 Okay. Hello, everyone. Okay. Please let me know who are there. So, in the today's class, we are going to discuss the problems of mainly the 11th portion numerical problems we'll discuss. Okay. First of all, our main focus will be on the problem practice in between if you require some concept that we'll discuss. Okay. So, I am starting with the first chapter, which is more concept, but I'm not doing exactly first chapter. It is the mixture of more concept than the concepts of equivalence and then n factor and all everything, normality and all. So, this chapter contains all the sum of these kind of concepts, the problem that we'll discuss. It's not only about moles and all. Right. So, what we'll do today, like whatever chapters we have in 11th class, because most of you are, you know, are more familiar with 12th portion. Now, since you are doing all these things from the past few months, okay. 11th class we have done in last year only. So, what we are trying to do today, I will just try to cover as many chapters as possible of 11th class today. All those chapters will discuss like 10 to 12 problems in every chapter, important problems and good problems, so that you'll go through with all the concepts. Okay. That is our objective today. So, like, let's start with, again, like I said, we'll solve problem mainly in this session. Okay. So, we'll start with problem. So, this is basically just as again, this is just a problem solving session. Okay. So, I'm starting with mole concept, right. And like I said, here we'll discuss all kinds of problem that is there in mole concept one, mole concept two concepts of equivalence and all. So, we'll start with a very basic question here. And the first question is how many moles of electron, how many moles of electron, moles of electron weigh one kilogram, one kg? How many moles of electron weigh one kilogram or one kg? Okay. Try to do this question and tell me the answer in the form of this. Avogato's number. Okay. I'll give you the option only so that you don't need to do those calculation and all. So, the first option we have in this question is the Avogato number itself and a second option is one by 9.108 into 10 to the power 31. Third option is 6.023 divided by 9.108 into 10 to the power 54. And the last option is one by 9.108 into 6.023 into 10 to the power 8. Tell me the answer. Is it C? All of you are getting C. All of you are getting C. Please check your answer. Don't do a silly mistake in all this kind of question. See, first of all, we have to find out what the number of moles of electron, which mass is one kg, right? So, we know what mass of one electron is what? Mass of an electron is 9.108 into 10 to the power minus 31 kg. Mass of one electron is this, right? This is the kg of, this is the mass of one electron, right? So, the number of electron, if I ask you, what is the number of electron in one kg? One kg, if you take in that how many electrons are present? That will be what? One by 9.108 into 10 to the power 31. This is the number of electrons in one kg. One kg number of electron is this, correct? Now, if we convert this into number of moles, then we can write what? 1 by 9.108 into 10 to the power 31. This whole thing divided by 6.023 into 10 to the power 23, right? Now, when you solve this, you see denominator, we have this into this and numerator by 10 to the power 8. Answer will be option D, okay? So, you see again, most of you made this mistake. You are getting answers C, but that is the option there, but the answer is not that, okay? So, don't do silly mistake, okay? Read out the question properly and then try to solve the question, okay? Now, next one you see, next question. The sulphate of a metal M contains 9.87% of M and the sulphate is isomorphous with ZnSO4.7H2O. Isomorphous with this. You have to find out the atomic weight of M, atomic weight of M. Options are 40.3, 36.3, 24.3 and 11.3. Do this and don't do silly mistake into this, okay? Okay, how many of you know what is isomorphous? This is a problem solving session, okay? We are not doing one particular topic, we are solving some problems into it. Okay, you see isomorphous means what? When we say, you see, it's just a definition, okay? See, when I say there is a sulphate of metal M, right? We can write down the formula as what? Like suppose if this information is not given, isomorphous, if this is not given, there's a general formula of the sulphate what we can write? We'll write simply MSO4 or any other thing we can write. AL2SO4 whole thrice, right? So since metal is mentioned here, so we cannot take the formula as M2SO4 whole thrice because AL2SO4 whole thrice, aluminium, we don't consider that as a metal, right? So whenever they say metal, it means we are generally talking about some like transition element like this or S block elements because those are metallic, those will have more metallic properties. Anyways, so what does this isomorphous means since this information is given and two substands are said to be isomorphous with each other when their molecular formula representation is same, right? Molecular formula representation is same or they have the same kind of crystals, same representation. So you see in this one ZNSO4.7S2O, right? This is the formula it is given and it is mentioned that the metal M, the sulphate of metal M is isomorphous with this, okay? This means what? The structure or the general formula of this sulphate of this metal M will write that will be MSO4.7S2. This is the general formula of this sulphate of this metal M since it is isomorphous with this, okay? This is the meaning of isomorphous. Same kind of formula representation, same shape will be, only the metal is different there. Now what is the answer tell me? Anyone? Tell me the answer. What happened? 24.3. Yes, 24.3 is correct option C. Okay, so what we'll do into this? First of all, we'll just assume the atomic weight of M of metal is M suppose, right? Now we can find out the molecular weight of this, right? That is M plus for sulphur it is 32, for oxygen it is 4 into 16 and then it is 7 into 18, right? So this gives you, the value will be what? M plus, if you add all these, it is 64, 32 plus 126, right? 126. When you add 32, it will get 222 here, 222. Now since what is the information given? In this substance, in this molecule, the percentage of metal is 9.87. So what we can write? The mass of the metal divided by the total mass of the compound into 100 is equals to 9.87. This is what the weight percentage we have. Okay, now when you solve this, you'll get M and that will come around 24.3 when you solve this. Clear? Okay, another question? Yeah, it's given in the question, contains 9.87 percentage of M, right? So this was a simple one. Next one you see, 0.5 moles of Bacl2, barium chloride are mixed with 0.2 moles of Na3PO4. And Na3PO4, the maximum number of moles Ba3PO4 whole twice can be formed is, what is the maximum number of moles of Ba3PO4 whole twice will form? I'm not giving you auction in this. Tell me the answer. What happened? 0.1. What about others? Tell me your answer. How did you do Vaishnavi? Condonia is getting 0.16. So what is the concept you applied here? Concept of limiting reagent, right Vaishnavi? Limiting reagent. Okay, yeah. That is what you have to do into this one. Okay, that's correct. You see, first of all, the reaction for this will be Bacl2 reacts with Na3PO4. And the product here it is Ba3PO4 whole twice plus NaCl, right? Plus NaCl. Now, when you balance this reaction, what is the balance reaction we get here? Na3PO4 we have, so NaCl3, so we'll write here Ba3, so 3 will write here, right? Then chlorine becomes 6. 6 chlorine we have here. Okay, so 2 will write here. And then 6 will write here. This is the balance reaction. Now, in this you see, the moles of these two are given, the reactant. It is 0.5 and this is what? 0.2. You see, the mole ratio is not same. It's not same, it is different, right? So whenever the mole ratio is different, so we have to find out the limiting reagent into this. Okay, if mole ratio is different and if number of moles are different. In both cases, limiting reagent concept will apply. Okay. So how do we find out limiting reagent? The given number of moles divided by the stoichiometric coefficient, given number of moles divided by the stoichiometric coefficient. It is 0.1 and it is obviously more than 0.1. Okay, so the value which is less that gives you the limiting reagent. So limiting reagent is Na3PO4 into this. Maximum moles of this we have to find out. See, whenever you have this term, maximum number of moles, then also you can understand we have to apply the concept of limiting reagent. Because maximum number of moles only possible when all these one of the reactant will react. Okay, when the reaction will complete or the reaction will stop, then only the maximum amount will get. Okay, so the reaction will complete or the reaction will stop when one of these reactant is not at all available here. That is only possible when the limiting reagent which is nothing but Na3PO4 gets consumed completely into this reaction. Okay, so we have to find out the moles of this. So simply unitary method you have to apply. And what is that? 2 moles gives 1, right, and we are using 0.2 moles. So the 0.2 moles gives what? 0.2 moles gives half of 0.2 which is nothing but 0.1 mole. So answer will be 0.1 mole. Okay, so most of you have done this question. Next question you see. See, these are three different types of questions we have done. Okay, so I'm trying to cover all those different types which is there in one particular chapter. Okay, so like this you figure it out. Like in one chapter, what all different different kinds of questions we may have. Okay, so that kind of framing of every chapter we have to do. Next question you see. A mixture of O and Fe2O3. You see it is also a different type we have. A mixture of Fe3O4. Fe3O4. When heated in air, when heated in air, constant weight, to constant weight, gains 5% in its weight. What is the percentage of, find the percentage of Fe3O4 present in the mixture. Percentage of Fe3O4 present in the mixture. Let me tell you one thing. This when heated in air to constant weight means the mixture of Fe3O4, just one hint I'm giving you. Fe3O4 and FeO, this mixture has a constant weight. Okay, whatever the weight we have here and here that is constant. And these two reacts with oxygen when heated in air. So that's what the thing is. Now you try to find out. The percentage of Fe3O4 present in the mixture. See the constant weight means what? See we have a mixture of Fe3O4 and FeO, right? So the mixture will have certain weight. So that weight will have some amount of Fe3O4 and FeO. That weight is constant. Means Fe3O4 and FeO will have some constant weight. That is not changing in this reaction. Basically this line they have given you to make you confused. Okay. You just what you consider, you have a mixture of this Fe3O4 and FeO and that mixture is heated in air. That is it. Then what happens? That you have to think. Is it close to 80? Yeah, it is close to 80 only. 79 point something we are getting. Correct? Done. Do I need to solve this? Okay, now you see. First of all, you see this mixture we have, right? And since the first thing you have to assume here is what? We have to find out percentage Fe3O4. So we assume the mass of this mixture is 100 gram. And out of this 100 gram, we have Fe3O4 X gram and rest is FeO, 100 minus X. And we have to find out this X only. Our question is, what is the mass? That is X only. So when this mixture is allowed to react in air, that is nothing but the reaction should be this. Fe3O4 reacts with O2, right? And it forms Fe2O3. When you balance this, it is 2 and it is 3, it should be 3, Fe2O3, you know, so it should be 3. And then here it will be half, 8 and 1, 9. Correct? Okay, this is the reaction we have. Now, since you see, we have this mole ratio of this reaction. Okay? And what we can say, we have X gram of this, right? So X gram of this is equals to how many moles that see? We have to find out the mass of this present. Okay? And since it is whatever the mass will get here, this plus the other reaction will be what? This FeO at the same time, this also reacts with O2 and this again forms Fe2O3. This actually you should know these two reactions. And when you balance this, we'll get 2 here and then we'll get half here. So this is the reaction. Now you see both this component of the mixture reacts with oxygen in air and forms Fe2O3. And that is how the mass is increasing. The increase in mass percent is given, that is 5%. So if we know the mass of it, suppose it is A and it is B. This A and B if we calculate, so what we can write, A plus B from the second condition is equals to what? The weight increases by 5%. Initially it was 100. So now it becomes what? 105. This is what we have to do. Right? Now what is the molecular mass of Fe3O4? What is the molecular mass of Fe3O4? It is 56 into 3 plus 16 into 4. And that will be, it is 64 and 164 and 168. So 164 plus 168, 200 something we are getting. So here we are getting 168 and 60. That will be 228 plus 4, 232 approximately. And since we have 2 into this, so what we can write here with this mass and this molar mass, what is the number of moles of Fe3O4 we have? X by 232 is the moles of this. Similarly we can find out the moles of this, 100 minus X by molar mass of FeO is 56 plus O and that is 56 plus O and that is 62. So 100 minus X by 62. If you are using one molecule of this and one molecule of this. But here in this question you see we are using two molecules of this. So here the number of moles we use, that will be equals to X divided by 2 into 232 because two moles we are using here. This is the number of moles. And here the number of moles will be 100 minus X divided by 2 into 62. This is the number of moles of FeO. Now the mole ratio will write what? Two moles gives 3. One moles gives 3 by 2. Two moles gives how much? Did you understand this? Next page I will write. You see the reaction is this, 2 Fe3O4 plus half of O2 gives Fe2O3. I will do this one first. Fe3O4 we have assumed X. So X by 2 into 232 moles of this we have. So what we can write? Two moles gives 3 moles of Fe2O3. So X by 2 into 232 gives what? 3 by 2 into X by 2 into 232 moles of Fe2O3. And that will be equals to the mass of Fe2O3 divided by molecular mass of this is what? Molecular mass of this you can find out 56 into 2 plus 16 into 3. It is a bit calculative, right? So this will put over here and that will be 16 into 3 is 48 and this is 112, so 160. So we will find out M from this, right? That will be 480 X divided by 2 into 24, 4 into this, so 8, 928. This is the mass of Fe2O3 we will get. Similarly, what we will do? We will write down the next reaction also. FeO plus O2 gives Fe2O3. And the reaction will be this. So this number of moles will be what? 100 by 100 minus X by 2 into 62. So this 2 moles gives 1. So 100 minus X by 172 it is. The molecular mass of this is 56 plus 16, 72. 144 gives 1 by 2 into 100 minus X by 144. And this will be equals to what? Again the mass of Fe2O3 divided by molar mass of this. That will be 56 into 2 plus 16 into 3. That will be again 160. So again you solve this M. You will get 80 divided by 144 into 100 minus X. So this is the mass of Fe2O3 we will obtain from this reaction. This is the mass of Fe2O3 we will obtain from this equation. Now according to the, yeah it is 72, correct, correct. Now according to the question, what we can write? When the reaction takes place, the mass increases by 5%. So 480 by 928 X divided by 80, 144, 100 minus X is equals to 105. You solve this for X and that is our answer. This is what you have done, Purvik. So in this question, all of you, in this question you can take some approximation because option I did not give you. Option is, one option is close to 74. Other option is close to 26. Next option is close to 21%. And the one option is close to 80%. So when you take some approximation, you'll get the answer here. So exact answer which is given here, which is 79.75% when you solve will get approximately this 79.75%. Is it clear all of you? So in this question, basically this reaction you should know and this is what the learning you should take. When this FeO reacts, it gives Fe2O3. When this Fe3O4 reacts, it also gives Fe2O3 in error. Sometimes what mistake we do that FeO reacts is going to be directly right here, Fe3O4. So this is what made, which is not true. So you must keep this in mind. Understood this one? So this question was based on percentage composition. Another type, percentage composition question. Only what you have to do in this kind of question, you have to frame the question according to the question. And then you have to solve those. A little bit of calculation will be there in this kind of question. Understood, fine, can we move on? Next question you see. Weight of 1 liter milk is 1.032 kg. It contains butter fat whose density is given and that is 865 kg per meter cube, the extent of 4% and this is given by volume, volume by volume. You have to find out the density of fat-free skimmed milk. I'll give you option in this. 1038.5 kg per meter cube, 1032.2 kg per meter cube, 997 kg per meter cube, kg per meter cube. Tell me the answer. Option A, 1038.5. Option C, option D. Who is getting D? Tell me. No one is getting D? Option A. Option A is correct. Option A is correct. This is again a simple question. We have to find out what the density of the fat-free skimmed milk. Means whatever the milk sample we have, from this milk sample which contains fat also, if you remove fat from this, then we'll get skimmed milk. For this only, we have to find out the density. And density is equals to what we know, mass by volume. So mass of this we have to find out. For this, we have to subtract the amount of fat present in this milk sample. Now, first thing is what you have to do. The weight of one litre of milk is given. So first thing is that you see the option everywhere we have kg per meter cube. So this litre you have to convert in kg per meter cube. So one litre we know, one litre is equals to what? 10 to the power minus 3 meter cube. It means we can say the mass of 10 to the power minus 3 meter cube milk is 1.032 kg, according to the question. So the mass of milk sample of one meter cube, that will be what? Will it be 1,000 by this? 1032 kg. So this is the mass of one meter cube. Now you see the density of butter fat is given, kg per meter cube, and 4% volume by volume we have. It means the volume of butter fat is 0.04 meter cube, because it is 4% out of 1 I am writing down, because here also I am taking the mass of one meter cube. This is the volume of the butter fat in one meter cube sample. So we know the volume, we know the density also butter fat that is 865 kg per meter cube. So what is the mass of this butter fat? If you know the mass of this butter fat, we can subtract it into the mass of the milk and we will get the mass of the skimmed milk. So mass of the fat will be what? Density into volume, 865 it is right, 865. So 865 into 0.04. This gives you the mass of the butter fat, right? Now since the volume of fat is 0.04, so volume of the skimmed milk will be what there? 1 minus 0.04, out of 1 the volume is this. So milk volume is 0.96. So now you see, we know the volume of the milk, we know the mass of the milk also. What is the mass of the milk? Of the skimmed milk. That will be what? The mass here, which is 1032 minus the mass of the fat. Okay, so when you solve this you will get just a second. Yeah, I got to tell me. Actually, God, I'm leaving tonight only. Okay, and I'm taking class now, okay? Yeah, okay, thank you. Okay, so this value when you solve, you'll get the mass of it, right? And that will be around 35 kg you'll get here. When you solve this, you'll get 35 kg. So 1032 minus 35 will be the mass of the skimmed milk, which is nothing but 997 kg. So the density here will be what? The mass 997 of the skimmed milk divided by the volume of the skimmed milk is 0.96. So when you solve this, you'll get option A, which is 1038.5 kg per meter cube. Clear? Okay, next one you see. Sea water contains 65 into 10 to the power minus 3 gram per liter of bromide ion, bromide ions. If all the bromide ion, all bromide ions are converted to BR2, then how much sea water is needed to prepare 1 kg BR2? Options are given 15.38 liter, 15.38 10 to the power 3 liter, 7.69 10 to the power 3 liter, 76.9 liter. Anyone? No. Should I solve it? Or still doing? Okay. Shares is getting option B. Atmesh is getting B again. Okay, you see. See, the sea water contains bromide ion, which is nothing but BR minus. And with BR minus, it is converting into BR2, right? So in this, we have to find out, we have to prepare 1 kg of BR2. So what is the amount of sea water we need for that? Okay. So obviously when the reaction is converting into, this is what, this is an oxidation reaction. Right? This is what it is getting oxidized. It is getting oxidized. Right? So what we can write the number of equivalents or equivalents of BR minus is equals to the equivalence of product that is BR2 forms, that is the law of equivalence. Equal equivalence reacts and equal equivalence forms also. What is the law of equivalence? When A and B reacts and suppose it forms AB, then the number of equivalents of A is equals to the number of equivalents of B, which is equals to the number of equivalents of AB. You know this concept of equivalence or you want me to discuss this? How do we calculate number of equivalents? Okay, we'll solve this question first, then we'll come back to this again. Okay? So we are just, from the law of equivalence, we are equating the number of equivalents of BR minus and BR2. Okay? So what we can write the weight of BR minus, number of equivalents is what? Mass divided by equivalent mass. Right? If for BR, the equivalent mass will be its atomic mass divided by n factor here. Here, there is only one negative charge, so n factor is 1, is equals to the mass of BR2. Right? Which is given in the question 1 kg of BR2 be required. 1 kg means what? 1000 gram. 1000 divided by, divided by the equivalent mass of BR2. That will be BR2 that atomic mass is 160 divided by its equivalent mass is 2. But obviously the reaction should be balanced, we should write down 2 BR minus here. Right? So when you solve this, you'll get, when you solve this, you'll get the mass of BR minus. So the W which is the mass of BR minus is nothing but 1000 gram. Right? And its density is also given. So the volume of water needed, volume needed will be what? This 1000 divided by its density 65 into 10 to the power minus 3. Okay? So when you solve this, you'll get 15.38 into 10 to the power 3 litre. This is the volume we required. Correct? Yeah, now you tell me, do you know the concept of equivalence? What is the formula of number of equivalence you tell me? How do we calculate number of equivalence? You know equivalent mass? Mass by equivalent mass, yes. But what is equivalent mass then? Yeah, that I know, mass by equivalent mass. Equivalent mass, you know. Right? One more formula we have, number of equivalences, normality into volume, Nv. And that's why you must have done some question in which we write down N1v1 is equals to N2v2. This kind of question you must have solved. Right? N1v1 is what? It is the number of equivalence of one substance. And this is the number of equivalence of another substance. And these two are reacting. So when they react, mass by N factor is equivalent mass. That's correct. Because the valence is also equivalent mass. That is also correct. Okay. So N1v1 equals to N2v2 when we write, nothing but we are equating the number of equivalence. Okay. Sometimes what happens if normality and volume is given, you just multiply these two, you'll get the number of equivalence of one compound. Sometimes they'll give you molarity. So we know the relation of normality and molarity. Right? So normality is equals to what we can write. It is molarity into N factor. Molarity into N factor. This is the another formula you must remember. So if molarity is given, an N factor you know, you can find out normality. Okay. And there is one more very important formula of the calculation of number of equivalence, which we often use when we solve the question of Mohl-Gohm's reaction. Okay. And that is the number of moles into its N factor. Number of moles into its N factor. Do you know this formula? The number of moles into N factor. Yes. So this must you keep in mind. This is very useful, this formula. Number of moles into N factor. So when you do not have any concentration term given, either normality or molarity, then with number of moles also we can find out the number of equivalence and we can equate that. Okay. So this is very important. Another question you see. So this is you see, this is the second type, another type of question in which equivalent concepts are equivalent we are using, which comes under more concept one. Next question you see. We have acidified KMNO4 oxidizes oxalic acid, acid 2 CO2, carbon dioxide. What is the volume? What is the volume? 10 to the power minus 4 molar KMNO4 required to completely oxidize, completely oxidize 0.5 liter, 5 liter of 10 to the power minus 2 molar oxalic acid in acidic medium. In acidic medium. Tell me this one. 10. The options are 125, 250, 220. Options are this. 125, 125, 0, 220. Rithvik is getting D. Ramchandran, tell me the answer. Ramchandran is also getting D. Kondanya. Vesnavi, did you get? Kondanya and Ramchandran, you know these guys, Saimi here, Rithvik, Vashnavi and all. Ramchandran and Kondanya, you know these guys. I'll solve Vashnavi just a second. Yeah, correct. Number of equivalence of KMNO4 is equals to the number of equivalence of CO2. You must have missed n factor. Ramchandran, you don't know. You know only Atmesh and Lahitya. Atmesh is there. Atmesh, Shreyash. Okay, you met only Atmesh and Lahitya. This Saimi here, Vashnavi, Rithvik, all these are from Rajinagar. Okay, what is the answer, Shreyash? Tell me. See, again the same thing you have to do. You have to equate the number of equivalence. The number of equivalence of KMNO4 should be equals to the number of equivalence of what? Oxalic acid, which is C2O4 2 minus. Right? And since you see here, molarity is given. So number of equivalence will write what? It is N1V1 is equals to N2V2V, right? And since molarity is given, so normality we can express in terms of molarity, which is the N factor into M1V1 is equals to N factor into M2V2. Now you see for oxalic acid, molarity is given and volume is given. For KMNO4 molarity is given, volume we have to find out. N factor we have to calculate. Right? So since you see the medium is acidic. So we know KMNO4 is nothing but MnO4 minus. In acidic medium it converts into Mn plus 2. This information you should have. That's why this part is a bit, I would not say difficult, but this part is very informative. Little bit of information if you do not have. You cannot do the question. Okay? This you have to memorize that in acidic medium this converts into plus 2. Right? So N factor for this will be what? The change in oxidation number. Here it is plus 7. So 7 minus 2 by 1. Change in oxidation number per molecule. So it is 5. Okay? N factor. For C2O4 2 minus, which always converts into CO2, 2CO2. Oxalic acid always converts into carbon dioxide. Okay? So in this case, you see the carbon oxide is balanced. To balance the charge we'll use what? 2 electron this side. So this is nothing but the N factor for this reaction. Right? So N factor for oxalic for KMNO4 is 5. So we'll write 5 into molarity of this is 10 to the power minus 4 into V1 we have to find out is equals to 2 into molarity is 10 to the power minus 2 into 0.5. So when you solve this, you will get V1 is equals to 20 liter. The answer is option T. Number of equivalence you have to equate and this N factor you must revise. Okay? N factor for various other, you know, reaction condition and acid and waste. N factor you must revise. How do we calculate N factor? You just make a note of all these and try to keep this in mind. Okay? Now just two, three more questions we'll discuss in this chapter and then we'll move on to the other chapter. There's two more problems we'll discuss. Okay? Last two problems for this chapter. Next one to write down. What volume of hydrogen gas what volume of hydrogen gas at 273 Kelvin and one atmospheric pressure will be consumed in obtaining 21.6 gram of elemental boron. Atomic mass for this is given 10.8 from the reduction of boron trichloride by hydrogen. Options are 89.6 litre, 67.2 litre, 44.8 litre, 22.4 litre. Option B. So I may have got option B. Okay, Ramchiran is getting B, Shreyas is getting C. Okay, most of you are getting option B. Okay, you see the, you have to find out the volume of hydrogen. So the reaction would be what? BCl3 boron trichloride react with hydrogen, right? So H2 gives elemental boron, so B plus HCl. So the balance reaction would be this, 2 and 2. Okay, now you see we know what, we know the atomic mass of boron, right? And its mass is also given. So with this data what you can find out the number of moles. So the number of moles of boron is what? 21.6 divided by 10.8 which is nothing but 2. So we are getting 2 moles of it, right? And corresponding to this, what is the mole of hydrogen we have? 3 moles. And you see the condition here it is what? 273 Kelvin and 1 atmospheric pressure, nothing but the condition for STB. And we know at STB 1 mole, the volume occupied by 1 mole is what? 22.4 liters, right? So if you do the number of moles of H2, number of moles of H2, that into 22.4 liters will be the answer, the volume of H2, correct? So we know 2 moles of boron we are getting and 2 moles corresponds to 3 moles of H2. So number of moles of H2 is equals to what? 3 moles with the associometry of the equation, right? Then the volume of H2 will be what? 1 mole is equals to 22.4 liters. So 3 mole is equals to 3 into 22.4 liters. 67.2 liters is the answer, okay? You see in this question the calculation becomes easier since we get the number of moles here that you are taking in the question that is equals to the associometry coefficient here. That's why the question becomes easier here. Now suppose if the data, if I give you, suppose the data is this, suppose if I take the mass of boron, I am making some small change into this question. The mass of elemental boron we are taking and that is equals to 27 gram, 27 gram. Then what would be the volume of H2 evolves into this? Everything is same, only instead of this 21.6, we are taking 27 gram. What will be the answer? If you take 27 gram here, I don't know, I have to calculate the continuum. 84 watt liter, okay? When the mass is 27 gram, then what is the number of moles of boron? That will be 27 divided by 10.8 and that gives you 2.5 approximately, 2.5 moles of boron, right? Now from this relation you see 2 moles of boron is equivalent to 3 mole of hydrogen because our objective is to find out the number of moles of hydrogen only. 2 moles is equivalent to 3. Since here we are getting 2 only, that's why directly the number of moles of H2 is 3 and we can calculate the volume here. But if the number of moles is not 2, then we have to calculate by symmetry methods. 2 moles requires 3. So 1 requires 3 by 2. So 2.5 requires what? 3.2 into 2.5 mole of H2, okay? Now this mole into 22.4 liter since the condition is at STP, right? This gives you the answer. Whatever the answer you will get here, whatever the calculation you do here, that will be the answer. So it will be 11.2 into 7.5 so approximately around 80 something you will get. 11.2 into 7.5. So 11 into 7 is 77, 77 plus 5.5, it's 82.5. Yeah, approximately 83, 84 you will get. Okay, the answer is that all. Understood? Your last question for this chapter you will write down. The mass of potassium dichromate crystal, potassium dichromate crystal required to oxidize, required to oxidize 750 centimeter cube of 0.6 molar Morse salt. Morse salt whose molecular mass is given 392 solution. Potassium dichromate molecular mass is also given, it is 294. This is the question here. Solve this one. Okay, solve this just a second. What happened? Are you done? Okay, see this. See, the mass of potassium dichromate crystal required to oxidize, this is the Morse salt, right? So we have to find out the mass of potassium dichromate crystal. So Morse salt, the formula is what? For Morse salt, it is FESO4.NH4 whole twice SO4.6H2. This is Morse salt. Potassium dichromate is K2CR2O7. Nothing is mentioned will take acetic medium only into this. Okay, so in this Morse salt, the only atom which will oxidize here is our only oxidizable part is this Fe. And this Fe plus 2 converts into Fe plus 3. Plus one electron releases. Oxidation of iron takes place when these two reacts. Okay. Potassium dichromate is nothing but CR2O7, 2 minus. And in acetic medium, this also you should know, in acetic medium it converts into CR2, sorry, 2CR3 plus in acetic medium. In basic medium, it goes into CR2O4 2 minus. CR2O4 2 minus, which is not required here, but this is the change we have in this reaction. CR2O7 2 minus will go into 2CR3 plus and then Fe2 plus will go into Fe3 plus. So now since this is, you know, here the oxidation number of chromium is plus 6. When you balance this reaction. Okay. So what will do? 7 oxygen we have. So we'll add here 7H2O. 7H2O to balance hydrogen will add what? 14H plus. All atoms are balanced now. Now we balance charge 2 minus 14 plus. We have 12 positive charge here and 6 positive charge here. So for to balance the charge, we have to add 6 electron this side. Okay. So this, the first half equation is this. Right. This is the first half equation. And the second half equation is this one. CR2O7 2 minus plus 14H plus. Plus 6 electron gives 2CR3 plus plus 7H2. Right. Now when we add these equations, because the net equation, if you have to write down, you have to add these two half equation. One is oxidation of other one reaction half. And when we add these two, this electron must get cancel. And to cancel this electron, we'll multiply this equation by 6. So the equation we get here is 6FE plus 2. Plus CR2O7 2 minus plus 14H plus. Gives FE plus 3 plus 2CR3 plus. Plus 7H2. This is the equation we get. Now, our objective is to find out the mass of potassium dichromate. So for mass will require what? The mole ratio. Mole ratio is 6 is to 1. Right. Mole ratio is 6 is to 1. Now the number of moles of, this more salt is given. Right. Which is 750 into 0.6. Number of millimoles since it is in centimeter q milliliter. So number of millimoles of more salt is what? 750 into 0.6. Right. 750 into 0.6 is the number of millimoles. If you convert this into moles, then this must be divided by 1000. Which is nothing but 0.45 mole will get here when you solve this. So moles of more salt we know. Now here you see from this equation. This is the information given in the question. From this relation what we can write. 6 moles of FE 2 plus. Of FE plus 2 is equals to 1 mole of CR2 or 7 2 minus. And here the moles we are getting what? Which is given the question is 4.45. So 6 moles is equivalent to 1. So 0.45 is equivalent to how much? 1 by 6 into 0.45. And when you solve this you will get 0.075 probably. Yeah 0.075. Okay moles of CR2 or 7 2 minus. Now since mass you have to find out so mass is equals to what? Number of moles into its molecular mass which is 294 and when you solve this you will get 22.05 gram approximately. This is the answer. So in this problem what you should keep in mind first of all the conversion. How this converts into some other ion? This conversion you should know. The first this information you should have if you do not know this information you should have solved this question. And when two substance are reacting so we must know the mole ratio of that in which mole ratio they are reacting. So for that purpose you have to balance the two reaction and add. So we will get the mole ratio here and then just unitary method we will apply. Is it clear? See the states of matter like this chapter we will finish now we have done it. Okay so we will not do any further question into this. In states of matter the gaseous state is not that much important. For example point of view. Solid state you must do. In gaseous state you must go through with A and B and compressibility thing. Just few two three questions we will see and then we will move on to another chapter. Okay so if you want to write or just write it down on your copy that for gaseous state the various molecular speeds formula VRMS V most prevalent V average that you must go through and then concept of A and B okay compressibility factor unit and all everything. Right these three four things you must revise for gaseous state. Okay and then the basic formula that you have because formula based question they might ask. Okay fine one question you write down we have done this solid state properly. Do you want to solve some question on to that also? Solid state. I guess it is not required because we have solved many questions in the class. Okay now you see the gaseous state will just solve a few problems three four problems in this. Okay first one a bubble of gas released released at the bottom of a lake at the bottom of a lake increases to times eight times its original volume its original volume when it reaches to when it reaches the surface the surface assuming that assuming that the atmospheric pressure equivalent to the pressure exerted by a column of water 10 meter high you have to find out the depth of the river of the lake what is the depth of the lake a b cnt option 90, 10, 70, 80 90 meter 70 meter 70 meter and 80 meter c how do you do this c is correct yes you have to apply a Boyle's law what about Rithvik Rithvik did you solve time here tell me okay we just have to apply p1 v1 is equals to p2 v2 okay so v2 is given that is 8 v so p1 is the atmospheric pressure is equals to p1 is equals to 8 p2 and p1 is equals to what we can write p2 is equals to p2 is equals to we can write because p2 is the pressure at the bottom suppose the lake is this here the pressure we have p2 and here it is p1 so p2 is the pressure at the bottom so this will be equals to atmospheric pressure plus the pressure because of the water column pressure because of water column for this you see one information is given this atmospheric pressure is the pressure because of water column so this is equals to times of something I have written here it is given that 10 meter no it is 10 meter of height right so this is this is because of 10 meter of height so this will be equals to what because of the 7 or 70 meter of height then only it will be 80 you see you can understand like this here we have p1 v1 and p2 v2 oh I have made one mistake p2 is equals to 8 p1 we should write correct so this is because of 10 meter of water column right and this is atmospheric pressure is p1 and this is the pressure because of water so this must be 7 p1 then only 7 p1 plus p1 will be equals to 8 p1 because of this we will get 7 p1 and p1 is the pressure because of 10 meter of water column so 7 p1 is the pressure because of 70 meter of water height like the depth of the water will be lake will be 70 meter and p1 will be option C easy one only poise so you have to apply ok next question you see in this that this is related with the vapor density see the question is too big but the solution is like only too like so the question is we have a mixture I will just give you data here we have a mixture of 204 and this is in the ratio of 2 is to 1 and 204 is to 202 is 2 is to 1 by volume right the ratio of 204 and 202 is to 1 by volume you have to find out the vapor density of the mixture what is the vapor density of the mixture then what happen 2.05 it is not the option option I will give you first option is 45.4 49.8 32.6 28.3 these are the 4 options 3 are gas 1 more mixture divided by volume what volume you have taken are you trying or should I solve it tell me others 3 are you ok option D 38.3 you are getting option D is correct ok you see what we will do into this since person takes by volume it is given so we assume the volume is 100 right whatever the unit but we assume the volume is 100 so what is the part of N204 into it it will be 100 of 100 of 2 by 3 and that of NO2 will be 100 of 1 by 3 which is volume of NO2 in this mixture will be 200 by 3 and the volume of NO2 will be 100 by 3 ok now since the mixture we have so what we can write the mass of if we balance the mass so mass of N204 plus mass of NO2 is equals to the mass of the mixture right so mass of N204 will be what it is density into volume plus of N204 plus density into volume of NO2 this is equals to density of the mixture we have to find out and volume we have assumed as 100 100 mixture is there 100 ml or whatever and you will take that 100 now density density is nothing but the vapor density of it it is half of the molecular mass of this right and molecular mass of this is what 92 so 92 divided by 2 is the vapor density of N204 into volume of N204 is 200 by 3 plus its molar mass is 46 so 46 by 2 100 by 3 is equals to density of the mixture which we have to find out in 200 now you can solve this 100 will get cancelled to 46 and here we have 23 so 1 by 3 will take common 46 this 2 also gets cancelled what is in 92 92 plus 23 is what 92 plus 20 is 112 plus 115 is equals to density of the mixture so that will be density of the mixture will be 38.3 whatever unit you will take that unit will take gram per ml, gram per liter whatever it is one more question will solve one last question see in this case it is still trapped as I told you they have generally asked the question of the ratio where you say speed rate of diffusion rate of diffusion z and a be related so those questions I have discussed already in the class so those are very basic questions I am not discussing all those you can go through on your own it is very basic thing some different question I am trying to give you this question you write down this is the last question for this chapter a mixture of NH3 and N2H4 a mixture of gas and N2H4 is placed in a sealed container is placed in a sealed container 300 Kelvin the total pressure is 0.5 atmospheric pressure is 0.5 atmospheric when the container is heated to when the container is heated to 1200 Kelvin both substance decomposes completely according to the reaction then decomposition takes place takes place and decomposition reaction is this NH3 gas converts into N2 and 3H2 another one is N2H4 gas converts into N2 gas and 2H2 after decomposition the total pressure Pt is equals to 4.5 atmospheric at 1200 Kelvin at this temperature the pressure is given what is the find the percentage of N2H4 gas the original mixture if you remember this kind of question I have discussed in the class 35% 40% 75% and then 25% what is the answer tell me you have to assume the ideal gas behavior here ideal gas behavior PV is equals to NRTA use number of moles initially you assume check what is the final number of moles final number of moles the dissociation is complete right the reaction is ok it is given that the both substance decompose completely according to the reaction ok so complete decomposition is there so what is the final number of moles of N2 and H2 from the from the reaction what about others tell me what happened tell me fast anyone Ramchiran, Kondinia Vaishnavi, Rithvik, Suresh where is Atmesh see when NH3 dissociates it gives total 4 moles of N2 and H2 right when NH3 dissociates it gives total 3 moles of H2 and 1 mole of N2 so total moles is 4 similarly for N2H4 the total number of moles is 3 see here see the first set of data is what P1 is given 0.5 atmospheric V you let it be as it is R you let it be temperature is given that is 300 Kelvin and you assume the number of moles of NH3 is N1 and that of N2H4 excuse me is N2 initially right so when you apply PV is equals to NRT so P1V N1 plus N2 into RT1 and that will be equals to the pressure is 0.5 V you let it be as it is N1 plus N2 R and T1 is 300 Kelvin this is 1 now you see when this dissociates this gives total 4 moles of this 3 moles of H2 and 1 moles of N2 this gives total 3 moles of N2N2 right so and this 4 moles is for 2 moles so 1 moles gives what 2 moles of this so what we can write for the final equation it becomes P2 is equals to 4.5 it is given V will be same R will be same T2 is 1200 Kelvin and for NH3 this N1 becomes 2N1 because 2 moles gives 4 moles 1 moles gives 2N1 right 2 gives 4 so 1 gives 2 initially we have N1 so gives 2N1 similarly N2H4 it gives 1 gives 4 so N2 gives 4N2 for N2H4 right now when you apply the PV is equals to NRT so 4.5 into V is equals to 2N1 sorry it is 3 not 4 see here it is 3 no so it is 3N2 so 2N1 plus 3N2 into R into 1200 Kelvin so this is equation 2 now when you divide 1 by 2 little side will get total 4 by 9 is equals to N1 plus N2 divided by 2N1 plus 3N2 and what we have to find out percentage of N2H4 in the original mixture so we have to find out N2 by N1 plus N2 into 100 this value you have to find out just to solve this equation you will get the answer we are taking 1 mole of NH3 and 1 mole of N2H4 decomposition takes place like this this is a balanced reaction this is using 2 moles but we are using here initially using 1 mole only 2 moles totally converts into 4 moles so 1 converts into 2 so N1 converts into 2N1 that is what 2N1 we have here similarly 1 converts into 3 so N2 converts into 3N2 now we will substitute here and then we will solve so when you solve this the answer let me check answer is answer is 25% is it option B is it it is D I think answer is 25% you just solve and check this okay so if you remember I have discussed one type of question there we have stopcock in the 2 vessel we have gas filled and then we just open that stopcock the gas mixed and then we calculate the pressure and all whatever the thing we have to find out but this is the different type and those 2 vessel you have which is connected to a small tube and then stopcock is there those 2 vessels may be at same temperature or different temperature so when they will at different temperature then we cannot add the volume over there if you remember I have done this kind of question in the class practice thing in the class I have done it okay and the vessel if it is at the same temperature then we can add the volume also and then we can solve it this is the third type of question we have where the gas dissociates okay temperature is different only but we are eating the mixture and the gas dissociates then how do we find out the final volume right this is no different types of question good question you must keep this type of question in mind how do we calculate final number of moles then only you can get stuck otherwise it is simple only only bbs goes to nrt you have to apply I hope it is clear right another question you see this one you solve correct maybe tell me the answer okay let others also do tell me the answer say may I repeat where is shares okay sign just after this question after this question wait we will take a break after this question say may I tell me the answer how do we know it is adiabatic okay you see I am explaining this first of all you see one by one you try to understand this an ideal gas in isothermally in thermally insulated vessel when thermally insulated vessel you have it means there is no exchange of heat thermally insulated means what means heat exchange is not possible right thermal insulation so heat exchange is not possible so obviously del q is equals to zero when del q is equals to zero so it is what adiabatic and when adiabatic is zero so this is the first option is correct yes thermal insulation means thermal means what the heat tendency to exchange heat that is the thermally thermal property we have basically to thermally insulated means what there is no exchange of heat okay now you see del q is equals to zero vessel at internal pressure P1 volume V1 absolute temperature T1 expands irreversibly so now when it is adiabatic process what we can write first thing PV to the power gamma is equals to constant and then what should be the answer right D but D is not the answer here why it is not the answer because this relation we can only write for reversible process adiabatic reversible process but here you see it is irreversibly so that's why I said that condition you must remember that what condition we are applying when under what condition will get the result so all those PV, VT, PT relation we have in adiabatic process that is all for reversible process not for irreversible process is it clear Ramchalan, Saimahir right so D is obviously not correct yes now next thing you see when zero external pressure we have so walk done is what always walk done is equals to minus P external into delta V we always write minus P external into delta V walk done we always write minus P external into delta V correct so when the P external pressure is zero so walk done is equals to zero right so when walk done is zero then what we can conclude there is no option related to walk done so we cannot do anything with this now from this first law of thermodynamics we know del U is equals to del Q plus delta W all other possibilities you have to check options one by one when the term comes into multiple correct so you have to cross check one all these options like one by one all these options you have to cross check right so we have done with A and D now we are trying to you know cross check this option D which is T1 is equals to T2 which is only possible when the change in internal energy is zero this is only possible when change in internal energy is zero because we know internal energy is a function of temperature only so at two different point internal energy will be sorry temperature if it is same it means the internal energy must be same means what del U must be zero is it possible here you see from first law of thermodynamics del Q and delta W is equals to zero here so del U is equals to zero eventually we are getting it means temperature is constant and when temperature is constant so we can write what T2 is equals to T1 so hence option B is also correct right now when temperature is constant so we can write what P1 V1 is equals to P2 V2 right and that is why option C is also correct right so A, B, C is the right answer here did you understand this all of you B and C is correct A is also correct okay so you must keep in mind that this is very good that's why they have given easily they have given you this thing that you can understand del Q is equals to zero so adiabatic so PV to the power gamma constant but this is only true when the process is reversible you must remember the derivation here adiabatic reversible process there we have done this derivation and why it is if suppose another thing I have explained irreversible and reversible process also right so for irreversible process the external pressure is constant so we cannot write this P1 P2 P1 and P2 here because these two are constant P1 and P2 right so for irreversible process this is not true irreversible process external pressure is not constant so we can write this okay so this derivation is to reversible process anyways so we will take a break here we will start at 5.30 correct okay I will send you the link again okay at 5.30 we will start I will send you the link again thank you