So this portion that we are going to do in this chapter, it is important for J mains, okay? Practical organic chemistry. See, in this particular topic, we are going to understand, like suppose we have an organic compound, we have a sample of organic compound given, right? And in this sample, water elements are present. Like do we have carbon present in this or not? Nitrogen present or not, halogens present or not, sulfur present or not. This kind of understanding that we are going to have in this. All these are practical methods, we'll do some practical methods are there, we have some reactions, we have some methods that we use in order to identify the presence of these elements in a given sample, okay? In a given sample, organic sample, correct? So how to do this, what methods we have and how to find out the amount of these elements present in the given sample of organic compounds, right? So this we also call it as qualitative analysis, okay? Qualitative, you must have seen this, name of the chapter is qualitative and quantitative analysis, quantitative analysis. So qualitative analysis is basically that detection, means whether these elements are present or not, that we'll find out by some reactions, by some methods, we'll find out the elements present or not. So it is a detection. Quantitative means estimation, like in a given sample of organic compound, what is the amount of carbon or nitrogen or sulfur present into this, okay? So detection and estimation. So we'll do both into this, how to find out the elements present or not, plus what is the method we use to find out the estimation. So estimation part is basically, we use the basic concepts of the first chapter that we did in grade 11, that is we use the concept of more concepts and basic concepts of chemistry. And we can do the calculation. So calculation part is not that difficult, okay? Now the first quantitative analysis we are going to do is for the detection of carbon and hydrogen. First point to write down, detection of carbon and hydrogen. So we won't write down the theory into this, like we take the sample, we add this, we'll get this and all. So each and every line we won't write, we'll do this in a simpler way. Like for example, what we do in the samples, you have the organic sample given. This is the organic sample given and they'll ask you to find out that whether the carbon is present in the sample or not, hydrogen is present in the sample or not, okay? If present, then what is the amount of carbon present in this, correct? Or hydrogen present in this. So what we do here, we'll take this organic sample and we mix copper oxide in this. C-U-O will mix into this sample and we heat it. This mixture then passes through and hydrous C-U-S-O-4 solution. We'll allow this to pass through and hydrous C-U-S-O-4 solution. So obviously these are the practical things. So all the paths are defined at what we need to do when, okay? So an hydrous C-U-S-O-4 is white in color. When this happens, so if the carbon atom is present here, then the carbon atom will get oxidized into CO2, one second. So why have we written an anhydrous C-U-S-O-4 solution? We don't mind. Sorry? So if it's an anhydrous C-U-S-O-4 solution. No, actually not solution, no. This mixture, no, will allow to pass through and hydrous C-U-S-O-4. Mainly solution you don't write, it's fine. We have a mixture of this. Okay, this sample mixture will allow to pass through and hydrous C-U-S-O-4 solution. Okay. So whatever carbon is present in the sample, right? It gets oxidized into CO2. And how do we get to know that it is happening, right? We have lime water, the entire apparatus, we have lime water present there, but lime water turns milky. That confirms the presence of, turns lime water milky. So this happens, it confirms the presence of carbon in the sample. If any hydrogen is present, this also gets oxidized and forms water, H2O, right? This water, once again, so this white color of anhydrous C-U-S-O-4 that you have, this water turns this color white into blue, correct? This method is called the copper oxide test because we are using copper oxide over here. So right on here, this water turns turns anhydrous C-U-S-O-4 into blue. The color changes from white to blue. So this confirms the presence of hydrogen over here. If you look at the reaction here, carbon with C-U-O and on this, the qualitative part, they won't ask any question. They'll ask you to find out the percentages of carbon present in the sample. So calculation, how do we do? That is here more important. Okay, so in this, you see, the reaction that involves in this, carbon plus C-U-O, it converts into two C-U-O, it converts into CO2 and copper, which reacts with this, converts into C-A-C-O-3, which turns the water milky, C-A-C-O-3 forms. Similarly, for hydrogen, you see two H plus C-U-O, it converts into H2O and copper gets reduced into C-U. This H2O with copper sulphate, white the color is, it converts into C-U-S-O-4, the hydrated form, dot 5 H2O, which is blue in color. So this color change confirms the presence of water, that is hydrogen, and if it turns milky, it is the carbon dioxide that is carbon. Okay, okay, so based on few informations, we can find out the amount of carbon or hydrogen present in the sample. That is the estimation of carbon and hydrogen, okay. So it depends upon what is the mass of CO2 we get, and what is the mass of H2O we get. In the book, if you see some book in which direct formula is given, the percentage of carbon is this, but we can do it this way also, because some data will be obviously given in the question based on that data you can find out. So write down the estimation, estimation of carbon and hydrogen in this sample. So we are just going to write down the general expression and based on the formula or the way that we calculate this, you can use that or methods you can use in order to solve the questions if you get this in the exam or test. So what we are assuming, we're assuming the mass of organic compound in which the carbon we need to find out. So mass of organic compound, organic compound I am assuming it as W-gram, for example. Okay, mass of CO2 that we obtained, mass of CO2 is suppose we have Y-gram and the mass of H2O we have is X-gram. This data we have, it will be given in the question. Obviously some data will be given because you're not going to do the practical in the exam. So data will be given in the question. Now what we can say here, 44 gram of CO2, carbon dioxide, right, contains, what is the mass of carbon in this? One mole of CO2 is 44 gram. Mass of carbon is 12 gram of C. So the mass of CO2 that we have, Y-gram contains how much of C? 12 by 44 into C. Sorry, why? Gram of carbon, because we are assuming all carbon is converted into carbon dioxide. When this reaction takes place, that is the understanding we have that all the carbon atom present in the sample, organic sample converts into carbon dioxide. So this is the relation we have. So this is the amount of carbon we get if the mass of CO2 is Y-gram, correct? So what is the percentage of CO2? Percentage of carbon, I'm sorry. What is the percentage of carbon? So this is divided by W into 100. Into 100, right? So mass of carbon, 12 divided by 44 into Y, divided by the total mass. We have assumed the organic compound, the mass is W. And since it is percentage calculation, so 100 we need to multiply. So the formula is 12 by 44 into Y by W into 100. I would say, keep it as it is, 12 and 44 only. It will be easier for you to memorize. 12 is the atomic mass of carbon, 44 is the molecular mass of CO2. And Y by W into 100. This is the percentage of C. Could you tell me the percentage of hydrogen present in the sample? So we'd 2 by 10 into X by W into 100. Right. Similarly, we can calculate the percentage of hydrogen present. So percentage of hydrogen present in the sample is 2 by 18 into X by W into 100. Next you see, first we'll do all these methods and then we'll see a couple of questions on this, right? Important ones we'll see. Okay. Second one is detection of nitrogen. Detection of nitrogen is the most important one. Okay. Neatmito, they have asked many times questions on this. What question I'll tell you, but it is very important, detection of nitrogen. If you look at the reaction here, one information we have already, that is the soda lime test. Soda lime test, it evolves ammonia, basically. All nitrogen converts into ammonia. So our basic concept based on this information, whatever organic compound we have the sample, organic compound, and we add soda lime. This reaction we have done already in organic and it gives ammonia. Nitrogen evolves in the form of ammonia. That is what the information we have. So we use this concept only in order to estimate nitrogen because whatever nitrogen is present here, it converts into ammonia, right? Okay. Now, in this, just a second. Yeah. So you see one thing we have. There's a test called LAWSAI, G-N-E-S, L-S-A-G-N-E-S test. So this test, it is not like it is done only for nitrogen. It is used to detect the presence of, write down all of you, used to detect the presence of, the presence of nitrogen, sulfur, halogen, phosphorus also to some extent. Yeah, you must have done this in the lab, yeah? Yes. Phosphorus in organic compounds. So question they will ask, this information you should know in organic compound, right? This is there. So it is not like this test is used only for nitrogen. Okay, first thing is that. So in this test, or in this less than this test or method, what happens? We'll have the sample, right? Let's search this. Okay, I'll go back. Copies. So in this, what we do, we allow this organic matter to fuse with sodium metal, right? So any metal will heat it at a high temperature, fusion basically, high temperature. And in this, the sodium metal melts. So basically this help when you allow this organic sample to react with sodium metal. The sodium metal combines with the various elements that is present in the organic sample. For example, you see, suppose our carbon or nitrogen present here. So it combines with carbon and nitrogen and convert into NaCN, sodium cyanide, ionic compound, okay? If sulfur is present, so it's reacts with sulfur, converts into sulfide, that is Na2S. Two Na plus S gives you this. If any hydrogen atom is present, Na plus X, it converts into halides, that is NaX. So this kind of reaction possible, depending upon what elements is present in the sample, organic sample, right? So this we use to convert into the ionic compound, covalent to ionic conversion happens in this, when we allow this to react with sodium, right? Now this compound that we get in this reaction, we extract this, right? We heat this, we get this, and then we filter it out. Or we can say extraction. We cool it down, and then we filter it, extract it. So after filtration, whatever sample you get, that we call it as sodium extract. The term we use, we call it as sodium extract. Or we also call it as lasagna extract, right? Sodium extract or lasagna extract. This is usually alkaline in nature, the sodium extract that we get. Usually alkaline, if it is not, then we add a bit of NaOH into this to make this alkaline. Usually alkaline, if not, then we add NaOH. Then we add NaOH. So if nitrogen is present into this, so this is common, till here it is common for all elements, nitrogen, sulfur, halogen, phosphorus, all elements, right? That we get fusion extract, sodium extract, or lasagna extract, okay? Now to test whether nitrogen is present in this or not. So after this, you write down test of nitrogen. In this only, test of nitrogen means till here, we have the common steps. Now specifically, if you try to find out whether nitrogen is present or not, what we do is see this. So whatever we have here, sodium extract, Na extract, we use to boil this with FeSO4 for nitrogen. We use to boil this with FeSO4, the compound that we get here, right? This compound, we acidified with concentrated H2SO4. Acidified with concentrated H2SO4. And when we do this, we get a blue color, Prussian, sorry, blue color we get here. We get blue color. And this blue color confirms the presence of nitrogen, confirms the presence of nitrogen. I'll write down the reaction also, first you copy this down. The reaction involved in this, reaction they won't ask you, I'll just write it down. Fe2 plus, plus, we have six CN minus, converts into FeCN6, four minus, Fe2 plus in presence of H2SO4 concentrated, converts into Fe3 plus. And this two combines FeCN6, four minus plus, Fe3 plus, converts into Fe4, FeCN6 plus, whichever you have ions, no, we'll get this. This complex gives blue color in the solution. So this blue color confirms the presence of nitrogen in this. So sodium extract that we are getting, it is common for other elements also, like, you know, halogens test we do, we do sulfur also, it is common for that. And then depending upon the element present, we use this, yeah, tell me. Shouldn't be thrice of FeCN6. Balanced reaction I haven't done. If you write down the ionic one, no. If you write down the balanced one, then it should be Fe4, Fe4 three plus. Fe4, FeCN6, the whole thrice. Like this here, Fe4 three plus, this three, four minus, then here also you should write down four, and this is three, isn't it? That's all. Correct? No. Yes. Balanced reaction is this. So you'll get this blue color. Now, this is the reaction, the reaction they won't ask you. So just to keep this in mind, we'll get blue color over here, which confirms the presence of nitrogen. Now, how do we estimate? Okay, lessen this test is this method is important. Two times they have asked this question in JEE, JEE mains, 2013 and 15. Whatever I remember, it is 2013 and 15 mains they have asked question on this. So nitrogen one is very important. Neat also, if you look at the previous question, or neat also they have asked the questions based on this many times. Two methods are there to find out the amount of hydrogen present, nitrogen present in this. Okay, that is gel dal method. Write down the next one, estimation of nitrogen. Heading you right down, estimation of nitrogen. Two methods we use here. One we have gel dal method. If you're going to revise this chapter in the last minute, this is one of the thing that you must revise. Okay, because you'll get questions on this. Okay, gel dal method. And another one is Dumas method for estimation of nitrogen. Sometimes they also ask this, which method we use for the estimation of nitrogen? Name they'll give you four different names. Dumas method we use, gel dal method we use, right? So three methods are very important. Two for nitrogen, gel dal and Dumas. One for halogens, that is carriers method. Okay, these three only important we have, for example. So what we do in gel dal method. First of all, one information you must have, write down. This method is not useful, all of you write down. This method is not useful for the compounds containing, for the compounds containing nitrogen in the ring, nitrogen in the ring, like pyridine, cunolin, okay, I'm repeating. This method is not useful for the compounds containing nitrogen in the ring, like pyridine, cunolin, comma, compound containing NO2, compound containing NO2, and compound containing diso group, N double bond N and the compound containing diso group. So all these things, if it is there, then gel dal method is not useful. Exactly on this information, need they have asked question twice or twice, okay? So this is important for need point of view. For G point of view also. So could you repeat that, I mean, net was very bad. I said, see gel dal method we cannot use, three types of compounds are there, for which we cannot use gel dal method. The compounds in which nitrogen is present in the ring, like pyridine is there, right? So pyridine, it is not useful. Cunolin, we cannot use gel dal method, comma, compound containing NO2, compound containing NO2 like nitroalkanes, gel dal method is not useful, and compound containing diso group, N double bond N. Thank you, sir. Yeah. So what we do in this method to estimate nitrogen here? See, we have the sample organic compound, right? And in this sample only, we are going to find out the amount of nitrogen present. So first of all, we use H2SO4, HCl also we can use, acid we use, known volume, known concentration. I am assuming the volume here as VML, and the concentration normality I'm assuming N1. This is known, known volume, known concentration. We get here ammonium sulphate, ammonium sulphate. There are nitrogens convert into ammonium sulphate, okay? NH4, whole twice SO4. Ammonium sulphate. Now, this ammonium sulphate is allowed to react with NaOH, NaOH, we'll add here. Then it converts into Na2SO4, Na2SO4, means in this sample we'll add NaOH, Na2SO4, then NH3, and then H2. Now, since we are using this H2SO4 in excess, right? Some amount of H2SO4 is also, you know, left behind. So that particular, that H2SO4 again neutralize this NH3. So in the next step, NH3 reacts with H2SO4, and it converts into ammonium sulphate, right? So finally, we'll get ammonium sulphate. So this is the reaction we have. Now, we are assuming since we have excess of acid here, so volume of acid left. Volume of acid left is Chotolavi, a small VML. This is the volume of acid left, okay? This NaOH that we're using here, it's concentration is also known. And do, this is the standard alkali we have, means concentration is known here, standard alkali. So the number of equivalents for acid here, which reacts, will be same as the number of equivalents of ammonium sulphate, and will be same as the number of equivalents of NaOH reacts, correct? Because they are reacting, we know number of equivalents of this equals to this equals to this equals to this, that are reacting, correct? So we can write here, the number of equivalents of H2SO4 would be N1 into volume of H2SO4, capital V is equals to the normality of this N2 into what we write, the volume we use over here, that is V, once again, and it's three we are getting, volume of, it's a cheek here, wait, this you let it be. Could you tell me, what is the volume of H2SO4 reacts in this, volume of H2SO4? See, this is left, initially it was this. So what is the volume consumes in the reaction? It is V minus V, right? So what we can say here, we have, since this is the standard NaOH, this is important, it is a standard NaOH here. A standard means what? It's normality is one normal, and its volume is also one liter. Standard concentration is this, right? Standard NaOH or standard alkali means concentration volume both is one, one normal and one liter. So what we can say here you see, with this information, that thousand centimeter cube means thousand ML of one normal, one normal NS3. Why NS3? Because one equivalent of NaOH gives one equivalent of NS3, one normal NS3 gives 14 gram of nitrogen, 14 gram of nitrogen. Number of equivalents will be equal, number of equivalents must be same if the compounds are reacting and the products are forming. The number of equivalents of reacting must be equal to the number of equivalents of product. Thousand ML, one normal NS3 gives 14 gram of nitrogen. You see, NaOH is present there in the middle. The left side we have S2SO4, and the final we have NS3. So this NS3 is the bridge we have in between. So we are relating the equivalence relation in all these reactions basically. So if you have S2SO4 volume, like suppose we have the volume of S2SO4 reacts is V minus V, initial minus final. If this ML of one normal NS3 we take, then the amount of nitrogen should be 14 divided by thousand into V minus V, right? If you have one normal, N normal then what happens? N normal then what happens? 14 divided by thousand into initial volume minus the volume of acid left into N normality, not N1, N the normality. So this is the amount of nitrogen present, amount of nitrogen present. Percentage, if you need to find out, percentage nitrogen would be the amount of nitrogen, 14 into V minus V into normality divided by 100. Achai, you did not mention it. Anyway, it's not a problem, thousand into one by W, the mass of sample I'm assuming into 100. This is the relation we have. Percentage nitrogen present in the sample. W is the mass of organic sample that we are taking. So with this formula we can find out the amount of nitrogen, percentage nitrogen. So why are we taking V minus V here? Because that is the volume of S2SO4 consumes in the reaction. Small V is the volume of acid left, yeah? Yeah, so. You can also take like this, you can also assume, okay, V is the amount that is reacting, consuming. So you'll have V1 over here, right? But when you get the question, they'll give you the volume of acid left and whatever like the understanding of expression you must have. V is the initial volume and V is the volume of acid left. So V minus V is volume react. So what is the use of NaOH tensor, the N2 of NaOH? NaOH, we are taking as a standard solution that is converting Na2SO4 into NH3, like ammonium sulfate into NH3. So all the ammonium sulfate, first we get ammonium sulfate and then we convert this into NH3 with the help of NaOH. And since NaOH is a standard solution, so its concentration and volume will be known. So we can equate the number of equations. Okay, yes. So in this case, the concentration of the acid or it's the concentration of acid, right? V, which concentration? So N, in the final example. N is the normality of acid, correct? Yeah. N is the normality of acid because we have equal equivalence, though. So if that is the normality, then what would be the amount of nitrogen we get? Yes, I got it. We'll do one questions on this. You will understand what's, let me finish this. Another matter of estimation of nitrogen we have. See, one more thing in the book, if you still simplify this, 100 and 100 will get canceled and instead of 14, the right here, 1.4. And N into V, it is a milli-equivalent, no. N into V, it is a milli-equivalent, normality into volume. Yes? Yes, sir. So in the formula, they also write 1.4 into, this is the volume is in ML and is a normality. It is a number of milli-equivalent of acid, which is taking part in the reaction. Millie-equivalent of acid, which is neutralized. 1.4 into milli-equivalent of acid divided by mass of the organic sample. That way also you can simplify this. It's ending. Okay, the another method of estimation of nitrogen we have, we call it as Duma's method. So in this method, right, Don, we heat organic compound with dry-cupric oxide. We heat organic compounds with dry-cupric oxide. Sir, could you repeat that one? In this method, we heat organic compound with dry-cupric oxide, C-U-O. So if suppose carbon is present in the sample, then it reacts with C-U-O, converts into CO2 and copper. Production of this oxidation of carbon. If any hydrogen is present, C-U-O, it converts into again copper and water. And if nitrogen is present, reacts with C-U-O, converts into N2 and we also get some oxides of nitrogen here, oxides of nitrogen, which is converted to N2, nitrogen converted to, converted to N2 by passing over copper gaze. This gaseous mixture, CO2, N2, right, all these gaseous mixture that you are getting, we collect this gaseous mixture over KOH, right? Or next time, gaseous mixture are collected over, over KOH, why we are using KOH? Because this KOH can dissolve all the gases except N2, right? Down, which dissolves all the gases except nitrogen, which dissolves all the gases, gases except nitrogen. And hence we can find out the amount of nitrogen present here. Okay, how, see, method is this. Now, how to calculate? See, nitrogen does not get dissolved in the KOH sample that you have. So it will accumulate there at the top. See, the arrangement, if you say, I'm not going to draw the entire arrangement, but this KOH is present in a beaker like this, right? In this, we have the KOH, where some arrangement is there, and KOH is present over here, right? And this is, we can easily find out the volume here. Measurement is there. So KOH is present over here, and all these gases that is present, we have the arrangement, with that arrangement, we allow this gas to pass through in this KOH solution like this. Roughly, I'm drawing this diagram. So all the gases will dissolve in this KOH, but since this nitrogen is not dissolved, the nitrogen will accumulate here at the top, correct? Here it will accumulate, right? And then we can find out how you see. Suppose you know the volume of this nitrogen, then what you can say? At STP, 22400 ml of N2 at STP is equals to what mole? One mole, and one mole is what gram? 28 gram of nitrogen, isn't it? Yes. If it is VML, suppose a volume of N2, we'll discuss how to find out the volume. If volume of N2 you are getting VML, then we can easily find out the amount of N2, right? That is 28 by 22400 into V, correct? Now, once you know this volume, you can find out the amount of nitrogen. So percentage amount of nitrogen is what? Percentage of nitrogen is equals to, what was the expression? 28 divided by 22400 into V divided by W, what is W? Into 100. W is the? Mass of organic samples. Mass of samples that we have, okay? Mass of samples. Now, so we are stuck here with the volume of N2. Once you know the volume, you can find out the percentage of nitrogen here. So how do we find out the volume over here, correct? So basically we have two different conditions, okay? One is, you see here, this will exert some pressure, right? This gas will exert some pressure, right? And we are doing this at some temperature, suppose T degrees Celsius, the entire process is taking place, right? And volume of gas we have V1, I'm assuming, N2 gas, which we need to find out. So if you compare, these are two different conditions, one at STP, and one condition is the experimental condition on which the experiment is taking place, right? Okay, so now this pressure of N2, we need to, this is a good pressure of this. If you apply P1 V1 by T1, atmospheric pressure we have in this direction, in this direction, and this compression, in this direction, the pressure of gas is P. So we'll have P plus P is equals to the pressure of N2, okay? So you see, here we'll have the open end here, this side, correct? We are trying to find out the pressure of this nitrogen, which is present over here, this pressure we need to find out, so that we'll have one experimental condition, some pressure, volume, and temperature will be there, at which the experiment is taking place. And we can always equate this with the STP, because the number of moles is same, so we can always equate this with the STP, so we can always write P1 V1 by T1, as it goes to P2 V2 by T2. So P2 V2 T2 we know, the pressure, volume, and temperature we know at STP. Here, we need to find out volume, pressure, and temperature we have, suppose temperature is T degree Celsius, so we can write down the value, because that will be given, right? So pressure of nitrogen here, it is what? The pressure exerted by nitrogen gas is equals to, which I'll write on this data, then you will understand. We have this, you see, experimental condition and the ideal condition, that is the STP we are assuming, just to understand the relation. So here the pressure is P2, we have 760 mmHg, here the temperature is T2, we have 273 Kelvin, volume is V2, and this volume we do not know. Here you see, pressure P1 is the total pressure, minus pressure due to the K-OVET solution that you have, aqua-stension the term we use for this, aqua-stension the term we use. V1, we can find out, okay, it's there in the vessel, so we can find out from that tube, temperature at any temperature, the 273 plus T Kelvin, temperature at which the reaction is taking place. So if you allow the relation here, if you apply this P1 V1 by T1 is equals to, we can write P2 V2 by T2. Now you don't have to worry for this, this will be given in the question. There's something that they would ask you to find out, all the details will be given, total pressure at which the reaction is taking place, the experimental condition this pressure will be given, aqua-stension of the water, the sample K-OVET you have, that will be given. So the pressure of N2 will be total pressure minus the pressure of the liquid, the K-OVET that you have, because K-OVET plus gas, both will exert the pressure, but we are considering only the pressure of nitrogen gas. So obviously the aqua-stension of liquid you need to subtract from the total pressure, correct? So this data will be given, pressure of N2 is this, volume will have that in the tube given temperature is this. If this volume will find out, right? And this volume, it's not like see, you cannot say that STP cell volume is 22.4 liter. We need to know that what number of moles we have, accordingly we can say the volume, right? If we say this is 22.4, if you're having confusion over here, why we are not taking 22.4? Because we do not know the number of moles of nitrogen, that is what we need to find out. So we'll find out V2, and then we can correspondingly, we can find out 22.4 liter is one mole. So this many volume is equals to how many moles? And then we can find out the mass, but that is not required because we just need to know this, we just need to know this volume. This volume will substitute here, and we'll find out the percentage of nitrogen. Did you get it? So V2, you solve from this, the volume of nitrogen that you get is equals to, we have 760 into temperature. What is the temperature we have? We have 273 plus T, T is the temperature at which the experiment is taking place, divided by T1 is the temperature at STP into the pressure we have, total pressure, I'll write on PT here, minus the aqua extension that we have, PT. All these values substitute, you'll get V2, V2 you substitute here, you'll get percentage of N2. Did you get it? If you want to see this, the pictorial diagram, I'll show you, just a second. So shouldn't V2 be the other way around? I've written it down, one by V2, okay? Yes, no? Heart T1 by PT, so I'll show you the diagram here, you see. So this volume we are getting from the experiment, just one second, just give me a second. Okay, so I'm getting some error, I'll show you here. You see this image, okay. So you see this, this is KOH, all these reactions takes place here in this vessel, CO2 will pass through, and this is the KOH we have, which can absorb CO2 which is present, the gas and species. All these will allow to pass through in this sample, CO2 will get dissolved, NO2 will accumulate over here. So with this reading, we can find out the volume of nitrogen which is present here, and this volume will be given in the question. Correct? This is the entire setup of this method we have, Duma's method. This question also they have asked many times that what we use in this process in order to absorb CO2, we use potassium hydroxide, okay? We'll do some questions on this, you see. Okay, this you don't have to draw. Just ignore this, yeah. The question is this, this one you try, both you see. Carrier method you let it be, we'll discuss carrier method and then we'll do this question. Question number six you try, done? Okay, what is the answer you got? W is given, yeah, D is correct. All of you have got D? So amount of acid absorbed is this 20 into one, 0.1, and the excess of acid required is this, any of which, this one amount of acid is also required. So amount that is reactive is two minus, we'll get 15 into this, that is 1.5, 0.5. 0.5 is the amount of acid or NH3 we can say which is required, right? So that 14 into, 0.5 is the amount of acid 14 into 0.5, 0.5 is N into V minus V. So 14 into 0.5 divided by W, right? W is 29.5 milligram, 10 to the minus three. Okay, see, I'll do this. See, evad ammonia was absorbed by 20 ml of 0.1 molar HCl. So first of all, it is HCl is monovalent. So it's molarity and normality, everything is equal, right? So milli-equivalent of acid absorbed, what we write? Absorbed, this would be 20 into 0.1, two milli-equivalent. Next is the milli-equivalent, the excess of acid required, this NaOH solution for complete utilization means the equivalent of acid must be equals to the equivalent of acid which is required to neutralize this. So we can directly write milli-equivalent of acid or excess acid or acid required, anything you can write, this would be equals to 15 into 0.1, that is 1.5 milli-equivalent. Again you see. Here it's 0.1 molar, not molar. It's a misprint. If molality is given, then density must be given, no? Without density, see, molality, you need to find out equivalent. So for equivalent, we require volume, we require what? We require volume. So if molality is given and density is not given, so you cannot convert this into molarity, are you getting it? If molality is correct, then density must be given, which is not given in the question. So I'm assuming this has molarity, there's a misprint you can see. If suppose molality is given and density is given, then we'll first first we'll convert this molality into molarity and then we have base city is one, acidity is one for this base, so we can find out normality accordingly. Means whatever molarity you get, that would be the normality and then you can find out the milli-equivalent of acid. So it is understood in this question, density is not given, so that should be molarity, not molality, okay? So this is the milli-equivalent of acid required. So this is the N into V initial walla, N into V final walla. The number of milli-equivalent here is two minus 1.5, so the percent is nitrogen would be, is it nitrogen? Nitrogen would be, we have this formula, if you remember, 14 into N into V minus small v divided by 1000 into W into 100. This is what we had discussed. So this would be 1.4 if you solve this and N into V minus V is nothing but the difference of this two, if you see, basically a more concept we are applying, two minus 1.5 because N into V is the number of milli-equivalent, number of milli-equivalent, two minus 0.15 is 0.5 divided by W is 29.5 milligram we have, 10 to the power minus three, this would be yours. This is what you did? Yes, sir. Yes, sir. Calculation is pretty easy. Just need to use the concept of the first chapter more concept, you'll get the answer from there. Okay, now the next is detection of halogen. So could you give one more problem? Carriers walla we'll see, we'll first finish the two and then whatever time is left we'll do problems on that. Okay? Okay, sir. So next is detection of halogen. Detection of halogen is the same thing, like I said, the sodium extract that you are getting, from that you need to change the method in order to identify halogen present or not. For nitrogen we have that particular method that we use, for halogen we use different thing. So sodium extract that you get here, this sodium extract, NA extract, this we use to allow this to react with halogen. Right, halogen, fusion takes place here and it converts into sodium halide, NAx. Now this sodium halide, we allow this to react with AGNO3, depending upon the result, we can conclude whether it is chloride, bromide or ionide. Okay, what is the, no, what is the reaction we have? So what we do, if it is NaCl, NaCl plus AGNO3, AGNO3, it converts into AGCl and NaNO3. This is the compound we get. Now this AGCl, it is white precipitate, PPT. It is white precipitate. And this is soluble in NH4OH, ammonium hydroxide. And this soluble in NH4OH, ammonium hydroxide, which indicates the presence of chloride, indicates the presence of, if it is bromide, NABR, plus AGNO3, we get AGBR plus NaNO3. The color of AGBR is dull yellow, dull yellow PPT we get. This is also soluble in NH4OH, in NH4OH. But this one is more soluble. So we write it as partly soluble. And color is also different. That indicates the presence of bromide. NAI, reaction you write down, NAI gives AGI over here. And AGI, the color is bright yellow and it is insoluble in NH4OH. Bright yellow, insoluble in NH4OH. So I did not write. This is how the direction of halogens we have, okay? Now estimation part of this, estimation is done by, is done by carrier method. So these three method only you need to know. Mostly I have seen in the past seven to eight years. They have asked only these three questions. Question based on gel dial, dumals, carriers, okay? These three methods only you need to understand. Carriers method is just a tube we have, which we call it as carriers tube, in that only all these reactions takes place. Very simple calculation we have here. Mass of organic compound is W. Mass of organic compound is W, gram we are assuming. The silver halide that forms, AGX, we are assuming this as X gram. So in AGX, what is the mass of halogen atom? Is it atomic mass? Okay, so AGX may, we'll have suppose M gram of halogen. M is the atomic mass. M is the atomic mass, okay? So AGX will have M gram. What is the mass of, okay, AGX we have this. So the mass of AGX, I should write down here, the mass of this, which is 108 plus M. Can we say this? This is the mass of silver halide we have. So in this mass, in this mass, the mass of halogen present is this. Can we say that? So whatever mass of AGX you get, which is X over here, so in X gram, what is the mass of halogen present? It is M by 108 plus M into X, this gram, yes, easy. So percentage of halogen is what? So that by W into 100. Yes, so it is M by MX by 108 plus M into W by 100 by W. This is the percentage of halogen. Should I go back? Sir, can you like go back to the previous answer? Yeah, just a second. So this is the, you see this question? Question number seven. Yes, sir. Okay. So all of you are getting option A. Yeah, that's right. So simple one, 250 milligram of an organic compound and 141 milligram of AGPR. So everything we know, W is 250 milligram, 141 is the milligram of AGPR, so X is given. So we know M, what is M? Mass of bromine is given, 80. So we have 80 into X is 141 milligrams into the power minus three divided by 80 plus 108, 188 into 100 and W is 250 into 10 to the power minus three into 100, yes, this is the expression, no? Yes. So this is how we can find out the percentage of halogen present in this. Now in the last one we have the estimation or the detection of sulfur here. It's also very much similar, see this? So in sulfur, what happens? Write down the heading, detection of sulfur. Write down just one point in this, write down a known mass of organic compound is heated with HNO3 and sodium peroxide, NNO2, HNO3 and sodium peroxide, NNO2, O2 in presence of BACL2, in presence of BACL2, correct? This is done in carrier steel, right? In this reaction, sulfur is oxidized to, sulfur is oxidized to H2SO4 and precipitated as BASO4, which is filtered and dried out. So precipitated BASO4 forms, right? So we'll filter it out, dry and we'll wait it. So we'll get the mass of BASO4. So method is simple, reaction is what? Given organic compound is mixed with HNO3 and sodium peroxide, NNO2, O2 in presence of BACL2. So the sulfur present in this sample, it oxidizes into H2SO4, which further precipitates in BASO4. So first, H2SO4 and then BASO4, HCl will go out. This will filter out and dry, so we'll get the weight of this component here. You can find out from this. Okay, so what is the molecular mass of BASO4? Suppose M is the molecular mass of BASO4, molecular mass of BASO4, correct? So in M, the mass of sulfur is present is the, what we say, atomic mass of sulfur, small ML right now. Is the atomic mass of sulfur? One sulfur atom we have atomic mass. So whatever weight we get over here of BASO4, we can say this weight is M dash. So in M dash, the amount of sulfur or the mass of sulfur we have M by molecular mass of BASO4, just to avoid the confusion into M dash. This is the experimental mass. 233. Yes, sir, 233. 233, okay. Yes, sir. So the atomic mass of sulfur is 32. So we can substitute here. It would be 32 divided by 233 into M dash, the mass of BASO4 that we get here. So percentage weight is what? Percentage of sulfur is this mass, 32 divided by 233 into M dash into 100 by W. This is a formula. So this is what we have for a test of the element present in the organic sample plus estimation what we can do. For JMAIN's portion, till here, you can say it is important. The next part of this, we will start in next class. We'll discuss, there are so many theories, we'll discuss important things there, right? But that part is not that important for JMAIN point of view. If you want, you can do it once. And afterwards, you should focus this part on, the later part that we are going to start. After JMAIN's, you can focus on this, right? Let's see this question first. One question we'll discuss. The question is, we have 0.4 gram of an organic compound. Organic compound, 0.4 gram. It is treated according to Jeldahl method, okay? Treated by the ammonia evolve I'll write it down the exact sentence, okay? The ammonia evolved, ammonia evolved was absorbed by, or absorbed in, in 50 ml of 0.5 molar H3PO4, H3PO4. The residual acid, residual acid require 30 ml of, 0.5 molar CAOH whole twice. Find percentage of N2 in the sample. Try this. Then N factor you must use. No, it's two less. 0.15, 0.75, it's two less for you. 29.1, 52.5, no, no, no. Yes, Akshit got the right answer. 70% is correct. See, this is the acid H3PO4 we are using. What is the N factor of H3PO4? Three. Three. N factor of this, CAOH whole twice? Just a second, guys. N factor is two. So what we can write, the melee equivalent of H3PO4, melee equivalent of acid which we are using initially, H3PO4 is 50 ml into molarity into its N factor. How much it is? 75, I guess. Is it 75? 10? 75, no. I think then the answer will also be changed. That is mistake. Okay. Let's see what is the answer. Just a second, 75. Okay. What is the melee equivalent of CAOH whole twice? Because the equal amount of acid also will be used, right? Of CAOH whole twice, this would be a 0.5 into two into two, correct? So we are getting 20. Is it 20? We have to... It's 20. 30 ml, we have no. 30 ml. 30 ml. This two is not there. Correct? Yes. So we have 30, yeah. 30 ml. 30 ml, I'm sorry. Okay. So this term Nv minus V is 75 minus 30, that is 45. So percentage nitrogen would be, I'll write down directly 1.4 into 45 divided by what? 0.4, correct? Is this correct? It was called 0.5. Yes, sir, that's what I'm getting. So what is the answer you are getting like this? So you get 35 into 300. 315 by 2, is it? Yes, sir, 315 by 2. How this value we are getting? We are getting more than 100, no, it's not possible. Are we getting more than 100? Yes, sir. Yes, sir. If it is, I think then there is a mistake in the question. It is S3PO4, okay. Consider this as S3PO3 then. Okay, I think there's a correction in the question. If it is S3PO3, then this would be two and this would be 50, correct? This would be 20. Then what is the answer we get? 1.4 into 20 divided by 0.4. What is this value? It's the same. They're getting 60 is something, correct? This is 70. 70, exactly. Yeah, I think there's a correction in that question. That's why the answer is given 70 only. I think there's a correction in that question because more than 100 is not possible, no. So it should be S3PO3, correct, see this. More than 100 is not possible. Basically concept, I think you understood how to do this. Many equivalent of, you know, as it will find out and then we'll find out the answer directly. So this is it for this chapter. Okay, half part we have done. The rest we'll start in the next class and we'll finish most probably. Okay guys, thank you. See you in the next class, take care, bye. Thank you, sir. Okay, sir.