47

I recently got a chance to attend a talk by someone who was working on developing analytical instrumentation on Mars. The interesting story is that the initial results by ion-selective electrode was that Mars soil is full of nitrates. Nobody knew on Earth that the nitrate ion selective electrode is far more responsive to perchlorate than nitrate. After ...


19

Your Question: Which "exotic salt" can lower water's freezing point by $\pu{-70 ^\circ C}$? Here is your "exotic compound" although it is not a salt by definition. It is a base: Aqua ammonia, also called ammoniacal liquor, ammonia liquor, or ammonia water, is produced by dissolving ammonia gas ($\ce{NH3}$) in water. The proper chemical name of aqua ammonia ...


15

According to [1, pp. 281–282], solution of sodium chloride $\ce{NaCl}$ prepared by dissolving 25 g of salt in 100 g of water has boiling point of 104.6 °C. Additional data is available in the following table for the aqueous solutions of common salts and bases. English transcription; column 1: Compound; columns 2–6: Concentration, g/100 g water — boiling ...


13

It's not a coincidence at all! If you do an online search for "derivation of osmotic pressure", you'll see how $R$ enters into the derivation. Indeed, that's one of the beauties of the van 't Hoff equation for osmotic pressure – it reveals that (under the modest simplifying assumptions of the van 't Hoff equation derivation) the osmotic pressure created ...


11

Magnesium perchlorate is far from unique. In fact, if you're willing to spend a little money at that hardware store you could pick up some calcium chloride, whose eutectic reaches about -50°C, not quite as low as magnesium perchlorate but still good enough to cover much of the temperature range on Mars. Hydrogen chloride, which becomes ionic upon ...


7

Well, you are certainly right in that the solubility of the gas will grow if you decrease the temperature, and so it will continue all the way down to the freezing point. That's when things get hairy. Ice is quite different from liquid water. It does not dissolve any $\ce{CO2}$ to speak of.(*) So pretty much all $\ce{CO2}$ will leave the solution and go to ...


6

Imagine a pure liquid. If part of it freezes, what happens to the rest of the solution? Well, it looks exactly the same since it's also pure, so nothing. However, this is not true for the system you've described. What happens if there's a solute mixed in liquid? When, now you're concentrating the solute in the remainder of the liquid, so you're creating an ...


5

I can see how this question is troubling. It don't think there is a solution. $K_b = \frac{RT_b^2M_w}{1000\Delta H_v}$ Where $T_b$ is boiling point, $\Delta H_v$ is the molar heat of vaporization, and $M_w$ is molecular weight of the solvent. (see Colligative Properties by W. R. Salzman) So you can see that knowing $K_b$ and $T_b$ is insufficient to ...


5

Here is a - hopefully intuitive - explanation based on kinetics: I start with a glass of water with ice cubes in it, sitting outside in winter on a day where the temperature happens to be equal to the freezing point of pure water. The system is at equilibrium, meaning that the rate of (solid) water molecules melting is equal to the rate of (liquid) water ...


5

It's not good notation on the TAs part. IUPAC defines a molecule as having to be electrically neutral. I see what they are intending, but solute particles would be more correct. This is why, as you noted, compounds that dissolve into ions produce a greater effect than the number of original molecules would suggest and molecules that aggregate in solution ...


4

In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property. A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature ...


4

No, in water the two solutions will not have the same boiling point. Ethanol is a volatile liquid with boiling point lower than water, and lowers the boiling point. Water forms an azeotrope with ethanol. There are positive and negative azeotropes, so when two or more volatile liquids are dissolved, you cannot easily predict whether boiling point will ...


4

I think that your first probabalistic argument is correct, i.e that as the solute interaction increases the effective number of them is reduced. In the second case, as the interaction become repulsive, the effective number of them cannot increase over the maximum number there. They are restricted in their configurational entropy because the repulsion means ...


4

The property of carbon that causes this is the ability to form rings and chains that are stable under atmospheric conditions. The only reason why ‘so many liquids contain carbon’ is basically confirmation bias. There are orders of magnitude more compounds known that contain carbon than those that do not. For the longest time in the history of chemistry, ...


4

It's a lie. Colligative properties do depend on the chemical nature of the solute and solvent - their interaction. The trick is: an ideal solution, in which the solute-solute intermolecular forces and solute-solvent interactions are more or less of the same character, has the property that colligative properties are dictated by composition given only in ...


4

When there is more than $1$ solute in the solution, the total elevation in boiling point ($\Delta T_\mathrm{b,total}$) or the depression in freezing point ($\Delta T_\mathrm{f,total}$) is defined as the sum of each individual value for each type of solute. What is Van't Hoff factor? According to Chem Libretexts, the Van't Hoff factor is defined as: $$i=\...


4

A more plausible explanation is that it is not water that is boiling. The "organic substances" and maybe the "nitrogenous substances" in the juice could include some more volatile components that may pass selectively into the gas phase at a lower temperature than most of the solution. Such a vapor would not be the pure organic or ...


3

A very simple, qualitative explanation: After your solute has dissolved, there are no more enthalpic effects to take into account. The solvation enthalpy is converted to a temperature change, and that's it. For colligative properties, the solute ideally has no significant own vapour pressure, does not precipitate, it just stays in solution. The solute ...


3

I'd like to specifically commment on this: According to the NCERT for Class XII, Part I, [pg. 46, para 3][1], In a pure liquid the entire surface is occupied by the molecules of the liquid. If a non-volatile solute is added to a solvent to give a solution [Fig. 2.4.(b)], the vapor pressure of the solution is solely from the solvent ...


2

While considering colligative properties we always assume solute to be non-volatile. So, it doesn't contribute to the vapor. The amount of vapor will only be due to solvent particles. So, in the problem we can use the following relation: $$ \Delta T = K_\mathrm b\cdot m $$ Initially $ \Delta T = 0.53\ \mathrm{^\circ C}$ so plugin the value of $K_\mathrm b$ ...


2

First of all, it is useful to envision an adiabatic process: add ice at the normal melting temperature to water or cold brine slightly above the MP in a perfectly insulated container (zero heat transfer to the outside) at constant pressure. What do you expect to happen? If the salt concentration is nil (pure water) then ice will melt and the temperature of ...


2

Phase change is not at constant temperature, phase equilibrium is [OP] it's a phase transition so the temperature of the ice should remain constant Here is a counter example: If you add ice cubes to hot water, the ice will melt, cooling down the hot water. In this system, there is no thermal equilibrium, so not all of the components are at the melting ...


2

A pressure cooker will allow boiling temperature to increase to 115 degrees or so by pressuring the cooking space. Its not quite salt, but you could find one in a kitchen. Commonly set to HIGH or LOW using a weight on the vent Setting LOW would be 0.4~0.55 bar and HIGH would be 0.9~1 bar so double atmospheric pressure. Readings are on top of the 1 bar ...


2

Vapour pressure is normally defined as an equilibrium phenomenon. In statistical mechanics terms it is the point where an equal number of molecules are leaving the liquid and entering the vapour and leaving the vapour and entering the liquid. This means that, however small the exposed surface, enough molecules have gone into the vapour phase to create the ...


2

The addition of one molecule (or salt) of solute produces $i$ effective solute particles. This defines the van't Hoff factor $i$, which encodes the strength of the solute with regards to colligative properties. As simple (idealized) examples in aqueous solution, the van't Hoff factor of $\ce{NaCl}$ is 2, as $\ce{NaCl}$ dissociates into the two solute ...


2

For a rough, back of the envelope, sanity check sort of calculation, your second approach is more appropriate than the first. (Though check your value for Avogadro's number.) This is something that should jump out at you if you keep track of units in your calculation. The $\ce{[H+]}$ in the equation should be measured in moles of hydrogen ion per liter, so ...


2

Freezing point depression is a colligative property, well defined by this Wikipedia article (emphasis mine): In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. This means the ...


2

You argument is largely based on enthalpy, whereas freezing point depression and boiling point elevation is (in the situation when the solute only weakly interacts with the solvent) mainly an entropic phenomenon. Because of the increased entropy by adding a solute (i.e. there are more configurations by having two types of molecules instead of one), the ...


2

Consider graphs of the vapor pressure of the liquid and of the solid, both as a function of temperature. Intuitively, both graphs increase with temperature, and intersect at the freezing point. By lowering the vapor pressure of the liquid, we lower the corresponding vapor pressure curve and hence the temperature at the point of intersection.


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