[mjc123]

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;
P_A = x_A*P⁰_A
where x_A is the mole fraction of A and P⁰_A is the vapour pressure of pure A. In your case, P⁰_A = 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?

[AdiDex]

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 .

[bingo95]

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;
P_A = x_A*P⁰_A
where x_A is the mole fraction of A and P⁰_A is the vapour pressure of pure A. In your case, P⁰_A = 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?

[AdiDex]

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.

[bingo95]

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.

[mjc123]

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?

[bingo95]

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..

[mjc123]

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?

[bingo95]

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?

[mjc123]

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?

[bingo95]

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.

[mjc123]

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?

[bingo95]

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.

[mjc123]

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!

[bingo95]

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!