvent molecules - not a very likely event! Suppose that in the pure solvent, 1 in molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time.

In an ideal solution, that would still be exactly the same proportion. Fewer would, of course, break away because there are now fewer solvent molecules on the surface - but of those that are on the surface, the same proportion still break away. If there were strong solvent-solute attractions, this proportion may be reduced to 1 in , or 1 in or whatever.

In any real solution of, say, a salt in water, there are strong attractions between the water molecules and the ions. That would tend to slow down the loss of water molecules from the surface.

However, if the solution is sufficiently dilute, there will be good-sized regions on the surface where you still have water molecules on their own. The solution will then approach ideal behaviour. According to Raoult's Law, you will double its partial vapor pressure.

If you triple the mole fraction, its partial vapor pressure will triple - and so on. In other words, the partial vapor pressure of A at a particular temperature is proportional to its mole fraction. If you plot a graph of the partial vapor pressure of A against its mole fraction, you will get a straight line. Now we'll do the same thing for B - except that we will plot it on the same set of axes. The mole fraction of B falls as A increases so the line will slope down rather than up.

As the mole fraction of B falls, its vapor pressure will fall at the same rate. Notice that the vapor pressure of pure B is higher than that of pure A. That means that molecules must break away more easily from the surface of B than of A. B is the more volatile liquid. To get the total vapor pressure of the mixture, you need to add the values for A and B together at each composition. The net effect of that is to give you a straight line as shown in the next diagram.

What do these two aspects imply about the boiling points of the two liquids? There are two ways of looking at the above question:. For two liquids at the same temperature, the liquid with the higher vapor pressure is the one with the lower boiling point.

We'll start with the boiling points of pure A and B. Since B has the higher vapor pressure, it will have the lower boiling point. If that is not obvious to you, go back and read the last section again!

For mixtures of A and B, you might perhaps have expected that their boiling points would form a straight line joining the two points we've already got. Not so! In fact, it turns out to be a curve. To make this diagram really useful and finally get to the phase diagram we've been heading towards , we are going to add another line. This second line will show the composition of the vapor over the top of any particular boiling liquid. If you boil a liquid mixture, you would expect to find that the more volatile substance escapes to form a vapor more easily than the less volatile one.

That means that in the case we've been talking about, you would expect to find a higher proportion of B the more volatile component in the vapor than in the liquid. You can discover this composition by condensing the vapor and analyzing it.

That would give you a point on the diagram. The diagram just shows what happens if you boil a particular mixture of A and B. Notice that the vapor over the top of the boiling liquid has a composition which is much richer in B - the more volatile component.

If you repeat this exercise with liquid mixtures of lots of different compositions, you can plot a second curve - a vapor composition line. The diagram is used in exactly the same way as it was built up. If you boil a liquid mixture, you can find out the temperature it boils at, and the composition of the vapor over the boiling liquid.

For example, in the next diagram, if you boil a liquid mixture C 1 , it will boil at a temperature T 1 and the vapor over the top of the boiling liquid will have the composition C 2.

All you have to do is to use the liquid composition curve to find the boiling point of the liquid, and then look at what the vapor composition would be at that temperature. Notice again that the vapor is much richer in the more volatile component B than the original liquid mixture was. Suppose that you collected and condensed the vapor over the top of the boiling liquid and reboiled it.

You would now be boiling a new liquid which had a composition C 2. That would boil at a new temperature T 2 , and the vapor over the top of it would have a composition C 3.