 Hello everyone, please type in your name. Please type in your name, all are there, then we'll start the class. Only me and you. What have we done last class? We were doing the concepts of equivalent, equivalent weight and all. Last class, we were doing concepts of equivalent. Have we discussed n factor? I think few questions, also we have discussed in this right, that copper oxides and all, few questions we have discussed already. So we have discussed about number of equivalence, equivalent weight and all, in various conditions for salt, acid and base. How do we calculate? Okay. Few conditions we'll see or few reactions we'll see in which how do we find out the number of the number of equivalence or the equivalent weight in reacting condition. Right. Number of equivalent weight in reacting condition. Stereo isomers, is it finished in the school? Areyman, have they finished stereo isomers in the school in Kormangala or in SSR? Let me know if they have done in Kormangala and SSR both. They are finished there? No, they haven't done that. That's why we are not starting stereo isomers, because we have to follow the curriculum of school. So I'm going with that topics only. Okay. Anytime we can do, we'll take two class for that, we can finish that. Anyway, so we'll continue with this few. Okay. One more thing, let me know. They have started Ionic equilibrium or what? They have started or not? Have they started Ionic equilibrium? Okay. They haven't started Ionic. Anyways, fine. So you see the first like I have already told you that there are various concentration term and a few of them we have discussed in the whole concept chapter, that is molality and molality. Okay. Now here that another concentration term we are going to discuss and that is normality. Okay. Fine. Fine just now. Okay. So now another concentration term we are going to see that is normality. Okay. Why it is like we have, why we are discussing this concentration term here because this involves the number of equivalents, right? To solve or to calculate normality of any solution, we should know the number of equivalents. So now we have discussed only, we have discussed number of equivalents already. Now we are considering this concentration term. Okay. So first of all, the definition of normality you write down, it is the number of equivalent normality is, it is the number of equivalent present in, present in one liter of solution, one liter of solution. Mathematical expression if you write normality, usually we express by capital N, right? So normality is equals to the number of equivalents, equivalents divided by volume of solution in liter, volume of solution in liter. Another way if I write, number of equivalents, volume of solution in milliliter and then we have to write down here thousand. Now, this number of equivalents into thousand is nothing but the number of milli equivalents. This divided by volume of solution in ml, right? So if you compare this normality and molarity term, the difference is what the difference we have here. In molarity, we take number of moles here, but in normality we are taking number of equivalents. That's a difference we have. Okay. So this is the expression, three different expression of, it's not different, same only, just we are changing the unit, right? Now, from this example also you see for any substance or solution, if you have to find out the number of equivalents, what you will write, you write normality into volume, okay, that is N into V. And this volume is nothing but this volume should be in liter, okay? But if this volume is in milliliter, then normality into volume gives you number of milli equivalents. So the number of milli equivalent is equals to normality into volume, and this volume must be in milliliter, okay? So this is the two formula we have that we use further. In ionic equilibrium also we use and we use this, right? So number of equivalents whenever you have to find out, because we know equal equivalents react, last class we have discussed few numericals on to that, we are equating the number of equivalents. So when you know the number or normality of one substance that is reacting with another substance, so we can equate the number of equivalents of A is equals to number of equivalents of B, and that we can write for A normality suppose NA volume is VA, so NA into VA, we can write NB into VB, which is nothing but N1 V1 is equals to N2 V2. When we write N1 V1 is equals to N2 V2, then we are equating the number of equivalents of the two substance, that's a simple meaning, correct? Now, here you see number of equivalents and milli equivalents we have. There are some other formula also, this is the one formula, all these three expression are related to each other, okay, they are not different, okay? Now, the another one you see, now the thing is what, this is also important expression we have. Suppose you have a sample, you have a sample of mass M, mass of the sample is suppose we have M, right? And this sample contains, and this sample contains X percentage of solute by mass, so basically it is mass percentage I am taking X percent, right? This means what, if you take 100 gram of the sample, whatever sample we are taking here, if you take 100 gram of the sample, the 100 gram of sample contains X gram of solute, that's a meaning there, X gram of solute, okay? X gram of solute, now in this you see, if we have to find out the mass of the solute present here, right? So we can say what, in 100 gram of sample, 100 gram of sample, X gram of solute is there, X gram of solute is present, right? So in M gram of sample, because we are taking the mass of the sample is M gram, this value, M gram. In M gram of sample, the mass of the solute will be X into M divided by 100, right? So if I write down the mass of solute present in this M gram of sample, that is equals to X into M divided by 100, right? One more thing we have to assume here, since we are going to find out the formula of normality, okay? So what we are assuming here also, the density of this sample is also D, is D, right? So if I write down the expression for density, density is equals to what we can write, that will be equals to mass of the sample divided by volume, mass is this M, capital M, mass of the sample divided by volume, okay? Now here one more thing you see, the volume we are taking here in ML, okay? So it means basically VML of the sample we have with a mass M density D, okay? Now here if I write down the formula of normality, the formula is what? It is the number of equivalent by volume of the solution or sample whatever it is. So volume is given in ML. So we'll write down here 1000. The number of equivalents we can write, it is mass divided by equivalent mass. Last class we have discussed, right? This into 1000 divided by volume in ML, this normality becomes mass of the sample we have already XM by 100. So we'll write X into M divided by 100 into 1000 will be as it is. And equivalent mass I'm taking here in the denominator that is E, equivalent mass, volume from this relation we can write down M by density D, right? Now when you solve this expression, what you'll get to see? This 100 gets canceled, right? Mass gets canceled, right? And what we can write here, the expression becomes X into D into 10 divided by E. So the formula of normality, the other formula we have that is X into D into 10 divided by equivalent mass of the substance. This is another formula of normality. Sometimes this is also useful. This formula is also useful, okay? Where X is what? X is the percentage mass of the solute, okay? Percentage mass of the solute. D is the density of the sample. E is the equivalent mass, okay? So this is the another form we have done. So how do we get this formula? The derivation is not important. This formula you just try to keep in mind and what are terms are there in this formula that also you must keep in mind, okay? Next one you see. Now the very important relation we have here, the relation between normality and molarity. Relation between normality and molarity. To establish the relation between normality and molarity, the formula of normality is number of equivalent volume in liter, volume of solution in liter. And the formula for molarity is number of moles volume of solution in liter. Now here you see what we can write here. Number of equivalents is what? Mass divided by equivalent mass. This into volume in liter. And equivalent mass what we can write? The mass will be as it is. Equivalent mass will be molecular mass divided by the n factor or valency factor into the, this is the valency factor, okay? So this mass and molar mass becomes what? Number of moles. So the expression we get here, n factor will go up, okay? So we'll get mass into n factor divided by volume. And here we have molar mass. Mass by molar mass, we know it is number of moles. So we can write moles divided by volume into n factor. And moles by volume is nothing but molarity. So this becomes what? Molarity into n factor. So relation of normality and molarity, we can write normality is equals to molarity into n factor of that substance, right? This relation is very important. Sometimes when you have to equate the number of equivalents for that you require, okay n factor we'll discuss. It is just the same thing that we have done, acidity and acidity, we'll discuss n factor after this. Okay, just give me a second. Okay, so normality into molarity into n factor. Why this is important? You see n factor is nothing but the valency factor. Suppose if you write S2, S4, right? So for this n factor is the number of replaceable H plus ion, n factor will be two, right? HCl if you write, n factor will be one, like this, NaOH if you write, n factor will be one in non-reacting condition, okay? I'll just give you a few data here. First of all, this relation you should know, why it is important because sometimes you have to equate number of equivalents. So for that normality is required, but normality will not be given in the question. They'll give you molarity. So with molarity and you know the n factor value, if you multiply these two you'll get normality and then normality into volume gives you number of equivalents, okay? So that is how we'll find out the relation of normality and molarity. Now a few examples I'll give you how this n factor you will calculate in reacting condition, okay? This kind of, you know, examples that we have already done in last class, okay? In reacting condition, how do we calculate n factor? You see right on the heading, equivalent weight, equivalent weight in reacting condition. Why I'm writing down equivalent weight here because we know equivalent weight E is equals to, it is a molecular weight divided by the n factor. So if you know the n factor, you can find out equivalent weight in a reacting condition, right? So n factor is this, we call it as n factor or we also call it as valency factor or valency factor. Now you see I write down a few reactions here. You see this reaction, H2SO4 plus NaOH gives Na2SO4 plus 2H2O. This is the reaction we have. Now the n factor of H2SO4 here, you see in this reaction, since H2SO4, see this kind, equivalent weight, what happens in reacting condition we are talking about. So equivalent weight of one substance may vary according to the reaction that is given, okay? See, I can also write down one more reaction here. This one you see, H2SO4 plus NaOH. But in this, the product is NaHSO4 plus H2O. The balance reaction will be 2 NaOH here. What is the balance reaction here? It is balanced already, I think. This one is balanced and this is a balanced reaction. Now in this reaction you see from H2SO4, H2SO4 has ability to donate 2H plus iron, right? But in these two reactions you see, here it is donating 2H plus iron and forms Na2SO4, all H plus is getting replaced. Here only one H plus is getting replaced, right? So in this case, what happens? The N factor of this is one, since only one H plus is getting replaced. But here, the N factor here it is what? It is two, okay? N factor here it is two NaOH, right? It is one because only one OH we have here. Here also N factor is one of NaOH, right? So if we have to find out the equivalent weight of H2SO4 in this reaction, the equivalent weight of H2SO4 in the first reaction, it will be its molecular weight divided by N factor. So that will be 98 is the molecular weight of H2SO4, N factor is two, so 49, equivalent mass. Here if you have to find out the equivalent mass of H2SO4, that will be its molecular mass divided by its N factor. Molecular mass is 98, N factor is one. So it is 98. Is it clear? So you see the equivalent weight of any substance depends upon the reaction condition. Okay, what is the reaction we have and what product we are getting according to that, the N factor will change and hence the equivalent mass will change. Tell me if you understand this. Okay, one more thing you see, just now we have stabilized the relation of normality and molarity. So normality is equals to molarity into N factor. This normality is number of equivalents divided by volume in liter. That will be equals to number of moles by volume in liter into N factor. So when you see this, you see this volume and volume terms gets canceled and the expression here we get is the number of equivalents is equals to the number of moles into N factor. So there are two different formula we have now to calculate the number of equivalents. One is what? One is number of equivalents is equals to normality into volume N into V. If normality and volume is not given and moles is given, so with moles also we can find out the number of equivalents that is number of moles into its N factor. Number of moles into its N factor. Understood? This is one formula, like I'm continuously repeating this thing that whenever two substance react, equal equivalents will react. So here with the help of this formula, what is the number of equivalent of S2 SO4 we have here if I ask you? Tell me, the number of equivalents of S2 SO4, the reaction is balanced, we know the respective moles of S2 SO4 and NaOH that is one and two. N factor you also know, what is the number of equivalent of S2 SO4? Tell me, what is the number of equivalents of S2 SO4? Yes, so number of equivalents of S2 SO4 will be what? Its N factor is two. Its number of moles is one here. So two into one is equals to two. Number of equivalents of NaOH if you see, N factor is one and its number of moles is two. So again you see equal equivalents are reacting. Similarly, you can find out the number of equivalents here also N factor is one and its number of moles is also one. So one here again N factor is one, number of moles is one, so one. So equal equivalents are reacting, correct? So with this formula, you see in this reaction, there is nothing called normality or molarity as given. Even volume is also not mentioned. Without all these term, normality, molarity or volume, we can also find out the number of equivalents with this formula which is N factor into number of moles, understood? Now another example you see, suppose a reaction I write Na2 CO3 plus HCl. HCl, this gives two NaCl plus H2O plus CO2. It will be two HCl here also, okay? So we know the N factor of HCl is one. The N factor of HCl is one. This N factor, we can find out, right? It is a salt. So what is the number of equivalents of HCl we have here? Number of equivalents of HCl will be equals to N factor into number of moles two. So here also, the number of moles of Na2 CO3 is one, right? So one into N factor of this should be equals to two, right? The number of equivalents here it is two which must be equal to this also. So number of equivalents is two. So N factor for Na2 CO3 is two again here. Anyways, it is a salt, right? So it dissociates like two Na plus plus CO3 two minus, right? So the two positive charge we have or two negative charge we have, so N factor is two. But like this we only do in non-reacting condition, right? In reacting condition, we have to consider the number of equivalents. Now like this also, one more reaction I can write and the reaction is this, Na2 CO3 plus HCl it gives NaHCO3 plus NaCl, plus NaCl. Now here it's N factor is one, number of equivalents is one. So for this also N factor is one, right? Number of equivalents also one. So if you calculate here the equivalent mass of Na2 CO3 that will be M by one which is the N factor and here it is M by two which is the N factor of Na2 CO3. Okay, calculation of N factor we have. So basically if you know N factor you can find out equivalent weight, okay? That is molecular mass or atomic mass by that. Now the second thing here and that is equivalent weight of oxidizing and reducing agent, reducing agent in a redox reaction. So in a redox reaction, how do we calculate the equivalent weight which is nothing but the N factor? So in this case, the N factor will be change in, change in oxidation number of oxidizing agent or reducing agent per molecule. We can also write in the another way that number of electrons oxidizing agent or reducing agent per molecule. This is what the formula we have, okay? Now, suppose if I write down few examples here the first one, and this is important, this you must remember, MnO4 minus converts into Mn plus two in acidic medium, Mn plus two in acidic medium. This you should know the conversion of the change in oxidation number of magnies in acidic medium or basic medium or alkaline or any neutral medium, okay? So in acidic medium it converts into Mn plus two, right? And here if I ask you, what is the oxidation number of magnies here? It is also plus seven, right? Plus seven, so N factor if you calculate, so that will be change in oxidation number that is seven minus two per molecule which is one only, so seven minus two is equals to five. So five is the oxidation number in acidic medium, right? So this is by the first definition I have written here, change in oxidation number of this, okay? We can also find out the change in the number of electron like you see, this is the half reaction. The reaction I have written here, I'm just giving you this one to understand, okay? MnO4 minus converts into Mn plus two. So this you can do for all other reactions. So in acidic medium when we balance this, what we do for oxygen we have here, so we'll add four H2O this side to balance hydrogen we'll add eight H plus this side, right? So here the change in oxidation number is seven and here the change is two, right? Seven to two we have to go, so five electron we have to add for this purpose. Now you see the number of electron exchange is five only per molecule because they have only one MnO4 minus. So n factor is equals to what? Five, which is the same thing that we are getting here. Both method you can apply and you can find out the number of the n factor of the molecule, correct? So this conversion of MnO4 minus into Mn plus two when acidic medium takes place in basic medium it converts into Mn plus six and in neutral medium it goes into Mn plus four. So this you must remember this is for neutral medium and this is for basic medium, right? This conversion plus seven to plus six plus seven to plus four plus seven to plus two according to the medium this you must remember. Next example, second one. Suppose I am taking Cr2O7 two minus. Cr2O7 two minus converts into Cr3 plus in acidic medium, in acidic medium. So this you must memorize again, okay? So in this since we have two chromium here so we'll balance this two chromium here. Again, we can find out the number of electron exchange by adding H2O and H plus and we can check, okay? And what another thing we can do? Here the oxidation state of chromium is what? Plus six to seven into two plus six here. So for this if you have to find out the n factor of this that will be total oxidation number here which is two and two plus six, right? Minus here, two and two plus three. So that will be and divided by one because we have one molecule. So that will be plus six or six simply. So n factor will be six here, okay? If Cr3 plus we have to find out, Cr3 plus. The change in electron is nothing but six only. So n factor for Cr3 plus is six divided by the number of molecules two, which is nothing but three. So we always find out n factor is a change in oxidation number per molecule. Is it clear? Is it clear? Tell me. Okay. So in acidic medium conversion is this. In basic medium, Cr2o7 two minus converts into Cr2o4 two minus. This is in basic medium, right? Again, you can find out change in oxidation number or you just balance the reaction according to the medium and then you find out what number of electron has been exchanged per molecule. That will be the answer, okay? Third one, oxalate ion. If I write down this reaction, C2o4 two minus, oxalate ion converts into C2o2, right? So here the oxidation state of carbon is plus four and here it is plus three for carbon, right? So change in oxidation number is what? Change in oxidation number is, here you see, two into three, double charges, two into three plus six, two into four plus eight. So n factor for this this value will always take positive, okay? Don't take negative value. So two into plus four, minus two into plus three. That will be eight minus six, two, n factor is two here. Now here you see n factor for CO2 will be what? That is two divided by two, one. You can also relate here that the number of equivalents of this should be equals to the number of equivalents of CO2, correct? So number of equivalents of this will be what? It will be n factor that is two into number of moles one that is two here. The number of equivalents will be what? Number of moles is two, n factor we have to find out and that is equals to two from this relation. So n factor is what? Which is also we are getting here. So anyway, you can do this, okay? So all these things you have to memorize like you write down n factor for H2O2. This is important, is always two. n factor for H2O2 is always two, okay? If the reaction is this, suppose I am written down two mnO4 minus and this converts into mn plus two and mn plus six. So in this case, you know, this condition is what? The same element, Magni is here, it is in plus seven oxidation number. The same element is going in change in oxidation number in two different states, plus two and plus six. So in this, we'll just find out the difference in oxidation number here. n factor for this will be, n factor for this will be, it is seven into two minus plus two plus six. 14 minus eight, it is six, right? And this six divided by what? This six we divide by two also because we have two molecules here. So we should write down here only divided by two. So six by two is three. n factor for this is three. Similarly, one important n factor calculation we have for FeC204, ferrous oxalate. FeC204 converts into Fe plus three plus two CO2, right? This is the reaction. So Fe here we have in plus two oxidation state and this is in plus six plus three, right? This is plus three and this is plus four, okay? How this reaction goes here, see? Total change in oxidation exchange in electrons we have to find out. Fe three plus plus one electron and then C204 two minus converts into two CO2, right? And this gives two electron because here we have two negative charge. This is neutral. We must have to electron this side, okay? So total electron exchange is what? When you add these two, the total electron exchange is three electron. So n factor for this will be three. This also you can memorize directly. n factor of FeC204 is three. Total exchange in electrons. Is it clear? Okay, now the last thing we'll see here, calculation of n factor in the case of disproportionation reaction. So in disproportionation reaction, how do we find out n factor? Suppose the reaction is this, Br2 plus OH minus and it gives Br minus plus BrO3 minus. This is the reaction we have. Can you balance this reaction? Tell me, what is the balanced reaction? I'll give you two, three minutes for this. Tell me these values because this is important. Like how to balance this. I have done this already in the class. Now tell me the value of A, B, C and D. Who are not there, Akash, Ahriman? And then tell me what is the answer? Akash, are you there? What about Ahriman? Okay, see to find out this A, B, C, D, we have to balance this equation, right? Disproportionation reaction. And you see here, first of all, this bromine here is in zero oxidation number. Here it is minus one and here it is plus five, right? From this bromine to here, this is oxidation, right? And this one is reduction, right? So the two half reaction is what? The first reaction is Br2 goes into Br minus. And the second one is Br2 will go into BrO3 minus, okay? So we have to balance this. So what we'll do? First of all, we'll balance all atoms other than oxygen and hydrogen. So this should be two here. This we should have two, right? Here now everything is balanced except charge that we'll do in the last, okay? And in the second one, you see, we have six oxygen here. To balance the six oxygen, we should add what? We should add six H2O this side. And when we add six H2O to balance hydrogen, we should add 12 H plus this side, right? Now we balance charge. So here we have total, zero charge. And here we have two negative and 12 positive. So 10 positive charge, right? Okay, so to make this 10 positive to zero, we have to add 10 electrons this side. Here if you see zero to two negative, so two electron this side, okay? Now when we add these two, the electrons must get canceled, right? We have to cancel out the electron when we add these two half reactions. To cancel out the electrons, we have to multiply this reaction by five and then we have to add these two. So these electrons and electrons gets canceled. Five BR2 plus one BR2. Here we have six BR2 plus six H2O. And this gives 10 BR minus plus two BRO three minus plus 12 H plus. So you see all atoms charge our balance. We have total, here we have total 12 negative 12 was zero discharge here, zero charge here and all atoms are balanced. So this is the balanced chemical reaction we have, okay? Now, the number of electrons that has been canceled out, the answer is what? 10 electrons is getting canceled, right? Two into five is 10, 10 electrons is getting canceled. It means that exchange, the number of electron exchange here, what we can write. Electron exchanged is the nothing but the number of electrons that cancel out, which is nothing but 10. So n factor for BR2 will be what? The number of electron exchanged per molecule, that will be five by three. Understood? Balance reaction is this and the electron exchanged divided by the number of molecules gives you n factor for this molecule, okay? Now there are another way also, like this you can balance the reaction and you can find out. One more possibility is what? Whatever the electron that has been exchanged in first reaction, this is first reaction and this is second reaction. Whatever the electron exchanged, suppose it is N1 and here the electron exchange is N2. So what we can find out, the N factor for this reaction will be N1 into N2 divided by N1 plus N2. N1 into N2 gives you 20 divided by 12. That will be five by three. Same value we are getting, okay? So this N1 and N2 are what? These are the number of electrons that is exchanged in the two half reaction, right? The total number of electron exchange is 10 because 10 electron is getting canceled but the number of electrons present in first half is two, second half is 10. So N1 is two, N2 is 10, N1 into N2 by N1 plus N2. This is the formula also we can apply to find out N factor. Is it clear? Okay. Now, you write down the question. Your question will solve. The question is, calculate the amount of H2O2 required to react with 316 gram of KMNO4 in acidic medium. Solve this. Today, Araman will finish this chapter. A few questions we'll discuss just now. This chapter is almost done and two more concepts we'll discuss. That is concept of oleum and volume strength of H2O2. And after that, probably we'll start this inic. Inic will not do today because I like, most of you are not there. Okay, then again, I have to do it in the class. So inic will not do, we'll see some questions then. Or if you have any doubt, you can ask me. We'll discuss that only. How was the exam? Akarsh, Araman, what about you? Akarsh, what about you? You tell me. Okay, then fine, Araman and Akarsh will finish this class by 6.30, even early also. We'll do some questions, we'll not start then any. Okay, then we'll not start any new chapter today. We'll just finish a few questions on this and then we'll leave the class, okay? You can go to study for UT tomorrow, for tomorrow's UT, okay? Tell me the answer of this question. 34 gram, is it 34? Araman, you check. Okay, you see. See, actually, the reaction is of KMNO4 and H2O2. You don't need to write down the complete reaction. Only these two are reacting. So what we can write, the concept we have, what? The number of equivalents of KMNO4 is because number of equivalent of H2O2, right? And we know in acidic medium, this KMNO4, which is nothing but MNO4 minus. KMNO4 is nothing but K plus and MNO4 minus. So MNO4 minus converts into what? MN plus two in acidic medium, this information we have already. So for this N factor we have calculated, which is five. And N factor for H2O2 is always two, I told you already. And if we, since both are reacting, so what is the concept we have? Equivalents or number of equivalents of KMNO4 should be equals to the number of equivalents of H2O2. This is what the concept we have studied. Equal equivalents react, correct? So number of equivalents of KMNO4 will be what? It is the number of moles, moles into N factor, because there's not normality or molarity given. Here also it is moles into N factor of H2O2. This is for H2O2, this is for KMNO4. So number of moles of KMNO4 will be what? The mass is given, 316 divided by molecular mass of KMNO4 is 158. When you count the molecular mass of KMNO4 it is 158 into N factor of KMNO4 we have calculated is five, is equals to moles we have to find out into N factor is two. So when you solve this, this will get canceled, these two will get canceled. Number of moles of H2O2 is equals to five, which is mass divided by molecular mass of H2O2 is 34. So mass is equals to five into 34, which is nothing but 170, that is the answer. Understood, another question you see, calculate the, calculate the number of moles of K2CR2O7, O7 required to oxidize, oxidize oxidize six mole of FeC2O4 ferrous oxalate, FeC2O4 in atheric medium. What happened? Three moles, yeah it's correct. Okay, just again you see we have to equate the number of equivalents. Okay, the two substants which are reacting here is K2CR2O7 so we can write the number of equivalent of K2CR2O7 is equals to the number of equivalent of FeC2O4, N factor we know number of equivalents is number of moles of K2CR2O7 we have to find out into N factor of K2CR2O7 is six, we have already calculated it in atheric medium. For N factor of FeC2O4 is three and number of moles of FeC2O4 is given six, so when you solve this number of moles of K2CR2O7 is three we'll get. Just we have to equate the number of equivalents that is it. Next we'll write down, calculate the normality of, calculate the normality of H2SO4 if three condition I have here, 49 gram very basic question formula based, 49 gram of it is present in 500 ml solution, 10 equivalents in 10 liter of solution. Tell me quickly, 98 gram in 100 ml of solution, one, one and 10, so 49 gram means of one equivalent, 10 equivalents in 10 liter it's 198 gram in 100 ml, so 100 into 10 will be there and 98 is 49, so it will be two, one and 20. The first one, 30 is the number of equivalent by volume of solution in ML if it is given then 1000 will be there in the numerator. The number of equivalents is what, mass divided by equivalent mass for H2SO4 it is 49 divided by 500 into 1000, so that will be two normal. Number of equivalents is 10, volume is 10 liter, so one normal, again 98 by 49 is the number of equivalents divided by 100 into 1000, so it is 20 normal. Yeah, what happened Akash? Akash will finish early, okay, only two concepts we'll discuss more, okay, then you can go. Is it fine? Only two concepts, 10, 15, 20 minutes maximum more. Akash is like what exam you have Akash tomorrow and Adema and you also. What exam you have, would you both have tomorrow? Okay, maths, English, English you know already. Akash, 10, 15 more minutes, we'll finish this. Write on concept of Oliam, concept of Oliam. See the formula of Oliam is H2S2O7, H2S2O7, which in another way we can also write H2SO4 plus SO3. This SO3, we call it as free sulfur trioxide, sulfur trioxide, free sulfur trioxide, okay. Now what happens when we dissolve this Oliam into water, right, see this Oliam contains few amount of H2SO4, but when we dissolve this water, so this free sulfur trioxide reacts with water, H2SO4 increases, amount of H2SO4 increases. So now you see how this reaction takes place, when you dissolve this Oliam into water, SO3 plus H2O, it converts what, it forms H2SO4, right. We actually do not know that what is the mass of this SO3 or free sulfur trioxide, right. But whatever the mass we have here, this mass will give some amount of H2SO4, some amounts we already have. So this plus this will be the total amount of H2SO4 in the aqua sample, aqua solution, right. So what we are assuming here you see, the concept of Oliam here, what we are assuming, suppose the Oliam sample we have here, suppose the Oliam sample we have here, H2SO7, we have 100 gram of this Oliam sample and out of 100 gram, we have SO3 and H2SO4. So if this SO3 is X gram, so H2SO4 will be 100 minus X gram. Now this X gram of SO3 is reacting with H2O. Since one mole of SO3 gives one mole of H2SO4, so we can write what number of moles of SO3 consumed is equals to the number of moles of H2SO4 forms. So number of moles of SO3 is what X divided by its molecular mass, right. That will be what, 80 for SO3 and that will be equals to the mass of H2SO4 that forms here and we do not know the mass of H2SO4 here, divided by the molecular mass of H2SO4, 98, right. So the mass of H2SO4 we get by this reaction is M of H2SO4 is equals to 98 by 88, X gram. This is a mass of H2SO4 we get from free sulfur trioxide when we dissolve into water, right. Now, since we have to find out the total mass of H2SO4, so the total mass H2SO4 is equals to, initially we have 100 minus X present and we also get 98 by 80 X mass of H2SO4 from this reaction when we dissolve this into water. So when you solve this, you'll get 100 plus 18X by 80 is the mass of H2SO4 you get, right. This is the total mass of H2SO4 when we dissolve this olium into water, formula, right. Must remember this formula. Now, the question that they ask here is this. When you see any olium bottle, it is there in the bottle. When you go to the lab, you can see this. You have a bottle there in which the olium is present and there is some labeling is there for that olium. Like they must have written like, suppose we have H2SO7, 100 gram of it and the bottle of this H2SO7, they label the olium like that, 109% suppose like this. Means the bottle, if you see in which the olium is kept on the bottle itself, on the bottle itself, they write this percent is 109%, 110%, 104% like that. It is, the bottle is labeled like this. Now, what is the meaning of this percent is that is labeled onto this bottle, olium bottle. The meaning of this is what, if 109% is written, I am explaining you with the help of this example. The meaning of this is what, that if you take, suppose this is 109% we have here in this bottle, if you take 100 gram sample of this olium, right. Means 100 gram of sample, sample what sample, this sample on the bottle on which 109% is written. 100 gram of this sample requires nine gram of H2O to react completely with, completely with free SO3, right. To react completely with free SO3. So when they write this and they ask you the question here they ask that what is the mass of, mass of free sulfur trioxide present in this sample, which is 109% labeled. The question is this. So what is the mass of free sulfur trioxide present the question is this, okay. Now this question they will ask, but the meaning of this you should know and what is the meaning that if you take 100 gram of this sample, this 100 gram requires nine gram of water to completely react with the free SO3 present. Now the reaction will be same, whatever the mass of free SO3 we have that reacts with H2O and forms H2SO4, right. We know the mass of H2O here, according to this data, this data, this gives the mass of H2O is nine gram. Mass of this we have to find out, so it is X gram. So we can equate the number of moles of these two. X by 80 is equals to nine by 18 and when you solve this you'll get X equals to 40 gram. This means in 109% of oleum sample, 40 gram of SO3 free sulfur trioxide is present. Free sulfur trioxide is present. So this is like I said, this one is the first definition or first kind of understanding of this labeling we have. The second thing what you can understand from this you can solve the question with this definition also. And the second thing is what, that this 109% also means that 100 gram of this sample when dissolved in water, when dissolved in water gives 109 gram of H2SO4. H2SO4 means if 109% is labeled, right? That means what you take the 100 grams of that sample, put into water, then the mass of H2SO4, the total mass that we have calculated in the previous slide, okay? The total mass of H2O we'll get is 109 gram. So if the bottle is labeled 105%, it means we'll get 105 gram of H2SO4 total. 110% means we'll get 110% of H2SO4. Then again, if you have to find out the mass of freeze SO3, so we know the total mass of H2SO4 is what, 100 plus 18x by 18. We have calculated this just now in the previous slide. This should be equals to 109 because 109% is labeled. So when you solve this again, you'll get x equals to 40 gram. So both way you can solve these kind of questions. Is it clear? Tell me yes or no. Now one more, this is one type of question they ask, where we have to find out the mass of SO3 present. Okay, similarly number of moles also they can ask. Okay, one more type of question you see on the disk as we are almost done. Find the percentage of strength, 20 gram of free SO3 is present. 20 gram of free SO3 present. See in this question, what we do, we try to compare with the previous example. Like when we write 109%, what is the meaning of this 100 gram of sample and nine gram of water is required to react completely with free salver trioxide. This gives you 109, okay? So here we have and this nine gram of water when you write down this reaction, SO3 and H2O gives H2O SO4. So this thing is always true that 18 gram of H2O reacts with 80 gram of SO3. Both are the molar mass of this. So for 18 gram, for 80 gram of SO3, we require 18 gram of water. So simply if you have to find out this percentage strength, percentage of this olium, we have to find out what mass of water is required to completely react with free salver trioxide, right? That is what we have to find out and that will add in 100, right? So 18 gram reacts with 80. So what we can write, 80 gram of sulfur trioxide reacts with 18 gram. So what is the question they have given? 20 gram. So one reacts with 18 by 80. So 20 reacts with 18 by 80 into 20. That gives you approximately 4.5 gram of water we require. Means this gram of water requires to react 20 gram with, to react 20 gram of SO3. So the percentage of this will reward 100 plus mass of water will add 4.5. So 104.5% of labelling will be there. Understood? Last thing we'll discuss for today, five more minutes, that is volume strength of H2O2, hydrogen peroxide. Right, done. Volume strength of H2O2 is what? It is the volume of oxygen. It is the volume of O2 gas evolved, evolved at STP, one liter sample of H2O2 being heated. If you heat one liter of sample of H2O2, H2O2, when you heat this, it forms H2O and half of O2, right? So what is the volume of O2 evolve at STP? That is the volume strength of H2O2, right? So from this equation you see, one mole gives half mole, right? So M mole, if you have, if molarity is there, right? If you're taking, okay, like you see, one mole of this gives half mole of O2 and we know molarity is what? It is the moles by volume. So when you take M moles, where M is the molarity, because you see the volume is one, when one liter of sample is heated, right? So this volume is one, so molarity is nothing but moles, right? So when M is the molarity, so when you take M moles of this, where M is the molarity we have, then this gives M by two mole of O2, where M is what? M is the molarity, you must write on this. M is the molarity. Now one mole we know, one mole at STP occupy 22.4 liter volume. So M by two mole occupies M by two into 22.4. So what we can write, 11.2 molar. So the formula, or we can write, only this you have to memorize, the volume is strength of H2O2. Definition is what? It is the volume of O2 evolve at STP, right? And in terms of molarity, the formula we can write, the volume is strength of H2O is 11.2 molar, M is the molarity, and molarity we know, normality into, normality is equals to molarity into N factor. So 11.2 into normality divided by N factor of H2O2 is two we already know, that will be equals to 5.6 normal. Any of these two formula you can memorize, 11.2 molar or 5.6 normal is the volume strength of H2O2. Is it clear? Yes or no, tell me, okay? Now, just one small thing. In the question you will see, they will write like this, 20 V of H2O2, this kind of question they'll write. Now, when you see like this 20 V, it means the volume is strength is 20. That is what the meaning we have. Understood? 10 V if they have written. It means the volume strength is 10, that is it. Okay, so this is it for this chapter. Okay, we'll not start any new chapter today. Okay, you can go and prepare for UT tomorrow, tomorrow's UT. Any doubt you have? If you have any doubt you can ask me, otherwise we'll wind up the class here only. Thank you.