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Offline JohanesK

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Vapour Pressure Question
« on: July 05, 2009, 03:26:13 AM »
I'm going through my tutorial notes and have just spent way too much time trying to figure out this question with no results:
"calculate the vapour pressure above a solution of 250.0 g of sucrose, C12H22O11, in 800.0 mL of water at 298 K. The vapour pressure and density of pure water under these conditions are 23.67 mm Hg and 0.997 g mL^-1 respectively".

Now I'm working with the formula P(solution) = P(sucrose)+P(H20)
so, P(solution) = P(sucrose)+23.67
I've found the mole fraction of sucrose but I can't figure out how to find the pressure so I have 2 unknowns in my final equation. I also don't know where the density comes into it all...

Can anybody explain what I'm doing wrong and the correct method?

Looking at the answer it's 23.37 mmHg

Cheers!

Offline Borek

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Re: Vapour Pressure Question
« Reply #1 on: July 05, 2009, 03:38:06 AM »
P(solution) = P(sucrose)*P(H20)

No such equation. No idea what you mean.

Quote
I've found the mole fraction of sucrose

Show how.

Quote
but I can't figure out how to find the pressure

Equation you have listed (the wrong one) contains three pressures. No idea which one you refer to.

Quote
I also don't know where the density comes into it all...

So how have you calculated molar fraction?
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Offline JohanesK

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Re: Vapour Pressure Question
« Reply #2 on: July 05, 2009, 04:18:23 AM »
No such equation. No idea what you mean.

P˚(A)= vapour pressure of pure A
P˚(B)= vapour pressure of pure B
P(A)= Partial pressure of A above the solution
P(B)= Partial pressure of B above the solution
X= mole fraction
P(T) = total vapour pressure of the solution

P(A)=X(A)*P˚(A)
P(B)=X(B)*P˚(B)

P(T)=P(A)+P(B) By Dalton's Law of partial pressures. That's what I was referring to, however I'm probably using the wrong equation as you said.

Edit: My bad, I had to correct the original post. The equation should have been + not *

Quote
Show how.

I could be wrong, but here are my calculations of the mole fraction:

number of moles of sucrose = 250.0g/(12.01*12 + 1.008*22 + 16*11) = 0.73 mol

To find mass of H2O:

m = density*volumte = 0.997*800 = 79.76 g

number of moles of H2O=79.76/180.16 = 4.43 mol

Mole fraction of Sucrose = 0.73/(0.73 + 4.43) = 0.14

Quote
Equation you have listed (the wrong one) contains three pressures. No idea which one you refer to.

The question asks for the vapour pressure above the solution. My equation must be wrong... Which one applies?

Quote
So how have you calculated molar fraction?

See above. I wasn't sure if the density played another role.

So that's about as far as I've gotten. Appreciate any help.

Offline Borek

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Re: Vapour Pressure Question
« Reply #3 on: July 05, 2009, 04:55:26 AM »
P(A)=X(A)*P˚(A)

That'll do - in general total pressure above the solution is sum of the pressures of water AND sucrose, but you can safely assume P˚(sucrose) to be 0.

m = density*volumte = 0.997*800 = 79.76 g

No.

Quote
number of moles of H2O=79.76/180.16 = 4.43 mol

No.

Change batteries in your calculator ;)

In general you are on the right track.
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Offline JohanesK

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Re: Vapour Pressure Question
« Reply #4 on: July 05, 2009, 05:47:25 AM »
  ;D Thanks for pointing that out -My calculator has a mind of it's own sometimes.
Here's the CORRECT calculations:

m = density*volume = 0.997*800 = 797.6 g

number of moles of H2O=797.6/18.016 = 44.27 mol H2O

Mole fraction of H2O = 44.27/(44.27 + 0.73) = 0.98

Now if P˚(sucrose) = 0 then P(sucrose) = 0, there's no pressure from sucrose so the final answer should be the pressure from H2O.
P(H2O) = P˚(H2O)*X(H2O) = 23.67 * 0.98 = 23.29 mmHg

I'm 0.08 off (double checked calculations). Could that just be a rounding error? Also, why do we assume sucrose's pressure is 0? Does that apply to all solutes in H2O? I'm a bit confused.

Offline Borek

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Re: Vapour Pressure Question
« Reply #5 on: July 05, 2009, 06:45:23 AM »
23.29 mmHg is the result I got as well.

why do we assume sucrose's pressure is 0? Does that apply to all solutes in H2O?

It applies to solutes that are either ionic or have a large molar mass and/or are solid at the temperature we are doing our calculations for. This is an approximation only, but it works quite well. In fact these pressures are not exactly zero, but they are usually so small they can be safely ignored. Sure, there are substances that can make an unpleasant surprise.
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