Colligative Properties - Boiling Point Elevation, Freezing Point Depression & Osmotic Pressure - Free Educational videos for Students in K-12

Colligative Properties - Boiling Point Elevation, Freezing Point Depression & Osmotic Pressure - By The Organic Chemistry Tutor

Transcript


00:0-1	in this video , we're going to talk about collective
00:02	properties . Now , you might be wondering what is
00:05	a collaborative property ? A collective property is a property
00:09	that is dependent on the concentration of solute particles and
00:15	not the identity of those particles . Now there's four
00:19	collective properties . We're going to talk about boiling point
00:21	elevation , freezing point , depression , osmotic pressure and
00:26	vapor pressure . So let's start with the first one
00:32	boiling point elevation . So what does that phrase tell
00:36	you ? Well , when you add solid particles to
00:41	a solution , the boiling point will elevate or it
00:45	will go up . Let's use water . As an
00:48	example here , pure water has a boiling point of
00:55	100 degrees Celsius at sea level . That is at
00:58	an atmosphere pressure of 1 80 M . The bullet
01:01	point does depend on pressure as you go up a
01:05	mountain , the boiling point of water decreases , but
01:08	at one ATM Let me do that again . The
01:13	boiling point is 100°C. . Now , what happens if
01:19	we add salt to water ? as you add salt
01:23	to water ? The boiling point of the solution goes
01:26	up . So is the solu water is the solvent
01:32	. But combined they make up the solution now ,
01:35	just to give you some numbers here . This is
01:39	going to be the morality of the solution and this
01:44	is going to be the boiling point . If you
01:47	use a one more , loan any cl Solution ,
01:51	if you have that , the boiling point is going
01:53	to be approximately one of 1°C. . If you have
01:58	a five M N hcl solution , the boiling point
02:01	will be approximately one of 5°C. . Technically it's like
02:06	one of 5.5 and if you use a 10 M
02:10	, any cl solution , the boiling point is going
02:13	to be 110 degrees Celsius . To calculate the boiling
02:20	point , you could use this formula . The change
02:22	in temperature is equal to K . B . Times
02:25	M . Times I K B . Is the boiling
02:30	point elevation constant . M . Is the morality i
02:33	is the event ha factor By the way . I
02:38	think I said one of 5.545M solution . It should
02:41	be like one of 5.1 . The KB for water
02:44	is .51 . And if the morality is five and
02:50	the vet ha factors to the reason why the benefactors
02:53	to is because when sodium chloride dissolves in water it
02:57	creates two solid particles and a plus and cl minus
03:01	. So when you multiply these three numbers out You're
03:04	gonna get 5.1 and that's the change in temperature .
03:07	So the original boiling point was 100 . That's the
03:10	bullet point is going to increase by 5.1 So it's
03:13	going to be one of 5.1 so that you can
03:17	see why boiling point elevation is a collaborative property .
03:21	It depends on the concentration of the Saudi particles .
03:26	As the concentration of the Saudi particles go up ,
03:30	the boiling point of the solution goes up . And
03:32	also it depends on the van Hoff factor as well
03:35	. The more ions that are dissolved in a solution
03:38	, the boiling point will go up as well .
03:40	Now the next type of collaborative property that you need
03:43	to be familiar with is something known as freezing point
03:47	depression . So think about what this expression tells you
03:54	. The word depression means something low , something that's
03:59	down . So as you add salt to water ,
04:02	the freezing point it goes down , it decreases the
04:06	formula for the freezing point depression , it's equal to
04:09	negative KF . Times the morality times the Van Hoff
04:15	factor . The negative sign is to help you to
04:18	see that the freezing point goes down . If you
04:21	don't use the negative sign , then KF is going
04:23	to have a negative value . So if you have
04:28	a positive sign , KF will be negative . In
04:31	the case of water , it's going to be negative
04:33	1.86 degrees Celsius per morality . By the way ,
04:40	morality is equal to the moles of the solute particles
04:47	divided by the kilograms of solvent . So that's how
04:52	you calculate the morality of the solution . For those
04:54	of you who might be wondering ? So as more
05:01	so is added to water , the freezing point ,
05:05	let's put FP for freezing point , it goes down
05:09	and this property is very useful , particularly in the
05:11	winter time . Perhaps , let's say for those of
05:13	you who live up north , you probably noticed that
05:16	during the wintertime so is added to the roads in
05:20	order to prevent freezing . And when you add salt
05:24	to water , it makes it difficult for the liquid
05:27	water molecules to freeze . And so that's why adding
05:31	salt to ice . It helps ice to melt because
05:33	the freezing point goes down and that's why . So
05:37	it's typically added to the roads in the winter time
05:40	, it makes it difficult for ice to form but
05:42	easy for ice to melt . Now let's add some
05:45	numbers to this situation . So let's see the relationship
05:50	between morality and freezing point . If the concentration is
05:56	zero , that is the concentration of the solvent ,
05:59	We have pure water water freezes at 0°C. . Now
06:04	if we have a one m sodium chloride solution ,
06:08	If you multiply negative 1.86 times one times the two
06:13	ions into sodium chloride solution , you're going to get
06:17	3.72 . But if you apply , the negative sign
06:20	will be negative 3.72 . So that means the freezing
06:24	point will decrease by 3.72 units from zero . So
06:28	the new freezing point , the solution will be negative
06:31	3.72°C. . If we were to increase the solute concentration
06:37	by a factor of five , The freezing point will
06:40	increase by a factor of five . That is the
06:42	change in the freezing point . So it's going to
06:44	be negative 18.6 degrees Celsius . Now let's say if
06:51	we were to use A 5M solution but of a
06:54	different substance A L . C . L . Three
06:57	. The identity of the substance doesn't matter . But
07:00	what matters is the quantity of the salute in the
07:03	substance , aluminium chloride Can break up into four ions
07:08	. We can get the aluminum three plus cat eye
07:12	. So that's one eye on and we can get
07:14	three chloride ions . So one formally unit generates four
07:19	ions , which means our I . Value is for
07:25	, So if you multiply negative 1.86 times five times
07:29	4 that will give you at freezing point . Or
07:34	the change in freezing point which will be negative 37.2°C.
07:39	. But starting from zero , the new freezing point
07:41	will be that value . So now you can see
07:45	why freezing point depression is a collective property because this
07:51	property depends on the concentration of solid particles . As
07:55	the concentration goes up , the freezing point depresses it
07:58	goes down . It doesn't depend on the identity of
08:02	the Saudi particles but it depends on the concentration .
08:05	For instance if I were to use a one m
08:08	solution of K . I . Both of these have
08:12	two ions per formal unit . The identity is different
08:16	but the concentration is the same . Thus the freezing
08:19	point will be the same and that's what makes it
08:23	a collective property . It depends on the concentration of
08:27	the solid particles and not the identity of the solid
08:30	particles . Now let's move on to the next collective
08:34	property . The next one is vapor pressure . This
08:39	property also depends on the concentration of the solid particles
08:46	. Now let's briefly talk about why the vapor pressure
08:49	of the solution is equal to the mole fraction of
08:53	the solvent , multiplied Bye , The vapor pressure of
08:59	the solving . Now , the more fraction of the
09:03	solvent , Let's say if we're using a saltwater solution
09:07	, water is gonna be the solvent . So this
09:09	is going to be the moles of H20 . Divided
09:12	by the total moles of the solution , which will
09:15	be the moles of the salt that we have .
09:18	Let's use sodium chloride , plus the moles of water
09:25	. So looking at this equation , we could see
09:27	that the vapor pressure of the solution depends on the
09:32	moles of solute particles . So if we were to
09:36	increase the moles of the solute particles , what's going
09:39	to happen to the vapor pressure will go up ?
09:42	Or will it go down ? Well , As we
09:45	increase the moles of solute particles , we are increasing
09:49	the value of the denominator of the fraction . A
09:55	fraction has a numerator and it has its nominator .
09:57	Whenever you increase the value of the denominator of a
10:00	fraction , the value of the whole fraction goes down
10:04	. There's an inverse relationship between the two . So
10:07	as we increase the moles of solute particles in the
10:11	solution , the mole fraction of the solvent , that
10:16	is this entire fraction goes down in value . As
10:21	the mole fraction of the solving goes down in value
10:25	, the vapor pressure of the solution goes down in
10:28	value . And so we see an inverse relationship between
10:31	solute concentration and the vapor pressure of the solution .
10:36	As the moles of the salary goes up , the
10:39	vapor pressure of the solution goes down , making it
10:41	a collective property . Now let's consider the 4th 1
10:46	. The 4th collective property we're going to talk about
10:48	today is known as osmotic pressure . Here's the formula
10:58	for osmotic pressure . It's pi is equal to M
11:02	. R . T . Times the benefactor I capital
11:06	M is the polarity are Is the gas constant ,
11:10	.08206 and its leaders times ATM divided by most times
11:17	Calvin . So because our has the units Calvin .
11:21	The temperature has to be in kelvin . Keep in
11:25	mind the Kelvin temperature is the C temperature plus 2
11:29	73.15 . The polarity is different than morality . Keep
11:36	my my clarity is the moles of solute divided by
11:40	the leaders of solution . And let's contrast that with
11:44	morality , morality is equal to the moles of solute
11:48	divided by the kilograms of solving . So both of
11:53	these quantities are different forms of concentration , but the
11:59	result is the same as we increase the concentration of
12:04	the solute particles . The osmotic pressure is affected .
12:08	In this case there is a direct relationship between the
12:10	two . The osmotic pressure goes up . So let's
12:17	review the four collaborative properties . Number one boiling point
12:20	elevation . As you increase the concentration of the solid
12:23	particles boiling point goes up . Number four Osmotic pressure
12:28	. As you increase similarity osmotic pressure goes up .
12:31	So for those two there's like a direct relationship Now
12:35	for number two and 3 freezing point depression and vapor
12:39	pressure . There's like an inverse relationship . If you
12:43	increase the solute concentration , the freezing point goes down
12:48	and if you increase the size of concentration the vapor
12:51	pressure goes down . So let's just write that so
12:54	you can see a summary . So as we increase
13:00	the concentration of the salyut or the salt , the
13:03	boiling point goes up and the osmotic pressure goes up
13:11	. The vapor pressure goes down . And What was
13:18	the other one freezing point ? The freezing point ,
13:23	He goes down as well . So that's the relationship
13:30	between the Saudi concentration and the four collective properties .
13:35	Now let's work on a few problems associated with boiling
13:37	point elevation and freezing point depression . So in this
13:42	example problem we have 20 g of sodium hydroxide dissolved
13:47	in 200 g of water . Our goal is to
13:50	calculate the boiling point of the solution and were given
13:55	the boiling point elevation constant K . B . It's
13:58	.51 . So what can we do here ? This
14:04	is the formula that we need the change in the
14:07	boiling point . It's going to be equal to K
14:10	. B . Time . Is the morality of the
14:14	solution times the Van Hoff factor , the delta T
14:19	. Is the difference between the boiling point of the
14:22	solution and the boiling point of pure water . So
14:33	I'm going to move this to the other side of
14:35	the equation . So the boiling point of the solution
14:41	is going to be the boiling point of pure water
14:46	plus KB times them times I . So we know
14:52	the value of K . B . And for the
14:56	van Hoff factor , sodium hydroxide consists of two ions
15:01	, the N . A . Plus ion and the
15:03	hydroxide ion . So when sodium hydroxide is dissolved ,
15:06	that one formal unit becomes two ions . Thus the
15:10	van Hoff factor is to for this problem . Now
15:14	, the only thing that we're missing is the morality
15:18	of the solution . Morality is moles of solute divided
15:23	by kilograms of solvent . So we're given two g
15:27	of the saloon 20 g of sodium hydroxide . What
15:30	we need to do is convert it to moles to
15:33	do that . We need the molar mass of any
15:35	ohh , so we need the periodic table . The
15:41	atomic weight of sodium Is 22.99 . Let me get
15:46	my periodic table just to make sure I have it
15:49	craft . And yes , that's it . 22.99 for
15:53	oxygen is 16 and for hydrogen it's one point 008
16:08	So this is 39 998 , which I'm going around
16:12	the 40 because 20 and 40 it's like just half
16:16	of each other . So one more of sodium hydroxide
16:23	has a mass of approximately 40 g of sodium hydroxide
16:30	. So the unit grams cancel and now we have
16:33	moles of solute . We need to divide this by
16:36	the kilograms of the solvent . The value is any
16:40	ohh , the solvent is water . So we need
16:42	to convert 200 g to kg . one kg Is
16:48	1000 g . So any time you need to convert
16:50	from grams to kilograms , simply divide by 1000 200
16:55	divided by 1000 . It's .2 . So we're going
16:58	to divide the moles of solute By .2 kg of
17:03	solving . So this here is going to give us
17:07	the morality of the solution . Now let's get rid
17:14	of that and let's Plug in the numbers . So
17:18	20 divided by 40 . That's 0.5 .5 , divided
17:22	by .2 Is 2.5 . So that is the morality
17:28	of the solution . Now let's fucking everything into this
17:31	formula . So we have the boiling point of water
17:37	or at least we know what it is . The
17:38	boiling point of pure water at atmospheric pressure of Let's
17:43	say , at sea level , at an atmosphere crush
17:45	of 1 80 M . That is 100°C. . Now
17:50	, KB is .51 with the units C per morality
17:57	Times a 2.5 molo solution Times of a factor of
18:03	two . So .51 Times 2.5 times two . That's
18:10	2.55 . And then once we add that to 100
18:14	, we're going to get the boiling point of the
18:16	solution , Which is 102 .55°C. . So that's how
18:23	you can calculate the boiling point of a solution when
18:27	you dissolve a soluble in water . Now let's move
18:31	on to the next problem number two determine the freezing
18:34	point of a solution . If 400 g of aluminum
18:38	chloride was dissolved in 1600 g of water And were
18:42	given the KF four . Water is 1.86 . So
18:50	the boiling point of the solution is going to be
18:53	the boiling point of the pure solvent , which is
18:56	water again . But this time is gonna be plus
19:00	the K . F . Value times the morality times
19:04	the benefactor . Now we're dealing with aluminum chloride when
19:09	he dissolved aluminum chloride in water , it's going to
19:13	break up into The aluminum three plus catomine . And
19:19	we're also going to get three chloride ions . So
19:23	notice that we get a total of four ions performing
19:28	the unit . Thus the vital factor is going to
19:31	be four in this example . Now let's calculate the
19:39	morality . So let's start with the grams of saw
19:45	you just like we did in the last problem .
19:49	So we need to convert grams of aluminum chloride into
19:52	marks which means we need the molar mass of A
19:56	. L . C . 03 So we have one
19:59	aluminum atom And three chlorine atoms . Using the periodic
20:03	table . The atomic weight of aluminium is 26.98 and
20:13	the atomic weight of chlorine Is 35.45 Times Street .
20:21	So let's perform the computation . So this gives us
20:30	a molar mass of 1:33 .33g promote . So this
20:36	tells us that one more of a L c L
20:39	three Equates to a mass of 133.33 gramps . Now
20:50	, our next step is to divide the moles of
20:53	sally by the kilograms of solvent . So we have
20:56	1600 g of water . If we divide 1600 5000
21:02	, It's going to give us 1.6 kg of solvent
21:08	. So remember the morality is equal to the moles
21:12	of solute divided by the kilograms of the solver .
21:19	So this is going to be 400 divided by 133.33
21:25	, Which is essentially three and three divided by 1.6
21:29	Gives us this answer . 1.875 . So that's the
21:34	morality of the solution . Now let's fucking everything to
21:40	get the final answer . So the boiling point ,
21:45	I mean rather the freezing point Of the solution is
21:49	going to be the freezing point of water which is
21:52	0°C plus chaos . Now , whenever you add a
21:58	salute to water , the freezing point goes down ,
22:01	the topic is freezing point depression . When you add
22:04	a solid to water , the border point goes up
22:07	. Does the expression born point elevation . So in
22:11	this problem the KF for water is going to be
22:13	negative because it depresses or decreases the freezing point .
22:18	So this is gonna be negative 1.86 times a morality
22:23	of 1.875 Times of a factor of four . Yeah
22:35	. And so we're gonna get negative 13.95 degrees Celsius
22:43	. So that is the freezing point of the solution
22:49	. Now let's move on to the last problem in
22:51	this video which are the following solutions has the highest
22:56	boiling point . Is it A B , C or
23:00	D ? Well , let's fight the equation . We
23:06	know that the change in temperature is going to be
23:08	equal to K . B times M times I now
23:12	K B is going to be constant for all of
23:16	the solutions . We're assuming that the solvent is the
23:18	same , most likely water . So the only part
23:21	that's different is the M . E . Product .
23:25	Because for each of the four solutions , I mean
23:29	the four answers , the morality is different and eventful
23:33	factor is different . So we got to find out
23:35	which of these has the highest semi product value .
23:40	So let's consider answer choice . A . The morality
23:45	is .35 and aluminium bromide has four ions , one
23:51	aluminum ion and three bromide ions . So if we
23:54	multiply .35 x four , That's gonna be 1.4 .
24:01	Now let's move on to answer choice B . The
24:04	morality is .75 but glucose doesn't ionized into ions .
24:11	Glucose is it's just a solid particle . That isn't
24:17	a nice . So it hasn't been a factor of
24:19	one . Now moving on part C . We have
24:27	them allow lee of 0.50 and calcium chloride history ions
24:32	CIA and the to see oh minus ions . So
24:36	.5 times three is 1.5 . Finally , as a
24:42	choice D . The Mulally is .8 , sodium chloride
24:46	breaks up into two ions , so 20.8 times two
24:49	is 1.6 . So therefore answer choice D . Is
24:54	the answer . Because we have the highest and my
24:57	value for that solution .

Summarizer

DESCRIPTION:

This chemistry video tutorial provides a basic introduction into colligative properties such as boiling point elevation, freezing point depression, osmotic pressure, and vapor pressure. It explains how to calculate the boiling point and freezing point of a solution as well as how to calculate the molar mass of a solute using osmotic pressure and freezing point depression.

OVERVIEW:

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