[MUSIC]
Lighthouse Scientific
Education presents
a lecture in the Mole series.
The topic; Molarity,
Molality and Mole Fraction.
The lecture is set up so
that these 3 sub-topics can be
viewed together or
individually as needed.
Material in this lecture
relies on an understanding
of the previous lectures
Atomic Mass and
Molecular Weight and
Percent Composition.
The discussion of these
important concentration scales
begins with an introduction to
the concept of concentration.
Concentration is covered
more in depth in its on lecture
in the 'Water and
Solution' series of lectures
but is brought up
here as a support
for concentration
scales that use the mole.
Some basic definitions
are offered along with an
introduction of the 3
mole based scales covered
in the lecture.
The first scale is the most
common of the three and
it is called molarity.
A setup of steps and examples
are offered to outline basic
calculations with this scales.
Molarity is also used
as a conversion factor.
That too will be reviewed.
The second scale is molality.
Steps and example of basic
calculations are provided.
Molality can also be used
as a conversion factor.
The last scale, and the least
common, is the mole fraction.
Basic calculations will be
presented with this scale.
Concentration and we
will limit the discussion
to concentration in liquids.
Some compounds dissolve in
liquids and form solutions.
Important terminology
includes the solute.
That is a compound which is
dissolved in the solution.
The dissolving liquid
is called the solvent.
Together, solute and
solvent are a solution.
The lecture on 'Concentrations
and Dilutions' in the
'Water and Solutions'
series covers these terms
more in depth.
The amount of solute per
volume of solution is
called the concentration.
Sometimes instead of
volume of solution it is
amount of solvent.
There are many different
concentration scales.
The ones of interest in this
lecture have the amount of
solute given in moles.
There are 3 common
concentration scales
that fit this description.
The first is molarity.
As a unit it is capital M.
Molarity is defined as the
number of moles of solute
dissolved in 1 liter of solution.
Capital M therefore stands
for moles per liter.
A similar scale is molality.
As a unit it is smaller case m.
Molality is defined as the
number of moles of solute
per kilogram of solvent.
Small case m therefore stands
for moles per kg.
The difference between
molarity and molality
is the first one uses volume
of the entire solution and
the second only considers
the mass of the solvent.
Both scales have their place.
The last scale is mole fraction.
It is given the unit 'chi'.
That is a Greek letter
that looks like a fancy x.
The definition of mole
fraction is moles of solute
over the total moles of
solution. It is indeed a
fraction with part
over the whole.
We will discuss each in turn.
Beginning with molarity.
That is the number of
moles of solute dissolved in
1 liter of solution;
moles per liter.
Written as an equation,
the moles of solute is the
numerator and the liters of
solution is the denominator.
Normally, division of the
numerical values attached
to the moles of solute
and the liters of solution
is done so that there is only
one numerical value
assigned to molarity.
That means the denominator
becomes 1 liter.
This will become
clearer when we get to
molarity calculations.
To familiarize ourselves with
this scale consider 1.5 M NaCl.
First of all, when actually
using this scale it is not
referred to as M or capital M.
In a concentration 'M' is
called molar, so the proper
way to call this concentration
is 1.5 molar NaCl.
Applying the definition of
molarity, 1.5 molar NaCl is
1.5 moles of NaCl in
1 liter of solution.
That holds even if there isn't
a whole liter of solution.
The scale is the proportion
of moles of solute
to volume of solution.
So half of a liter or 0.5
liters of 1.5 molar NaCl
will only have
half as much NaCl.
Dividing 0.75 by 0.5
will get that 1.5.
Here 3.0 molar NaCl.
By definition it is 3.0 moles
of NaCl in 1 liter of solution.
Half a liter of this
solution would have half
as much solute. 1.5
divided by 0.5 gives 3.
Concentration is proportion
not absolute amounts.
One more point to bring
up before moving into
calculating the molarity.
A solution can have
more than 1 solute.
Each solute will
have its own molarity.
Each solute will use just
the value of its own moles
in the calculation.
They will all share
the volume of solution.
Calculating molarity
is an essential skill
in the study of chemistry.
It is not one of the more
difficult calculations
as can be seen with
these 3 basic steps.
Begin by calculating moles.
Convert mass values
of solute into moles.
That usually involves
molecular or formula weights
as conversion factors.
The second step is to get the
volume of solution in liters.
Sometimes the volume is given
in milliliters or even mass.
Use your conversion skills to
get those values into liters.
Finally, divide moles from
step 1 by volume from step 2.
That'll do it. Alright, let's
put these steps to work with
a couple example problems.
Example 1; 2.19 g of
NaCl dissolve in water
to make a 750.0 ml solution.
What is the molarity?
Step 1 has us calculate
the moles of solute.
But which substance
is the solute?
Is it the NaCl or
is it the water?
If we revisit the definition
of solute and solvent we find
the dissolved component
is called the solute
and the dissolving liquid
is called the solvent.
From the problem it is
seen that the NaCl is
dissolved (solute). It is
dissolved in water (solvent).
Notice that the problem
does not say that there are
750 ml of water. It says that
there are 750 ml of solution.
The actual amount of water
will be less since the salt is
also taking up volume.
Be sure to keep solution
and solvent straight
So, it is the solute NaCl
that needs to be converted
in to moles. And there
are 2.19 g of it.
Converting grams into
moles requires a molecular
or formula weight
of the compound.
At this point the student
should be relatively comfortable
with getting such values so
we will skip the calculation
and just use the value.
Inserting the formula weight
of sodium chloride into this
equation with an orientation
that allows grams of NaCl
to cancel will produce
a mole value of 0.0375.
Step 2 is to get volume
of solution into liters.
Currently, it is in milliliters.
The conversion into liters
is one that many students
can do in their heads.
We will use a metric
conversion factor
that relates mls to liters
and has an orientation that
allows mls to cancel.
The math produces 0.750 L.
The last step is to divide
the mole value obtained in
step 1 by the volume
value obtained in step 2.
0.0375 moles of NaCl divided
by 0.750 L of solution.
The molarity of the
NaCl is 0.0500 molar.
A check of sig figs in the
problem verifies that having
3 sig figs (500) in
the molarity is correct.
That's the process.
A second example asks what's
the molarity of a solution
prepared by mixing 6.07
g potassium nitrate, KNO3,
in water to a volume of 240 ml.
The wording 'to a volume'
is there to tell us that
water is added until the
entire solution has
a volume of 240 ml.
Returning to the definition
of solute and solvent and
reviewing the problem has
the potassium nitrate being
mixed in as the solute and
the water as the solvent.
Onto calculating
moles for the solute.
There are 6.07 g
of potassium nitrate
and they needs to be
converted into moles.
The conversion factor
is a formula weight,
101.10 grams per mole,
with an orientation that
allows grams of potassium
nitrate to cancel out.
The math will produce
0.00600 moles of KNO3
The 2 zeroes after the 6
are significant figures.
Step 2 has us get the volume
of solution into liters.
Like the previous problem the
volume is given in milliliters.
And like the previous problem
it uses a metric conversion
with an orientation that
allows mls to cancel.
The math produces 0.240 L. By
inspection of this calculation
we can reason out that moving
from mls to liters is moving
3 decimal place to the left.
We finish up by dividing the
mole value obtained in step 1
by the volume value
obtained in step 2.
0.00600 moles of potassium
nitrate divided by 0.240 L
of solution gives a molar
concentration of 0.0250 molar.
That's the number value.
The molarity has
3 significant figures as does
numerical values
in the problem.
Staying with molarity but
moving into conversion factors.
Molarity can be used to
move between moles of solute
and volume of the solution.
Molarity, as defined,
has 3 variables. Given values
for 2 of the variable the
3rd can be solved for.
As a demonstration of
the principle consider the
arrangement of units under the
heading of current units as in
have or given units
and desired units
as in wanted or
asked for units.
The conversion that begins
with liters of solutions and
ends with moles of solute
requires a solution molarity
in the orientation that allows
the units liters to cancel
out leaving the desired units
of moles of solute.
The two given values are
volume of solution
and the molarity.
The third value, moles
of solute, is solved for.
Flipping the arrangement
around so that moles of solute
are the current unit and
liters of solution are the
desired unit will still require
molarity as the conversion
factor but with moles of
solute in the denominator so
that units cancel out
leaving the desired units of
liters of solution.
The two given values are
moles of solute and the molarity.
The third value,
volume of solution,
is solved for.
These principles are very
much like the principles
that underlie molecular
weight as a conversion factor
between grams and
moles of a compound.
Let's see these
principles in work.
How many moles of nitric acid,
HNO3, are present in 0.250 L
of 0.75 M HNO3 solution?
Is this a conversion problem?
Well, we should take our time
and consider the problem first.
We can do that with
a set up to find
the known and unknowns.
Looking at the problem
it can be seen that
given is 0.250 L of
the acid solution.
Make a note of that.
And that the solution
has a molarity of 0.75.
Make a note of that.
What the question asks for
is moles of nitric acid.
That is an unknown.
The relationship between
these values is current or
given units of the 0.250 L
and desired or wanted units
of moles of nitric acid.
This is indeed a
conversion problem.
It is a conversion
problem of one aspect
of the solution into another.
The question to ask is what
conversion factor relates
the volume of a solution
to the moles of solute?
It is the molarity.
But written in this form,
with capital M, it
is not that helpful.
It looks and feels more like a
conversion factor when
written in its definition form.
0.75 molar nitric acid is
0.75 moles of nitric acid per
1 liter of solution.
Fortunately, this orientation is
the correct one for canceling
liter units out and leaving
mole units.
Some calculator math
and 0.250 L of a
0.75 molar nitric acid
is shown to contain 0.1875
moles of nitric acid.
A quick check of sig. figs.
has us trimming that value down
to 2 significant figures.
Okay, to expand on the
problem solving using
molarity we will keep this
same problem but insert
a minor alteration to one
of the values in the problem.
Say that instead of 0.250 L
of solution the problem gave
250 ml of the solution.
This is an expansion
because the volume units in
molarity are in
liters and not mls.
But we've come across
the issue before when
calculating molarity.
There we converted our mls
of solution into
liters of solution.
The metric conversion factor
is used. Units cancel out
and the conversion into
0.250 liters is done.
The amount of 250 mls
was selected because the
converted value fits right
into the earlier solution.
So, the only real change
with this new value is
an additional
conversion before the
molarity conversion.
Can the two conversion
be combined into one
dimensional analysis equation?
Of course they can.
Convert the starting current
units of 250 mls into liters
with the metric
conversion factor.
Verify that units of mls
cancels leaving units of liters.
Convert those units into moles
of solute with the molarity
written in definition form.
Verify that unit liters
cancel out leaving unit
moles and then do the math.
The same 0.19 moles of
nitric acid should come out.
A second problem is
probably in order.
0.10 moles of sulfuric
acid, H2SO4, makes how many
liters of 0.050
molar H2SO4 solution?
As with the previous problem
a setup that searches the
problem for knowns and
unknowns is a helpful step.
In this problem the given
is 0.10 moles of the solute
sulfuric acid and
the concentration is
0.050 molar sulfuric acid.
The unknown in this question
is the liters of solution.
The relationship
between these values is
given or current units of the
0.10 moles of sulfuric acid
and desired or wanted
units of liters of solution.
Yes, we have another
conversion problem.
Given this relationship what
is the conversion factor that
should be inserted between
the moles of solute and the
unknown volume of solution?
It is the molarity.
Does this orientation
of molarity allow for
the canceling out of
the units of moles?
No. it doesn't. It will
be necessary to flip the
orientation so that the
moles are in the denominator
which does allow canceling
of moles and leaves the
desired unit of liters.
Doing the math
produces a value of 2.0
liters to 2 sig figs.
This is the core of a
mole to volume conversion
using molarity.
Such problems can be a bit
more complicated though.
For example the moles of
sulfuric acid could be given
as grams of sulfuric acid.
Essentially, this is
the same problem with
an addition step.
What is that step?
Well, molarity can only
convert between moles
and volume so the gram
value is going to have to be
converted in to moles.
There will be 2 conversions
in the problem if grams
are given instead of moles.
What is the conversion
factor that allows grams
to be converted into moles?
It is the molecular weight of
the acid. 98.1 g of sulfuric acid
is 1 moles of sulfuric acid.
Is this orientation of the
molecular weight correct
for the conversion?
No it isn't. Grams
need to cancel out so the
denominator of the
conversion factor has to have
grams of sulfuric acid.
The molecular weight
needs to be flipped.
Now grams can cancel
out and the math yields
0.10 moles of sulfuric acid.
Just like the value in
the original problem.
This calculated mole value
can now be inserted into the
core solution giving the
requested value of 2.0 L.
These two conversions can
be combined into a single
dimensional analysis equation.
The given or starting or
current units in this problem
are the grams of sulfuric acid
which will need to be
converted into moles.
Insert the first conversion
factor, the flipped
molecular weight and
cross out gram units.
That leaves moles which the
second conversion tells us
uses the flip of the molarity
as a conversion factor.
Verify that units moles
cancel out. Get the calculator
and punch a final
value of 2.0 L.
This modified example 2
is a very common homework
molarity problem.
The next concentration scale
we are covering is molality.
It is not a common scale but
its usefulness can be seen with
the limitation of the scale
we just covered; Molarity.
Yes it is very easy to
confuse these two names.
They only differ by one letter.
Never the less,
the problem with molarity
is that the volume can be
temperature sensitive.
That is many solutions
expand with
increasing temperature
(volume gets larger).
If volume increases
with temperature,
the denominator of the
fraction gets larger
making the overall
fraction smaller.
That means the molarity
decreases and this is so
even though the same mass of
solute and solvent are used!
The amount of matter
does not change but the
concentration does.
The makes molarity a less than
optimal concentration
scale for some processes.
Especially those process
that are accompanied by a
temperature change.
Consider molality.
It is the number of moles of
solute per kilogram of solvent;
mole per kg.
Importantly, the mass
(of the solvent) is not
temperature sensitive.
Molality does not vary
with temperature like
molarity does. This
makes molality the
optimal concentration
scale for some process.
Especially those that
are accompanied by a
temperature change.
It is also called molal
concentration. Molal is used
with molality like molar
is used with molarity.
Molality is used when dealing
with colligative property.
Such properties include
boiling point elevation
and freezing point depression.
Colligative properties
are covered in their own
lecture in the 'Water
and Solution' series.
In that lecture being able
to calculate molalities is
an expectation.
Which is convenient since
that is our next topic.
Calculating molality is very
similar to calculating molarity.
Begin by calculating
moles of solute.
Convert mass values
of solute into moles.
Then, get the mass, or weight,
of the solvent and that needs
to be in kilograms.
Solvents are usually measured
by volume so some sort of
conversion will probably
be necessary.
A convenient relationship
exists for the weight and
volume of water which is
the most common solvent
found in our studies.
Between the temperatures
of 4 degrees and
40 degrees Celsius, to
2 significant figures,
1.0 kg of water is
equal to 1.0 L of water.
That allows the following
conversion factors.
Moving between weight and
volume of water in this
temperature range is
very straightforward.
The last step is to take the
moles of solute from step 1
and divide them by the
kilograms of solvent in step 2.
Two examples will help
solidify the process.
The first example asks if
87.7 g of sodium chloride,
NaCl, is added to 0.500
kg of water, what is the
molality of the solution?
Step 1 in getting a molality
is to get moles of solute.
That requires a recognition
of the solute and
solvent in the solution.
NaCl, the solute, is a
salt added and dissolved
in water, the solvent.
The solute is
given as 87.7 grams.
The objective is to
convert that into moles.
The conversion factor
is the formula weight of
NaCl; 58.44 g per mole.
Note that the formula weight
is oriented so that grams
are in the denominator
allowing units to cancel.
87.7 divided by 58.44 is 1.50.
There are 1.50 moles
of sodium chloride
in 87.7 grams of
sodium chloride.
Now to getting the
weight of the solvent.
Conveniently, the weight
is already given in
kilograms; 0.500 kg. It is
in the correct units so there
is nothing more
to do in this step.
Molality is gotten by dividing
the moles of solute from step 1
by the kg of solvent in step 2.
1.50 moles of NaCl
divided by 0.500 kg of
water giving a molal of 3.00.
Not too bad. This is the core
of a molality calculation.
Variations to this calculation
include the solvent being
given in units of volume.
Liquids are usually measured
by volume so more often than
not that is the unit presented.
The change to the core
calculation will be in step 2,
getting the weight
of the solvent.
Now the starting value
here is the volume of water.
As was discussed when
outlining these rules there is a
straightforward conversion of
liters of water to kg of water.
That did come with a
temperature range caveat.
The temperature of the
solvent needs to be included
to make this conversion.
Now units can be canceled
and a final value obtained.
The 0.500 kg of water
is the same value the
problem originally provided so
the step 3 calculation
is the same with the same
3.00 molal answer.
A second example says
that a 4.5 g sugar cube
(sucrose; C12 H22
O11) is dissolved in
350 ml of lemonade (at 4 oC).
What is the molality of the
sweetened lemonade?
Begin by getting the get
moles of solute. A review
of the problem shows the
sugar to be the solute (it
is getting dissolved) and
the lemonade as the solvent.
Sugar is given a mass of
4.5g and that needs to be
converted into mole. The
molecular weight of sucrose
is given in the problem
as 342.2 g per mole.
The orientation of the
molecular weight is such that
grams cancel out.
The division of 4.5 by 342.2
gives a mole value 0.0132.
In this problem that
is the easy part. A bit
more difficult is getting the
weight of the solvent.
It is given in mls so a
conversion into kg is
necessary. Is there a
convenient volume to mass
conversion for lemonade?
Not really. But lemonade
is mostly water and the other
component, lemon juice,
is also mostly water.
Given that, we will say that
volume of very dilute solutions
is almost equal to
volume of the solvent.
It will not be too big of
a stretch to consider the
lemonade as essentially
all water which does have
a nice conversion.
There are temperature
constraints to this
conversion and a check of
the problem shows that the
conversion is acceptable
to 2 significant figures.
A second issue with
the 350 mls is that the
conversion factor uses liters
as its volume units not mls.
It there a conversion factor
that gets mls into liters?
Yes, the standard
metric conversion.
These two conversions can be
done in separate steps or in
a dimensional analysis
equation as shown here.
Canceling liter leaves
kg units and a bit of math
gets the mass of lemonade as
0.35 kg to 2 significant figures.
The molality calculation is
moles of solute from step 1
divided by the kg
of solvent in step 2.
0.0132 moles of sucrose
divided by 0.35 kg
of water or lemonade.
That gives a solution
of 0.377 molal sugar.
Notice, however, that
the given solvent mass
is to 2 significant figures.
The final answer must also
be limited to 2 sig. figs.
The molality is 0.38.
Molality can be used
to move between moles of
the solutes and mass (kg )
of solvent much in the same
manner as molarity
allowed the conversation
between moles of solute
and liters of solution.
Using the same headings
of current units and
desired units, conversions
between kg of solvent
and moles of solute
requires a molality in
an orientation that has
the kg units cancel out
leaving the desired
units of moles of solute.
Flipping that current and
desired units around gives
an arrangement where moles
of solute are converted
into kg of solvent
A flip of the molality is
the conversion factor here
since units of moles of
solute cancel out leaving the
desired units of kg of solvent.
An example problem will
emphasis just how similar
these principles are to those
used with molarity and
molecular weight conversions.
The example asks "How
many kg of water is needed
to make a 1.50 m
solution from 0.500 moles
of potassium chloride, KCl,?"
To be sure that this is a
conversion problem
a review of the
problem is in order.
Known for the problem
is the amount of potassium
chloride in moles.
Also known is the
molality of the solution.
It is 1.50 molal.
What is unknown is the kg
of solvent in the solution.
The relationship between
known and unknown is
current units of 0.500
moles of potassium chloride
and desired kg of solvent.
Clearly a conversion problem.
The other known quantity
from the set-up is the molal
and it will serve as the
conversion factor.
However, the orientation
as written, 1.50 moles
of potassium chloride over
1 kg of water, is not correct.
The current units of moles
do not cancel out.
That is an easy fix though.
Flip the orientation and
the mole units cancels out
leaving units of kg.
Some calculator work and
0.500 moles of potassium
chloride dissolve in 0.333 kg
of water. And that will
make a 1.50 molal solution.
This is a core
molal conversion.
A modification to this strategy
would occur is the question
asked not for kg of water but
the much more common unit of
liters of water. Since volume
is temperature sensitive
a temperature
needs to be given.
Here it is given as 22
degrees Celsius which
falls with the temperature
range that allows the
kg of water to liters
of water conversion
we used earlier (which
means we can use it again).
The core conversion gives kg of
water and that is the starting
value for the second
conversion: kg to liters.
The conversion factor needed
here will have the current
unit kg in the denominator
and desired units of liters
in the numerator.
Starting units cancel and a
very simple math step give the
final answer 0.33 L of water.
The 2 significant figure answer
comes from the 2 sig. fig. limit
of the kg to liter conversion.
For completion sake we
should combine these
two conversions into
a single dimensional analysis
equation. It starts with
the original given value of
moles of potassium chloride
and aims at the final
value or desired units
of liters of water.
Like the equation below
the staring unit of
moles will be converted
into the units of kg of water
with a flip of the given molal
as validated by the
moles canceling out.
The new current unit
of kg of water will be
converted into liters of water
with the conversion factor
of the above equation.
Multiplying all the numerators
together and dividing
all the denominators will
produce that same value
of 0.33 liters of water.
The logic of conversions
are essentially the same.
The more we see of them
the more the logic sinks in.
The third mole based
concentration scale
is the mole fraction.
It is the ratio of moles of
one compound to the
total number of moles
of the solution or mixture.
It is given the symbol chi which
is a lower-case Greek letter.
And like all fractions it is
the ratio of part to whole.
The values in this fraction,
however, are strictly moles.
If we consider a mixture that
just contains substance 'A'
and substance 'B',
the mole fraction of A
is moles of A over moles
of A plus moles of B.
That might be easier to
follow in fraction form.
The part, moles of
A, over the whole,
moles of A and moles B.
Additionally, the mole fraction
of B is moles of B over
moles of A plus moles of B.
The part, moles of B, over
the whole, moles of A plus
moles of B.
Because the sum of the
parts must equal the whole,
the mole fraction of A plus
the mole fraction
of B equals 1.
Mathematically that can be
shown by putting the mole
fractions in fraction form
and since these two fractions
have the same denominator
add the numerators and
keep the denominator.
It follows that if the
numerator and the denominator
have the same value
the fraction is 1.
Anything divided
by itself is 1.
Mole fraction of A plus
mole fraction of B is 1.
This equation tells us
that if we know one of the
two mole fractions we
can calculate the other.
If we subtract the term mole
fraction of B from both sides
we get a new equation solved
for mole fraction of A.
An equation that is solved
for mole fraction of B
is obtained in a similar manner.
These are very useful
equations in 2 part mixtures.
Many students may never be
faced with a mole fraction
situation that has more
than 2 parts. But just so that
we know, for mixtures with 3
or more parts the statement
that the sum of the parts
equals the whole still
holds and shown here is the
generic mathematical
equation that accompanies
that statement. On the left
are all the mole fractions
for any number of parts and
on the right is the sum of 1.
The mole fraction for any
one parts can be isolated
mathematically by subtracting
all the other parts from
both sides. This is just
an expansion of how we
generated the two equations
at the bottom of the screen.
A specific and common
example of mole fractions
involves solutions that
are made of a single solute
(substance A) and a
solvent (substance B).
The mole fraction of the
solute is the moles of solute
over moles of solute
plus moles of solvent.
The mole fraction of the
solvent is moles of solvent
over moles of solute
plus moles of solvent.
The mole fraction of the
solute plus the mole fraction
of the solvent will be
equal to 1. Sum of the parts
still equals the whole.
Mole fractions are used for
specific purposes like when
dealing with the partial
pressure of a gas. Actually
calculating mole fractions
is not the involved.
Mole fractions can be
determined in a 3 step process.
Start by calculating the
moles for all components in
the mixture or solution.
This will usually
involve the formula
weight of the compounds
as conversion factors.
From there get the
total number of moles
by summing all the moles
calculated in step 1. That
is the whole in the fraction.
To get the mole fraction
of interest, that is of
any one substance, divide
moles of interest (step 1)
by total moles (step 2).
Once the moles have been
determined in step 1
the rest of the process
is basic arithmetic.
A couple examples
will prove that.
First example: 333.0 g
of water is needed to make
a 1.50 m solution from 0.500
moles of potassium chloride.
What is the mole fraction
of the potassium chloride?
What is the mole
fraction of the water?
This problem can be
considered a follow up
question to the example
problem covered in molality
as a conversion
factor since it uses
the same compounds
and same values.
Beginning with the potassium
chloride's mole fraction
but sharing the first 2 steps
with water's mole fraction.
Calculate the moles
for all components.
The potassium chloride was
nicely provided in mole form
so that compound is set.
The 333.0 grams of water
will need to be converted
into moles of water.
What do we use to
complete this conversion?
It is the molecular weight of
water; 18.02 grams per mole.
The orientation of the
molecular weight is such that
grams of water cancel out.
Doing the division math gets
18.48 moles of H2O.
Step 1 is done and that
is the bulk of the work.
Step 2 is the arithmetic step
of adding all the mole values
together to get a whole.
That is 0.500 moles of KCl
plus 18.48 grams of
H2O. A good way to get
the number of significant
figures correct is to line up
the decimal points and note
the last digit of significance.
The sum is 18.98 moles.
Clearly there is much more
water than potassium chloride
and we should expect the
respective mole fractions to
show that great difference.
To get the mole fraction
of the compound of interest,
potassium chloride, take the
moles of potassium chloride
and divide them by the
total moles from step 2.
Part divided by the whole.
The math comes to 0.0263.
The mole fraction is
usually given as a decimal.
We'll keep track of that value
as we move to the second part
of the question; the
mole fraction of water.
The nice part about this
calculation is that step 1 and 2
are identical to those for
chi KCl. All that is left to do
is divide the moles of
interest by the total moles.
That is the 18.48 moles
of water divided by
18.98 total moles.
The fraction looks like
it's pretty close to
1 and it is; 0.9737.
Let's keep track of that value
and double check our answer.
Earlier, it was said that the
mole fraction of the solute
plus the mole fraction of
the solvent should equal 1.
If we did our calculations
correctly then 0.0263 + 0.9737
should be equal to or
close to 1; and it is.
The parts sum to the whole.
There is another way
to could come up with
the mole fraction of water
after we calculated the mole
fraction of potassium chloride.
It starts with the same
summation equation and then
algebraically solves for the
mole fraction of the solvent.
The resultant equation
has the calculated value
on the right hand
side of the equation.
Plugging those values in and
solving the math gives the
same value as was calculated
using the 3 step process.
Both ways work if the
math is done right.
Circumstance will
show which is easier.
Our final example
says that a solution contains
10.0 g pentane (C5H12),
10.0 g hexane (C6H14) and
10.0 g benzene (C6H6).
We are asked to find the
mole fraction of pentane?
The first thing to note is
that there are 3 substances,
all liquids, in this equation.
It is still solved the same
way as the previous
two substance example.
Our first task, and the one
that involves the most work is
to get mole values for all
components or substances.
Beginning with the
10.0 g of pentane.
To speed matters along
the molecular weights of
the organic liquids
will be provided.
The molecular weight of
pentane is 72.15 g per mole.
The orientation of
the molecular weight
has the current units of
grams in the denominator
so that units cancel out.
The division
of the numbers produces
0.139 moles of pentane.
The 10.0 g of the hexane
are converted the same way.
The molecular weight is
86.18 g per mole of hexane
it's inserted such that
grams cancel out and a value
of 0.116 moles of
hexane is produced.
The 3rd substance, benzene,
is converted into mole with
the molecular weight
of 78.11 grams per mole.
10 grams of benzene is
0.128 moles of benzene.
The hard work of the problem
is done and nothing but
arithmetic awaits us.
Totaling the moles is addition.
The values have been lined
up on their decimal points
and each has the same
number significant figures.
The sum or total number
of moles is 0.383.
Pentane is our substance of
interest and its mole fraction
is calculated by getting
its moles from step 1
and dividing that by the
total number of moles 0.383.
The mole fraction of pentane,
in decimal form, is 0.363.
Is it correct?
Well, the best way to verify
the accuracy of the
calculation is with the sum of
mole fractions
equals 1 equation.
That would require us to
determine the mole fractions
of the other two organic liquids.
Since it is relatively
straightforward to calculate
those mole fractions once we
had determine one of the mole
fractions we should do that.
Let's keep track of this value
and find the mole fraction
of hexane.
Steps 1 and 2 are
identical and getting the
mole fraction of interest
has us take the number of
moles hexane from
step 1 and divide that
by the total number
of moles 0.383.
Hexane's mole
fraction is 0.303.
We'll save this value and
move on to the calculation
of the mole fraction of
benzene. Again, there is
no need to repeat
steps 1 and steps 2.
The mole fraction of benzene
has the moles of
benzene divided by the
total moles for a
decimal value of 0.334.
Adding that value to the
other mole fractions puts us in
the position to verify whether
or not we have correctly
calculated the mole fractions
for these 3 organic liquids.
Returning to the sum of the
parts equation we see that
adding the 3 mole fraction
together should get a value of
1, or very close to that.
This result lends great
confidence that the
mole fraction of pentane
was correctly determined.
That ends our coverage
of concentrations
containing mole values.
Recapping the lecture.
Some compounds dissolve in
liquids and form solutions.
The dissolved compound
is called the solute
The dissolving liquid
is called the solvent.
More detail is provided
on solutes and solvents in
the 'Solution and
Solubility' lecture.
The first of the 3 concentration
scales involving moles
is molarity, capital M. It is
the number of moles of solute
dissolved in 1 liter
of solution (mole/L).
Molarity is also called
molar and is a very important
scale in chemistry.
As a conversion factor,
molarity can be used to
move between moles of solute
and volume of the solution.
One of the limitations of
molarity is that volume can
be temperature sensitive.
That is the molarity can
change with temperature.
That is not the case for the
second concentration scale
molality , lower case m. It is
the number of moles of solute
per kg of solvent (mole/kg).
Both the numerator and
denominator are based
on weight which is not
temperature sensitive.
Molality is also called
molal concentration.
Molality is used when dealing
with colligative property
The third concentration scale
is called the mole fraction.
It is the moles of solute over
the total moles of solution
or mixture.
It is given the lower-case
Greek letter chi as a symbol.
Since it is a fraction it follows
that the sum of ALL of
the parts equals the whole.
That means adding all the
mole fractions together has
to equal 1. And that
concludes our lecture.
While we covered 3
concentration scales molarity
is the most common scale
used in chemistry and
the one that should
get your focus.
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