Colligative Properties Video Lecture | Physical Chemistry

hey you guys this is mr. Millington
today we're going to learn about
colligative properties so what are
colligative properties well it says
right here that colligative properties
are properties of a solution example
being boiling point freezing point vapor
pressure and osmotic pressure that
depend on the amount of dissolved solute
particles in the solution so what does
that mean let's suppose we have two
beakers here and each one of these
beakers has some sort of dissolved
solute in it this one has some salt and
this one has some sugar and we know from
prior experience that when you put salt
and water and sugar and water they
dissolve you stir them up and they
dissolve right well what happens to the
boiling point or what happens to the
freezing point of aqueous solutions when
you place them in water well what we're
going to learn is that when you place a
dissolved solute in water it's going to
increase it's going to increase the
boiling point of that solution and it's
going to decrease
it's going to decrease so the yeah the
freezing point of this solution here is
going to decrease so normally we know
that water has a freezing point of zero
degrees Celsius right the freezing point
of h2o is normally zero degrees Celsius
assuming we are at sea level the boiling
point of water at sea level or at one
atmosphere pressure is 100 degrees
Celsius so if we take a look here when
we dissolve something in water either
salt or sugar or anything for that
matter it's going to increase the
boiling point so this solution is no
longer going to boil at a hundred
degrees Celsius it's going to boil a
little bit higher and how much higher
depends on how much of this stuff we put
in it and same thing with the freezing
point the freezing point of water is
normally zero degrees Celsius however
once we start dissolving stuff in it
whether
be salt or sugar ionic compounds or
polar covalent compounds the freezing
point is going to decrease so this
solution is no longer going to freeze at
zero degrees Celsius it might freeze at
negative five degrees Celsius or
negative six degrees Celsius depending
on how much dissolve solute is in the
solution so colligative properties are
properties of a solution that depend on
the amount of dissolved solute particles
in the solution okay so the boiling
point and freezing points of these guys
really depends not on the chemical
makeup of these substances but how many
dissolved particles are in the solution
the more dissolved particles you have
the higher the boiling point and the
lower the freezing point okay so it's
important to understand that colligative
properties do not depend on the type of
chemical species present rather they
depend on how many dissolved particles
are in that solution and in this little
video we're going to focus our attention
on boiling point and freezing point but
understand that colligative properties
are properties of a solution such as
vapor pressure and osmotic pressure that
depend on the amount of dissolved solute
particles in the solution so let's take
a look at a couple of examples here all
right so we have three beakers of water
here and in this beaker here we have
some sugar that we're going to dissolve
in here we're going to put some sugar in
here and we're going to stir it up and
we're going to dissolve dissolve it and
we're going to throw some sodium
chloride or table salt and here we're
going to stir it up and it's going to
dissolve it last but not least we're
going to put some potassium sulphide in
here we're going to stir it up and it's
going to dissolve okay and the question
that I want to ask is which one of these
will have the greatest difference in
boiling point and which one of these
guys will have the greatest difference
in freezing point well to answer that
question what we first need to look at
is whether or not the substance is polar
or ionic that plays a big role in
determining this question or answering
this question if we take a look at sugar
we talked about sugar earlier this unit
as being polar
it is a polar molecule so when I put
sugar in water it does not dissociate
it's not going to break apart so when
you put this substance in water it stays
intact
it still dissolves because of the
hydrogen bonds between the oxygens and
hydrogen's here however it is not going
to dissociate okay so that is what polar
compounds or polar molecules will look
like when you put them in water
these two guys this salt here in this
salt here we learned from a prior or
earlier video that when you place these
in water they dissociate they break
apart into the ions that make them up so
when I put this salt in water it's going
to break apart into a bunch of sodium
ions and a bunch of chloride ions and
when I put this salt in water because
it's ionic it's also going to break
apart or dissociate into the ions that
make it up it's going to dissociate into
a bunch of potassium ions and a bunch of
sulfide ions okay but let's take a look
at what happens let's suppose I have 1
mole of NaCl and I put this in this
water right here it's going to dissolve
and dissociate and when it dissociates
what ends up happening is it produces
two moles of dissolved particles whereas
over here if I have one mole of sugar
and I dissolve this in water I still end
up with 1 mole of dissolved particles ok
so understand that concept when
something dissociates it breaks apart so
one mole of salt is going to end up
producing two moles of dissolved
particles and if we take a look at this
right here at potassium sulfide if I put
this in water it's going to dissociate
and if I have 1 mole of this it's going
to dissociate and produce 3 moles of
dissolved particles the two right here
the two moles of potassium ions and the
1 mole of sulfide ions okay so this is
an important concept to understand so if
I were to ask you which one of these
guys or which one of these solutions
here is going to have the greatest
effect on boiling point and the freezing
point of the solution it's going to be
this because there are three moles of
dissolved particles which one will have
the least effect on its boiling point or
freezing point it's going to be the
sugar because there's only one mole of
dissolved particles so it's an important
concept to understand that if something
dissociates if an ionic compound
dissolves it's going to dissociate it's
going to break apart it's going to
produce more moles and therefore it's
going to have more of an effect on the
boiling point and freezing point of that
solution compared to a polar substance
so let's take a look at how we can
calculate changes in boiling points and
changes in freezing points depending on
the amount of dissolved substances in
that solution all right if we take a
look at this example here or if we take
a look at this formula here what we're
looking at is we are looking at a
formula that allows us to determine
changes in a solutions boiling point all
right so to calculate the change in the
boiling point of a solution it's simple
we take the boiling point constant of
water which is 0.5 1/2 degrees Celsius
per this lowercase M here which means
molality of the solution which we
learned in an earlier video we then
multiply it by the molality of the
solution and then we multiply it by this
right here I is the vant Hoff factor you
can think of this as the number of
particles the substance will dissociate
into and we're going to use this
specifically for ionic compounds or
electrolytes so let's look at this
example and hopefully you catch on in
this example here of a 1.5 mol au
aqueous solution of sodium chloride
we're going to assume we're at sea level
so pressure doesn't come into play here
so to calculate the change we want to
calculate I'm sorry we first want to
khaki or we want to calculate the
boiling point of this solution in order
to do that we must find the change in
the boiling point temperature so to do
that here to get Delta T to get the
change in the boiling point
of this solution here we take since
we're dealing with water here it says
it's aqueous right what says it's
aqueous we take this boiling point
constant of water is 0.5 1/2 degrees
Celsius over lowercase M which stands
for molality we're then going to
multiply this by the molality of the
solution well it says right here the
molality of the solution is this right
here the molality is 1.5 m and last but
not least we need to multiply this by
the vant Hoff factor and let's think
about this if I put this in water it's
going to dissociate and if it
dissociates it's going to break apart
into two different ions right so the
vant Hoff factor here is going to be 2
so we'll multiply this by 2 and we'll
get our calculator out here and we end
up with 1.5 degrees Celsius so this
isn't our final answer this is just the
change in the solutions boiling point
all right we know that water normally
boils the boiling point of water is
normally a hundred degrees Celsius at
sea level and so we need to add we need
to add this 1 point 5 to it and we will
end up with 101 point five degrees
Celsius so what will the boiling point
be of a one point five molar
aqueous solution of NaCl assuming that
we're at sea level it looks like this
water or this solution this solution of
saltwater is going to boil at a
temperature of 101 point five right
the boiling point has been changed by a
temperature of one point five degrees
Celsius let's take a look at this
example here
what if we have the same molality here
we have a solution of salt that has the
same
ality and we're asked to calculate the
boiling point of it except this time
take a look we have a different type of
salt we have calcium chloride we're
going to assume that we're at sea level
so once again to get the change in the
solutions boiling point we take the
boiling point constant 0.85 1/2 degrees
Celsius per mole ality we're going to
multiply this by the molality which is
1.5 and last but not least we need to
think about this what does the vant Hoff
factor going to be here if we put this
in water here it's going to dissociate
all of these ions are going to
dissociate and we should end up with 3
ions right we have one calcium ion here
and we have two chloride ions here that
it's going to dissociate into so we're
going to multiply this by the vant Hoff
factor which is 3 I'll put this in our
calculator 0.5 1/2 times 1.5 times 3 and
we end up with 2.3 so the solutions
boiling point is going to be changed by
2.3 degrees Celsius we need to now add
this to the boiling point of water at
sea level which we know is a hundred
degrees and we will end up with the
boiling point of this solution equaling
102.3 degrees Celsius
okay so the final boiling point of this
solution is going to be 100 and 2.3
degrees Celsius let's take a look at a
couple more examples using freezing
point okay if we take a look here this
is the formula that we're going to use
to find the freezing point of a solution
okay so to get the freezing point or to
get the change in the freezing point of
an aqueous solution we take the freezing
point constant of water which is one
point eight six degrees Celsius per
Mille ality times the molality of the
solution times the event hot factor so
this is very similar to what we just
were working on so let's take a look at
this example here
calculate the freezing point 0.5 molar
sugar water solution
assuming you're at sea level so in this
problem here we want to know what it's
going to freeze at this solution so we
must first calculate the change in the
freezing point to do this we take the
freezing point constant which is one
point eight six degrees Celsius per
Mille allottee which it says right here
times the molality of the solution zero
point five times the vant Hoff factor if
you take a look we're dealing with sugar
sugar is polar so it's not going to
dissociate so one mole of sugar is going
to produce one mole of dissolved
particles so we'll just multiply this by
one we get our calculator out one point
eight six divided by I'm sorry one point
eight six times twenty-five and we end
up with zero point nine three or zero
point nine if we're using the correct
number of sig figs zero point nine
degrees Celsius so this is just the
change in the freezing point we know
water freezes at their freezing point of
water is equal to zero degrees Celsius
we're now going to subtract write this
is a depression that means to lower so
we're going to subtract this point nine
here
from zero and it looks like the freezing
point of this solution is going to be
0.9 degrees Celsius let's take a look at
this example we have the same mole ality
however now we're dealing with saltwater
so which is going to lower the freezing
point more sugar water solution or a
salt water solution well you can imagine
because the vant Hoff factor here is 1
and you know it's going to be 2 here
that salt will in fact decrease the
freezing point so let's take a look here
in order to figure this out we must
first calculate the change in the
freezing point so we take 1 point 8 6
times the molality of the solution which
is 0.5 times the vant Hoff factor if you
put this in water it's going to
dissociate and produce two different two
different particles right it's going to
dissociate into sodium ions and chloride
ions so two different particles and if
we put this in our calculator we should
end up with one point eight six or if we
use the correct number of sig figs Apple
just will just will ignore the sig figs
for this problem we'll just say one
point eight six so this isn't our final
answer to get the freezing point of this
solution here we will take the freezing
point of water which we know at sea
level to be zero degrees Celsius and
we'll subtract one point eight six and
we will end up with negative one point
eight six degrees Celsius whoops it
should be negative one point eight six
degrees Celsius okay so which one is
going to lower the freezing point more
sugar water or so
water it looks like salt water is going
to in fact or the salt is going to lower
the freezing point more and in fact most
of the salt in the United States is
actually used on icy roads because it
lowers the freezing point of that water
prevents that that water from freezing
and prevents many accidents so this or
these are colligative properties and I
hope you found this helpful