Brandon Snider

I am faced with the question: The vapor pressure of pure ethanol at 60 °C is 0.459 atm. Raoult's Law predicts that a solution prepared by dissolving 10.0 mmol naphthalene (nonvolatile) in 90.0 mmol ethanol will have a vapor pressure of __________ atm. If I convert the milimolar to molar I get: 0.01 mol Nap. 0.09 mol Eth. and 0.10 mol total Thus the mole fractions should be: X_Nap. = 0.01 / 0.10 = 0.10 X_ Eth. = 0.09 / 0.10 = 0.90 Which according to Raoult's Law Vapor Pressure of Soln. = Sum of Vapor Partial pressures 0.459 atm * X_Nap. -{0.10} or 0.459 * 0.10 = 0.0459 Partial Pressure Nap. 0.459 atm * X_Eth. -{0.90} or 0.459 * 0.90 = 0.4131 Partial Pressure Eth. I sum the two and get: 0.459. Which is consequently the same as the pure vapor pressure. I am assuming that this is related to the fact that naphthalene is supposed to be non-volatile. My problem is that this matches none of the answer choices. Any help appreciated. My exam is in 20 mins. Answer Choices: 0.498 0.0918 0.367 0.413 0.790 *Chem software says that 0.413 is the answer. Not sure how.

I'm just having a hard time because I thought that it was illegal to move RT in that way... I really need to take math again. So (P)(V) = (n)®(T) (P) = (n)RT / V = (P/1) = [(n/1)(RT/1)] / V So how is it possible to move the RT... Got it sorry. Just been a long time now. Dividing by RT is just like multiplying both sides by (1/RT) which is perfectly sound. Appreciate the help. It takes some thought.... really need to do more math though.

I have a chamistry final tomorrow and was trying to study. I my professor's notes, I found an altered form of the ideal gas equation PV=nRT. It's been quite a while since I've had a good math course, so I'm wondering if someone here can show me how PV=nRT is rearranged to give n/v = P/RT I'm having this problem because I don't remember some basic math tricks. Been a while. I'd like to understand how this equation is derived. Thanks in advance for the help.

Oops. Sorry. Didn't read carefully. Thought you said "driving." My mistake.

This may be a bad statement, but I'm almost curious to find a way to drop 33 feet quickly to experience this... Live in the mountains at the moment so...EDIT: Also, how much pain and damage are we talking? Not wanting to kill myself. Also, really appreciate both answers. They are really insightful. Edit: Also, D H, would you mind explaining what you mean by pressure doesn't work that way? What I am confused about is the fact that Gravity should be causing atmospheric pressure. It is one thing to assume that there is no gravity, and that the pressure inside of and outside of a system are equal, but why does pressure make us lighter if (our internal pressure = environmental pressure) and then throw atmospheric pressure caused by gravity into the mix. I'm sorry I'm just a little bit confused because the way that I understood atmospheric pressure is that gravity pulls the mass of atmospheric gasses towards earth. Sorry if my intent/question in ^ that wall of text isn't very clear.

Given that I am in a second semester general chemistry course, atmospheric pressure has been discussed. I was wondering why we do not feel the effects of the atmosphere's intense pressure? Following this: http://suite101.com/article/weight-of-earths-atmosphere-a56021 and a previous calculation that I did on my own, the atmosphereic pressure is actually rather high (in terms of pounds.) Why do we not feel this? I understand that, given evolutionary theory, our bodies may have adapted to this standard pressure, but I feel as though there may be a rather large disparity between atmospheric pressure at the coast and pressure at 3,333 ft. Given that the pressure is high in terms of pounds (being about 14.7 PSI) why do we not feel the change? I feel like we should notice having more than 14 pounds of pressure off of our bodies. Sorry if this question seems elementary or stupid. Thanks