There are several ways of stating Raoult's Law, and you tend to use slightly different versions depending on the situation you are talking about. You can use the simplified definition in the box below in the case of a single volatile liquid (the solvent) and a non-volatile solute. In equation form, this reads:  In this equation, Po is the vapour pressure of the pure solvent at a particular temperature. xsolv is the mole fraction of the solvent. That is exactly what it says it is - the fraction of the total number of moles present which is solvent. You calculate this using: Suppose you had a solution containing 10 moles of water and 0.1 moles of sugar. The total number of moles is therefore 10.1 The mole fraction of the water is: A simple explanation of why Raoult's Law works 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. Now suppose you added enough solute so that the solvent molecules only occupied 50% of the surface of the solution. A certain fraction of the solvent molecules will have enough energy to escape from the surface (say, 1 in 1000 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. Limitations on Raoult's Law Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law. Features of an ideal solution 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 1000 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 2000, or 1 in 5000 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. 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 (beyond the scope of this site). 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. There was a good reason for that! What matters isn't actually the number of moles of substance that you put into the solution, but the number of moles of particles formed. For each mole of sodium chloride dissolved, you get 1 mole of sodium ions and 1 mole of chloride ions - in other words, you get twice the number of moles of particles as of original salt. So, if you added 0.1 moles of sodium chloride, there would actually be 0.2 moles of particles in the solution - and that's the figure you would have to use in the mole fraction calculation. Unless you think carefully about it, Raoult's Law only works for solutes which don't change their nature when they dissolve. For example, they mustn't ionise or associate (in other words, if you put in substance A, it mustn't form A2 in solution). If it does either of these things, you have to treat Raoult's law with great care. A knowledge of Latin can help us to better understand the English language—specifically, scientific terminology. Consider the word "colligative". Where did that come from? If you know a little Latin, you know that it comes from two Latin words meaning "to tie together". A colligative property is a property of a solution that depends only on the number of solute particles dissolved in the solution, and not on their identity. Recall that the vapor pressure of a liquid is determined by how easily its molecules are able to escape the surface of the liquid and enter the gaseous phase. When a liquid evaporates easily, it will have a relatively large number of its molecules in the gas phase and thus will have a high vapor pressure. Liquids that do not evaporate easily have a lower vapor pressure. The figure below shows the surface of a pure solvent compared to a solution. In the picture on the left, the surface is entirely occupied by liquid molecules, some of which will evaporate and form a vapor pressure. On the right, a nonvolatile solute has been dissolved into the solvent. Nonvolatile means that the solute itself has little tendency to evaporate. Because some of the surface is now occupied by solute particles, there is less room for solvent molecules. This results in less solvent being able to evaporate. The addition of a nonvolatile solute results in a lowering of the vapor pressure of the solvent. The lowering of the vapor pressure depends on the number of solute particles that have been dissolved. The chemical nature of the solute is not important because the vapor pressure is merely a physical property of the solvent. The only requirement is that the solute was only dissolved and that it did not undergo a chemical reaction with the solvent. While the chemical nature of the solute is not a factor, it is necessary to take into account whether the solute is an electrolyte or a nonelectrolyte. Recall that ionic compounds are strong electrolytes, and thus dissociate into ions when they dissolve. This results in a larger number of dissolved particles. For example, consider two different solutions of equal concentration: one is made from the ionic compound sodium chloride, while the other is made from the molecular compound glucose. The following equations show what happens when these solutions dissolve. \[\begin{array}{ll} \ce{NaCl} \left( s \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) & 2 \: \text{dissolved particles} \\ \ce{C_6H_{12}O_6} \left( s \right) \rightarrow \ce{C_6H_{12}O_6} \left( aq \right) & 1 \: \text{dissolved particle} \end{array}\nonumber \] The sodium chloride dissociates into two ions, while the glucose does not dissociate. Therefore, equal concentrations of each solution will result in twice as many dissolved particles in the case of the sodium chloride. The vapor pressure of the sodium chloride solution will be lowered twice the amount of the glucose solution. Summary
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When a non volatile solute is dissolved in a solvent the vapour pressure of the solution is lowered?
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