Mole concets (NEET PYQs)

Indiziert: 2026-05-15

[00:00:00]It will come in this. Correct.

[00:00:08]Yes. So dear students, today we are going to talk about NEET previous year questions.

[00:00:14]And this time that we have got right now, yes, it is fine.

[00:00:21]We will use this time that we have got. Until any new information about NEET comes, we are going to prepare ourselves better.

[00:00:29]We will try to solve all the previous year questions and from today we are going to start solving previous year questions of each chapter daily, as many as we can. All the children who are joining the live, they must send their name in the comment box and in the chat box and if you have any doubt, any question is coming again, you want to ask, then you can ask that too.

[00:00:58]Ok? So those who are joining the class, getting added, they should send their names so that I can also know which students are present in the class. And today what we are going to start with is we are going to start with questions on mole concept. The basic concept is such a topic that many children are afraid of it, but you see that questions come and such questions come which we can solve very easily.

[00:01:25]If we use our brain a little, then the questions revolve around NCERT and if we talk about the level of NEET, then the questions revolve around NCERT and we can approach it very easily.

[00:01:38]So please also tell me whether the voices of those present are coming out properly?

[00:01:59]And those who are present can also send their names to me. The live has started.

[00:02:13]So now let us talk about the previous year questions of NEET.

[00:02:16]Okay, yes, okay, okay, lock, Akash, Kakad, okay, okay, okay, okay, sound, okay, Akash, everything is fine, visibility, sound, everything is fine, let's start now. So see, basically the questions that have been coming, like the first question which was asked in NEET was at STP ok ok ok, it is fine. At STP the density of CCL4 vapors in grams per liter will be nearest to. They have asked you what will be the density at STP?

[00:03:15]liters per gram of CCL4 vapors. So look if we talk about One Mole CCL4 vapors. If we talk about one mole, read the question again for a second. They are asking what is the density of CCL4 vapors in grams per liter at STP. What have you been asked? Density. See, if we talk about one mole of CCL4 vapors, then what will be the molecular mass of one mole? First of all, CCL4. If we look at the molecular mass, what is 12 + chlorine, brother? 30 35.5 * 4 + 12 When you calculate this, it will come to approximately 154 grams and what is the volume of one mole at STP? 22.4 litres. You have read that if we talk about one mole, then what is the volume of one mole at STP? 22.4 litres. So now you see from here we have been asked that at SP the density of CCL for vapors in grams per liter. So let us do one thing. From here we can calculate the density.

[00:04:55]What is density brother? Weight by Volume. A simple formula has to be applied.

[00:05:01]What is density? Weight by Volume.

[00:05:04]And in which do we have to take the weight?

[00:05:06]In the village. And what should the volume be taken in?

[00:05:09]What will we do here in liters? 154 / 22.4 and when you calculate this then your answer will be 6.87, you have to do the calculation. I have said this earlier also that whenever we take classes, you have to do the calculation part because you do it in the exam also. I am going to tell you how to approach the question. The second question is, we have a mixture of gases, what do we have in it?

[00:05:43]One second in which we have H2, O2 and what is the ratio given to us? 1:4 ratio is given and how is that given? Weight by Weight W / W So what is the molar ratio of the two gases in the mixture? Molar ratio in the mixture has been asked. So look brother, we are given in the question that how much is WH2 / WO2? This is 1/4. Weight wise ratio is given.

[00:06:15]Now what do we have to do in this? We have to find their ratio mole wise. So what do we have to remove from them? Number of moles. And what is the formula for number of moles? If we look here, we write NH2 NO2, what is the number of moles, brother? Given mass upon molecular mass. So if the mass of H2 is one, what is the molecular mass of H2, brother?

[00:06:47]What will be the mass of two O2? 4 / What is molecular mass? 32 So from here, when you solve this, what value will you get?

[00:07:03]This will become 1/2 1/8. How much will this cost?

[00:07:07]4:1 Clear? So what will be the ratio? 4:1 What is your answer? Are you getting the point?

[00:07:17]These types of questions come. Third question if we see what volume of oxygen gas measured at 0 degree Celsius and 1 atm is needed to burn completely 1 litre of propane gas measured under the same conditions.

[00:07:43]Now if we look at this question.

[00:07:46]Now if we look here carefully, read it again. What is the question given? What volume of oxygen gas measured at 0° C 1 atm pressure is needed to burn completely 1 litre of propane gas. Now see, if you write the reaction of combustion of propane gas, then see what reaction will we write?

[00:08:11]Questions about combustion reactions come from stoichiometry. Keep in mind, what method have I told you to write the reaction of Combus, that make as much carbon as there is, carbon dioxide plus half of the hydrogen, you make water, so what happens is four, you have H2O, now you balance the oxygen, 3 * 2 = 6 and 4 ≠ 10 means what happens here is 5O2.

[00:08:41]Now if we read the question, it is saying how much volume of oxygen gas measured at 0° Celsius, 1 atm is needed to burn completely 1 litre of propane gas under the same conditions. So see, if we look at the mass of the equation that we have written here, from this equation we need 5 litres of oxygen.

[00:09:09]If we look at it according to the law of Galvak law and if we look at it according to the Galvak law, then we understand that 5 litres of oxygen is required for 1 litre of propane. So what will be the answer from here?

[00:09:25]5 liters. What will be the answer for fourth? C. Did you understand the question?

[00:09:31]One question is missing in this. Question number three. We are doing this also first.

[00:09:38]Now see what is there in the third question? If you want any question repeated or want to ask it again, you can ask. Now we are talking about question number three. If the Alvedro number changes from 6.022 * 10 to the power of 23 to 6.00 * 10 to the power of 20, it will become Alvedro number.

[00:10:02]So they are asking whether there will be a change?

[00:10:04]So if we look at this and if we see the important thing in it, which is the mass of 1 mole of carbon, that will change. Because if we see, if you change 6.022 to the power of 10 23, then the mass of 1 mole of carbon will change, how much will it become from 12 grams to 12 milligrams, so what will be the answer to this important thing, the rest of the units like gram etc. will not change, the ratio of chemical species will not change in the balance equation, the ratio of elements to each other in a compound will also not change, but what is the mass of one mole that we read now? 12 grams. And if you make this change, how much will it cost? 12 mg. So it will be a change.

[00:10:53]Got it?

[00:10:56]After this, the next question we have is fifth.

[00:11:02]Fifthly, we have given the molecular weight of O2 and SO2 are given at 15° Celsius and 150 mm pressure. We have given that 1 litre of O2 contains N molecules capital N molecules. So, what is the number of molecules we have in 2 litres of SO2 under the same condition of temperature and pressure. So see what will you do in this now? If we talk about the number of molecules of water here or if we look here, what am I talking about if we don't have water?

[00:11:48]No water. If we talk here according to Avogadro's Law, if we go here then see what Avogadro's Law tells us? That equal volumes of all gases contain equal numbers of molecules. If we have the same temperature and pressure, what will be the equal number of equal volumes? Molecule. So if there are n molecules of one gas in 1 litre, then how many will there be in 2 litres? 2n, so brother, what will be the answer for fifth? If there are n in 1 litre, how many will there be in 2 litres?

[00:12:20]What will be the answer to 2n and the fifth? C Clear Got It?

[00:12:30]Now let us talk about question number six.

[00:12:34]You see six. The number of water molecules is maximum in. These questions come up a lot. You will ask the number of water molecules. We will ask you the number of atoms.

[00:12:45]Who has more atoms? Whose atoms are less?

[00:12:48]So if we look here at the number of water molecules. And the first one we've got is 18 molecules. The second one is given as 1.8 grams. What needs to be removed first? Number of moles. So 1.8/18 * Na is 18 grams of water in it. So you are finding 18/18 * Na number of molecules. And what are you using along with it? Given Mass / Molecular Mass.

[00:13:16]Given here 18 moles. So this will be directly 18 * Na. So now you see, the maximum value that is coming from these is that those who are present live, please send your name because some children have sent messages that the online class is not running, those who are connected live, please let me know because some children have just sent messages that perhaps some issues are coming up regarding the class. The class is not running live.

[00:14:14]I am also checking it. Is the class going on smoothly or are any issues coming up?

[00:14:47]Those who are present live, please tell us. I am also checking. Rest I am checking once because there were messages from some children on WhatsApp that the class is not going on live.

[00:15:21]All ok ok, okay then let's start the next question, if all is ok then let's start the class, very good, see what we have to find out here children, number of moles and how many moles will be there from this number of moles? 1 mole. How many moles will it cost? 0 points. Who has the most value? D K. So there are 18 moles. So which one has the most particles? 18 * Na.

[00:16:03]So when you guys solve this, please solve 18 * NA.

[00:16:45]Ok? So, these must be your maximum particles. When you multiply 18 * Na, what answer will you get for six? D. Now let us come to the next question. The next question we have is question number seven. See, in this he has given that there is an element x, the following isotopic composition is given to you. The percentage composition of these isotopes has been given and the average atomic mass has been asked.

[00:17:15]If you open NCERT, then near atomic mass in NCERT they have written average atomic mass. How do you calculate average atomic mass? So it has a simple formula.

[00:17:28]Mass of isotope one * its percent abundance plus mass of isotope two * its percent abundance and so on divided by total abundance. So if we try to solve this here.

[00:17:44]If we find its average atomic mass, then see what we will do here? First of all, what is its atomic mass given to us?

[00:17:59]How much have you given in 200 abundance? 90 plus 199 * How much is Abundance? Ate, I am writing it again. I did not come. I have arrived. Ok. Similarly, how much abundance has been given plus 202? 2% whatever the percentage is, divide it by the total percentage, what will be the sum of 100, when you solve this, what will you get, the average atomic mass, so the average atomic mass that will come in seventh will be approximately 200 am 199 something, when you solve it accurately, then what will be the approximate 200 am question number 8, if we see, it is counted among the good questions.

[00:19:01]If we talk about question number 8, see, specific volume is given. A cylindrical virus particle is 6.02 * 10 to the power -2cc in grams with a radius of 10 angstro and a length of 10 angstro. The values ​​of Na have been given, the value is given. And what have you been asked? Molecular weight of the virus. So if we look at question number eight, specific volume means volume of 1 gram. You have a given in the question. So what is the specific volume word given, basically? Volume of 1 gram of cylindrical virus. This is given to you 6.02 * 10 to the power of -2, I am writing it again.

[00:20:03]So the specific volume is the volume of 1 gram of cylindrical wires.

[00:20:14]How much have you given me, children? 6.02 * grams to the power of 10 -2 cc. This is given to you which is the volume of 1 gram of virus.

[00:20:27]Now the radius is also given to you, the length is also given. So we have to remove the volume of virus.

[00:20:36]This becomes πr² l volume of virus, how much will it be when you extract it? πr² l So do you know the value of pi? 22 / 7 Radius If you look at the question, how much is given to you? 7 Angstro means this will be 7 * 10 to the power - 8 cm.

[00:21:06]And in the same way, what is the length?

[00:21:10]10 angstroms. So what will happen to this also?

[00:21:12]10 * 10 to the power of -8 centimeters.

[00:21:16]What will you take away from this? Volume. So what will you get from here? Volume.

[00:21:23]Once you calculate the volume, you will calculate the weight of one virus particle.

[00:21:29]What will happen to the weight of one virus particle, brother?

[00:21:35]So you will divide the volume of virus that has come to you by the specific volume.

[00:21:45]So the volumes of both will cancel out. And where will you get the answer?

[00:21:51]In grams. Do it, calculate and tell me the answer, everyone, what was your answer?

[00:21:57]Reply quickly after finding and calculating the answer.

[00:22:45]Everyone who is connected to Jo Jo Live.

[00:22:48]Yes, this answer has been sent by some child Sumit.

[00:22:52]What is Sumit's answer to this? D. So, Sumit gave the correct answer to the question about eight. What will be the answer to this? D.

[00:23:04]15.4 kg per mole. Ok? Ok. This is the correct answer. Very good Sumit. Now let's come to the next question. Let the rest of the people also keep doing it.

[00:23:15]Now let's come to question number nine. See, in nine again there is a question about average atomic mass. will that do. In the same way, it was just explained that its mass of isotope one into its percent abundance plus mass of isotope one into its percent abundance and divided by the total abundance. So whatever expression you will make here, I am writing it here also, the average atomic mass, brother, will be 10 * 19 + 11 * 81 / 100, clear?

[00:23:58]So in this way your question will be solved. The question is visible, proper visibility is proper.

[00:24:06]After this comes the next question. The next question is question number 10.

[00:24:12]You were asked about the highest number of helium atoms, see, I had said that such questions come. Yes, this is a direct formula based question. Yes absolutely Sumit and such questions come. So the focus has to be on not letting the formula based questions that come directly go wrong.

[00:24:29]They must be right for us.

[00:24:31]That's why I am making you practice the questions.

[00:24:33]So that the question gets resolved once and for all. The highest number of helium atoms is the same as 4 moles, so 4 * na.

[00:24:45]Now see, the four unified masses of helium are given.

[00:24:48]Then given is 4 grams of helium. So if we look here in 10ths, how many are there in 4 grams of helium? One mole. So what would 4 moles of helium be here? The highest.

[00:25:01]What will be the highest? 4 moles of helium.

[00:25:06]Yes. Ok. Piyush.

[00:25:09]Ok. 10th will have A. That's perfect.

[00:25:13]Now let us come to question number 11.

[00:25:25]What is the weight of one mole of carbon atoms given below? 12 grams. The number of atoms in it is equal to two. So brother, I have directly asked you the definition of price.

[00:25:37]What did you ask so directly in 11th?

[00:25:40]What will be the answer to the definition of mole? A Because what is the definition of mole?

[00:25:47]What is the mass of one mole? Amount of any substance which contains at many particles as there are number of atoms present in 12 grams of C12 isotope.

[00:26:00]Yes, of course.

[00:26:02]After this we come to question number 12.

[00:26:06]Maximum number of atoms is back. So what should you do whenever the number of atoms comes? The number of moles has to be calculated. First, multiply the mole by the Alberro number and then multiply it by the number of atoms of that particular type in a molecule or the total, whichever is in question.

[00:26:25]Like here is magnesium so what will happen into? Forest. But if for O2 I do N * Na and how many oxygen atoms will there be here? Two.

[00:26:36]Or if there is lithium, there will be forest here. Here there will be a multiplication of one. So you will calculate all this and see which answer is coming. After this the question number is 13.

[00:26:52]The number of moles of hydrogen molecules required to produce 20 moles of ammonia through the Haber process.

[00:27:01]I think all of you must remember Haber Process. Those who are preparing for NEET, it is expected from you that you must remember Haber Process.

[00:27:10]What is the process? n2 + 32 gives 2NH3.

[00:27:17]This is Haber Process. Now in Haber Process, they are talking about number of moles of hydrogen molecules. They have asked for producing 20 moles of ammonia. So look, if we talk about question number 13, if we have to produce one mole of ammonia, if I look here, how much ammonia is there? Two moles. If I want to produce one mole of ammonia, how many hydrogens do I need to produce 1 mole of ammonia? 3/2 mole of hydrogen. Did you understand? Where did I do it from?

[00:27:53]And you see it here? Yes, okay.

[00:27:56]So how many do we need here? 3/2 mole of hydrogen to 1 mole of ammonia. But what the question is telling us is how many moles of ammonia are there? 20 moles of ammonia. So brother, how much will it be?

[00:28:11]Yes, stoichiometry will happen. Simple Piyush suddenly became * 20 30. Ok? So these types of questions also come. If we remove this.

[00:28:28]Now let's come to the next question.

[00:28:33]Next is in v case the number of same same question. What is the maximum number of molecules in it? So you will solve this. I have told you the method. It contains 18 ml water.

[00:28:47]How to find the number of moles of water if its volume is given? So the given volume is at STP divided by 22.4, so from here you will take it in ml, 22400, so when you solve question number 14, then mass or you can take it directly from here, 18 ml is given, STP is not given, if STP was given then you would have gone by that approach, it is given normal, what is the density of water, you would have taken approximate one, we have 18 ml, so what is 18, grams and if it is 18 grams then how many moles are there? Forest. And if there is one mole then how many molecules are there brother?

[00:29:29]What would be the molecules in Na and all the rest? will be less than Na. If you look at this also, there will be less here also.

[00:29:35]What is the ultimate answer?

[00:29:37]A of 14. clear?

[00:29:41]Now let us come to question number 15. Suppose there are two elements. One is x and one is y. When these are combined, two compounds are formed.

[00:29:51]x y2 and x3 y2 They are saying that when what is the weight of 0.1 mole xy2? 10 grams.

[00:30:03]And what is the weight of 0.05 mole x3y2? 9 grams. So whose atomic weight do you want to tell? of x and y. So how will you do question number 15? So what is the number of moles? Given Mass / Molecular Mass. What is the number of moles brother? Given Mass / Molecular Mass. Let us consider the mass of x. x is.

[00:30:36]What value of y did we take?

[00:30:38]y So brother, if we look at the first one, it is 0.1 mole given.

[00:30:42]Whose given is it? of xy2. So what is the number of moles?

[00:30:50]How much is the given month?

[00:30:54]What will be its molecular mass in 10 upon brother?

[00:30:58]x + 2y clear?

[00:31:02]Make another equation in the same way. 0.05 mole is given. It is 10 grams. It's not by 10 brother. Its 9 grams 9 grams by x3/2 3x + 2y two equations. There are two unknowns.

[00:31:23]Solve it quickly and after solving it, tell me what values ​​of x and y are coming in it? Do 15 quickly. There is mathematics, there is calculation. Tell me what are the values ​​of x and y coming out to be?

[00:31:39]Question number 15 Everyone please hurry up who is present?

[00:32:01]Salim Khatun, you children, just keep your mind calm during this time and first of all do not panic, whatever has happened has happened, whatever is happening is happening for everyone, you just have to practice new questions, Salim Salima Salima Hatun, you children, just have to practice questions, just nothing new, if there are any topics which you feel you need to do better in, you can work on them. So, in the 30 days that you have left, practice only questions.

[00:32:38]Ask as many doubts as possible. Ask questions.

[00:32:41]Solve all the previous year questions.

[00:32:44]And you should keep your main focus on Biology and Chemistry.

[00:32:49]Do previous year questions Salima. As many as you can. We will take this class in TTS. This class will be conducted in TS and per day we are going to do questions on one topic. This class will be held daily at around 5.15 in TS.

[00:33:04]So you can join this also. Yes Sumit has sent the answer.

[00:33:12]Sumit, which answer did you send? B 30 20 and send some quick answers.

[00:33:21]Yes.

[00:33:35]Question number 15 Salima, please ask the question with us and send the answers quickly.

[00:33:45]Question number 15 Sumit, there is a correction in it. Please check your answers again in question number 15. The calculation is mathematics, there is some mistake in the calculation.

[00:34:03]Do question number 15 and as I told you, you have made these two equations.

[00:34:10]From here, find the values ​​of x and y. The value of x and y is C C is the right answer Piyush C is the right answer is absolutely correct. C is the right answer. So brother, do the calculation. I am speaking again.

[00:34:34]Physical paper will come. Calculations will come. She came this time also. Then you will be worried. You will have to leave the question. So keep doing the calculations. Do not be light on calculations.

[00:34:44]Yes CC, that's absolutely right Piyush. very good.

[00:34:47]Let's move on to the next question. Now the next question is question number 16.

[00:34:59]16 is the V check maximum number of molecules.

[00:35:02]Same question happened. We have asked many questions.

[00:35:04]You will do it. What needs to be done?

[00:35:06]Given mass up molecular mass.

[00:35:16]Find the number of moles from the given mass up molecular mass and directly ask for the molecule, if it is not asked for the atom then what should be done into it, na, yes Sumit, you should not make this slippery mistake, take special care of it, yes, then we will do this in this, this is an easy question, ok ok Salima, now let's come to the next question, this is everyone's given mass up molecular mass *, na, now let's come to question number 17. See, in 17 he is talking about the number of atoms in 0.1 mole of triatomic gas.

[00:35:59]He said triatomic gas. So tri atomic means that the gas is some A3 type gas. Meaning how many atoms are there in it? There are three atoms. They may be the same or they may be different. It will depend.

[00:36:14]But we are assuming that there are three atoms. It is a triatomic gas and the number of atoms is asked. So what will you do? Mole is given to you. I told you what is multiplied in the mole? of the Agodro number. And into which one will you multiply? Number of atoms. Calculate it. Your answer will come.

[00:36:33]You have to do the calculations and you must practice them. Don't shy away from calculations.

[00:36:38]So the answer to question 17 will be B. That's perfect. Okay Piyush Salima very good. Question number: What is its answer? B.

[00:36:51]Now let us come to question number 18. Volume Occupied by One Molecule of Water. The density of water has been given to you as 18. And if density has been given, how much has it been given?

[00:37:05]Forest. And since you have asked about volume, what is density equal to brother? Mass Up Vol.

[00:37:14]Simple what will we do? What will the volume be? Mass/Density.

[00:37:21]So volume occupied by one molecule of water if we were asked.

[00:37:27]So what does volume equals to? Mass up density.

[00:37:32]So how much mass is there? 18 Now this question of yours will be wrong for some children who will not use their brain, this is a good question, a small mistake but the question itself will be wrong, brother, if you are keeping the mass as 18 grams then it will be the volume of 1 mole, what is 18 grams of water per mole, so we have kept 18 here but this volume that we have got, what is the volume of this per mole. So from here, this 18 ml volume that you got or this 18 cm volume that you got, how did this volume come?

[00:38:17]But mole. But he did not ask for the price.

[00:38:21]What you have been asked is the volume of one molecule of water.

[00:38:26]So what do you do to extract one molecule? Now you do here that volume of one molecule of water.

[00:38:38]So brother, if one molecule is of water.

[00:38:42]How much is this volume? Of one mole. Yes, CC means mole only. It is ml. cc means ml milliliter. So 18 cm. It's done but at a price. With whom should we divide this?

[00:38:54]from the Agodro number. With whom should we divide it?

[00:39:02]Divide the mole by 18 cm by the Odedro number 6.022 * 10 to the power 23, the answer will come from here, did you understand, where will the answer come from, from here will come the volume of one molecule of water, yes divided by Na, very good Salima, by which should it be divided, by Na, by Odedro number, yes Piyush Piyush's answer has also come, what is the answer to the 18th? B will come. But please check its answer calculation again. What's coming? Is B coming?

[00:39:43]Check the calculation once.

[00:39:47]After this let us move on to the next question.

[00:39:49]Again that's the maximum number of. Now see, whatever question will come in moles in your NEET paper, it will definitely revolve around the number of atoms, okay, yes C is the answer, okay, then which add maximum number of molecules, then the same thing has to be done, can we calculate moles by one molecule and then calculate its mass, see what will happen, Sumit? If you calculate moles from one molecule then that method will not be your length as you are taking the number of molecules. One molecule divided by 6.022 * 10 to the power of 23 will increase the calculation error. So the easy method has to be approached. The answer comes in both ways.

[00:40:43]Now if you find that easy then you can go with that also. But the main thing is which calculations we have to reduce.

[00:40:51]Keep this in mind that it is a power hour, so generally you Bio students are understanding that mistakes happen, so you should not make mistakes, otherwise if you have practice then you can learn from that, now let's come to the next question, which change is the maximum number of molecules, you will definitely do this question, I am not telling you the maximum number of molecules, it will only waste time in this, we have done this many times, now let's come to question number 21, good question assuming fully decomposed, what have you given us in 21, fully decomposed? The Volume of CO2 Released at HTP on Heating Barium Carbonate. So, when you are decomposing barium carbonate for the first time, you will have to write its reaction. And what will be its reaction, son? What will happen to the BaO+ CO2 reaction? BaO + CO2 Now read the question. Fully decomposed. The volume of CO2 released on heating at HTP is asked to us 9.85 grams. The atomic mass of barium is also given.

[00:42:06]Stoichiometry is the best question. This Devil Sahib, the answer for 12 is three. Child, which answer have you sent? Please tell me about 21 once.

[00:42:25]Yes. So now see, when you solve on 21st, yes the third answer will come. Yours is absolutely correct.

[00:42:32]But what do you have to do? First of all you should understand the method. I am talking to those who have not been able to do stoichiometry.

[00:42:41]What to do first?

[00:42:45]What can you calculate for barium carbonate? Molecular mass.

[00:42:47]Because stoichiometry tells you how many moles of a substance are being produced from one mole of it? 1 mole. We will use stoichiometry.

[00:42:57]How many moles of CO2 are produced from 1 mole of barium carbonate? 1 mole. So when you calculate the molecular mass of barium carbonate, it will come out to be approximately 197 grams. So if we see from here that 1 mole of it is being produced from 197 grams and what is the volume of 1 mole of any gas at STP? 22.4 litres. This is what has been asked of us in terms of volume.

[00:43:23]So how much will it be for one? 22.4 /197 and 9.85. So the expression will be 22.4 / 197 * 9.85, you have to calculate this. I am writing it here clearly so that you can do it 22.4 * 9.85 / 197, this will be your answer.

[00:43:56]So the question of stoichiometry is easy.

[00:43:58]Catching that Stychometry Questions.

[00:44:04]okay brother? Let us now move on to the next question.

[00:44:09]Next question is 22.

[00:44:12]Look, read question number 22.

[00:44:19]What is 22 saying to us? Hemoglobin contains 0.33 4% of iron by weight. Iron is given.

[00:44:28]How much is iron by weight? 0.334%.

[00:44:30]The molecular weight of hemoglobin is approximately given. Number of iron atoms present in one molecule of hemoglobin. So, if we talk about question number 22, what can we do if we want to find the number of iron atoms? Molecular weight of hemoglobin * Percentage of iron. What can we do? We can write here. If you want, you can write down the number of iron atoms and note down how you will calculate it. The simple formula will be molecular weight of haemoglobin * percentage of iron, whatever is given to you in the question, divided by 100 * atomic weight of iron.

[00:45:24]If you use this formula here, then if I use this formula, then what is the mass of molecular weight of hemoglobin given to us?

[00:45:33]How much percentage of iron is given to us? 334 / 100 * What is the atomic weight of iron? 56 If you don't know then it is given in the question.

[00:45:44]You solve this. From here you will get its number of iron atoms. Do it quickly.

[00:45:59]And this is not only about iron, you can ask this about any medicine, any other medicine also because the question does come and it is not necessary that the question will come only on iron. A question related to some polymer or some medicine based question came.

[00:46:23]Ok? Done. Did you get the answer?

[00:46:26]Tell me the answer to this quickly about the iron one. 22 of c. Ok.

[00:46:30]Ok. very good. Four will come in this. Let's come to the next question on 23. Look, let's make some space here to write.

[00:46:45]20 come to the question of three. You are being told the weight of one molecule of a compound and what is the compound given? C6 H12 weight of one molecule is asked. Yes C of 22 is absolutely correct Sumit.

[00:47:08]Now look, we have been asked about this molecule and for this molecule we are being asked about the weight of one molecule. So what do you do whenever you calculate the weight of one molecule?

[00:47:19]Find the molecular weight and divide it by what? from the Agodro number.

[00:47:25]So what will be its molecular weight here? 12 * 60 + 122 is the molecular weight. The weight of one mole is.

[00:47:44]Divide this with whom? From the Hevobedro number.

[00:47:47]Do.

[00:47:49]So what will you get from here? The weight of one molecule of it. So, if you have ever been asked about the mass of a molecule or the mass of an atom, then by what do you divide the molecular mass or the molar mass? from the Agodro number. Keep this in mind.

[00:48:09]Questions will keep coming on this also. Next question is Na is the Agodro number. Then number of valence electrons. See, electrons also ask. Then the number of valence electrons in 4.2 grams of nitride ion.

[00:48:26]What is nitride? n-3 nitro nitride in one unit ion yes, okay whatever the answer is.

[00:48:36]How many electrons are there in one unit ion of nitride?

[00:48:39]How many electrons does nitrogen have? Seven and three added. What is 7 3 to the minus? 10 electrons. Well then?

[00:48:49]So what should you remove first? Number of moles. So 4.2 up now see whether you add electrons or remove electrons there is no effect on the mass means the atomic mass of nitrogen is going to be 14 only, there will be no change in it if we talk in terms of 24 so what will we do simply that brother 4.2 / 14 is how much it is, 0.3 moles now take out the number of ions in this, 0.3 * Na and by what will you multiply them?

[00:49:30]From 10. If yes then you will get your answer.

[00:49:35]Yeah, okay views.

[00:49:37]After this we do 25. The number of oxygen atoms is asked in these 4.4 grams of CO2, so we will solve this question.

[00:49:54]Now see, in this the oxygen atom asked is only 4.4/ 44 * Na * 2 because how many oxygen atoms are there in one molecule? There are two.

[00:50:06]Keep this in mind. You have asked how many oxygen atoms are there in CO2?

[00:50:21]Now let's come to question number 26, number of gram molecule of oxygen gram molecule asked 26 now see a big child remains confused about gram molecule gram of molecule so see it is always mole only, like it is written gram molecule of gram atom of if off is coming later then it represents mole only, so gram molecule has been asked from us, is there any off between these, if not then what is this? I am asking you about the number of moles.

[00:50:55]Keep this in mind. What are gram atom, gram ion, gram molecule, gram electron?

[00:51:06]Number of moles is being asked.

[00:51:09]Because if this was mass, what would you have written?

[00:51:11]Gram of molecule. So many grams of molecules. Off if it comes in the middle, what are these? Wet. But if off is not coming in the middle then it means what is it?

[00:51:23]Number of molecules. That number of moles. So brother, I have asked about the number of moles in so many molecules of oxygen. So first find out its value. Just a while ago, Sumit had told the method that if the number is given, then divide it by the add number.

[00:51:47]What will you get from here? Number of moles. And the number of moles that will come Yes, 25 was A. Ok? The number of moles of CO that come from here will be 10 moles.

[00:52:02]And who will have only 10 moles? of oxygen.

[00:52:05]Because one mole in one molecule CO is one mole in one mole CO?

[00:52:11]Oxygen.

[00:52:13]clear?

[00:52:14]Now let us come to question number 27.

[00:52:22]1 ml in 27 cc means milliliter. So don't worry that it is cc. I read ml. cc means centimeter and 1 cm. What is a cube equal to?

[00:52:34]1 ml. So 1 ml N2O at NTP cont. So you will be able to do it.

[00:52:45]27 Do it quickly.

[00:53:01]Tell me which answer is correct in this? Use your brain quickly in 27.

[00:53:09]Atom has asked us. 1 ml of N2O 27 Do it fast Sumit Argal D is perfect. So how will we do it? See, if we talk about 22400 ml at STP, how many molecules are there? 6.022 * 10 to the power 23 is the same number of molecules, so what is there in so many ml? Molecule.

[00:53:57]So how many will be there in 1 ml? So 1 ml will be 6.022 10 to the power 23/22400 molecules, which means this is correct. Then if you see the atom in it, by how much will you multiply it? From three. So this will also be correct.

[00:54:17]Then if you look at electrons, how many electrons will there be in N2O?

[00:54:22]How many are in nitrogen? 7 * 2 = 14.

[00:54:25]And your oxygen is gone. It's 22 now. If you multiply 22, you will get this answer. Meaning all the answers were correct.

[00:54:36]Now let us come to percent composition.

[00:54:39]This is a very easy topic. The tough topic has been covered.

[00:54:43]Now this is an easy topic. Percent composition. Let's try some of its questions as well. There is little time. So let's do the questions now. Look, then I will give you the PDF of this PD or if you have any previous year question book then put its questions from that, whatever remains, we will do them the day after tomorrow, after this you will have to ask a few questions, so there will be a few less, you will be able to do it quickly, if there is one less than before, a compound x contains 32% a 20% of b whose remaining percentage is c, see what you have to do for the empirical formula? Make a simple table.

[00:55:29]This is any empirical formula question. First of all stop. I'll make this a little bit smaller. So you will be able to see it easily.

[00:55:41]See, whenever a question on empirical formula comes, what should you do? I want to make a table.

[00:55:47]Simple C.

[00:55:50]First of all write in this table what are the elements? So what was there in this question? x, not a, b and c.

[00:56:05]Ok? Then write their percentage here. So sir, how much percent of A is given to you? 32. How much is given by B? 20.

[00:56:16]Whatever the remainder of C is. So this is 52, meaning we will subtract 48 from 100. Second, after the percentage, now write these, whatever the atomic weight is here, it will be given to you in the question.

[00:56:31]How much is given to you here.

[00:56:36]If we look at how much A has given in this?

[00:56:39]64, B's 40 and C's 32. Then find out their atomic ratio. Take it out like a simple mole. Don't do anything.

[00:56:50]Consider this percentage as mass.

[00:56:53]Divided by atomic mass. So you will have to do some calculations here - 32/64 20 / 40 48 / 30.

[00:57:03]Find out what the values ​​are coming out to be.

[00:57:14]This is coming out to be only 0.5. This is also 0.5 and this is 1.5, these have to be made into whole numbers. Divide everyone with whom? From two. So these two became one.

[00:57:25]Meaning how much is A? Forest. B How much is it? Forest. And how much is C? Three. So what will be the answer in this?

[00:57:36]B ABC 3 of 28 is absolutely correct Piyush very good, all the other children are able to do ABC 3, so we have completed this question, I am doing one last question, it is an organic compound, I am zooming in on it a little, question number 29 is an organic compound in which 78% is carbon and the remaining is hydrogen. It is simple. Please tell me the right option for the empirical formula. Quickly answer question number 29. This should not take much time for you.

[00:58:27]29 Do it fast.

[00:58:30]What is 78%? What is carbon remaining?

[00:58:34]How much hydrogen is left? 22 Do it fast. Is the answer coming for 29?

[00:58:43]Tell me quickly.

[00:58:54]Sumit with percentage poverty. Ok? So in the next class I will ask questions related to percentage. Sumit, when you come tomorrow morning, you can also come offline, so ask me offline. Otherwise, if you want to ask in this class, I will get you some questions related to percentage tomorrow.

[00:59:12]Okay Sumit Argal, so this was a live class for some of you children who are already associated. Sometimes the classes would get disconnected. So I would like to ask, yes Piyush has sent the answer CH3 is absolutely correct. very good. Piyush, how is your experience of this class today? You have taken classes with me before also.

[00:59:37]I am talking about online. So the online experience here or the rest of the children who have taken classes before.

[00:59:45]So how was this experience compared to the previous one?

[00:59:52]Respond quickly to live classes.

[01:00:11]Good, okay, let's stop here today.

[01:00:16]We had said that we have done up to 29 questions.

[01:00:20]Okay, after this we will meet day after tomorrow. Okay, okay, voice is also fine. Okay, okay Sumit. Okay, okay, okay, children, bye, good night, revise and those who are coming to the class tomorrow morning, please revise offline and come.

[01:00:44]I will ask you. Do the remaining questions and daily we will keep doing these questions here in TTS. Thank you, thank you and from next time the class will start at 8:15. The class on the next turn will start at 8:15. I will share the link in the group and those who are not from the group, please take some care.

[01:01:07]Good night, good night children, nine people were present.