# Tag Info

1

The key to understanding this type of question is that the equilibrium (saturation) vapor pressure of a pure liquid is to a first approximation (that is, assuming ideal behavior) independent of the presence of other gases, such as (in this case) oxygen. Therefore, you can think of $\pu{355 Torr}$ as being always the pressure of the vapor when in equilibrium ...

2

Use the ebullioscopic equation (the first equation) in this Wikipedia article, $$\Delta T = K_\mathrm{b} m$$ First solve for $m_\mathrm{init}$, the initial molality of urea. Second figure out at what molality $m_\mathrm{fin}$ the boiling point is elevated by $\Delta T = \pu{1.5 °C}.$ Since the amount of urea is constant, the % change in the mass of water ...

0

I'm a little late to the party here, but wanted to chime in. The link you posted isn't working anymore, but you appear to be making too many assumptions about the system being measured. Nothing about the measurement of the volume change implies there is necessarily an open space above the liquid for vapor to form. Imagine an open beaker of ethanol with a ...

0

I know I'm really late to the party, but I was interested in this as well. Essentially, it appears that water and ethanol form a solution that has some structure to it, not unlike micellar aggregation. The OH end of the ethanol molecules aggregate with each other and the polar water, which creates a packing structure that has negative excess volume. ...

1

Asking why When you ask why, you want to know about causality. If I ask "why does the cold pack show a decrease in temperature" and the answer is "because the reaction is endothermic", this might be considered a tautology. After all, endothermic means that energy is needed, and this energy can come from the surrounding, lowering the temperature. Why does ...

0

Let’s do a thought experiment to see what is going on. Combine 500 mL of pure H2O and 500 grams of ice, each at 0 C, in a perfectly insulated container. Then add 300 grams of NaCl (also at 0 C). At first, not all the NaCl will dissolve, because NaCl is only soluble 26% in water at 0 C. (The temperature of the water will decrease slightly because of the ...

-2

No. Unfortunately we seem to all be forgetting a fundamental property of ice - I kicked myself when I realized it. Think of a large chunk of ice taken out of your home freezer, sitting in a bucket, floating in its own melt-water. All the liquid water surrounding that ice is at $\pu{0 ^\circ C}$. The extreme surface of that ice, that is exposed to the ...

-1

Raoult's Law is only valid when the solute is non-volatile. Therefore, you cannot apply it.

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Zinc is an amphoteric metal which means it can dissolve both in acidic and basic conditions. However, neutral pH is the transition point. Please see this article Recovery of Rare Earths from Waste Cathode Ray Tube (CRT) Phosphor Powder by Selective Sulfation Roasting and Water Leaching on its behavior in water and eg. I believe the reason is that in eg there ...

0

In order for the $\ce{Ca(OH)2}$ to dissolve, it should become solvated, which requires it to bind some of the OH- ions in molten $\ce{KOH},$ leaving some $\ce{K+}$ relatively bare. We already know that $\ce{Ca(OH)2}$ can be dehydrated easily, while $\ce{KOH}$ cannot (it boils at 1325 °C). The heat of hydration of $\ce{K2O}$ is much greater than the heat of ...

2

IUPAC “Green Book” recommends to use $\mathrm{sln}$ for denoting a solution in general [1, p. 54], referring to earlier Recommendations 1981. Appendix No. IV to Manual of Symbols and Terminology for Physicochemical Quantities and Units [2, pp. 1240–1242]. This has been extensively covered in the following posts: What is the standard way to denote physical ...

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At low concentrations for which the ideal bp elevation/fp depression expressions usually apply, molarity is linearly proportional to molality, therefore the statements are equivalent. This webpage explains nicely why low concentrations are important: Raoult's law only works for low concentration solutions. Why? Well, in order for our approximation to ...

2

Solution 1 Let's assume there is 1 mole total of solvent and solute. With a mole fraction of 0.2 solute, there is 0.2 mole of solute, the rest (0.8 mole) is benzene. We can calculate the mass of 0.8 mole benzene from the molar mass, and express it in kilograms: $$m_\mathrm{benzene} = n_\mathrm{benzene} \cdot M_\mathrm{benzene}$$ = \pu{0.8 mol} \cdot \pu{...

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