Decreasing Vapor Pressure Q

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If you have a molecule that dissociates into 3 particles vs. one that dissociates into 2 particles. The one that weighs more will cause a much more drastic decrease in vapor pressure?

But if you have 10 g of a substance, then one that weighs the least will cause a bigger change in vapor pressure because it will become more moles?

Can someone explain molecular weight and dissociation to decrease vapor pressure in a way that I'll understand?

 

[deleted329605]

So for vapor pressure, it decreases when you add impurities to a solution causing new because the impurities create new intermolecular interactions cause a greater amount of energy to break them. For these types of problems, the identity of the impurity is not important, vapor pressure depression is a colligative property. However, you must concern yourself with two things. First the van't hoff factor (i) which determines the number of solute particles that your impurity dissolves into. For glucose C6H12O6 it would be 1 because it doesn't dissociate. For NaCl it would be 2, MgF2, 3. Then you have to figure in the actually amount you have. You can get this from the grams of substance and the molar mass of the impurity. So essentially if you have equal masses of impurity with equal van't hoff, the one with the smaller molar mass will have the greater vapor pressure depression simply because there are more moles.

I don't know the vapor pressure equation of the top of my head, but if you look at other equations that are dependent on colligative properites, you see that things like boiling point elevation and freezing point depression are usually based on "i" and moles in some shape or form (molality when T changes, Molarity when T is constant b/c increasing T increases V which changes Molarity).