 So this experiment right down is used to calculate the size of a molecule. So I am going to introduce a method in which you can use to find out the size of oilic acid molecule. What is oilic acid? You know that? Some organic acid? Yeah. Is this the one that you found in olives? No. Could be actually. I am not sure about it. So this is one of the organic acid. It has two COOH groups. So COOH is right. Oilic acid is C18H34 right now. This is the oilic acid. So why do you have taken such big molecule? Because it will be bigger. It will be slightly bigger to deal with. If you are taking a very small molecule, then it becomes impractical for us to use this method to find the size of it. So we have used the bigger molecule in this experiment. So what you do is that you dissolve 1 centimeter of oilic acid into 20 meter cube of water. Why we have used not 20 meter cube of water in fact 19 so that the total solution is 20. Or you can say you are dissolving 1 centimeter cube of oilic acid into water so that it makes a total of 20 meter cube of solution. Now tell me, can you answer like why we are using oilic acid molecule? Why not some other organic molecule which is big? It is soluble in water. It is polar. It is a polar. So you can get soluble, it is soluble in the water. Sir, it is 1 centimeter. 1 centimeter. 1 centimeter. So you need a cube of water. Yes. Then you have a solution like this. Suppose this is 20 mm of this solution. After that you have seen the dropper. Yes. Have you ever used dropper to fill the ink pen? Yes. So you have this dropper. You press the dropper, put it inside. You take some solution inside the dropper. Then you have an empty test tube or whatever and you drop n drops of this. You are dropping n drops of this solution over here. So there are n drops of the solution. Getting it? Yes. Understood. These are n drops of this solution. Now what you are doing is, you are again diluting it up to 20 ml of mark. You are putting water so that it becomes 20 ml of solution. Understood? Yes. Okay. It becomes 20 ml of solution. So what is the total dilution over here? Suppose 1 drop. 1 by 20. 1 by 4. 1 drop has a volume of V. So how much of total volume of this solution inside this? n into V. Okay. Which was anyway diluted up to 20. This divided by 20. So this is the, let us say, meter cube of the Olig assay divided by the total volume of the solution. This is how much it has diluted. And V divided by 20 by 20. 1. So you have diluted it by 20 times initially. Yes. Then again you diluted 20 times. So total dilution is 1 by 400. And into n into V. Yes. The n into V drops. Okay. This experiment is slightly weird but then this is how it is conducted. It is a need to do like this. We will first conduct the experiment. Then we will come back and understand why we are doing it like this. Okay. So this is the procedure. First you diluted to the 20 meter cube. Then you take n drops in empty beaker and again diluted to 20 meter cube. Okay. Alright. Now, now what you are doing is the volume of n drops of the solution is n into V cube. And amount of oleic acid is n V divided by 20 into 20. Yes. Okay. Now this volume n V divided by 20 into 20. You are taking an vessel with larger cross section area which has water in it. Okay. And then you are spreading some powder. That powder is called. Why? Because it makes it so complicated. Is that powder? Lycopodium powder. This is the only chapter in physics. You have to remember a lot of things. But that is very easy. Is that an indicator? Is that a basic powder or indicator? First write down. We will come back to that. Lycopodium powder you are spreading on the surface of the water. Lycopodium powder is an organic powder. It will not get, let's say, it will not get soluble in the water. It will be on the surface. Okay. But there is a porous boundary between the powder and the water. Are you getting it? We are not using a membrane sort of thing that if I create a light, if there is a lycopodium powder over here, okay. If I put anything on the lycopodium powder, it can touch the water also. It is a porous boundary. Okay. Now on that lycopodium powder you are pouring this. Okay. So what will happen? It will dissolve in the organic solvent. So lycopodium powder. The lycopodium powder will hold the organic part of the oleic acid. Okay. And water will try to hold the polar part of the oleic acid. Okay. So lycopodium powder, you can say it is used as inhibitor for oleic acid to get dissolved in the water itself. Okay. Understood? So we are spreading this much volume onto the surface of lycopodium powder. What will happen that there is a water also? Lycopodium, sorry, the oleic acid is only nV divided by 400. Rest all is water. That will go inside the water itself. Lycopodium powder will not hold the water. Are you getting it? Only this much portion. Suppose you take n into V drops. And you take this much, which is n into V volume. Only this much will be held by the lycopodium powder. Out of n into V, only this much is the oleic acid. Only remaining will get dissolved in the water. But shouldn't the oleic acid hold the water which was previously dissolved in it? No, it's not like that. Oleic acid is dissolved in water, correct. But then water gets dissolved with water, better than water getting dissolved with oleic acid. Okay. So this water will try to form a hydrogen bond with water in this volume of the solution and pull it down. And lycopodium powder will hold this much oleic acid on the surface. Getting it? And it will spread across, let's say, uniformly. And it's a bigger area. Bigger area is there. So it will spread as much as possible. Okay? The assumption here is, the greatest assumption is that the surface is so large that it will spread as unimolecular layer. Are you getting it? You will spread it, the lycopodium powder, or you will spread it till it will stop spreading. Getting it? And when it will stop spreading, then there is a unimolecular layer. One molecule here, one molecule there, like this. Suppose there is one more molecule at the top. It can further spread or not. If this layer can come and sit adjacent to it, it can further spread. But after this spreading is not possible. Okay? So this is the biggest assumption. And that is why it is called just estimation. It is not called calculation. It's just a rough estimation. Okay? So using a big beaker, you have spread the oleic acid in such a manner that on the surface, unimolecular layer is there. Okay? Now, how to calculate the diameter of the molecule? Can you tell me that? Suppose the area of the beaker is known as A. Can you do it? Get the expression. Don't talk in air. Tell me what is the diameter of the molecule. Understood? How to go about this? No idea? What is the total volume of oleic acid? How much it is? This much is the volume. Okay? So write down Nv divided by 20 to 20. This is the volume of the oleic acid which is spread on the surface. Okay? We know the area of cross section of the beaker. Okay? This should be equal to area of cross section of the beaker into thickness of the layer. Can I say this thickness is the diameter? That is the answer. This, if you divide this by the area of the beaker, you get the diameter of the molecule. Thickness of the oleic acid. Thickness of the oleic acid on the surface. If you assume it to be monomolecular layer, is the diameter of it? See, this is, all of you understood, this is the volume. This is the volume of oleic acid which is spread. Any doubt here? This is the volume of oleic acid which is spread. And in how much area it is spread? Area of A. So this should be equal to area A into thickness of the layer. Right? This is a monomolecular layer. So thickness should be the diameter of the molecule itself. So area into thickness should be equal to this. Because it is oleic acid that is spread across. Getting it? So that is what we have done. We have equated area into thickness with this tree to get the thickness. And that thickness is your size of the molecule itself. I can understand your pain. It is not very intuitive. But then this is I think a very nice method to come up with roughly the size of the molecule. It is a rough estimation but still you don't get any simpler method than this without using any instrument. So is it practically possible to do it? Yeah. So how do you reach the monomolecular layer? What? How do you know when you reach the monomolecular layer? If it stops spreading. Can you see it spreading? There can be... You can see the layer breaking. It is like an oil. It is an organic thing. Won't you be able to see the oil on the surface of the water? Or you can just take a large group and put one drop of oleic acid until it stops spreading. Yeah, I didn't understand. Why do you have to dilute the oleic acid? No, why? Guys, stop talking. Why do we have to dilute it so much? To reduce the amount of oleic acid is one. But what you are saying is why not just take one drop of the oleic acid and put it over there? Yeah. It is much harder to concentrate. First is that it will be concentrated only in the one drop, all of it. So it becomes difficult to spread it. So if you have so many drops, you can put the drops all around and then it can spread easily. It is nothing to do with, let us say, science of it. Why can't we do it? Definitely you can do it. You can do it on the paper. But when it comes to practically, there are some practical issues. Are you getting it? You don't need to even dilute. You can take one molecule, put it at one place, take another molecule, some hand, put it at another place. But practically not possible. Fine? So all these practicals, they are not talking about idealistic scenario. Okay? Even equations like force to mass emigration is not very accurate. Okay? When you are trying to apply in practical world, things changes. And this is how it is, practically. In fact, whatever science you are learning, you are done with the entire science by the time you complete the class 12. In the engineering, you are learning the practical aspects of it. Okay? There comes a lot of difficulty. Okay? I hope you understand this thing. Okay? This is a favorite question in school exams, to devise a method to estimate the small distance. And this is what it is all about.