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Topic: Vapor Pressure  (Read 5050 times)

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Offline big

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Vapor Pressure
« on: November 17, 2010, 06:53:24 PM »
Assuming ideal behavior, which of the following aqueous solutions will have the lowers vapor pressure at 25°C?
a. 0.800 molal C12H22O11
b. 6.00 g of urea (NH2CONH2) dissolved in 100. g of H2O
c. 5.00 g of NaCl dissolved in 100. g of H2O
d. χC6H12O6 = 0.0200

so I might be wrong in my calculations, but otherwise I have that the mole fractions of water in each case is:
a. 0.200
b. 0.9823
c. 0.98482
d. 0.98

but then after that, how do you tell what the vapor pressure is?

Offline Borek

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Re: Vapor Pressure
« Reply #1 on: November 18, 2010, 02:46:25 AM »
What laws describing pressure over ideal solution do you know?
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Offline opti384

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Re: Vapor Pressure
« Reply #2 on: November 21, 2010, 07:22:23 AM »
The vapor pressure of a solution will be lower than the pure solution and that magnitude depends on the molal concentration.

Offline big

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Re: Vapor Pressure
« Reply #3 on: November 22, 2010, 05:25:56 PM »
well there's raoult's law, which is that p=xsolv*P°psolv. So is the answer A? but if I remember correctly, I think the answer was supposed to be D or something? does the van't hoff factor have anything to do with this problem?




Offline opti384

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Re: Vapor Pressure
« Reply #4 on: November 24, 2010, 10:03:40 AM »
does the van't hoff factor have anything to do with this problem?

Of course it does.

Offline big

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Re: Vapor Pressure
« Reply #5 on: November 28, 2010, 01:46:02 PM »
I figured out where I went wrong - I calculated the wrong mole fraction of water for answer A, but now I know how to get D, thanks  ;D

Offline big

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Re: Vapor Pressure
« Reply #6 on: January 22, 2011, 06:24:47 PM »
does the van't hoff factor have anything to do with this problem?

Of course it does.

My apologies for reviving this, but what exactly does the van't hoff factor have to do with raoult's law? I don't see the van't hoff factor anywhere in raoult's law.

Offline Borek

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Re: Vapor Pressure
« Reply #7 on: January 22, 2011, 07:31:04 PM »
It is hidden in the way molar fraction is calculated.
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Offline big

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Re: Vapor Pressure
« Reply #8 on: January 24, 2011, 09:21:51 PM »
Oh, that makes sense. So when calculating the molar fraction of NaCl in part c, do you just multiply by two because it dissociates into two ions?

Offline opti384

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Re: Vapor Pressure
« Reply #9 on: January 24, 2011, 11:24:16 PM »
Yes.

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