Raoult's Law , vapor pressure lowering?

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How does Raoult's law apply to solids dissolved in liquids. like NaCl dissolved in water
how would we be able to determine how much did the vapor pressure decrease in our solution?

Usually if we have two liquids we can use each of the individual pressures of the liquids and multiply by Molar percentage and then add each partial pressure to get the vapor pressure for all the solution

Do solids have a vapor pressure in order for us to use that equation?

I would think it does apply to small amounts of dissolved solids. Not sure though, Kaplan only says:

Raoult's law holds only when the attraction between the molecules of the different components of the mixture = the attraction b/w the molecules of any one component in its pure state.

The NaOH bond and the NaCl bonds are both ionic bonds, so they are equal strength attractions.



Edit: Found a page that had more info:

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.

http://www.chemguide.co.uk/physical/...ultnonvol.html

Oh i got it ..
Thanks for the link made it clear
So to calculate the change in vapor pressure we just use the
molar percentage of solute * pure vapor pressure of solvent
Then we subtract it from pure vapor pressure
Excellent !!