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vixen

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I absolutely hate gen chem. Just a venting thread for those of us who are studying. I never learned those Ksp equations ever...its like I'm learning them for the first time...you know those ones....if Ksp for MX3, then Ksp=27x^4. Do you just memorize these equations? Desidentist, if you are online sometime today, IM me, I have a question. Thanks.
 

DesiDentist

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seema,

I'm on SDN right now so you can post questions you have.

DesiDentist
 

portlander

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Vixen,

If you have an chem questions, I might be able to help. I am studying for the DAT as well, and I hate gruding up old quantative analysis stuff. I've done MORE than my share of titrations, so I'd say I'm pretty seasoned in equilibria and solution chemistry.

If you want to share any advice, I would also appreciate it!!

Carmel
 

vixen

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ok I'm posting a question....I didn't get it when I was studying, and am not fully understand the answer...so first lets see if anyone gets it:

The osmotic pressure at STP of a sol'n made from 1L of NaCl (aq) containing 117g of NaCl is:

A) 44.77 atm
B) 48.87 atm
C) 89.54 atm
D) 117 atm

Carmel, you can PM me if you want... :)
 

ItsGavinC

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Just a quick glance at this question since I'm in the middle of studying for midterms.

The equation is very simple: Osmotic pressure = MRT (Molarity x R constant x T)

So 117 g of NaCl is equal to 2 moles (roughly 58g of NaCl per mole). 1 L is given in the problem, so there is a 2M solution. R = .08 and the temperature under STP = 273K. Therefore:

Osmotic pressure = 2 x .08 x 273 which = 44.xx atm

(A) should be the correct answer.

Note: if you confuse STP (273K) with standard conditions (298K) you'll get the wrong answer. Perhaps this question is meant to test on that concept, since the question itself is so simple.

-G
 

vixen

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That's what I thought...here's what the answer says....its choice C because....

The osmotic pressure (II) of a solution is given by II=MRT. At STP, T=273K, R=8.2x10^-2. To determine Molarity, find the formula weight of NaCl from the periodic table (58.5).

117g/L divided by 58.5 g/mol = 2 mols of undissociated NaCl/L

BUT since NaCl is a strong electrolyte, it dissociates in aq sol'n and there are actually 4 moles of particles per L of sol'n, (i.e. 2 mols of Na+ and 2mols of Cl-) so....

II= (4mol)(8.2x10^-2)(273)
=89.54

Remember that colligative properties depend upon the # of particles, not their identitiy.

I understood this problem, until the BUT part....I would never have guessed that...I hope I don't get a problem like this one on there...nice try GavinC, exactly what I thought! :rolleyes: +pissed+
 

JP2005

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its all about keeping the equation balanced...

2 NaCl ---> Na + Cl is unbalanced

2 NaCl ---> 2 Na + 2 Cl is balanced......

i took MCATS almost 2 years ago in April Y2K (and you dont gotta deal with this crap in med school), but i think thats the concept they were probably testing...... if you know how to write the balanced chem equation of a substance that completely dissociates...thank god chem days are over......
 

ItsGavinC

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Oh yeah, that's the correct answer also :)! When you've already taken the test one tends to forget uncessary information. Kapan likes to drive home the colligative property functions.

Actually, I had an nearly identical problem on my exam -- so it's good to remember this stuff when you need to. I'm just glad that I don't need to.

Good luck, seema.

-G
 
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