Raoult’s Law only works for solutes which don’t change their nature when they dissolve, unless you think carefully about it.
They mustn’t ionise or associate. Needless to say, the diagram also includes the melting and boiling pure points water from the original phase diagram for pure water. Now we are finally in a position to see what effect a non volatile solute has on the melting and freezing solution points. Besides, look at what happens when you draw in the 1 atmosphere pressure line which allows you to measure the melting and boiling points.
In any real solution, say, a salt in water, there are strong attractions between the water molecules and the ions. In an ideal solution, that would still be exactly identical proportion. That would tend to slow down water loss molecules from the surface. Notice that if the solution is sufficiently dilute, there going to be good sized regions on the surface where you still have water molecules on their own. Suppose that in the pure solvent, 1 in 1000 molecules had enough energy to overcome the intermolecular forces and break away from the surface in any given time. The solution will then approach ideal behaviour.
If you reduce solvent number molecules on the surface, you are going to reduce the number which can escape in any given time.
It may well stick, if a solvent molecule in the vapour hits a tad of surface occupied by the solute particles. It won’t make any difference to molecules ability in the vapour to stick to the surface again. There are obviously attractions between solvent and solute otherwise you wouldn’t have a solution first and foremost.
There is another thing that you have to be careful of if you are going to do any calculations on Raoult’s Law. There was a great reason for that! Moles number of particles formed, what matters isn’tain’twas not actually moles number of substance that you put into the solution. You may have noticed in the little calculation about mole fraction further up the page, that I used sugar as a solute rather than salt. For each sodium mole chloride dissolved, you get 1 sodium mole ions and 1 mole of chloride ions -in other words, you get twice moles number of particles as of original salt.
There isn’tis notwas not any problem about them returning, this net effect is that when equilibrium is established, there may be fewer solvent molecules in the vapour phase -it is less likely that they are going to break away.
In an ideal solution, it takes exactly energy same amount for a solvent molecule to break away from surface of the solution the surface as it did in the pure solvent. Attraction forces between solvent and solute are exactly nearly identical to between the original solvent molecules -not a very likely event!
If this is questions first set you have done, please read the introductory page before you start. That point is the system triple point -an unique set of temperature and pressure conditions at which it is possible to get solid, liquid and vapour all in equilibrium with each other at identical time. You will need to use the BACK BUTTON on your browser to come back here afterwards. Of course, you will see that the point at which the liquidvapour equilibrium curve meets the solid vapour curve has moved, if you look closely at the last diagram.
This page deals with Raoult’s Law and how it applies to solutions in which the solute is ‘nonvolatile’ -for example, a solution of salt in water.
You can use the simplified definition in the box below in a single case volatile liquid and a nonvolatile solute. For instance, there are several ways of stating Raoult’s Law, and you tend to use slightly different versions relying on the situation you are talking about.
It is also a melting system point -although not the normal melting point because the pressure isn’tis notain’t 1 atmosphere, since the triple point has ‘solid liquid’ equilibrium present. You could prove to yourself that doesn’t affect what we are looking at by re drawing all these diagrams with that slope particular line changed. Besides, for most solvents, these slope forwards whereas the water line slopes backwards. The only difference is in the solidliquid slope equilibrium lines.
It is also the curve showing temperature effect on the saturated vapour water pressure, it can be thought of as the effect of pressure on the boiling point of the water.
Level Because I am aiming at, I’m just going to look at the simple way. These two looking ways at identical line are discussed briefly in a note about half way down the page about phase diagrams. Some information can be found easily by going on the web. There are two explaining ways why Raoult’s Law works -a simple visual way, and a more sophisticated way based on entropy.
The effect of Raoult’s Law is that the saturated vapour pressure of a solution is should be lower than pure that solvent at any particular temperature. The next diagram shows the phase diagram for pure water in the region around its normal melting and boiling points. The 1 atmosphere line shows the conditions for measuring the normal melting and boiling points. That has important effects on the solvent phase diagram.
That saturated vapour pressure is what you get when a liquid is in a sealed container.
That saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where particles number breaking away from the surface is exactly pretty much similar to the number sticking on to the surface again. An equilibrium is set up where particles number breaking away from the surface is exactly similar to the number sticking on to the surface again.