General Chemistry Question Thread

[QofQuimica]

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All users may post questions about MCAT, DAT, OAT, or PCAT general chemistry here. We will answer the questions as soon as we reasonably can. If you would like to know what general chemistry topics appear on the MCAT, you should check the MCAT Student Manual (http://www.aamc.org/students/mcat/studentmanual/start.htm)

Acceptable topics:
-general, MCAT-level gen chem.
-particular MCAT-level gen chem problems, whether your own or from study material
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-how best to tackle the MCAT physical sciences section

Unacceptable topics:
-actual MCAT questions or passages, or close paraphrasings thereof
-anything you know to be beyond the scope of the MCAT

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If you really know your gen chem, I can use your help. If you are willing to help answer questions on this thread, please let me know. Here are the current members of the General Chemistry Team:

-QofQuimica (thread moderator): I have my M.S. in organic chemistry and I'm currently finishing my Ph.D., also in organic chemistry. I have several years of university general chemistry TA teaching experience. In addition, I teach general chemistry classes through Kaplan for their MCAT, DAT, OAT, and PCAT courses. On the MCAT, I scored 14 on PS, 43 overall.

-Learfan: Learfan has his Ph.D. in organic chemistry and several years worth of industrial chemistry experience. He scored 13 on the PS section of the MCAT, and 36 overall.

-Sparky Man: Sparky Man has his Ph.D. in physical chemistry. He scored 14 on the PS section of the MCAT, and 36 overall.

-GCT: GCT scored in the 99th percentile on the PCAT. He has also taught introductory physics and general chemistry.

 

[jace's mom]

Working at In 'N Out in Southern California is better. If you become manager in one of the franchises, you earn 90K and upwards.

But can I afford California on 90K? 🙂

IIIA - VIIA, I'm pretty sure, refer to groups (columns) 13-17 on the periodic table (the column that starts with Boron through the column that starts with Flourine). The transition metals are all "B" elements. So if we're talking elements in IIIA-VIIA (group) beyond period II (row), that would begin at Al.

 

[heymanooh1]

But can I afford California on 90K? 🙂

IIIA - VIIA, I'm pretty sure, refer to groups (columns) 13-17 on the periodic table (the column that starts with Boron through the column that starts with Flourine). The transition metals are all "B" elements. So if we're talking elements in IIIA-VIIA (group) beyond period II (row), that would begin at Al.

Huh the periodic table I downloaded from somewhere has it wrong then

 

[jace's mom]

I asked one of the Chemistry teachers at school today, and she said the old periodic table used to be numbered that way (transition metals numbered IIIA-VIIA), but they've since renumbered it. You can find a new copy at:

http://www.swansontec.com/selement.htm

(Mods, I apologize if I'm not supposed to post links.)

What a great discussion! Until now, I didn't know there were other versions of the table. I would think the new one is the one to go by for the test, though, because that's the version that was provided in my Kaplan prep book.

 

[alex1231]

What is the defenition of the pure solvent for Keq, delta H and delta S.
Thank you

 

[alex1231]

What is the defenition of the pure solvent for Keq, delta H and delta S.
Thank you

 

[QofQuimica]

Answers B and D are out because the possible ranges of m1 don't correlate with the respective value of l given. That leaves choices A and C. Answer C doesn't work because from the given value of l =3 this designates occupation of an f-orbital. The problem with answer C is that l = n - one which is not possible for the value of n given and filling in f-orbitals is possible only when n = 4 or greater. That leaves answer A.

The last electron should have quantum numbers n=3, 1 = one, m1 can either be -1, 0, or 1 and ms = +/- 1/2 since we don't know which p-orbital is filled and whether in a spin up/down fashion.

For simplicity's sake let's say that the quantum #s for last electron to be filled in aluminum is: n = 3, l = one, m1 = -1, ms = +1/2. This designates the ground state electron configuration in aluminum. Now answer A is possible because it is a set of quantum #s for an electronically excited state of aluminum i.e. the lone p-orbital electron (l = one) goes in to the next highest available orbital which is 3d (l = 2) in this case. Choices B, C, D are not possible even if they were excited states for the reasons given in the first paragraph.

Also, rows IIIA-VIIA on the periodic table refers to transition metal/d-block elements which aluminum is not.

This explanation is right on. The question doesn't ask for the ground state of aluminum. It asks for a possible electron configuration for that last electron. 🙂

 

[QofQuimica]

I asked one of the Chemistry teachers at school today, and she said the old periodic table used to be numbered that way (transition metals numbered IIIA-VIIA), but they've since renumbered it. You can find a new copy at:

http://www.swansontec.com/selement.htm

(Mods, I apologize if I'm not supposed to post links.)

What a great discussion! Until now, I didn't know there were other versions of the table. I would think the new one is the one to go by for the test, though, because that's the version that was provided in my Kaplan prep book.

Posting links is fine. You are correct that the main group elements are generally referred to as the As and the transition elements are the Bs. I hadn't ever seen a backward periodic table like hey's before either.

 

[QofQuimica]

What is the defenition of the pure solvent for Keq, delta H and delta S.
Thank you

I'm not sure I understand your question, but I think you are asking how to recognize a pure solvent that wouldn't be included in a Keq expression? If so, the answer is you can often recognize them from the equation itself. This is because the pure solids and liquids have (s) or (l) after them, while the dissolved substances and gases have (aq) or (g) after them. (Pure solids don't get included in the Keq expression either, but dissolved substances and gases do.)

 

[alex1231]

OK
Thanks

 

[heymanooh1]

I thought I had finally gotten a handle on this but a question on a practice test proved me wrong.

Take Ti[22], the solutions say the electrons from 3d^2 will participate in a bond. Those are the valence electrons.

Now I looked up valence electrons and it was defined as electrons in the outermost shell. .. Wouldn't the electrons in the outermost shell be 4s because the principle quantum number of 4 is bigger than 3?

I'm pretty sure this is simple but I can't get it. Please help Q~ 🙂

Both the 3d and the 4s electrons are valence electrons for titanium. You can't just go by the principle quantum number when you talk about transition metals like this, because the 3d and 4s orbitals are very close together in energy. In fact, they are close enough that 4s actually fills first before 3d does. (4s is slightly lower in energy than 3d.) Confusing the issue is the fact that once they are both filled with electrons, the 4s orbital becomes higher in energy than the 3d orbital is. So, the 4s fills before the 3d does, and the 4s empties before the 3d does also.

I have to disagree with the explanation that the 4s orbital is energetically lower than the 3d orbital. Clearly the 4s orbital is higher in energy because if neutral Titanium is singly ionized to a +1 oxidation state, the first electron removed is from the 4s orbital. This has been proven experimentally.

The real reason why Titanium fills in the 4s orbital first rather than the 3d can attributed to the penetration effect. Experimentally, when a "radial distribution v. distance from nucleus" for electrons is plotted, the 4s electron spends significanly more time i.e. penetrates closer to the nucleus compared to the 3d electron. Because of this effect, the 4s orbital is "preferred" first when filling in the orbitals for a transition metal such as Titanium. Also, the radial distribution plot confirms that the 4s electron in Titanium spends more time further from Titanium's nucleus compared to the 3d electron. This correlates with the experimental fact that the first ionization energy of neutral Titanium is due to the removal of the higher energy 4s electron and not the lower energy 3d electron. The consequence of Titanium being singly ionized to +1 is that the remaining 4s electron will drop down to the 3d orbital to give the following electron configuration Ti +1: [Ar] 3d3. If 4s were energetically lower than 3d, then the electron configuration for Ti +1 would be: [Ar] 4s1 3d2. This hasn't been shown experimentally.

 

[QofQuimica]

I have to disagree with the explanation that the 4s orbital is energetically lower than the 3d orbital. Clearly the 4s orbital is higher in energy because if neutral Titanium is singly ionized to a +1 oxidation state, the first electron removed is from the 4s orbital. This has been proven experimentally.

The real reason why Titanium fills in the 4s orbital first rather than the 3d can attributed to the penetration effect. Experimentally, when a "radial distribution v. distance from nucleus" for electrons is plotted, the 4s electron spends significanly more time i.e. penetrates closer to the nucleus compared to the 3d electron. Because of this effect, the 4s orbital is "preferred" first when filling in the orbitals for a transition metal such as Titanium. Also, the radial distribution plot confirms that the 4s electron in Titanium spends more time further from Titanium's nucleus compared to the 3d electron. This correlates with the experimental fact that the first ionization energy of neutral Titanium is due to the removal of the higher energy 4s electron and not the lower energy 3d electron. The consequence of Titanium being singly ionized to +1 is that the remaining 4s electron will drop down to the 3d orbital to give the following electron configuration Ti +1: [Ar] 3d3. If 4s were energetically lower than 3d, then the electron configuration for Ti +1 would be: [Ar] 4s1 3d2. This hasn't been shown experimentally.

I won't argue with your explanation being more technically correct. However, it is also WAY beyond the scope of what anyone would be expected to know from any freshman-level chemistry course. As such, it is also way beyond the scope of MCAT-level chemistry. I guarantee you that a very large percentage of the students using this subforum do not understand what you just wrote. This is always a tough call, but sometimes it is necessary to simplify explanations and lose a bit of accuracy in order to make these answers useful to the askers and avoid having students get too caught up in the extraneous details. It is important to always consider your audience and taylor your response to them, rather than expecting them to be at your level. If they get the general concept that the 4s and 3d orbitals are close together in energy and that this sometimes leads to deviations from the Aufbau filling order, that is enough for many of them. Those students who want to actually understand why these things happen are the ones who take p. chem; when they do, they find out that nearly everything we learned back in gen chem isn't strictly accurate. 😉

 

[mcat_study]

hi,
i have a question about maxwell-boltzman distribution curve.

at higher temperatures more molecules are moving at higher speeds.

But at the time, Kaplan book says that at higher temperature, FEWER molecules will move at exactly new AVERAGE SPEED, because more molecules move at a greater range of speeds. does it still mean that more molecules have greater speeds, it is just average speed is rare????

i really do not understand this? please help, someone!!!
thanks 🙂

 

[estairella]

hi,
i have a question about maxwell-boltzman distribution curve.

at higher temperatures more molecules are moving at higher speeds.

But at the time, Kaplan book says that at higher temperature, FEWER molecules will move at exactly new AVERAGE SPEED, because more molecules move at a greater range of speeds. does it still mean that more molecules have greater speeds, it is just average speed is rare????

i really do not understand this? please help, someone!!!
thanks 🙂

I assume you mean either "exactly" or "near" instead of "exactly new".

Let's go with exactly, though of course, there is no molecule that will be going at EXACTLY the average speed.

What you said is correct. If you imagine a distribution of molecular speeds for any given temperature, the higher the temperature, the larger the range of speeds. But the area of the distribution has to stay the same - you have the same # of molecules.. so as a consequence, the distribution flattens out. At the higher average speed, the height will be lower than at the previous lower average speed, and height corresponds to frequency of molecules... I.E. there are fewer molecules travelling at exactly the average speed as Kaplan suggests.

For these kinds of things, its sometimes easy to think about the extremes.

Think about absolute zero. By definition, no molecule has any kinetic energy, so EVERY molecule is at the "average" speed of zero. Increasing above that will mean some are above average and some are below average. You can extrapolate from there that the more you move away from absolute zero, the larger the distribution of speeds.

 

[mcat_study]

I assume you mean either "exactly" or "near" instead of "exactly new".

Let's go with exactly, though of course, there is no molecule that will be going at EXACTLY the average speed.

What you said is correct. If you imagine a distribution of molecular speeds for any given temperature, the higher the temperature, the larger the range of speeds. But the area of the distribution has to stay the same - you have the same # of molecules.. so as a consequence, the distribution flattens out. At the higher average speed, the height will be lower than at the previous lower average speed, and height corresponds to frequency of molecules... I.E. there are fewer molecules travelling at exactly the average speed as Kaplan suggests.

For these kinds of things, its sometimes easy to think about the extremes.

Think about absolute zero. By definition, no molecule has any kinetic energy, so EVERY molecule is at the "average" speed of zero. Increasing above that will mean some are above average and some are below average. You can extrapolate from there that the more you move away from absolute zero, the larger the distribution of speeds.

thank you very much 🙂

 

[chem neub]

when 5.8g of sodium sulfate is dissolved in 150g of H2O in a calorimeter the water temp. rises from 21.5C to 27.6C. Find the molar heat of hydration for sodium sulfate. ( assumen no heat is lost to the surroundings)
I apologize if this is posted in a bad place but it would be nic for someone to answer this before monday Dec. 4 thanks

 

[spicedmanna]

when 5.8g of sodium sulfate is dissolved in 150g of H2O in a calorimeter the water temp. rises from 21.5C to 27.6C. Find the molar heat of hydration for sodium sulfate. ( assumen no heat is lost to the surroundings)
I apologize if this is posted in a bad place but it would be nic for someone to answer this before monday Dec. 4 thanks

Hi, there. I'm not an expert in Thermodynamics, but this question seems like a simple application of constant pressure calorimetry:

q = mc(deltaT), where q is the amount of heat gained or lost by the water, c is the specific heat of water (4190 J/Kg * C), and deltaT is the change in temperature of the water. If q > 0, then the water absorbed heat, but if q < 0, then the water lost heat.

So for this problem:

q = (0.150 Kg of water)(4160 J/Kg*C)(27.6 - 21.5) = 3.81 kJ

Now we need to find the number of moles of sodium sulfate dissolved (assuming we are talking about anhydrous sodium sulfate):

Na2SO4 = 2(23 g/mol) + (32.1 g/mol) + 4(16.0 g/mol) = 142.1 g/mol

number of moles of Na2SO4 = 5.8g/(142.1 g/mol) = 4.0816E-2 mol

Now for the final answer:

q/mol = (3.8064 Kj)/(4.0816E-2 mol) = 93.1 kJ/mol

So, the amount of heat absorbed by the water per mole of Na2SO4 dissolved is approximately 93.1 kJ/mol. Thus, by convention, the molar heat of dissolution (deltaHsoln) would be a negative number since the dissolution process is exothermic (the water gained heat): -93.1 kJ/mol.

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Well, that was the heat of dissolution for anhydrous sodium sulfate... If you are looking for that, then, you are done. To find the heat of hydration for sodium sulfate, however, you need to know that it forms a hydrate in the presence of water: (Na2SO4)*(10H2O).

The definition of the heat of hydration is the difference in enthalpy between the hydrate and it's anhydrous salt (http://www.rit.edu/~axlsch/classes/schp445/Hydrate%20lab.pdf). According to what I found on that site, you can obtain it using Hess' Law:

find deltaH for: (Na2SO4)*(10H2O) (s) + n H2O ---> Na2SO4 (aq) + (10 + n) H2O (l)
find deltaH for: Na2SO4 (s) + (10 + n) H2O ---> Na2SO4 (aq) + (10 + n) H2O (l)

According to Hess' Law, the heat of hydration is:

deltaH(Hydration) = deltaH(Na2SO4) - deltaH[(Na2SO4)*(10H2O)]

So, from the first part of the exercise, we have deltaH(Na2SO4), which is -93.1 kJ/mol. Now, we need the deltaH[(Na2SO4)*(10H2O)]. If you don't have that, you will need to look it up.

Hope this helps; it's my best guess for the answer. If anyone else understands this process better than me, please feel free to make any corrections and or other suggestions.

 

[commuter9]

I'm having some trouble w/ solvents and solutions etc (everytime I think I have it figured out, I get a problem wrong and then I'm back to the drawing board).

Here is the question:
If 11 g of CaCl2 is added to 100ml of water, what is the molarity of the solution?
A. 5M
B. 9M
C. .9M
D Can't be determined?

The answer is D because they say that in this case water is a solvent not a solution (M= moles/L). The explanation goes on to say that if you add 11g of salt to 100 mL of water, you are ending up w/ more than 100mL of solution since the salt would take up room. But what about a problem that says you have 5M glucose? Doesn't the glucose take up space in this case? I feel like I'm missing something that is key (and easy).

 

[QofQuimica]

I'm having some trouble w/ solvents and solutions etc (everytime I think I have it figured out, I get a problem wrong and then I'm back to the drawing board).

Here is the question:
If 11 g of CaCl2 is added to 100ml of water, what is the molarity of the solution?
A. 5M
B. 9M
C. .9M
D Can't be determined?

The answer is D because they say that in this case water is a solvent not a solution (M= moles/L). The explanation goes on to say that if you add 11g of salt to 100 mL of water, you are ending up w/ more than 100mL of solution since the salt would take up room. But what about a problem that says you have 5M glucose? Doesn't the glucose take up space in this case? I feel like I'm missing something that is key (and easy).

Ok, first of all, for MCAT purposes you can definitely round the molarity off and say that the volume of the salt is so small that it doesn't greatly affect the volume of the entire solution. In other words, you can assume that the volume of the solvent is nearly equal to the volume of the entire solution. We round off like that all of the time for dilute solutions, and I think that's why you're feeling confused. But the problem here is that they are wanting you to calculate the EXACT molarity. To do that, you need to know the volume of BOTH solvent and solute, because technically you are supposed to divide the moles of solute by the total volume of the entire solution to get molarity. We know that water is the solvent and calcium chloride is the solute. You can get the moles of calcium chloride by dividing its mass by the MW. But, you cannot get the volume of calcium chloride from the info given, and if you want to get the exact molarity, you must divide the moles of calcium chloride by the TOTAL volume (volume of water + volume of calcium chloride), not just the solvent volume. Again, on the MCAT, just divide your moles of solute by the volume of solvent and that will almost always be a close enough approximation.

In your second example, no, you wouldn't have to factor in the volume of the glucose. Since the glucose is already in solution, its volume has already been accounted for. In other words, if you have one liter of a five molar solution, some of that one liter is occupied by glucose, while most of it is occupied by water. So, the only time this issue comes up is when you are changing the volume, as by adding a solute to a solvent where you wind up with a new volume of total solution.

 

[tik-tik-clock]

An ethanol and water solution can be distilled to, at best, a 95% ethanol solution, which boils at 78°C. What can be said about this mixture?

It is a minimum-boiling azeotrope with water as the most volatile component.
It is a minimum-boiling azeotrope with ethanol as the most volatile component.
It is a maximum-boiling azeotrope with water as the most volatile component.
It is a maximum-boiling azeotrope with ethanol as the most volatile component

-----------------------------------------------------------------------
Answer: THe answer is B, I understand that the ethanol is the most volatile substance out of both water and ethanol, but I dont understand how the mixture turns out to be a minimum boiling azeotrope.
A minimum boiling azeotrope is the one in which both the components in the mixture have a lower boiling point than the mixture. They give us in the passage that the booiling point of the mixture is 73 degrees, but gives no information about their individual boiling points, so how am I supposed to know whether the BPs of the individual components are lower than the solution or not

 

[QofQuimica]

An ethanol and water solution can be distilled to, at best, a 95% ethanol solution, which boils at 78°C. What can be said about this mixture?

It is a minimum-boiling azeotrope with water as the most volatile component.
It is a minimum-boiling azeotrope with ethanol as the most volatile component.
It is a maximum-boiling azeotrope with water as the most volatile component.
It is a maximum-boiling azeotrope with ethanol as the most volatile component

-----------------------------------------------------------------------
Answer: THe answer is B, I understand that the ethanol is the most volatile substance out of both water and ethanol, but I dont understand how the mixture turns out to be a minimum boiling azeotrope.
A minimum boiling azeotrope is the one in which both the components in the mixture have a lower boiling point than the mixture. They give us in the passage that the booiling point of the mixture is 73 degrees, but gives no information about their individual boiling points, so how am I supposed to know whether the BPs of the individual components are lower than the solution or not

Well, hopefully you know that water boils at 100 degrees, right? So if the azeotrope boils at 78 degrees, it's lower than the BP of water, and you know it has to be a minimum boiling azeotrope even if you don't know the BP of ethanol (which is slightly above 78 degrees). 🙂

 

[tik-tik-clock]

Thanks Q! 🙂

I think I had got the definition of azetrope wrong , the first time around

so it is the solution that has a lower BP than the individual components forming the solution, right?

 

[QofQuimica]

Thanks Q! 🙂

I think I had got the definition of azetrope wrong , the first time around

so it is the solution that has a lower BP than the individual components forming the solution, right?

Right. I looked up the exact BPs for absolute ethanol versus 95% and it's a difference of like 0.2 degrees. But the ethanol is slightly higher. Anyway, this problem is actually a good lesson in not panicking. It doesn't matter if you know the BP of ethanol as long as you use what you do know, which is the BP of water. BTW, if you have to guess, most azeotropes are minimum boiling. Not that you're ever likely to see a question about one again. 😛

 

[commuter9]

Ok, first of all, for MCAT purposes you can definitely round the molarity off and say that the volume of the salt is so small that it doesn't greatly affect the volume of the entire solution. In other words, you can assume that the volume of the solvent is nearly equal to the volume of the entire solution. We round off like that all of the time for dilute solutions, and I think that's why you're feeling confused. But the problem here is that they are wanting you to calculate the EXACT molarity. To do that, you need to know the volume of BOTH solvent and solute, because technically you are supposed to divide the moles of solute by the total volume of the entire solution to get molarity. We know that water is the solvent and calcium chloride is the solute. You can get the moles of calcium chloride by dividing its mass by the MW. But, you cannot get the volume of calcium chloride from the info given, and if you want to get the exact molarity, you must divide the moles of calcium chloride by the TOTAL volume (volume of water + volume of calcium chloride), not just the solvent volume. Again, on the MCAT, just divide your moles of solute by the volume of solvent and that will almost always be a close enough approximation.

In your second example, no, you wouldn't have to factor in the volume of the glucose. Since the glucose is already in solution, its volume has already been accounted for. In other words, if you have one liter of a five molar solution, some of that one liter is occupied by glucose, while most of it is occupied by water. So, the only time this issue comes up is when you are changing the volume, as by adding a solute to a solvent where you wind up with a new volume of total solution.

Thanks Q. So what you are saying is that if I'm asked the exact molarity (and usually MCAT we aren't asked that) then I need the volume of the the solute, but otherwise it is okay to just assume that the solute doesn't take up a lot of space.

 

[QofQuimica]

Thanks Q. So what you are saying is that if I'm asked the exact molarity (and usually MCAT we aren't asked that) then I need the volume of the the solute, but otherwise it is okay to just assume that the solute doesn't take up a lot of space.

Right. Technically you are supposed to divide by the total volume, which is the sum of the volumes of the solute plus the solvent. But, since the volume of the solvent is generally much larger than the volume of the solute, you can usually neglect the volume of the solute and say that the volume of solution is approximately equal to the volume of the solvent.

 

[commuter9]

Right. Technically you are supposed to divide by the total volume, which is the sum of the volumes of the solute plus the solvent. But, since the volume of the solvent is generally much larger than the volume of the solute, you can usually neglect the volume of the solute and say that the volume of solution is approximately equal to the volume of the solvent.

Thanks Q!!!

 

[RAD11]

Hi can someone please tell me if I have my concepts correct. Thanks so much in advance.

Why does vapor pressure of a liquid increase with increasing temperature?

Vapor pressure increases with temperature because @ a higher temperature the molecules have more kinetic energy to break free from the liquid and go into the gas phase.

Btw, is vapor pressure the "pressure" that is holding the molecules down in the liquid phase?

Why is vaporization an endothermic process?

It is an endothermic process because breaking the bonds of the molecules requires energy (whereas forming a bond requires a release of energy).

 

[heymanooh1]

Hi can someone please tell me if I have my concepts correct. Thanks so much in advance.

Why does vapor pressure of a liquid increase with increasing temperature?

Vapor pressure increases with temperature because @ a higher temperature the molecules have more kinetic energy to break free from the liquid and go into the gas phase.

Btw, is vapor pressure the "pressure" that is holding the molecules down in the liquid phase?

Why is vaporization an endothermic process?

It is an endothermic process because breaking the bonds of the molecules requires energy (whereas forming a bond requires a release of energy).

Your summary to the first question is correct. Mathematically, we can see that the vapor pressure of a gas is proportional to temperature as given in the ideal gas equation, PgasV =nRT. The kinetic energy is essentially the sum of vibrational, translational and rotational energies that allows a molecule to "escape" from the liquid phase at a given temperature.

Vapor pressure is the pressure exerted by a gas in the volume it occupies over its liquid phase. The vapor pressure isn't necessarily the pressure that "holds" molecules down in the liquid phase. You might be thinking that if there's a beaker of water at STP, then it's the atmospheric pressure is what keeps it in the liquid phase. This isn't the case since there will be a distribution of water molecules that have enough energy to go into the vapor phase. The water molecules that are in the vapor phase will have a gas pressure that will be less than the atmospheric pressure. Look into a general Chemistry textbook to see that liquid water has different vapor pressures at different temperatures.

In the 2nd question, you need to modify your explanation i.e. bonds are not being broken , ionic or covalent, but inter-molecular intearctions such as hydrogen bonds, dipole-dipole or van der Waals. From the perspective of the system, the liquid phase, vaporization is endothermic since a net input of energy is required to bring the liquid molecules into the gas phase. Think of the input energy as a "kick" to the liquid molecules. In order to get that "kick", there must be an energy input, such as heat for example, to go from the liquid to the gas phase.

 

[RAD11]

Your summary to the first question is correct. Mathematically, we can see that the vapor pressure of a gas is proportional to temperature as given in the ideal gas equation, PgasV =nRT. The kinetic energy is essentially the sum of vibrational, translational and rotational energies that allows a molecule to "escape" from the liquid phase at a given temperature.

Vapor pressure is the pressure exerted by a gas in the volume it occupies over its liquid phase. The vapor pressure isn't necessarily the pressure that "holds" molecules down in the liquid phase. You might be thinking that if there's a beaker of water at STP, then it's the atmospheric pressure is what keeps it in the liquid phase. This isn't the case since there will be a distribution of water molecules that have enough energy to go into the vapor phase. The water molecules that are in the vapor phase will have a gas pressure that will be less than the atmospheric pressure. Look into a general Chemistry textbook to see that liquid water has different vapor pressures at different temperatures.

In the 2nd question, you need to modify your explanation i.e. bonds are not being broken , ionic or covalent, but inter-molecular intearctions such as hydrogen bonds, dipole-dipole or van der Waals. From the perspective of the system, the liquid phase, vaporization is endothermic since a net input of energy is required to bring the liquid molecules into the gas phase. Think of the input energy as a "kick" to the liquid molecules. In order to get that "kick", there must be an energy input, such as heat for example, to go from the liquid to the gas phase.

Hey thanks for the response! Your answer to my 2nd question is very clear to me now, but I'm still a little if-y on my 1st question.

By looking at the equation PV=nRT, I definitely see the direct relationship between pressure and temperature; however, I'm just trying get a deeper understanding (or maybe I'm dwelling on it too much 😀)

So is vapor pressure part of the atmospheric pressure? Okay so when the temperature increases, the KE of molecules in the liquid increases, thus allowing these molecules escape into the atmosphere at a faster rate. Since the molecules are are essentially being transferred from liquid to gas phase, more molecules can now be found in the gas phase, increasing vapor pressure?

 

[heymanooh1]

So is vapor pressure part of the atmospheric pressure? Okay so when the temperature increases, the KE of molecules in the liquid increases, thus allowing these molecules escape into the atmosphere at a faster rate. Since the molecules are are essentially being transferred from liquid to gas phase, more molecules can now be found in the gas phase, increasing vapor pressure?

Yeah that's basically it but watch out on how you define "the atmosphere".Vapor pressure is not necessarily part of atmospheric pressure. One can create are an atmosphere consisting of different components such as methane gas, sulfur gas, carbon monoxide etc. and place it over liquid water at 25 degrees Celsius. Before equilibrium is reached, let's say the total pressure of the "atmosphere"/different gases above liquid water = 0.6 atm. As equilibrium is established, the "atmospheric" pressure will eventually rise above 0.6 atm because liquid water develops a vapor pressure, adding to the initial value of 0.6 atm.

To clarify a bit more. Fill a flask with water at a given temperature, evacuate and seal it from atmospheric/external pressure. Initially, the vapor pressure above the liquid will be zero. Letting the flask equilibrate at the given temperature, there will be a distribution of liquid molecules that can go into the vapor phase because there's sufficient energy. At the same time there will be gas molecules losing energy and returning to the liquid phase. Eventually these two rates equal each other, thus establishing equilibrium. The resulting vapor pressure above liquid water will be a different value from atmospheric because it has been sealed off.

Don't confuse the vapor pressure of a liquid being equal to atmospheric/ external pressure. This is true when a liquid has reached it boiling point temperature. The boiling point temperature of a liquid depends on atmospheric/external pressure and the substance(s) that makes up the liquid. For example, at a sea level pressure of 1 atm, the boiling point of water is 100 degrees Celsius, therefore the vapor pressure of water = 1 atm. At a higher altitude in Denver, CO, the boiling point of liquid water will be 94 degrees Celsius because the external/atmospheric pressure will be lower (~0.8 atm). The vapor pressure of boiling water at this elevation will also be ~0.8 atm. Now if you measure the vapor pressures of liquid water at both elevations at 25 degrees Celsius, they will be the same since the distribution of water molecules in the vapor phase will be identical and yet the atmospheric pressure is different.

The linear relationship(s) between (vapor) pressure and temperature are valid over a given region as observed on their respective phase diagrams and dependent on the substance.

 

[RAD11]

heymanooh1,

I get it now!!! Thank you so much for taking your time to explain these concepts to me. 🙂

 

[lisichka]

hello,
i am confused about one of those B.P elevation problems-->in one of kaplan homeworks please help-->here it is:
what is B.P. temperature of a 0.5 molal Na2SO4 solution in water at 1 atm, Kb=0.512 C/m
and them use Kb*molal=change in T-->0.521*0.5=0.256
but.....
why they don't take into consideration Vant Hoff factor(i) which should be 3 in thic case, since sodium sulphate is soluble??? change T=i*Kb*molal
am i missing something?
😎

 

[heymanooh1]

hello,
i am confused about one of those B.P elevation problems-->in one of kaplan homeworks please help-->here it is:
what is B.P. temperature of a 0.5 molal Na2SO4 solution in water at 1 atm, Kb=0.512 C/m
and them use Kb*molal=change in T-->0.521*0.5=0.256
but.....
why they don't take into consideration Vant Hoff factor(i) which should be 3 in thic case, since sodium sulphate is soluble??? change T=i*Kb*molal
am i missing something?
😎

The question is asking for the boiling point temperature of the electrolytic solution not the change in temperature, which the van't Hoff equation calculates for electrolytic solutions.

boiling point temperature of electrolytic solution = (boiling point temperature of pure solvent) + (change in boiling temperature).

 

[lisichka]

The question is asking for the boiling point temperature of the electrolytic solution not the change in temperature, which the van't Hoff equation calculates for electrolytic solutions.

boiling point temperature of electrolytic solution = (boiling point temperature of pure solvent) + (change in boiling temperature).

no i realize that, so it would be 100+0.2smth=100.2
but if you consider vant hoff factor in temp. change i (3)Kbm=0.7, so final temp should be 100.7

my question is why they did not consider i in temp change, and therefore final temp?

 

[heymanooh1]

no i realize that, so it would be 100+0.2smth=100.2
but if you consider vant hoff factor in temp. change i (3)Kbm=0.7, so final temp should be 100.7

my question is why they did not consider i in temp change, and therefore final temp?

I think it's because you have to assume that it's an ideal solution where the ions don't interact with each other, unless information in the passage indicates it's not ideal. i in the van't Hoff equation is essentially empirically determined because it takes into account interactions between the ions in solution.

 

[gridiron]

hello,
i am confused about one of those B.P elevation problems-->in one of kaplan homeworks please help-->here it is:
what is B.P. temperature of a 0.5 molal Na2SO4 solution in water at 1 atm, Kb=0.512 C/m
and them use Kb*molal=change in T-->0.521*0.5=0.256
but.....
why they don't take into consideration Vant Hoff factor(i) which should be 3 in thic case, since sodium sulphate is soluble??? change T=i*Kb*molal
am i missing something?
😎

Hey! You must take into account solubility rules. Here is a helpful site with the solubility rules you will need for the MCAT: http://www.ausetute.com.au/solrules.html.

The question you pose is interesting. Sodium sulphate, according to solubility rules, behaves as an electrolyte in water. Therefore, it should have an dissociation factor of 3. You then would use the boiling point elevation equation for an electrolytic solution to calculate the boiling point of the solvent in the solution:

Tb -100 = iKbm

Where you should get 100.768 C. I believe this is correct, but let Q affirm this.

 

[QofQuimica]

Hey! You must take into account solubility rules. Here is a helpful site with the solubility rules you will need for the MCAT: http://www.ausetute.com.au/solrules.html.

The question you pose is interesting. Sodium sulphate, according to solubility rules, behaves as an electrolyte in water. Therefore, it should have an dissociation factor of 3. You then would use the boiling point elevation equation for an electrolytic solution to calculate the boiling point of the solvent in the solution:

Tb -100 = iKbm

Where you should get 100.768 C. I believe this is correct, but let Q affirm this.

I agree with you. I think that the answer key is wrong here. There is no reason to assume that sodium sulfate in water would not dissociate; it's a completely soluble salt. If you start with 0.5 mol Na2SO4/kg, you will wind up with 1.5 moles of ions/kg.

lisichka, is 100.8 one of the answer choices? And just out of curiosity, what did the explanation to the problem tell you was the reason for choosing 100.3?

 

[akinf]

Hi

I am currently taking an intro. inorganic chemistry course and prepping for the exam. We talked a little about solubility. The first time I wrote the MCAT, I didn't memorize the solbility rules because I didn't feel I needed to. It never came up on the MCAT, so maybe I got lucky. However, I was wondering, as a general guide, as the difference in size of the cation and anion increase, solubility increases. Is this a safe general rule to go with, or would actually memorizing the rules be better.

The reason that was suggested to us is based on lattice enthalpy and hydration enthalpy. I probably don't need to know that for the MCAT, but I figure this might be a better way to remember the rules since I got the feeling that MCAT never really went into anything too specific in the PS section.

 

[QofQuimica]

Hi

I am currently taking an intro. inorganic chemistry course and prepping for the exam. We talked a little about solubility. The first time I wrote the MCAT, I didn't memorize the solbility rules because I didn't feel I needed to. It never came up on the MCAT, so maybe I got lucky. However, I was wondering, as a general guide, as the difference in size of the cation and anion increase, solubility increases. Is this a safe general rule to go with, or would actually memorizing the rules be better.

The reason that was suggested to us is based on lattice enthalpy and hydration enthalpy. I probably don't need to know that for the MCAT, but I figure this might be a better way to remember the rules since I got the feeling that MCAT never really went into anything too specific in the PS section.

There is no mention of the solubility rules on the AAMC study guide list. See page 19 for the solubility topics that are covered.

 

[lisichka]

I agree with you. I think that the answer key is wrong here. There is no reason to assume that sodium sulfate in water would not dissociate; it's a completely soluble salt. If you start with 0.5 mol Na2SO4/kg, you will wind up with 1.5 moles of ions/kg.

lisichka, is 100.8 one of the answer choices? And just out of curiosity, what did the explanation to the problem tell you was the reason for choosing 100.3?

hi QofQuimica,
no they do not have 100.8 as a choice, only 100.2 and some other much higher or lower numbers like 200 C or so. so i chose 100.2 as closest 🙁

as an explanation they had a formula Kb*molal=Temp change but no i included and no mention of ionic solubility. strange

well, i guess it is a mistake then? 🙄

 

[QofQuimica]

hi QofQuimica,
no they do not have 100.8 as a choice, only 100.2 and some other much higher or lower numbers like 200 C or so. so i chose 100.2 as closest 🙁

as an explanation they had a formula Kb*molal=Temp change but no i included and no mention of ionic solubility. strange

well, i guess it is a mistake then? 🙄

Ok, well in that case, pretend like sodium sulfate is glucose, and then you're fine. 😉

Yeah, I think it is a mistake. You and BME are both right that sodium sulfate should dissociate into three moles of particles for every one mole of salt. There should be a way for you to report mistakes in the practice materials in your account online; if you think about it the next time you're on, please do this so that other students will know about this problem. And good job catching that. 🙂

 

[lisichka]

Ok, well in that case, pretend like sodium sulfate is glucose, and then you're fine. 😉

Yeah, I think it is a mistake. You and BME are both right that sodium sulfate should dissociate into three moles of particles for every one mole of salt. There should be a way for you to report mistakes in the practice materials in your account online; if you think about it the next time you're on, please do this so that other students will know about this problem. And good job catching that. 🙂

yep i will let kaplan know,
thanx

 

[commuter9]

I have two Acid/Base questions.

EK:
Which of the following is the weakest base?
A. H-
B.Na2O
C.N^3-
D.OH-

The answer is D. I thought OH- was a strong base?

Which of the following explains why the measured pH of a solution represents an equilibrium condition?
A. The delta G of an acid-base reaction is negative
B. The delta H of an acid-base reaction is negative
C. The delta S of an acid-base reaction is positive
D. Proton transfer reactions take place very fast.

The answer is D. The explanation says that a,b,c are related to thermo and so thermo doesn't effect rate. I know that thermo doesn't affect rate, but where are they asking for rate? They are asking about equilibrium conditions and that makes me think of delta G.

 

[heymanooh1]

I have two Acid/Base questions.

EK:
Which of the following is the weakest base?
A. H-
B.Na2O
C.N^3-
D.OH-

The answer is D. I thought OH- was a strong base?

Which of the following explains why the measured pH of a solution represents an equilibrium condition?
A. The delta G of an acid-base reaction is negative
B. The delta H of an acid-base reaction is negative
C. The delta S of an acid-base reaction is positive
D. Proton transfer reactions take place very fast.

The answer is D. The explanation says that a,b,c are related to thermo and so thermo doesn't effect rate. I know that thermo doesn't affect rate, but where are they asking for rate? They are asking about equilibrium conditions and that makes me think of delta G.

One way to approach this problem would be to look at the conjugate acid of the base such that you end up with a neutral molecule.
H- ----> H2 (hydrogen)
O^2- ----> H2O (water)
N^3- ----> NH3 (ammonia)
OH- ----> H2O (water)

Notice for water, two protons need to be removed to get O^2-. Comparatively speaking, this makes O^2- a stronger base than OH-. Now going back to the conjugate acids H2 and NH3. Ask yourself if these are weaker acids compared to H2O, keeping in mind that three protons need to be removed from HN3. If the acid is weaker than H2O, then the conjugate base must be stronger. One hint that can help you out in determining if H- is a stronger base/acid than OH- comes from Organic Chemistry. If you recall, hydride salts, such as lithium aluminum hydride, are much stronger reducing agents/stronger bases than sodium hydroxide. Now comparing NH3 to H2O, ammonia is a stronger base than H2O since the free electron pair is less localized to nitrogen because it has smaller electronegativity compared to oxygen, therefore making NH3 a better base.

I'll answer the other question in more detail, I'm just too lazy right now. One way you can reason out the answer is to define what the dissociation constant Ka for acids is, define pH and also derive the Henderson-Hasselbach equation. Notice that for the HH equation, the terms delta G/H/S don't show up.

 

[commuter9]

Hi heymanooh,
I'm a little confused. First of all why is the conjugate acid of Na2O H2O? Wouldn't it be Na2OH? Also the conjugate acid of N^3- is ammonia? I didn't realize that (is that because you need 3 protons to cancel out the charge? I thought that a conjugate acid is the addition of only one proton.

One way to approach this problem would be to look at the conjugate acid of the base such that you end up with a neutral molecule.
H- ----> H2 (hydrogen)
O^2- ----> H2O (water)
N^3- ----> NH3 (ammonia)
OH- ----> H2O (water)

Notice for water, two protons need to be removed to get O^2-. Comparatively speaking, this makes O^2- a stronger base than OH-. Now going back to the conjugate acids H2 and NH3. Ask yourself if these are weaker acids compared to H2O, keeping in mind that three protons need to be removed from HN3. If the acid is weaker than H2O, then the conjugate base must be stronger. One hint that can help you out in determining if H- is a stronger base/acid than OH- comes from Organic Chemistry. If you recall, hydride salts, such as lithium aluminum hydride, are much stronger reducing agents/stronger bases than sodium hydroxide. Now comparing NH3 to H2O, ammonia is a stronger base than H2O since the free electron pair is less localized to nitrogen because it has smaller electronegativity compared to oxygen, therefore making NH3 a better base.

I'll answer the other question in more detail, I'm just too lazy right now. One way you can reason out the answer is to define what the dissociation constant Ka for acids is, define pH and also derive the Henderson-Hasselbach equation. Notice that for the HH equation, the terms delta G/H/S don't show up.

 

[heymanooh1]

Hi heymanooh,
I'm a little confused. First of all why is the conjugate acid of Na2O H2O? Wouldn't it be Na2OH? Also the conjugate acid of N^3- is ammonia? I didn't realize that (is that because you need 3 protons to cancel out the charge? I thought that a conjugate acid is the addition of only one proton.

Na2O is an ionic compound, so its dissociation in aqueous solution is 2 Na+ and O^-2. Na2OH is also a salt with an overall charge of -1 since there are two + charges from ionic sodium and -1 from hydroxide. Technically, the conjugate acid of O^-2 is OH-.

Yes, I realize that you make the conjugate acid from the base by adding one proton. The purpose of putting NH3 in there is to work with a more familiar molecule and to realize that by removing subsequent protons from NH3, the resulting molecule is a stronger base than the previous one. To be specific, base strength from strong to weak goes: N^3- > HN^2- > H2N-1 > NH3. The same reasoning holds with H2O.

Now if you want to justify the answer to double check, look in a general chemistry textbook for the Ka and/or Kb values of water, ammonia and molecular hydrogen. From the data you'll be able to figure out which will be the stronger/weaker conjugate acid/base.

In essence, the question was asking for the basicities of the molecules given relative to each other.

 

[rcd]

1. 1M CH3COOH exists as:

a) H+(aq), CH3COO-(aq), CH3COOH(aq)
a) H+(aq), CH3COO-(aq), CH3COOH(l)

2. T/F
CH3COOH (non-dissociated) can be dissolved in a solution without dissociating.

 

[QofQuimica]

I have two Acid/Base questions.

EK:
Which of the following is the weakest base?
A. H-
B.Na2O
C.N^3-
D.OH-

The answer is D. I thought OH- was a strong base?

OH- *is* a strong base, but so are all of the others. B and D are related to one another by the following equilibrium:

H2O <-> OH- <-> O2-

Thus, O2- is a stronger base than OH-. In fact, O2- is the conjugate base of OH-.

H- is a very strong base, much stronger than OH-. When we compare pKas, OH- has a pKa of about 16, but H- is about 35. Keep in mind that a one unit change in pKa is a change in *10 times* the number of protons. Thus, the difference between 35 and 16 is a difference of 1 x 10^19 protons. That is huge! N^3- is the conjugate base of HN^2-. I don't even know what the pKa of that is, but it's going to be really huge. N^3- is related to NH3 as follows:

NH3 <-> NH2- <-> NH^2- <-> N^3-

The first equilibrium between NH3 and NH2- has a pKa of about 37. The others will all be even higher. N^3- would therefore be a very, very strong base.

Which of the following explains why the measured pH of a solution represents an equilibrium condition?
A. The delta G of an acid-base reaction is negative
B. The delta H of an acid-base reaction is negative
C. The delta S of an acid-base reaction is positive
D. Proton transfer reactions take place very fast.

The answer is D. The explanation says that a,b,c are related to thermo and so thermo doesn't effect rate. I know that thermo doesn't affect rate, but where are they asking for rate? They are asking about equilibrium conditions and that makes me think of delta G.

Basically what this question is asking you is how you know that when you measure the pH, you are at equilibrium and not under a non-equilibrium condition. Kinetics tells you how fast the reaction will reach equilibrium. If the reaction were occurring slowly, you wouldn't get an equilibrium value (K) when you measured pH if you didn't wait long enough after mixing the acid and base together; instead, you'd get a value called Q, which is the non-equilibrium ratio of the products (ionized acid) to the reactant (unionized acid). So the reason why your measurement of the pH is the K value and not the Q value is because the proton transfer reaction happens so quickly.

I would agree with you that this question wasn't very clear. On the MCAT, there should not be any ambiguity as far as what you are being asked is concerned.

 

[QofQuimica]

1. 1M CH3COOH exists as:

a) H+(aq), CH3COO-(aq), CH3COOH(aq)
a) H+(aq), CH3COO-(aq), CH3COOH(l)

2. T/F
CH3COOH (non-dissociated) can be dissolved in a solution without dissociating.

What do you think? We'll help you if you need it, but only after you try it yourself first.

 

[rcd]

What do you think? We'll help you if you need it, but only after you try it yourself first.

😀 CH3COOH is polar, so it dissolves in water and therefore exists as aqueous even when not dissociated.

So, the answer to my first question is a, more specifically the top a.

The answer to my second question is T.

How'd I do?

 

[QofQuimica]

😀 CH3COOH is polar, so it dissolves in water and therefore exists as aqueous even when not dissociated.

So, the answer to my first question is a, more specifically the top a.

The answer to my second question is T.

How'd I do?

I concur. 🙂