When you mix two liquids 50/50, and one has a freezing point of 20 Celcius and the other has a freezing point of 0 Celcius, does the new liquid have a freezing point of 10 Celcius?
In: 6
Not necessarily, no. It will follow some curve based on concentration. Here’s the curve for an alcohol/water solution:
https://www.engineeringtoolbox.com/amp/ethanol-water-d_989.html
Based on personal experience, it also depends on how well the liquids are/can be mixed. Some friends left a cheap vodka bottle outside, and it separated into ‘ice’ and ‘alcohol’.
And I’ve always wanted to find out how that happened…!
Metal alloys typically have lower melting points than the individual pure metals. Lead = 328C, Tin = 232C, Lead-Tin Solder 183C
It’s way weirder than that.
For example: standard car antifreeze is made of ethylene glycol. Pure ethylene glycol freezes at -13c.
A 50-50 mix of water and ethylene glycol freezes at… -37c!
Mix the two and the freezing point is *MUCH lower than the freezing point of either of the two components of the mixture!*
So freezing points aren’t all that predictable. There’s a curve based on the concentration but that curve won’t necessarily stay between the two starting points.
Only in an ideal world. “Ideal” means “if everything worked perfectly without special rules for the specific thing”, in a way. In an ideal world, a mixture will have a property which varies directly in proportion to the relative contents of each item.
A two-end-member ideal mixture such as this would show a straight line in the value of the property (melting point in this example) and the fractional proportion of the two components. The fraction is in terms of atoms, not weight or mass or volume. How many little balls are present and what fraction are type A and what fraction are type B. In an ideal system, the balls (the particles) do not behave differently depending upon size or shape.
An ideal system is usually only true in reality when one of the two components is at a tiny concentration (variations of Henry and Raoult laws). Few solutions or mixtures behave ideally in actual fact. the two different compounds react to each other differently from how they react to themselves, either interacting more than expected or repelling more than expected.
So, even for an ideal mixture or solution, how you define that idea of 50/50 matters: 50 ml of one thing and 50 ml of another will not be 50% of each in fractional percent except when the molar volume (the number of atoms per unit volume) are the same. If you measure 50/50 by mass, then density comes into play and 50% mass will not be 50% atomic fraction.
The ideal system assumes perfect behavior, that all particles behave identically without regard to size or composition. Real systems differ from that.
Depends on the chemicals involved I believe. Sometimes their reaction can create a new molecule that can be different than 10, like 5 degrees for example.
**ELI5:** No. Boiling points and freezing points are based on the amount of stuff in the more prominent liquid. The more stuff dissolved, the lower the freezing point will be. In your case of two liquids, the liquid that was present in lesser amount would be considered to be “dissolved” in the greater amount liquid. (assuming they’re miscible.)
**ELI’m Older:** Boiling and Freezing points (along with the oft forgotten about vapor pressure) are colligative properties. The equation that determines these things is pretty much all the same.
DT = K * m * i or, in words “The change in temperature of the freezing point (compared to the freezing point of the pure liquid) is the proportionality constant for that liquid (which you can look up in a table) times the concentration of whatever you’re dissolving in that liquid (in molality) times the van’t hoff factor for that thing(or how much the thing you’re dissolving splits apart.)
Basically, ignoring the constant because…. it’s a constant, the freezing point will decrease if you A: dissolve MORE of something or B: dissolve something that physically splits apart more.
For example. 1 molal CaCl2 will have a LOWER freezing point than 1 m NaCl because CaCl2 splits apart into 3 things (one Ca^(2+) ion and 2 Cl^- ions). But 2 m NaCl will have a LOWER freezing point than 1 m CaCl2 because there is literally 4 molal of “stuff” in solution (2 molal Na^+ and 2 Molal Cl^-) where as for 1 m CaCl2 there is only 3 molal of “stuff” in solution (1 molal Ca^(2+) and 2 molal Cl^-.)
Basically, colligative properties only depend on the amount of “stuff” in solution. They do not particularly care what that stuff is.
The van’t hoff factor is the way we get away from “ideal” materials. IDEALLY, 1 M (yes I’m switching to molarity) NaCl would produce 2 molar of “stuff” in solution. 1 molar Na and 1 Molar Cl. But in practice that’s not what we see. The more “stuff” that’s in solution (so the higher the concentration) the less more “stuff” wants to go into solution. So the van’t hoff factor for NaCl is like 1.9 instead of the idealized 2 because, on average, you’ll get 1.9 “things” when you dissolve NaCl in water (instead of the 2 like we’d expect.)
Worth noting that yes, van’t hoff factors DO change with concentration, and the 1.9 above is measured at 0.05 Molar (which is a pretty low concentration.)
**As for WHY?** Well, colligative properties are determined by intermolecular forces. (Forces that occur BETWEEN molecules.) The stronger the forces, the higher the boiling point/lower the freezing will be (because the molecules LITERALLY “hold onto” each other harder therefore making boiling more difficult and making rearranging into a solid more difficult.) When you dissolve something into something else, you often incur STRONGER intermolecular forces such as ion-dipole (which are the strongest intermolecular forces). That’s why it’s not just the pure liquids’ temps averaged. You literally have stronger forces now. The sum of the parts is greater than that of the whole is one way to put it. With the pure liquid you have no way to get ion-dipole forces, but when you dissolve something in it, you do!
Latest Answers