Free Chemistry NEET Video Lectures for Solutions

00:00:01 [Music] [Applause] [Music] Now we are going to discuss solubility. So let's see. We have a solute A which we are being dissolved in solvent B. So we add a little bit of A in solvent B and then we shake the flask vigorously. Okay. Until this A is dissolved completely. If it is going to dissolve I mean there might be a case that we keep on shaking the J and it's never going to dissolve. There are sol uh uh solute which is not going to dissolve in a given solvent B. So okay but let's assume in this case A is going to

00:>00:57 dissolve completely in B. Now we keep on adding A little by little and keep shaking the flask and keep on repeating until we reach a point if we add a little bit more of a it doesn't dissolve in B. So if we add any more of A, it's not going to dissolve in B. It's just simply going to pascate out. It's just going to settle at the bottom of the flask. So, we have reached the maximum concentration of solute A in uh solvent B. So, the maximum amount of solute A which dissolve in solvent B without

00:01:45 precipitating out in the is the solubility of solute A in solvent B. So basically this is the maximum concentration attainable of solute A in solvent B that is called solubility. So there are salt which is very highly soluble and there are salt which is very low has a very low solubility in a given uh solvent. For example in water we know it is highly soluble. In fact, little less than 400 g of Ncl can be dissolved in one liter of water. On the other hand, we have silver chloride. Only a few mg around 1.9 mg of AgCl can

00:02:45 be dissolved in one liter of water. If I add anything more than 1.9 mg of AgCl in water in one liter of water, it will simply precipitate out. It will simply settle at the bottom of the flask. The solubility depends upon the nature of solute and uh solvent. It depends upon the interaction between uh among the solute molecule between solvent molecules and between most importantly between solute and solvent molecules. Okay, let's try to understand it. Let's see briefly. Okay, so we have already

00:03:32 seen NaCCl. So NaCCl and sugar they easily readily dissolve in water but another example anthraine does not dissolve in water at all but it readily dissolve in benium. So if you try to understand the interaction NaCl is ionic is goes into the solution and dissociate as ion it becomes Na plus and Cl minus get solvated by water which is polar and that's how it has a very high solubility sugar it has hydrogen attached to oxygen and that leads to hydrogen bonding. Again, water has lots of hydrogen bonding

00:04:40 and so sugar becomes part of this hydrogen bonding network and that's how sugar get dissolved in water. Enthosin it is nothing but a hydrocarbon. No ionic character, no hydrogen bonding. So this there is no mechanism which will help him to dissolve in water. It's basically non-polar benzene. This is also non-polar. It's exactly like empasin. Both of them are hydrocarbons. So this empas dissolve in benzene but doesn't dissolve in water. So we can conclude this polar solute dissolve in polar solvent and

00:05:31 non-polar solute dissolve in non-polar solvent. We can say like dissolves like when solute is added to solvent to solvent the concentration of solute increases and that is called dissolution. But at the same time another process is going on the solute molecules or solutes are going to strike heat uh the other solute solid and in this process there might be some crystallization and they comes out of the uh solution and that is called precipitation. And these two process are happening all the time. So we are we are having solute

00:06:31 plus solvent and I'm we are getting solution. So in the forward direction the dissolution is happening and in the backward direction precipitation is happening. And then there's a dynamic equilibrium. So at the rate at which solute goes into the solvent or the rate the sol the sol solute comes out of the solution when the two rate becomes equal then we have equilibrium and as we keep on adding solute we reach to a uh stage when solute going to the solvent is no longer physible and we have no

00:07:20 more uh dissolution and at that time we reach is to the maximum concentration of solute possible and that is called saturated solution. Saturated solution is when the solute has reached its maximum concentration we cannot add any more solute in the solution. If we add more solute the solid solute in the uh solvent or the solution it will simply not dissolve. It will simply come out from the solution. At this stage this is we can say that solution has saturated uh with the solute cannot add we cannot

00:08:10 increase the concentration of solute anymore. Okay. So now we have to discuss the effect. So this is a reaction. We can treat it as a reaction and it should follow what is called leatalia principle. Leatalia principle. Um so we can study the effect of pressure and temperature and temperature on this reaction on this dynamical equilibrium. Okay. Let's say solute is we let's discuss first uh solid and liquid where sol solute is solid and solvent is liquid and what would be the effect of pressure

00:09:07 there is not going to be much effect of pressure because most of solid and solvent is usually very highly incompressible. So if we increase the pressure there's no any change in the volume change in of the reaction of the reaction material. So there's not going to be much change u in the direction of the reaction due to the pressure change. But temperature change would have a lot of effect if there's a enthalpy is involved delta H. If this reaction is exothermic that is delta H is uh negative

00:09:53 that means as the rea reaction goes forward the heat is liberated. We are getting energy. So increase in the temperature uh will uh take the reaction in a backward direction because it want to minimize the uh the uh the effect of temp uh we are having due the temperature increase as the if the reaction goes forward the delta h is eliminated the system has more heat. So the temperature would increase but the reset says the reaction will move in a direction where the stress can be minimized. So it

00:10:30 has to move in a direction where the heat is absorbed by if we increase the temperature. So it's clear that for delta for exothermic reaction reaction will go backward will move in the backward direction. From the same logic we can now can conclude if delta H is greater than zero that need uh energy to move the reaction uh energy is absorbed to move the reaction forward. React is endothermic then the reaction will move forward. Okay. Now that's that about a solid dissolved in liquid. Okay. Let's now

00:11:32 consider uh a gaseous solute is being dissolved in a solvent liquid solvent. Now same uh observation can be made by Lisa principle. If the reaction is delta is exothermic that is delta h is less than zero reaction goes backward. If delta h is greater than zero reaction is endothermic reaction goes forward. But in this special case let's conditioner consider a which is gaseous goes to a liquid. This is a condensation. We know to going from liquid to gas that is basically boiling we need energy. So

00:12:20 going from gaseous to liquid the gas energy will be evolved. So this reaction is exothermic and delta H is less than zero. So we can say as the temperature is increase solution the reaction will go backward that is uh the solubility of gas gas in the solution decreases as we increase the temperature. Okay. Now let's think about the pressure. So this is my vessel where I have my solution. In this case we have the sol solute particle and this is the solute particle in a gaseous phase. And this is a closed flask and I have

00:13:06 some pressure pressure applied on the system. Now if uh pressure is increased what is going to happen the solution being highly incompressible. It doesn't compress but gas will compress. And now I have more gaseous particle stored in a smaller volume. So what the listener level will say to reduce the stress we have to go in a direction uh that it can reduce the stress. So of course this solute particle which has concentration has increased in this region even though the concentration in this remain constant

00:14:03 because it's not compressible. So these will start going to the solution to reduce the effect of stress and it's very clear that as we increase the pressure solubility of a gaseous solute increases in solution and this has been formulated is given a mathematical expression for this cons the solubility of gaseous particle in uh concentration of gaseous solute in the solution by Henry. And the law says P is equal to KH X where P is the pressure of the solute over the solution. X is the concentration of solute in solution.

00:15:02 So as the pressure increase the mole fraction the concentration of solute increases and it's clear we can put it the other way around P / KH higher the Hendry constant lesser the solubility of gaseous solute in solution okay let's next discuss vapor pressure of liquid liquid solution. So let's first consider the pure liquid. Let's say I have a liquid A. Put it in a closed flask. So I have introduced some liquid over here. This is a volatile liquid. So at the this given temperature some of

00:16:07 the this uh liquid molecules is going to escape and they are going to go into this empty space. Now these gaseous molecule of this same liquid is moving around and they are going to hit back the surface and some of them can start going back into the liquid form. So there's a some kind of a uh process is going on where sol solution uh liquid a goes to gaseous and then this gasous comes back. So there's a reaction going forward and reaction coming backward and there's a dynamic equilibrium

00:16:54 and at the equilibrium the pressure this liquid uh this gaseous molecule of that same liquid A uh is going to exert some pressure on this surface and that pressure is called the vapor pressure and that we put zero due to this pure And same thing is going to if we have another uh evacuated flask and there we put this uh liquid B. Same thing is going to happen. is going to fill up this empty space and then again we are going to have a dynamical equilibrium and this gaseous molecule is going to exert

00:17:44 the pre exert pressure on the surface and this is going to call PB0 the vapor pressure of pure B pure liquid B. Okay. And of course in this I have considered this is evacuated flask. So only pressure is going to be due to this uh gaseous molecule of B. And same thing here the pressure is going to be due to only due to only the gaseous molecule of liquid A. Now let's mix them together. Now we are going to mix them together. And now we have a + b here. And same thing a + b. We have a and solution of a and b in

00:18:40 liquid form and uh gaseous a and b in gaseous form. And now there should be some relationship between this uh what is the pressure exerted by a what is the pressure exerted by b and what is the total pressure t. So this is a partial pressure due to a partial pressure to b and this is the total pressure. That's how I can classify uh the gaseous uh phase. But what about the liquid phase? liquid phase I can classify by mole fraction of a mole fraction of b. So once we have classified what is the

00:19:23 relationship between uh these quantity and this is given by R's law says that this P A is proportional to X A P B proportional to XB. We also know when x a = 1 that is when we have a pure uh liquid a pa is equal to pa = p a zero and xb = 1 then we have PB is equal to PB 0. So we get the proportionality constant immediately and I can write this equation in this form. So let's look at this in a in this diagram. What I'm plotting is the pressure partial pressure of A and B. This is a mole fraction. So over here we have a

00:20:53 mole fraction uh X A =0 and X A is equal to 1. And same way over here we have XB =0 and XB = 1. So when we have a pure B the partial uh partial pressure due to partial pressure of B is going to be PB0 and partial pressure of A is going to be zero. What about when we have a pure liquid A? Then the partial pressure of A is going to be due to the P A zero and partial pressure of B is going to be zero. And you can one can see that this is a straight line. As I change XA, PA is going to linearly change. So if I

00:21:53 am going to change X a from zero to 1 the partial pressure of A is going to increase linearly. Sorry this is try to make a straight line. Okay. And same thing when the partial pressure when we have a pure liquid B. So as the part uh concentration of B is increasing the partial pressure of B also will increase uh would have a straight line and we get this straight line. What about the total pressure? Okay. The total pressure we can calculate from this equation. So total pressure is p a 0 x a plus pb 0 xb

00:22:49 and I can uh substitute xb is equal to 1 - x a so we get p a 0 x a plus pb 0 1 - x a so I at PB 0 plus PA A0 minus PB 0 X a. So this is again a linear function with respect to concentration in A. And of course we are going to get this straight line which is really a sum of these two line. This is the P total. This is uh a uh this is concern partial pressure of A. This is a partial pressure of B and this is P total which is P A plus PB. Okay. So that's how the solution would be. Let's do a uh example given in the

00:24:06 book. Let me read it out. Let me clear the blackboard first. Okay. Let's do the example from the book. It says the example is as follows. Vapor pressure of chloroform and dchloromthane at 298 Kelvin at 200 mm Ag and 450 mm Ag respectively. Okay. plus. So we have chloroform CH CL3 let's call that A we have dchloromthane CH2 CL2 let's call that B and we are given vapor pressure of these of the pure components of P P 0. So P A0 is going to be uh 200 mm Ag and PB0 is going to be 415 mm HG.

00:25:18 Okay. Okay. Now number one. Calculate the vapor pressure of the solution prepared by mixing 25.5 g of CCL3. So we are mixing 25.5 g of this and 40 g of uh sol uh sol uh liquid B and we are asked the vapor pressure of A and B and the total vapor pressure. Okay. Okay. So what is the for now? Let's go back to the basic and the formula is P A0 P A is equal to X A PA A zero. PB is XB PB 0. Now P A Z and PB 0 is given. This is PA 0. This is PB 0. And so I need mole fraction X A and XB. How the X A

00:26:16 and XB is defined? X A is N A / N A + NB and same way I can calculate XB and also I can calculate XB simply by 1 minus X A. Now to calculate NA I need the molecular weight. Okay. So let's see what is the molecular weight of molecular weight. Molecular weight of CL3 is 119.5 and molecular weight of uh CH2 CL2 is 85 g per mole and then we can calculate NA number of moles of E. So that would be 25.5 gide by 109.5 g per mole. That will give me moles of CCL3 that is a. So let's do that 25.5 divided by

00:27:28 119.5 and answer we are going to get is 0.213 213 and same thing we need to do for B that is dividing 40 by 85 40 / 85 and answer I'm going to get is 0.470 470 and now I need to calculate simply X a mole fraction so it's all given there NA is 0 213 / NA 0.213 plus 0.470. So using calculator 213 divided by 683 I get answer is 3 0.312 and of course using this formula 1 minus 312 I at 886. 688. Now I have all the information required. The PA would be given by XA we have calculated 312 multiply by uh PA0 which

00:28:58 is given right here 200. So I get 62.4 4 and the PB okay XB is given here 688 ult* 415 answer I'm going to get is 688 * 41 5 answer is 285 5 mm mg and of course the total pressure would be simply sum of these two quantities. So the total pressure is going to be.9 4 347.9. So that is the total pressure. So we have calculated part a. Now it's asking mole fraction of each component in vapor phase. mole fraction of each component in vapor phase. Now we know this ideal gas law we are going to apply

00:30:16 that PV is equal to N R T. Okay. And if it is a partial pressure then P A N A RT. So we need to calculate mole fraction of A and B in vapor phase. So mole fraction would be let's call that capital X a that will be okay I have already used N so let's put uh vapor okay over here so N A in vapor divide by total moles N A in vapor plus NB in vapor and if I Substitute NA from here. The answer would be simply I'll get PA / PA + PB. And I have all this information. PA is given right here. Uh P and PA A plus PB

00:31:27 is nothing but P total. So I get mole fraction of A in vapor phase and that would be 62.4. 4 / by uh 347.9. So let's see 62.4 divided by 347.9. So.179 and what about XB? That would be 1 - X A. So I get 1 2 8.821. Okay. So we have calculated mole fraction of A and mole fraction of B in solution. mole fraction of A and mole fraction of B in vapor. One thing to notice is this the vapor has become richer in B. See in the liquid phase the mole fraction of B was 688. Now the mole fraction of B in

00:32:50 vapor phase has become 821. And that is has something to do with the volatility. And you can see V is has a higher vapor pressure means it has a higher tendency to go into vapor phase. It's it has higher volatility. So it is a volatile compound more volatile compound than A. So it has higher tendency to go into the vapor more than A. So it is going to richer in vapor phase than A. Okay. Okay. Let's do one more problem from the index question. Okay. Let let me clean the blackboard first.

00:33:40 The vapor pressure of pure liquid A and B are 450 and 700 mm hg respectively. So we have A and B and P 0. The P A 0 and PB0 is 450 and 700 mm AG. 4 50 and 700 mm hg respectively. So vapor pressure of pure A is 450 mm hg. Vapor pressure of pure B is 700 mm hg. Now at of course at 350 kel find out the composition of liquid mixture of if total vapor pressure is 600 mm hg. If pt is 600 mm hg find the composition of of liquid mixer also find the composition of vapor phase. Okay. Now let's first assume

00:34:52 in the liquid phase the mole fraction of A is X A. Then of course the mole fraction of B is going to be 1 minus XA. Okay. The vapor pressure once I have mole fraction assumed I can write the vapor pressure of A and B in terms of XA. So that is going to be 450X. The vapor pressure of A is 450 XA and the vapor pressure of B is going to be 700 1 - XA and the question tells me the total pressure. So that is this plus this equal to this. So I have a one equation that is 450 x a + 700 1 - x a one equation one unknown solve it and we

00:35:51 have the answer. So let's try to solve it. What we'll get 450 x a + 700 - 700 x a. So that is uh 700 minus 250 x a bringing other side dividing by 250 I'm going to get x a is equal to 100 divid by 250. So that is 04. So we have found that x a is 04. So pressure the uh partial pressure of a is going to be simply 450 into 04 that is going to be 180 and uh partial pressure of b is going to be 700 * 1 - x that is 6 or is going to be 420. So we can check the total pressure is 180 + 420 that is 600 mm energy. Now in

00:37:07 the part B the same question is also find the composition of vapor phase and in the last example we have already seen the composition of vapor phase. We can calculate the part from the partial pressure. The composition of vapor phase is X of vapor phase. Okay, let me clear the blackboard a little. So composition of in vapor phase of A is simply partial pressure of A divided by total pressure. And answer is straightforward divide by 600 and I get.3. I can calculate uh compos the mole fraction of B in vapor pressure either

00:38:04 by 1 minus.3 and I can try that one also just for checking 420 divid by total pressure that is 600.7 I get same answer by using both the method and I have completed this task starting from the from the very basic principle okay so next topic which we are going to discuss is Vapor pressure of vapor pressure of sir of solution of solution of solid and liquid. solid in liquid. Okay, let's examine the Rod's law one more time. What it says? It says P A the partial pressure of one component let's say component A is equal to

00:39:14 X A PA A zero and same thing PB is equal to XB PB 0. So if I have a solid solid what is the vapor pressure over it for most of the solid it's going to be negligible this solid is usually not volatile I'm considering that there are some solid which of course have a uh they evaporate and they would have some vapor pressure or for most of the solid there's no vapor So if I leave this solid, it's not going to evaporate. If I if I leave let's say alcohol or liquid, it's going to pretty

00:40:03 much evaporate after some time. But if I leave the solid in let's say the stable or metal object, it's not going to evaporate in my lifetime. So of course there are some solid which will evaporate but most of them not. In such cases the PB0 let's say this is uh A is the uh uh solvent B is the solute solid solute then in that case PB0 is going to be zero. So this is going to be zero. So Rod's law in general it says any solution for any solution the partial vapor pressure of each volatile

00:40:48 component in the solution is directly proportional to its mole fraction. In this case, the vapor pressure is going to be due to only the solvent A because it's going to evaporate in time. But B the solute is not going to uh uh not going to give any component. It's going to be the pressure is going to be zero. So all the pressure is going to come from the partial pressure of A. Okay. And this can be understood also by this following diagram. So we have solvent pure solvent and it's going it

00:41:34 is in a closed evacuated flask. So in this flask we have only this uh a this uh solvent a and it's going to evaporate and it's going to fill in filled up this whole. So this is a gas and this is liquid of a and the uh equilibrium which is going to happen is from this surface there are some gaseous molecule which is going to extract the surface and going into the solution and from the surface some of them is going to have enough energy to escape. Okay. Now if I add a uh solute to this what is happening? There are some

00:42:25 molecule of solution of the solvent and there are some molecule of solute. So the concentration of solvent molecule at the surface has decreased. Of course solute is not going to evaporate. That's what we have considered. These are not volatile at all. That is the assumption here which is true in most of the cases. But the solvent is going to evaporate but there's a lesser number of solvent uh molecule at the surface. So the evaporation has decreased. The moment we have added the solute

00:43:07 the concentration of solvent molecule at the surface has decreased but this is has not changed. Let's consider not the concentration of or partial pressure of A has not changed. So it's going to strike at the same rate. So evaporation rate has decreased but the condensation rate has not changed. In that case the there's going to be more condensation than the evaporation and as condensation has increased. So partial pressure of A is going to decrease and we are going to reach to a new equilibrium

00:43:45 where the partial pressure of A is going to be less than what it was. And that can be understood again from this because since in the pure solvent there's no solute XA is one. So PA is equal to PA0 and as the XA has become less than one PA has become less than PA0. And the exact relation is given over there. Okay. Next we are going to discuss ideal solution and non ideal solution. So what makes a solution ideal? So if a solution follow the ROS's law over the whole range of concentration then this

00:44:39 solution is called ideal solution. But question is what makes them to follow the rolls law and that's where comes the two very important quantity those are delta V of mixing and delta H the enthalpy of mixing and they should be zero. So if I mix, let's assume if I mix one liter of uh solvent A plus 2 L of solvent B. If the total volume after mixing is 3 liter 3 liter of solution then it follows the condition one and the second condition is if no heat is evolved during the mixing that is if the

00:45:42 temperature doesn't change uh after the mixing then it is a ideal solution. I mean during the chemistry experiment I'm sure you must have observed if I mix water with concentrated H2SO4 the mixture becomes quite warm that is delta H of mixing is non zero it's a uh exothermic reaction if the total volume is greater than or less than 3 liter or there's a uh heat is involved then this solution is non ideal. Okay, let's try to understand uh this with a simple diagram. Let's see if

00:46:28 I have a uh uh solvent a in a uh container and on the this uh vapor and liquid boundary zone. Right now I have only uh solvent uh A. Now to that I'm going to add B. But before hand let's try to understand uh what is happening. So there's interaction between A and A. If it's a pure uh uh solvent A. Now I'm going to add solvent B. So some of the A uh molecule at the boundary will be replaced by another solvent B. It could be solvent or it could be solute. If you are mixing let's say 50 50% of water and ethanol

00:47:29 then these are two solvent which we are mixing. If I'm mixing let's say sodium chloride in water then sodium chloride is going to be solute and water is a solvent. So it's a uh so now so there is a solvent A and somewhere in between is solvent B it could happen uh very often if there concent depending on the concentration now initially in the pure solvent A had an interaction between A and A. Now once the another uh solvent or solute is added B then we also have a interaction A and B and they could have an

00:48:18 interaction B and B depending upon the concentration. Now this A and A interaction could be stronger than AB interaction. What would happen? So initially right instead of this there was A and A and that in interaction is stronger stronger than AB interaction. So A and A interaction is stronger and AB interaction is weaker. So we have replaced a stronger interaction with a weaker interaction. Now this molecule A is lesser stable. So it has a higher tendency to go into vapor phase. It requires a lesser amount of energy,

00:49:25 lesser push to go into the vapor phase. So what would happen? It's become uh the vapor pressure will increase. So okay, let's let me draw uh this diagram which I have drawn earlier. So that is uh so we what we are changing is mole fraction of a over here x a is 1 over here x a is zero. So vapor pressure of A is going to be this. I say this is P A0. And same thing if XB over here is zero. XB is over here one. And we get another line. This is vapor pressure of B. This we have done earlier. And this is PB0.

00:50:15 And the total pressure is going to be this. This is ideal solution. But now AA interaction is stronger than AB interaction and the B molecule has interrupted this stronger interaction and replaced with a weaker interaction. Now this A can go into the vapor phase easily and that's by increasing the total pressure or individual vapor pressure. So now we have a what is called positive deviation. So if AA interaction is stronger than AB interaction then we have positive deviation. Okay. And other way around if

00:51:03 AA interaction is weaker than AB interaction then we have negative deviation. So in this diagram for a positive deviation the total vapor pressure has changed into the positive direction and same with the individual component also would change in the positive direction and for a negative deviation where interaction between A and B is stronger than A and A, we are going to have a division in another direction and that would lead to negative division. Okay, let's look at the example. Ethanol and acetone.

00:52:03 Ethanol C2H5 O acetone CH3 COH3. So this molecule in a uh solution has a lot of hydrogen bonding in a liquid phase. It's going to have a lot of hydrogen. There's a polar hydrogen is available there. So it's going to lead to hydrogen bonding and there's no such interaction in acetone. So now when there's only ethanol present there's a strong A and A interaction due to hydrogen bonding. And now when I add acetone to it this hydrogen bonding network is interrupted. So this molecule becomes molecule A that is

00:53:06 ethanol becomes lesser stable. and it has higher tendency to go into the vapor phase and it leads to positive deviation. Let's look at the other example that is between chloroform and acetone. So acetone uh that is CH3 COH3 and chloroform that is CL3 H. Now we can see that there is no hydrogen bonding within acetone or within chloroform. But once we put them together there's hydrogen bonding between this uh oxygen and hydrogen as there's a very powerful electron withdrawing group over

00:54:09 here. it will withdraw all the electron density making it is quite polar and now they can have a U hydrogen bonding. So that will lead to a stronger interaction. So there's a stronger interaction between A and B and as we have discussed that will lead to negative deviation. So, so we can see by looking at the interaction between between the components we can kind of guess that in which direction the uh vapor pressure would change whether it's going to be a negative deviation or positive deviation. Okay. When we have a

00:54:58 very large deviation in binary solution then they form what is called edotropes. Okay. So we have seen earlier the vapor phase is usually richer in richer in more volatile solution richer in more volatile component. And using this property, one can deise a way to separate uh these two components. Just by heating the solvent solution, collecting the vapor which is richer in volatile component uh condensing it back again, heating it and the vapor which we are going to get from this condenser is going to be even richer in volatile

00:56:06 component. And we if we keep on doing it we will get we should be able to separate out the two components. But is a special kind of solution where the liquid phase and the vapor phase has same concentration. Same concentration of liquid phase, liquid phase and vapor phase. And it's clear if there's no difference between concentration of uh liquid phase and vapor phase. If I just collect the vapor phase, it's going to give me uh the condensate is going to give me the same concentration as liquid phase and

00:57:00 it cannot be uh I cannot separate A and from B. So we have already discussed weaker interaction between two component compared to interaction between uh component A and or component B leads to positive deviation that is it has higher vapor pressure then it will be if I calculate the vapor pressure using rolls law we know if higher the vapor pressure lower the boiling point higher vapor pressure means more volatile uh component more volatile lower the boiling point so in this case weaker interaction ction positive deviation

00:58:05 that means higher uh vapor pressure than calculated by Ro's law and it will leads to minimum boiling edotrop and similarly if we have a stronger interaction between A and B compared to interaction between A and A or B and B then we have a negative deviation that is the vapor pressure is lower than what it will be if calculated using R's law then we are going to get maximum boiling adotropes and the example for such kind of a binary solutions are if I have 95% ethanol by volume in water it forms

00:59:20 minimum boiling edotropes and similarly if I have 68% HNO3 by weight in water then it forms maximum boiling edotropes. So that's where we stop. Thank you very much.