Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: bingo95 on September 08, 2014, 08:01:00 AM
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Hello,
I need help to calculate something.
In an enclosed chamber with the volume 100.9 cm^3, I have added 200 µl (0.2 ml) of a solution containing the chemical X. The concentration of the solution with chemical X is 0.05M and 0.5M (yes I have done this twice). The vapor pressure of chemical X is 400 Pa.
What is the concentration of chemical X in the gas phase of the enclosed chamber?
This is not some random test, this is real and I need your help guys. Thank you very much.
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Why two numbers? 0.05M and 0.5M??
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Why two numbers? 0.05M and 0.5M??
Because I did this experiment 2 times using different concentrations.
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Let's start from the very beginning: what is the concentration definition?
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Let's start from the very beginning: what is the concentration definition?
Do you mean "M"? M is mol/l.
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Do you mean "M"? M is mol/l.
No idea what kind of concentration you want to calculate, but if you go for molarity...
It is number of moles per volume. How many moles of the substance, what volume of the chamber?
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I think you should use Raoult's law to find partial pressure of that chemical X...
Then Use the Dalton's law of partial pressure to get mole fraction of chemical X in the gaseous phase....
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This is a tough one, because if you apply Raoult's Law unthinkingly, you will get the wrong answer. Because the volume of solution is so small compared to that of the gas, that answer would imply the amount of X in the gas phase is more than all the X initially present in solution. You have to realise that as X goes into the gas phase to establish an equilibrium vapour pressure, the concentration of X in solution will decrease, so you can't use the initial concentration in Raoult's Law. There is, however, enough information given to solve the problem.
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This is a tough one, because if you apply Raoult's Law unthinkingly, you will get the wrong answer. Because the volume of solution is so small compared to that of the gas, that answer would imply the amount of X in the gas phase is more than all the X initially present in solution. You have to realise that as X goes into the gas phase to establish an equilibrium vapour pressure, the concentration of X in solution will decrease, so you can't use the initial concentration in Raoult's Law. There is, however, enough information given to solve the problem.
Dear mjc123, thank you for your comments, they are very enlightening! Could you please show me how to solve the problem? I appreciate all help from everyone guys, thank you so much.
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Could you please show me how to solve the problem?
It is up to you to solve, we don't do work for others.
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Well, for a start, can you show that you can use Raoult's law (without correction) to give the "simple" answer? Once you can do that, we can think about modifying it.
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Well, for a start, can you show that you can use Raoult's law (without correction) to give the "simple" answer? Once you can do that, we can think about modifying it.
Raoult's law states that mole fraction is proportional with vapor pressure. The mole fraction is for instance 0.5M, then the vapor pressure is already known and 400Pa. There is a very simple equation for Raoult's law that does not take into account diffusion into a set volume. I must admit that based on these clues I can not advance.
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Well, for a start, can you show that you can use Raoult's law (without correction) to give the "simple" answer? Once you can do that, we can think about modifying it.
Raoult's law states that mole fraction is proportional with vapor pressure. The mole fraction is for instance 0.5M, then the vapor pressure is already known and 400Pa. There is a very simple equation for Raoult's law that does not take into account diffusion into a set volume. I must admit that based on these clues I can not advance.
No diffusion etc. is needed if you only want the equilibrium solution. Those are for studying the transient.
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Well, for a start, can you show that you can use Raoult's law (without correction) to give the "simple" answer? Once you can do that, we can think about modifying it.
Raoult's law states that mole fraction is proportional with vapor pressure. The mole fraction is for instance 0.5M, then the vapor pressure is already known and 400Pa. There is a very simple equation for Raoult's law that does not take into account diffusion into a set volume. I must admit that based on these clues I can not advance.
No diffusion etc. is needed if you only want the equilibrium solution. Those are for studying the transient.
Hello curiouscat, and thank you for your comment. I appreciate your time, but this is a bit out of my field, so I am not able to proceed even with your clue here.
"Equilibrium solution" and "the transient" mean what exactly? Thank you in advance.
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Equilibrium = what happens when you wait long enough
Transient= How you get there
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Raoult's law states that mole fraction is proportional with vapor pressure. The mole fraction is for instance 0.5M, then the vapor pressure is already known and 400Pa.
There appear to be a couple of misconceptions here. Raoult's law says that the actual vapour pressure of A above a solution is proportional to the mole fraction of A in the solution;
PA = xA*P0A
where xA is the mole fraction of A and P0A is the vapour pressure of pure A. In your case, P0A = 400 Pa. This is a constant and is not proportional to mole fraction.
Second, 0.5M is not a mole fraction, it is a molar concentration. Do you know the definition of mole fraction? How would you calculate it in this case?
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First of tell methat are you conducting this experiment....?
Or is this just a numerical...??
If this is a experiment , then if it is possible to measure the vapour pressure of pure Element X and pure solvent , Then please find it .
Or if Element X is non-volatile then measure the vapour pressure of pure solvent only .
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Raoult's law states that mole fraction is proportional with vapor pressure. The mole fraction is for instance 0.5M, then the vapor pressure is already known and 400Pa.
There appear to be a couple of misconceptions here. Raoult's law says that the actual vapour pressure of A above a solution is proportional to the mole fraction of A in the solution;
PA = xA*P0A
where xA is the mole fraction of A and P0A is the vapour pressure of pure A. In your case, P0A = 400 Pa. This is a constant and is not proportional to mole fraction.
Second, 0.5M is not a mole fraction, it is a molar concentration. Do you know the definition of mole fraction? How would you calculate it in this case?
Maybe the molar fraction would be the mol of chemical X divided by the mol of the solvent in which X is diluted? Chemical X was diluted in an oil without a known chemical formula. How should I proceed with that? Maybe we should proceed pretending it was water. Is it correct by the way, regarding mole fraction?
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Assum that you have 1 litre of water = 1 Kg water ( You can asume density = 1 g/ml since dilute solution )
So now you know the moles of element X in 1 litre ( Which is equal to molarity)
And you know the moles of water in 1 Kg ( Equal to 1000÷18)
So find the mole fraction.
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Assum that you have 1 litre of water = 1 Kg water ( You can asume density = 1 g/ml since dilute solution )
So now you know the moles of element X in 1 litre ( Which is equal to molarity)
And you know the moles of water in 1 Kg ( Equal to 1000÷18)
So find the mole fraction.
So the mole fraction should be:
moles of X equals 0.5 mol
moles of water equals 55.49 mol
total number of mols equals 55.99 mol
mole fraction equals 0.5 mol / 55.99 mol = 0.0089
Raoults law states that P1 = X * P, where X is the mole fraction and P is the vapor pressure, which in this case gives 0.0089*400 = 3.57. Now, what does this 3.57 mean exactly? Im not sure what unit it is. Is this correct so far by the way?
Thank you for your help so far.
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That's basically correct; in fact if you added 0.5 mol X to 1L water you would probably have a volume a bit more than 1L, so in a 0.5 mol/L solution there would be a little less than 55.49 mol water. But as a first approximation it's good enough. However, it would give a wrong answer if in fact your solvent is not water but oil. You can proceed with the water calculation if you wish, to get practice with the method. If you want to get a better approximation to your experimental situation, I would suggest using a model compound like, say, dodecane (molecular weight 170.33, density 0.75 g/mL) to represent the solvent.
Mole fraction is a dimensionless quantity, so P1 has the same units as P, i.e. Pa. This gives you the pressure of X vapour in the container (assuming, for the moment, that there is a large enough volume of solution, so the solution concentration doesn't change).
Now we have the gas equation of state, PV = nRT. Can you see how to manipulate this to get an expression for the molar concentration of X in the gas phase?
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That's basically correct; in fact if you added 0.5 mol X to 1L water you would probably have a volume a bit more than 1L, so in a 0.5 mol/L solution there would be a little less than 55.49 mol water. But as a first approximation it's good enough. However, it would give a wrong answer if in fact your solvent is not water but oil. You can proceed with the water calculation if you wish, to get practice with the method. If you want to get a better approximation to your experimental situation, I would suggest using a model compound like, say, dodecane (molecular weight 170.33, density 0.75 g/mL) to represent the solvent.
Mole fraction is a dimensionless quantity, so P1 has the same units as P, i.e. Pa. This gives you the pressure of X vapour in the container (assuming, for the moment, that there is a large enough volume of solution, so the solution concentration doesn't change).
Now we have the gas equation of state, PV = nRT. Can you see how to manipulate this to get an expression for the molar concentration of X in the gas phase?
Alright, lets go for the oil you suggested. Is it correct that:
X = 0.5 moles
oil = 4.4 moles
total = 4.9 moles
mole fraction of X = 0.5/4.9= 0.102
Anyway, then I should move on to using the gas equation of state.
P = pressure
V = volume
n = moles
R = a constant
T = temperature
Where does mole fraction fit in here? P = 1 atm. V = 100.9 cm^3. n = the fraction? R = 8.315. T = room temperature = 21 celsius degrees. I feel really stupid not seeing what I should solve for..
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Mole fraction doesn't fit in because you've already used it (or should have) to calculate P, the pressure of X vapour. Not 1 atm! (Strictly V = 100.7 cm3 if you subtract the volume of liquid (200 µL), but the correction is minimal in this case, and we're of necessity only being approximate.) n is the number of moles of X vapour in the volume V. T is the absolute temperature, in K.
Now you want the concentration of X vapour. What is the expression for that? How can you relate it to the other quantities?
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Mole fraction doesn't fit in because you've already used it (or should have) to calculate P, the pressure of X vapour. Not 1 atm! (Strictly V = 100.7 cm3 if you subtract the volume of liquid (200 µL), but the correction is minimal in this case, and we're of necessity only being approximate.) n is the number of moles of X vapour in the volume V. T is the absolute temperature, in K.
Now you want the concentration of X vapour. What is the expression for that? How can you relate it to the other quantities?
Alright I think Im getting closer to a minimal understanding here now. We have to use PV=nRT and solve for n, n=PV/RT
n = 40.8*100.7 / 8.315*294.15 = 1.68 mol
Is this correct?
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If you start with 200 µL of a 0.5M solution, do you think you could get 1.68 mol out of it?
You have to be careful with units. With all the other terms in SI units (P in Pa, T in K, R in J/mol/K) V must be in m3. To convert from cm3 to m3 you must divide by 106, so n = 1.68 x 10-6mol.
And do you know what? I've just realised I made a similar mistake - I confused L with m3 and overestimated the amount of X in the vapour (by a factor of 1000 rather than 1000000). So to a first approximation you don't need to correct the solution concentration for X lost to the vapour phase (calculate the number of moles in the solution and compare it to that in the vapour).
Now can you get what you originally wanted, the concentration of X in the vapour?
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If you start with 200 µL of a 0.5M solution, do you think you could get 1.68 mol out of it?
You have to be careful with units. With all the other terms in SI units (P in Pa, T in K, R in J/mol/K) V must be in m3. To convert from cm3 to m3 you must divide by 106, so n = 1.68 x 10-6mol.
And do you know what? I've just realised I made a similar mistake - I confused L with m3 and overestimated the amount of X in the vapour (by a factor of 1000 rather than 1000000). So to a first approximation you don't need to correct the solution concentration for X lost to the vapour phase (calculate the number of moles in the solution and compare it to that in the vapour).
Now can you get what you originally wanted, the concentration of X in the vapour?
Dear mjc123,
Again thank you for your patience and help. I'm sorry that this is going at an annoyingly slow pace.
If I understood things correctly now, it should be like this:
P = 40.8 Pa
V = 0.0001007 m^3
T = 294.15
R = 8.315
n = (40.8*0.0001007) / (294.15*8.315) = 1.68*10^-6 mol
This is what you just wrote, and what you wrote after that was a bit confusing to me. Where exactly did you make a mistake? Which number does that affect?
Thank you again.
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When I wrote
This is a tough one, because if you apply Raoult's Law unthinkingly, you will get the wrong answer. Because the volume of solution is so small compared to that of the gas, that answer would imply the amount of X in the gas phase is more than all the X initially present in solution. You have to realise that as X goes into the gas phase to establish an equilibrium vapour pressure, the concentration of X in solution will decrease, so you can't use the initial concentration in Raoult's Law. There is, however, enough information given to solve the problem.
I had worked out the answer (so I thought), and it seemed the gas contained more moles of X than were present in the original solution. That was because I had used a value for the volume in L rather than m3 in the gas equation, so got n wrong by a factor of 1000. My mistake doesn't affect any of your numbers - but it shows how careful you have to be with units, even us old hands can get it wrong.
Now, you have 1.68 x 10-6 moles in a volume of 100.7 mL. What is the concentration?
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When I wrote
This is a tough one, because if you apply Raoult's Law unthinkingly, you will get the wrong answer. Because the volume of solution is so small compared to that of the gas, that answer would imply the amount of X in the gas phase is more than all the X initially present in solution. You have to realise that as X goes into the gas phase to establish an equilibrium vapour pressure, the concentration of X in solution will decrease, so you can't use the initial concentration in Raoult's Law. There is, however, enough information given to solve the problem.
I had worked out the answer (so I thought), and it seemed the gas contained more moles of X than were present in the original solution. That was because I had used a value for the volume in L rather than m3 in the gas equation, so got n wrong by a factor of 1000. My mistake doesn't affect any of your numbers - but it shows how careful you have to be with units, even us old hands can get it wrong.
Now, you have 1.68 x 10-6 moles in a volume of 100.7 mL. What is the concentration?
Alright, I see. Thank you for clarifying.
The concentration should then be 1.67*10^-5 M, or 1.67*10^-2 mM, right?
Just to summarize this now:
1) Find the mole fraction
2) Use the mole fraction and the vapor pressure to find P
3) Use the ideal gas law to find n
4) Calculate M using n and the volume
Any adjustments that should be taking into account?
Thanks again for the great help.
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Yes, that's right for the concentration. (now the other one...)
These steps are OK if you know the gas volume, as in your case. Sometimes (often in exam or textbook questions) the volumes of liquid and/or gas are not specified. You can still work out the concentration by rearranging the gas law as n/V = P/RT, without knowing n or V explicitly.
Where you do know the volumes, it is worth checking to see whether the amount of X in solution decreases significantly. In this case, the initial amount of X is 0.5M * 0.2 mL = 1 x 10-4mol - so losing 1.68 x 10-6 mol to the gas phase does not change this significantly.
But I can't emphasise enough, for anything you do in science, be careful about units!
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Thank you very much for following my thread and helping me out despite the snail-paced learning curve. I will make sure to follow your advice regarding units from now on.
Thank you again mjc123 and all other people who contributed!