DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
Mole fraction of KCI in Benzene and increase in boling point
Hi all, if you guys don't mind, I'd really appreciate some help with this question:
What mole fraction of KCl in Benzene would make the boiling point increase by 3.7degrees C?
Not sure where to even begin or what to do.
Benzene has a boiling point of +2.53
[Edited on 30-6-2015 by DesertFox]
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
First of all can you share whether this is a homework problem or not?
Here's the problem I have with this. I assume this is a homework question okay. So KCl probably is not soluble in benzene to any meaningful extent, so
it probably would be impossible to even do that experiment.
The question is basically about "colligative properties" ( https://en.wikipedia.org/wiki/Colligative_properties)
Say for the sake of homework, it is soluble and infinitely so or in the range needed to answer the question. Then do we ignore that KCl will not
dissociate into the respective ion's inside of a non-polar solvent such as benzene? This changes the answer of the question by a factor of two.
If you want a pretty simple explaination of the subject at hand I found this cute little website - http://www.chemprofessor.com/colligative.htm
I can gaurantee you Kahn academy has a video on this subject on youtube, its a classic general chemistry type question.
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
Quote: Originally posted by smaerd  | First of all can you share whether this is a homework problem or not?
Here's the problem I have with this. I assume this is a homework question okay. So KCl probably is not soluble in benzene to any meaningful extent, so
it probably would be impossible to even do that experiment.
The question is basically about "colligative properties" ( https://en.wikipedia.org/wiki/Colligative_properties)
Say for the sake of homework, it is soluble and infinitely so or in the range needed to answer the question. Then do we ignore that KCl will not
dissociate into the respective ion's inside of a non-polar solvent such as benzene? This changes the answer of the question by a factor of two.
If you want a pretty simple explaination of the subject at hand I found this cute little website - http://www.chemprofessor.com/colligative.htm
I can gaurantee you Kahn academy has a video on this subject on youtube, its a classic general chemistry type question. |
I read all of that and that in the link and it did nothing to help me.
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
You didn't answer my first question.
Okay, I'll help explain it.
Boiling point elevation is a colligative property. Meaning to elevate the boiling point of a liquid by dissolving something, it doesn't matter what
that something is, just how much of it there is.
So we know the equation - ΔT = i * Kb * m
Where, ΔT is the change of the temperature. i is the number of ions which occur when something is dolvated(ex: NaCl -> Na+ + Cl- (i=2); CaCl2
-> Ca2+ + 2Cl- (i=3)), and Kb is the constant of boiling point elevation, and m is the molality of the the thing dissolved.
We want to solve for m to get the concentration so rearrange the equation...
m = ΔT/(i * Kb)
Cool let's subsitute what we know. So we know ΔT = + 2.53 from the question. Let's assume KCl does not dissociate in benzene so i = 1.0. According
to wikipedia, Kb for benzene is 2.53 (https://en.wikipedia.org/wiki/Ebullioscopic_constant) so that's convenient.
m = 2.53/(1.0*2.53)
m = 1 molal or 1 mole KCl/kg of benzene
So now we need the mole fraction... You know I'll leave that for you to do. The conversion is a good excercise.
Anyways, if your teacher assumes that KCl (s) -> K+ (l) + Cl- (l) then you need to use i = 2. It shouldn't be like this. But I have no idea its
obviously not a real world question so you never know.
Does this all make sense?
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
This is also why people dump salt all over the roads when it's icey. The salt raises the melting point of the water so it turned to a liquid instead
of a slippery solid. I guess I should also ask you this, which would you prefer to use to melt ice on a drive-way or a road.
NaCl
or
CaCl2
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
I already know how to solve for m, that's the easiest part. What I don't know is how the hell do I go from solving (m) to actually getting a mole
fraction?
3.7 deg. Celsius = (2) (m) (2.53)
m = .731
What I know is:
Mole fraction of KCI = (moles of KCI) / (total moles)
and
m = (moles of solute) / (kg of benzene)
But I have absolutely no idea how to even start or what to do.
[Edited on 1-7-2015 by DesertFox]
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
oh, I had no idea that's where you were stuck. in the original post it sounded like you didn't understand that part of the problem.
So what is mole fraction? Moles X / (Moles X + Moles Solvent)
What is molality? Moles X/ Mass of Solvent
Where this gets a little interesting for alot of general chemistry students in my experience is this. We have to assume some arbitrary amount of
solvent. So lets do that. Let's say we have 1kg of benzene.
MW Benzene = 78.11 g/mol, so we know that we have 12.80mol of benzene.
Great. So what was our molality, you got 0.731mole KCl for every kg of benzene. I'm not going to check your work, because you assumed i = 2, etc those
are your choices.
so for every 0.731mole of KCl we have 12.8mole of benzene
So the mole fraction: 0.731/(0.731 + 12.8) = 0.0540 mole fraction.
Does that make sense?
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by smaerd | oh, I had no idea that's where you were stuck. in the original post it sounded like you didn't understand that part of the problem.
So what is mole fraction? Moles X / (Moles X + Moles Solvent)
What is molality? Moles X/ Mass of Solvent
Where this gets a little interesting for alot of general chemistry students in my experience is this. We have to assume some arbitrary amount of
solvent. So lets do that. Let's say we have 1kg of benzene.
MW Benzene = 78.11 g/mol, so we know that we have 12.80mol of benzene.
Great. So what was our molality, you got 0.731mole KCl for every kg of benzene. I'm not going to check your work, because you assumed i = 2, etc those
are your choices.
so for every 0.731mole of KCl we have 12.8mole of benzene
So the mole fraction: 0.731/(0.731 + 12.8) = 0.0540 mole fraction.
Does that make sense? |
Yeah the professor said to use 2 for i when the elements come from column 1a and 7a in the periodic table.
Okay so I am supposed to just invent a random amount for Benzene? Does it matter what amount (in kg) I use or is one specifically recommended?
By the way, yes I understood how you explained it. I'm guessing the fraction will always be the same regardless of what amount of benzene I choose?
[Edited on 1-7-2015 by DesertFox]
[Edited on 1-7-2015 by DesertFox]
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
it doesn't neccesairily matter what number you use. To be really brief always choose 1kg for this conversion it's way easier. If you use 1kg then you
don't have to resolve the molality. What I mean is this.
If I choose 100g of benzene then I have 1.28 mole of benzene
But what does that mean for the molality?
Do some "dimensional analysis"/unit conversion
0.731Mol KCl/1000g benzene * 100g benzene = 0.0731mole KCl for ever 1.28mole of benzene
Thus the mole fraction: 0.0731/(0.0731+1.28) = 0.054 mole fraction
See it's an extra step. Use 1kg and you don't have to worry about it. Just grab the moles of your solution, and do the fraction(&sum).
Hopefully this didn't confuse you
edit - mathematically you can think about it like this. You are multiplying the numerator and denominator of the mole fraction by a constant which
cancels out. The proportion/ratio of solvent to solute is the same if the math is done correctly.
[Edited on 1-7-2015 by smaerd]
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by smaerd | it doesn't neccesairily matter what number you use. To be really brief always choose 1kg for this conversion it's way easier. If you use 1kg then you
don't have to resolve the molality. What I mean is this.
If I choose 100g of benzene then I have 1.28 mole of benzene
But what does that mean for the molality?
Do some "dimensional analysis"/unit conversion
0.731Mol KCl/1000g benzene * 100g benzene = 0.0731mole KCl for ever 1.28mole of benzene
Thus the mole fraction: 0.0731/(0.0731+1.28) = 0.054 mole fraction
See it's an extra step. Use 1kg and you don't have to worry about it. Just grab the moles of your solution, and do the fraction(&sum).
Hopefully this didn't confuse you
|
Awesome. I was breaking my head over here only to realize how simple the question was once you explained how to do it.
|
|
|
smaerd
International Hazard
Posts: 1262
Registered: 23-1-2010
Member Is Offline
Mood: hmm...
|
|
DesertFox - No worries. I've been there many times before. You know, it's usually the really easy things that confuse us the most. One thing that
helps me when I find myself doing the same thing over and over again is taking a break. Then returning to the problem and outlining what do I need to
know, what do I know, and thinking about it more broadly.
Keep practicing :]
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
Back with a couple other questions althought these relate to electrons, wavelength, ionization energy, and such.
Okay so the question states: How much energy will it take to remove only one electron on the surface of Nickel? If you hit the surface of
Nickel with a photon that has a wavelength of 90 nm, what will the wavelength of that ejected electron be?
What I know:
Nickel has an ionization energy of 757 according to my textbook.
90 nanometers = 4.36332313 × 10^-13 meters per second
Other than that, I do not know how to start, or which equations I should use.
|
|
|
blargish
Hazard to Others
Posts: 166
Registered: 25-9-2013
Location: Canada
Member Is Offline
Mood: Mode Push
|
|
Recall that ionization energy is the energy required to remove one mole of electrons from one mol of gaseous atoms. Thus, finding the
energy required to remove one electron from one (Ni) atom merely requires a simple scaling operation.
As for the photon question, the energy of the incident photon is found by E = hc / λ, where h is the planck constant, c is the speed
of light, and λ is the wavelength of the incident photon. That energy is going to be divided between going towards the ionization energy and the
kinetic energy of the ejected photoelectron (so a simple subtraction operation with the value obtained in the first part of the question gives you the
EK of the electron). Once you have the EK of the electron, you can substitute into the equation EK =
mv<sup>2</sup> / 2 and solve for the velocity of the electron (I assume that you are given the mass of an electron). Then
substitute into the de Broglie wavelength equation λelectron = h / mv and solve for the wavelength of the electron.
Alternatively, you can directly find the momentum of the electron by recognizing that EK = p<sup>2</sup> / 2m,
and you can isolate p (p = sqrt(EK(2m)) and plug that into the simplified de Brogie wavelength equation:
λelectron = h / p (since p = mv). This omits the need to calculate the velocity of the electron.
[Edited on 1-7-2015 by blargish]
BLaRgISH
|
|
|
blogfast25
International Hazard
Posts: 10562
Registered: 3-2-2008
Location: Neverland
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by DesertFox |
What I know:
Nickel has an ionization energy of 757 according to my textbook.
90 nanometers = 4.36332313 × 10^-13 meters per second
|
757 what? Lemons? Potatoes?(*)
1 nm (nanometer) = 1 x 10<sup>-9</sup> m (0.000,000,001 meter)
(*) 737 kJ/mol (kJ mol<sup>-1</sup> , 1<sup>st</sup>
ionisation energy of Ni.
[Edited on 1-7-2015 by blogfast25]
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
Okay so going off of the ionization energies found in my textbook:
(757,000 Joules) / (6.022 x 10<sup>23</sup> avogrado) = 1.25705746 x 10 <sup>-18</sup>
and then
E= (6.63 x 10 <sup>-34</sup>j/s) (3 x 10 <sup>8</sup> m/s) / (9.0 x 10<sup>-8</sup> meters)
which got me: 2.21 x 10 <sup>-34</sup> joules?
So now I subtract (1.25705746 x 10 <sup>-18</sup>) - (2.21 x 10 <sup>-34</sup> joules?) ??
[Edited on 1-7-2015 by DesertFox]
|
|
|
unionised
International Hazard
Posts: 5126
Registered: 1-11-2003
Location: UK
Member Is Offline
Mood: No Mood
|
|
https://en.wikipedia.org/wiki/Work_function
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
We never used work functions in class.
|
|
|
blargish
Hazard to Others
Posts: 166
Registered: 25-9-2013
Location: Canada
Member Is Offline
Mood: Mode Push
|
|
| Quote: Originally posted by DesertFox |
and then
E= (6.63 x 10 <sup>-34</sup>j/s) (3 x 10 <sup>8</sup> m/s) / (9.0 x 10<sup>-8</sup> meters)
which got me: 2.21 x 10 <sup>-34</sup> joules?
[Edited on 1-7-2015 by DesertFox] |
Take another look at your calculation there.
Also, in the context of this question, the work function and ionization energy of the metal can be considered the same thing IIRC
BLaRgISH
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
| Quote: Originally posted by blargish | | Quote: Originally posted by DesertFox |
and then
E= (6.63 x 10 <sup>-34</sup>j/s) (3 x 10 <sup>8</sup> m/s) / (9.0 x 10<sup>-8</sup> meters)
which got me: 2.21 x 10 <sup>-34</sup> joules?
[Edited on 1-7-2015 by DesertFox] |
Take another look at your calculation there.
Also, in the context of this question, the work function and ionization energy of the metal can be considered the same thing IIRC
|
I don't understand what I did wrong?
|
|
|
blargish
Hazard to Others
Posts: 166
Registered: 25-9-2013
Location: Canada
Member Is Offline
Mood: Mode Push
|
|
Make sure you are dividing by 9.0 x 10<sup>-8</sup> metres
[Edited on 1-7-2015 by blargish]
BLaRgISH
|
|
|
DesertFox
Harmless
Posts: 10
Registered: 29-6-2015
Member Is Offline
Mood: No Mood
|
|
Okay I ended up getting E= 2.21 x 10 <sup>-18</sup>
So now I subtract (2.21 x 10 <sup>-18</sup> joules) - (1.25705746 x 10 <sup>-18</sup> ?
which got me: 9.5294254 x 10 <sup>-19</sup>
Do I substitute this value for (Kinetic Energy) = mv<sup>2</sup> / 2 ?
[Edited on 1-7-2015 by DesertFox]
|
|
|
blargish
Hazard to Others
Posts: 166
Registered: 25-9-2013
Location: Canada
Member Is Offline
Mood: Mode Push
|
|
Yes, think about it. The energy of the incident photon is 2.21e-18 J. Part of this energy is going to go towards removing the electron from the atom
(the value for the ionization energy of a single electron that you previously calculated), and the rest is going to go towards the kinetic energy of
that electron. This is more or less the photoelectric effect.
Since you are trying to find the wavelength of the electron, you can use the value of its kinetic energy to find its velocity (assuming that you are
given the mass of the electron) with EK = mv<sup>2</sup> / 2. Hence, its de Broglie wavelength can be calculated with the
obtained values using the equation I outlined in a previous post.
BLaRgISH
|
|
|
![[*]](images/xpblue/default_icon.gif){kind=icon}
