Raoult’s Law tells us that if we dissolve something, “anything,” in that liquid (so our liquid’s now a solvent), our solution will have a lower vapor pressure (vapor pressure depression). And this will lead directly to boiling point elevation because more heat is required to make more molecules evaporate so you raise the vapor pressure. These are a couple of colligative properties, so they depend on the number of particles dissolved, so we have to take Van’t Hoff factor (i) into account. Let’s de-jargon this a bit…

In order to evaporate (go from a liquid to a gas) a molecule:

  1. has to be at the air-liquid interface (this is like a lottery – more players (particles) less chance of winning (being at surface))
  2. has to have enough energy to break away from intermolecular attractions

When molecules do evaporate, they whiz around randomly, banging into things & exerting a pressure on remaining liquid, walls of container, other molecules in the air, etc. This creates pressure, which we call vapor pressure.

↑ evaporation leads to ↑ molecules in the air and ↑ vapor pressure

At & above a temperature called the boiling point (b.p.), this vapor pressure’s strong enough to push away whatever external pressure there is & remain a gas. But that’s a sort of “average” thing and even below the boiling point there will still be some molecules with enough energy to break free if they’re at the surface. So you get evaporation even when things aren’t boiling. more here: https://bit.ly/boilingpointvaporization

In an open system, these molecules can whiz far away, so the solution can continue to evaporate until it dries out (liquid nitrogen evaporates really quickly, so we go through a LOT of it!)

But in a closed system (like if you put a lid on the container) the air will become “full.” Molecules in the air have fewer places to go, so they keep colliding with the surface of liquid & getting “sucked back in” (condense back from gas to liquid). As a result, you eventually reach a saturation vapor pressure equilibrium where rate of evaporation (liquid → gas) = the rate of condensation (liquid ← gas). This is a dynamic equilibrium, so molecules are still moving between these phases, but there’s no longer a net movement.

For any temperature, each pure liquid has a unique saturation vapor pressure that depends on how tightly its molecules are held together (if they’re held together tightly it’s harder to escape so they have a lower vapor pressure). 

But when you add in other stuff, you change this vapor pressure. Raoult’s law tells us that if we dissolve something, “anything,” in that liquid (so our liquid’s now a solvent), our solution will have a lower vapor pressure (vapor pressure depression). Why? Because the solute (dissolved thing)…

  1. takes up some of the valuable real estate near the surface, where molecules have a chance of escaping from &
  2. increases the entropy (randomness) of the solution so there’s less benefit to becoming a gas

And these effects are proportional to how much stuff (solute) you add. In other words, vapor pressure depression is a colligative property, meaning it depends only on the number of particles, not the identity of those particles. More non-solvent particles leads to less evaporation leads to lower vapor pressure.

↑ solute leads to ↓ evaporation and ↓ vapor pressure

This is something x-ray crystallographers think about a lot and take advantage of when we do vapor diffusion… More on this here: https://bit.ly/gettingcrystals

So we just need to know what proportion of all the particles (solvent & solute) are solvent. This is just the mole fraction (χ) of the solvent in terms of particles (moles of solvent particles divided by total moles of particles) http://bit.ly/2DRc1mO

The vapor pressure of the solution (at equilibrium) (Psolution) will equal the vapor pressure of pure solvent (P⁰solvent) times the mole fraction (χ) of that solvent (in terms of particles):  

Psolution = (χsolvent)(P⁰solvent)

⚠️ I’m stressing the “particles” part so much because forgetting it can be disastrous…

  • non-electrolytes like sucrose don’t break down any further when you dissolve them (because they’re held together by strong covalent bonds
  • but electrolytes like NaCl DO – they dissociate into their component ions (charged particles). Thus, the # of dissolved particles is greater than the # “solutes” you put in. How many times more is its Van’t Hoff factor (i).
    • For example, table salt, NaCl, breaks into 2 ions (Na⁺ & Cl⁻) so its i=2. So if you dissolve the same number of “sucroses” as “NaCls” you get 2X the # of dissolved particles from the NaCl, so the mole fraction of the solvent ↓ more, so you get ↓ evaporation & an even ↓ vapor pressure

⚠️ Another common error to avoid – make sure you solve for solvent! For colligative properties, solvent’s “all” that matters so it gets the spotlight. 

This is assuming your solutes are non-volative (i.e. they don’t evaporate easily, so they’re stuck in the liquid phase). If you have solutes that are volatile, then they contribute to the vapor pressure in proportion to their amount. So the final vapor pressure will equal the sum of their individual vapor pressures (we call the contribution of each volatile component the partial pressure).

Psolution =  (χsolvent)(P⁰solvent) +  (χvolatile solute t 1)(P⁰volatile solute 1) + (χvolatile solute 2)(P⁰volatile component 2) … for as many volatile components as you have

you might seen this written as Ptotal = ∑(χi)(Pi)

∑ just means sum over (add em all up!)

But, this is all if your solution is acting ideally… And, let’s keep things real – your actual solution is NOT IDEAL! 

To summarize what we talked about already, Raoult’s law says that dissolving something, *anything*, in a liquid, will lower the tendency of that liquid to evaporate. When less evaporation occurs, there are fewer of that molecule in the air so there’s lower vapor pressure. And Raoult’s law says this vapor pressure depression is directly proportional to the number of dissolved particles. e.g. if you were to draw a graph of solvent concentration vs. vapor pressure you’d get a straight line

But if you were to actually perform the experiment to get those values and plot them, your line probably wouldn’t be straight.  If your line bulges up, we call it positive deviation & if it sags down we call it negative deviation

These deviations occur because the solute can interact with the solvent. Raoult’s law is only strictly true for ideal solutions. In an ideal solution, solvent-solvent interactions are the same strength as solvent-solute interactions – they’re basically interchangeable (the solute acts more like a generic “placeholder” -> so adding solute “dilutes” the solvent so there’s less solvent at the air-liquid surface (where it has a chance to evaporate) -> but, for the water that is there, it doesn’t make it any easier or harder to break free from surrounding molecules 

Positive deviations mean that the solute doesn’t lower the vapor pressure as much as you’d expect (i.e. it’s easier for solvent molecules to escape than Raoult’s law says it “should be”). It occurs when solvent-solute interactions are WEAKER than the solvent-solvent interactions – this can happen when you dissolve things with differences in polarity

– e.g. ethanol + water: water’s more polar so it’d rather bind other water molecules 

Negative deviations mean that the solute lowers the vapor pressure LESS than you’d expect (i.e. it’s harder for solvent molecules to escape than Raoult’s law says it “should be”. It occurs when solvent-solute interactions are stronger than the solvent-solvent interactions

Thought exercise: imagine a chain. Each link in the chain represents a solvent molecule. Now take that chain and, between each of the original chain links, put in a new chain link, identical to the original. Your chain is just as strong, so it’s no easier to break apart, but if you were to randomly choose a link to try to remove, you only have a 50% chance of it being one of the originals.

Now imagine that instead of identical links, you added weaker links. Now the entire chain’s weaker and easier to break apart. You still have a lower chance of selecting one of the originals but if you do, you have a better chance of actually being able to remove it. This is analogous to positive deviation

If instead of weaker links you added stronger links, the entire chain would be stronger, so it would be harder to break apart. So in addition to making it harder to “be selected” it’s harder to remove it even if you do select it. This is analogous to negative deviation.

Bottom line: solutes change vapor pressure but not always exactly the way you calculate

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