DENA LEGGETT: Hi, and welcome so much to AP Daily
practice sessions with multiple choice.
My students call me Doc Dena, and I'm here
to help you prepare for that big day.
Let's jump right into some multiple choice questions.
This is a good one because it has a little nuance.
You want to make sure you're reading carefully.
So when you're doing these, you want
to make sure you're circling and highlighting
important information.
And we have two volumes, and we're asked for a new molarity.
The reason I added this one because we're not
asking for the molarity of magnesium chloride.
We're asked for the concentration of the chloride
ion within.
So, we get to do a boom-boom, because we added volume
to volume.
Remember that total volume isn't 400.
It's the sum of the two.
And that will give us our magnesium chloride molarity.
Don't grab onto the a yet, because we're not done.
We need to convert magnesium chloride to moles
of chloride over that liter.
So our answer is actually 0.4 we get twice as much
for the molarity of chloride as magnesium chloride.
All right.
This next one, I hopped in because my students
get tripped up by some of the information in this question.
So we have mass and molar mass, so we know we can get to moles.
And it asks us to find out what the molarity is
of this solution.
And it asks us to find out what the molarity is
in a 50 milliliters sample.
That's the trigger that gets to my kids.
We have 200 milliliters.
We did not add volume to volume.
We're taking out a particular volume.
We're taking out 50 mLs.
The concentration stays the same.
We never use that 50 mLs in the calculation at all.
We just need the molarity of the original 200 mLs.
Don't be tempted to boom-boom.
So we convert our milligrams to grams,
mass to moles, use molar mass.
Now, we've got the moles you divide by the volume.
And we have our molarity.
That trick is, sometimes there's extra information.
You might not always need it.
So read carefully.
This is according to VSEPR model.
I have a decrease in bond angles.
And the question asked me to account for that.
Now, my phrase is, when in doubt, draw it out.
VSEPR hybridization polarity, you
want to draw some Lewis dot structures.
In this case, we can start out by getting rid
of some distractors.
That may simplify our work.
So VSEPR stands for "Valence Shell Electron Pair Repulsion."
So our answer must have to do with some electrons that
are repelling.
It's not about strengths of bonds.
It's not about the size of the central atom.
Because the arrangement is around the central atom.
And it's not due to the electronegativity.
Now, mind you, have this been about polarity,
that electronegativity could have had an impact.
Instead, it's about increasing the number of unpaired-- shared
pairs of electrons.
Now, if you had to draw those Lewis dot,
you want to take that time.
And you would find that carbon doesn't
have any non-bonded pairs on the carbon, nitrogen has one,
the oxygen and water has two.
And that's why the bond angles progressively decrease.
And this one, we have nitrogen molecules
absorb ultraviolet light, but not visible.
I2 absorbs visible and not ultraviolet,
which best explains it.
Now, hopefully your teacher went through the electromagnetic
spectrum, and what ranges that we
use to study different properties
and movements in molecules.
So the first one talks about vibration of molecules.
And we're in the visible and ultraviolet region,
and vibrations occur in the infrared region.
So that's lower region.
That's not going to help us.
More energy is required to remove an electron from an I2
then N2.
That's in the x-ray or gamma.
Taking those rate, electrons off takes a lot of energy.
Visible does not produce transitions
between electronic energy levels.
Well, if you've done atomic emission spectroscopy,
you would know that excitation and relaxation
occur in the violet and visible ranges.
That's probably our answer.
But I want to take a look at D, just to make sure.
D is talking about molecular masses.
We would study those, possibly, using mass spectrometry.
And at the AP level, we really apply that to isotopes.
So that ones out too.
So, our first inclination was right.
Visible and UV is commonly used for excitation relaxation
in electronic transitions.
Now, this next one has quite a bit of information.
We've got three molecules, differing sizes,
same temperature, got some different pressure here,
and the question says that we're going
to increase the pressure at constant temperature
and monitor condensation.
So which gas condenses first?
Methane, ethane, or butane?
So we're going to need to evaluate.
If we're doing condensation, we're
going to be looking at a phase change, gas
to liquid, which means we need to evaluate intermolecular
forces.
So if we increase the pressure, we're
going to be decreasing the distance between the molecules.
Now, as molecules start to get closer and closer and closer
and closer together, they start to feel attractions.
So we start to increase the attractive forces
between molecules.
Now, the strongest attraction is going to condense first.
They're going to feel that attraction,
and boom, they're going to hop from gas
to liquid very quickly.
So let's evaluate which one of these
has the strongest attraction.
All three are non-polar covalent molecules.
All three have only London dispersion intermolecular
forces between neighboring molecules.
As you increase the size, you increase polarizability.
That increases the strength of the London dispersion.
So that means that butane, our biggest molecule,
has the strongest London dispersion,
and so, would condense first.
This next one shows chromatography-- paper
chromatography.
We want to look at what can be concluded about the dye.
So it asks about molar mass, intermolecular forces,
attraction to stationary phase versus mobile phase.
My guess is that it's showing us that mobile phase, it's showing
us the stationary phase, it's probably C or D.
But I want to make sure you understand this question.
The paper is stationary.
It doesn't move.
Stationary phase.
The solvent is the mobile phase.
It's going to move up the paper.
Read questions carefully, because sometimes,
the mobile phase is polar, and sometimes the mobile phase
is non-polar.
We're not given any indication of that.
OK?
So we want to look at the comparative attraction
between the solvent as the dye moves up to the solvent front.
So, the more the dye moves with that solvent
as it goes up to the end where the solvent front is,
the more attracted it is to that solvent.
This technique is not separating based on molar mass--
there are those that do, but paper chromatography doesn't.
We don't have any indication of weak intermolecular forces.
We can't even talk about weaker versus stronger,
because we're not told what the polarity of the solvent front
is.
So that's out.
So now, what we have to do is look at,
does it have a weaker attraction for the stationary
and a stronger for the molar mobile phase as c says?
Or does it have a stronger phase for the stationary
and weaker for the mobile?
Well, if it was more attracted to the stationary,
it would stay stationary.
If it's more attracted to the mobile phase,
it's going to be mobile with the mobile phase.
And our dye moved pretty far.
It was very mobile with the mobile phase.
And therefore, c is our choice.
This next one has to do with intermolecular forces.
We have two molecules, nonane, and you don't have
to name the one on the right.
What we have to look at is that the one on the left
is all carbons and hydrogens.
It is non-polar with only London dispersion.
The one on the right has fluorine,
which makes it polar with a London dispersion
and dipole-dipole.
Now, it tells us in the question--
so we're going to have four choices there.
They say they have nearly identical molar masses,
but no nonane has a greater boiling point.
That's kind of surprising.
Because nonane is London dispersion,
and the one on the right, the 2,3,4-trifluoropentane is
polar.
It has dipole.
So this is surprising that the nonane
has the higher boiling point.
So we've got to figure out why.
Well, right away, we've got to get rid of that misconception.
Molecular compounds are breaking intermolecular forces
between neighboring molecules.
We are not breaking bonds.
So we want to get that out of there.
Now, the second one says C-F is more polar than C-H. True
statement.
The problem is, that would argue for the higher boiling point
for the 2,3,4-trifluoropentane, and that's not what we see.
So that's out.
Carbon chains are longer in nonane.
What that does is it allows those chains to nestle up
against and get close to one another.
And that maximizes their London dispersion, decrease distance,
increase London dispersion.
So I think that's the answer.
But I'm going to make sure by reading the last distractor.
It says, the carbon chains are farther apart in nonane.
Now, we just said Coulomb's law says
that if you increase the distance,
you decrease attraction.
We want to decrease the distance.
It's clear they are more attractive.
And that could be explained by a decreased distance,
not the increased.
So d is out.
Our answer is c.
Always evaluate all possibilities
before you jump on and answer too quickly.
OK.
This is comparing boiling points of some ionic compounds.
And it very specifically says, based on Coulomb's attractions.
So if we're going to be talking about Coulomb's attractions,
we want to talk about charge, and we
want to talk about distance.
Increase charge, decrease distance, increased attraction.
So I like to evaluate charge first.
It is typically the most important contributing factor.
So if we compare NaCl to MgO, MgO's the clear winner.
+2 cation, -2 anion, to +1, -1.
So we're going to make sure that we don't pick those ones that
place NaCl as having a stronger attraction,
hence, higher boiling point than magnesium oxide.
Now, we're down to having to deal with distance.
They give us the ionic radii, and both of the two,
NaF and NaCl, have the same cation.
So what we have to focus on is the anion radius.
Since fluoride is smaller, it has a shorter distance
between our cation and anion.
Shorter distance means increased attraction.
And that means that NaCl will have a lower
boiling point than the NaF.
NaF is going to be stronger.
Wow.
I know that was fast.
I hope you paused as needed, rewind as needed.
I'm glad you joined us.
If you need more information, there's
the QR code to so much review that can help you.
Thanks so much for giving me your time as you
review for this test.