Prohaar CEE 2027 ChemistryAHSEC Assam Board | Chemistry Concept Revision Class 1

[00:00:00]I mean so this is a concept building class okay so we will try to cover up some of the small thing okay I mean just topic that is from class 11 again okay that is from class 11 again small topic easy Level you have to clear your basic first.

[00:00:42]I don't know.

[00:00:48]Okay.

[00:00:50]Yes.

[00:00:58]We will provide it. Okay.

[00:01:08]Okay. And those who people are watching me from WhatsApp. Hi everyone. Good evening. Okay.

[00:01:14]WhatsApp.

[00:01:26]Okay. And once more one thing people watching me from YouTube.

[00:01:33]So then yes if you're coming for study then please download P lens application.

[00:01:42]poll type of questions participate.

[00:01:46]Okay.

[00:01:48]Hello.

[00:01:50]Hi. Good evening. Okay. Let's start the class. Okay.

[00:01:56]Okay.

[00:02:00]What you people know? Okay. About the thing. Yes. Probably right. But yeah, let's go through it and check it. Okay. So the topic that we are going to cover in today's class is what mole concept. Okay. Mole concept question.

[00:02:20]Okay.

[00:02:23]Stoometry and what stochometry calculation. And if you ask me say why you have chosen only this topic two topic because yeah it is time time consuming. Okay.

[00:02:34]Right. at the same time.

[00:02:41]Okay.

[00:02:46]Right.

[00:02:57]Right. So this is that is the reason that is the simple reason why I have chosen this particular topic. Okay.

[00:03:08]Yes. Okay.

[00:03:11]So, uh before starting Yes. What exactly mole is right?

[00:03:24]Mole basically it defines what the amount of substance right very simple example to you people.

[00:03:57]Right.

[00:04:00]One of Okay. specific amount specific 6.022 22 into 10^ 23 right anything rightificle when I'm saying one of Right.

[00:05:04]I hope right one is basically what this number. I mean simply one mole is basically what this number but Why are from where we got this right?

[00:05:50]Hi. Hi. Hi. Hi.

[00:05:55]Right.

[00:06:15]Let's say we having something in my right hand. Okay. And I'm saying it is one mole.

[00:06:36]Hello. Hello.

[00:06:48]That is the main question.

[00:06:53]Okay.

[00:06:58]12 g of carbon 12 isotope. You know carbon right? carbon carbon 12 isotop 12 g or particle carbon or atom carbon 12 g what is the number Number of carbon atom number one when I having the same amount of substance which is exactly equal to the number of carbon atom that present in 12 g of what? Carbon 12 is that equal to what? That is equal to what? Carbon 12. Okay. That means 12 g of what? Carbon 12.

[00:08:30]Okay.

[00:08:45]12 g of carbon 12.023.

[00:08:59]Okay.

[00:09:01]to that means this number is equal to the number of carbon atom that present in what 12 g of carbon 12 is okay uh so an hello hello s27 Sorry.

[00:09:40]Okay.

[00:09:45]Good evening. Good evening.

[00:09:48]Yes.

[00:09:50]DB homeworks.

[00:09:53]Yeah, you can think about it. Okay.

[00:09:57]Yes. Then yes, that's it.

[00:10:13]Right. Okay. So then this class is helpful for you.

[00:10:21]Okay.

[00:10:23]Okay.

[00:10:25]Right. Okay. Okay. Okay. Okay. Yes.

[00:10:29]So in SI system basically and introduce as what the seven base quantity for the amount of substance right to mole basically meaning I have already start discussed it in the beginning right to mole basically amounted okay amount of substance and mole is basically M it is the what symbol of SI unit of what amount of substance and one Exactly contains what? This number of what? Elementary entites. So this number of elementary entitities is simply the number of carbon atom that is present in what? That is present in what? 12 g of carbon 12 isotope.

[00:11:19]Yes. I mean yes definitely definitely definitely.

[00:11:34]So this number is what fixed numerical value okay of the avagadro constant when expressed in the unit per mole and it is called what a do number. So when we talk about one mole so I just already explained to you right people. So have a look here. You can see it here that a mole is an amount of substance with which consist of as many entities as there are atoms in 12 g of what? Carbon 12 is okay to 12bon.

[00:12:17]Okay. One more meaning I problem Z7 level.

[00:12:37]So now problem B 25 mass of carbon 12 was calculated by mass spectrometer and pound to be equal to 1.98 1.992648 into 10us 3 okay partism So the mass of C12 of a carbon 12 mass 1.992648 into 10us 23 g that is the mass of what that is the mass of what one carbon 12 atom one so we need to find out what that 12 g carbon 12. Okay, that is what we need to find out. What that is 12 g of carbon 12 isotop.

[00:14:16]Okay everyone.

[00:14:25]Okay. So therefore the number of atoms in it is equal to what? 12 g right divided by what? This that means mass of what? One carbon 12 atom. That means what we have to do is that we have to find out that in 12 g of carbon 12 is what exactly the number of carbon 12 atom we are having. Okay.

[00:14:49]So that is equal to what this 6.022 1367 into 10 the^ 23 atom. And this is what we call one mole. And this particular quantity is what we call one mole. You say you say I hope it is clear right simply you people have to know that rightated relations.

[00:15:33]That's what you people need to know.

[00:15:35]Okay, okay, let's have a look here. There is some relation, right? We used to call what?

[00:15:44]Molar mass, right? Molar mass. So, that is what mass of mass of what?

[00:15:53]One mole of the substance, isn't it? Mass of one of the substance mass Right.

[00:16:14]Uh, egg mall.

[00:16:38]That's it.

[00:16:41]Okay.

[00:16:56]One CH4.

[00:17:00]If I say one of CH4, it is one CH4. Right? 1 CH4 equal to 1 CH4 equal to what? In terms of mass in terms of mass, how could I define it?

[00:17:17]Okay.

[00:17:18]Okay.

[00:17:27]G of CH4 to one CH4.

[00:17:31]16 g of CH4.

[00:17:34]Correct.

[00:17:35]Next it is can I say like this that one CH4 means one CH4 means key meaning 1 M CH4 meaning that 6.022 022 into 10 to the power 23 number of CH4 meaning go by yes sir honey to 1 mole CH4 means 16 g in terms of mass but what do we know if we talk about one mole CH4 that means is in terms of number of CH4 molecule A2 meaning that means indirectly that means indirectly what mass of 6.02 222 into 10 ^ 23 number of CH4 equal to what 16 g can we say like this that mass of this number of CH4 equal to 60 16 g everyone is it correct? So look this is how we have to look after the relation so that you know we can answer it very easily.

[00:18:52]Okay.

[00:18:57]Is it clear that how we can you know find out the relation right number of that means mass of this number of CH4 g of CH4. So this is how you people have to you know find out the relation.

[00:19:29]Okay.

[00:19:32]Yes.

[00:19:33]Sorry volume.

[00:19:45]Okay. Especially gases per kilodar gases. Simple simple volume of volume of one mole of the gas and STP.

[00:20:05]We are considering at what at standard temperature and pressure.

[00:20:15]iot at STP. at have a look here. So at STP volume of 1 mole of gas equal to 22.4 L consider okay 22.4 4 liter consider if this is the relation how molar mass then number and one concept is related okay that is what volume have a look here my next let's have a look here I'm saying you one mole of CH4 gas to in terms of mass in terms of mass how I can define I can define 16 g of CH4 right next what next one mole of CH4 gas equal to 22.4 4 liter of CH4 in terms of volume. This is in terms of what mass right in terms of volume.

[00:22:16]Okay.

[00:22:23]1 of CH4 gas equal to 6.022 into 10 to the power 23. So in terms of number of molecule in terms of what number of molecule if if I'm supposed to explain these things relation 1 mole of CH4 gas equal to what 16 g of CH4 then that means what 16 g of CH4 occupies 22.4 liter of volume.

[00:23:10]Can I say like this one of CH4 gas 16 g of CH4 16 g of CH4 16 g of CH4 22.4 Liter volume.

[00:23:43]So things are related as like this.

[00:23:48]Okay. As things are as related like I I can look after like this also that I I can say like this that 6.022 222 into 10 ^ 23 number of CH4 CH4 gas equal to 16 g of CH4.

[00:24:12]Yes. That mean this number of CH4 get means this in terms of mass in terms of mass correct in terms of mass of the occupies occupies 22 this much of volume you will say yes if this is so that means can we say like this that 6.022 222 into 10 ^ 23 number of CH4 gas occupies occupies what 22.4 4 liter volume right briangu not ignoring but where I'm ignoring you.

[00:25:08]Okay.

[00:25:14]Okay.

[00:25:36]They are what? Related to each other.

[00:25:38]They are what? Related to each other.

[00:25:40]Okay.

[00:25:45]Probably next either Monday.

[00:25:50]Okay.

[00:25:52]Yes.

[00:26:02]Avid Aloj.

[00:26:16]Okay. So yes uh these are the few things that you people need to remember. Okay.

[00:26:23]H Okay. Next it is. So there are some calculations you know related to what mole right? Okay relations related to concept. Okay.

[00:26:38]Okay.

[00:26:44]For any given amount of substance we can calculate value. You people know right.

[00:27:00]Okay.

[00:27:06]If you know amount certain amount of substance is given then wre So [snorts] W is what your given amount of substance but given mass. Okay.

[00:27:38]N= M say for for any given amount of substance we can calculate the number of atom and what electrons.

[00:27:56]Okay.

[00:28:10]Okay. Number of number of atoms per entity per molecule.

[00:28:23]Next number of molec22 number of value number of electron number of electron. Okay.

[00:30:57]Par entity.

[00:31:09]So yes, righteous entity can be what?

[00:31:27]Okay.

[00:31:30]Okay.

[00:31:32]With respect to this given amount, we need to calculate what?

[00:31:38]A once you calculate mole you can calculate what? You can calculate what?

[00:31:50]Number of atom right number of molecule right but number of electron. Okay.

[00:31:59]Number of proton.

[00:32:11]atom like H2O.

[00:32:34]Okay.

[00:32:40]So whatever it is asked to calculate it depends on that. Okay.

[00:32:48]Right.

[00:32:51]Yes. I'm giving some time.

[00:33:34]Yesh.

[00:33:40]Okay.

[00:34:16]Done.

[00:34:30]Done.

[00:34:36]Okay.

[00:34:47]The total number of proton in g of what?

[00:34:50]Calcium carbonate.

[00:34:55]Ca3. Right.

[00:35:00]massate.

[00:35:03]So you need to calculate massium 40 + 12 + 48 right 48.

[00:35:26]So given mass what is the given mass given mass equal to 10 g. So therefore mole so mo number of moles so 10 by 100 correct 0.1 correct so this is the value we are having right 0.1 and one total number of proton. Total number of proton. So what you have to do is that first of all you have to calculate the proton. So number of proton Ca3. So proton basically is equal to what? Atomic number. Proton is equal to what? Atomic number. Atomic number to proton.

[00:36:28]So yeah atomic number 20 plus carbon six plus oxygen oxygen 8 24 that means what 50 so therefore number of proton equal to mole number of mole basically into 6.0 22 into 10 to the power 23 that is what I get number right then what into proton total proton 50 number of proton present in the entity that is 50.1 into 6.022 222 into 10^ 23 into 50.1 50 to 5 into 6.022 into 10^ 23 simple calculation 30 11 into 10 the power 23 decimal so 3.011 into 10 the power 24 3.011 011 into 10 24 that means this 3.01 right option A correct answer option correct answer 50% B 50% uh D Okay.

[00:38:16]Well, key everyone batch guys.

[00:38:30]Rahan Vanita would sor a no ain amid aloc.

[00:38:58]Okay. So this is how it is supposed to be solved. Okay.

[00:39:07]Yes. So these are the type of questions that you that may be asked in examinal number of electron.

[00:39:30]Okay. Next question.

[00:39:34]Yes.

[00:39:37]Another question.

[00:39:42]Okay.

[00:39:45]Mandhu. Okay. There is one new student.

[00:39:48]Okay. Clear.

[00:39:52]Simple question right. The mass of 11.2 L of ammonia relation.

[00:39:59]Volume mass mole. So they're related to each other right? So on that particular basis this is the question direct question to both right that's it.

[00:40:34]Okay. Dejra. Okay.

[00:40:42]Dej.

[00:40:51]Yes. Done everyone.

[00:41:02]to download. Okay.

[00:41:22]17.

[00:41:30]Okay. Okay. Right. Mass of what? 11.2 L.

[00:41:36]Remember 11.2 L ammonia. What is the amount of ammonia gas we having in 11.2 L? That is what we require. Right?

[00:41:50]If I say one mole of ammonia NH3 one mole of NH3 guess equal to what? In terms of volume. So if I say like this volume of let's clear it.

[00:42:07]If I say like this that is a general thing now that volume of one mole of NH3 gas at STP ganabh if it is 1 mole then at STP what will be the volume so you people will you people will probably say sir it is 22.4 liter think about it think about it right 22.4 4 L. So 22.4 L now. Okay. Okay. That is that is what 22.4 liter of NH3 gas at STP equal to what?

[00:42:51]What is the mass of NH3? That is 17 g.

[00:42:54]So 22.4 L 1 M to 1 M 17. Correct.

[00:43:01]Correct.

[00:43:02]Ultimately 11.2 that means half mole that is therefore 11.2 liter of NH3 gas at STP probably I should write mass now that is mass of okay mass of 11.2 liter of what ammonia at STP to 17 into 11.2 by what 22.4 4 to ultimately 17 by 2 that means what 8.5 think about it everyone 8.5 okay 11.2 two half.

[00:43:49]Okay.

[00:44:11]Option A is what? Correct answer here.

[00:44:15]He is not bad 83. Yeah. 33%. Okay. GU Kalita A Rupin Madua Mo. Okay. Ba.

[00:44:32]Okay. Very good.

[00:44:38]Relations.

[00:44:41]Okay.

[00:44:45]Okay.

[00:44:49]So, yes. Next.

[00:45:20]Combined interesting engineering college chapter Chemistry class 11 first first chapter Mr. S option number two okay question let's let's try to find out the question the number of water molecules okay the number of water molecules in a drop In a drop of water weighing this is so simple that means basically the mass of water is given to you ultimately right. Huh?

[00:46:20]Did you get the thing?

[00:46:24]So according to question the mass of water equal to what? 0.05 g.

[00:46:37]Therefore mole equal to what? Mole equal to given mass by water or mar mass you know right? 18 isn't it? It will look like this.

[00:46:52]It will look like this. Huh? So what you need to do? You need to find out the number of molecules. So therefore now what we can do? What we can do now?

[00:47:04]Number of water molecule. So number of water molecule equal to what? Mole into what?

[00:47:12]Na. That's it. This is what you have to do. Correct. So mole value 0.05 by 18 into 6.022 into 10^ 23. That's it.

[00:47:30]So it will come out as 18 into 6.022 into minus to 21.

[00:47:40]So 30.11 that is 0 into 10 the power 21 by 18 to 2 obviously right 3 10 36 10 by 6 1 something ultimately so 1 30 right So 3 10 36 right. So 10 one time go to 6 to decimal four 6 36 yeah 1.6 something into 10 the power 21 answer the option say yes that is the correct answer 1.672 672 into 10^ 21 this is the option probably yes J yeah you're correct Mr. is yes you are also correct good option B correct answer okay okay moishitan okay very good so isn't it was too easy Right.

[00:49:07]Right.

[00:49:17]Just just trying to you know go through with the thing right just simply discuss. Okay everyone it is very simple discussion that is going through okay so in details definitely okay yes because we are also having a limited period of time for the class right so yeah this is all about more concept okay next stochometry stochometry okay stochometry and what stochometric calculation so this is one of the important topic okay if you're preparing for uh you know if you are preparing for competitive exam then these are the main two topic okay exactly July definitely.

[00:50:39]Okay.

[00:50:42]Yes. Stoometry and stochometry calculation.

[00:50:47]When we talk about stochometric coefficient, right? Whenever we having a reaction related basically refers refers to the quantitative study of what reactants and product involve in a chemical reaction and this stochometric value talks about number of thing. Okay, it can talk about volume also. It can talk about mass also. It can talk about number of molecule also. It can talk about number of atom also. So it give us lot of information. Okay, it give us what lot of information. Okay.

[00:51:49]Now uh yeah somewhere I have definitely written it. So the coefficient of the balanced chemical equation are called what? Isometric coefficient. Okay. And isometric coefficient replaces what the number of moles molecules of what?

[00:52:02]Reactant and product in a balanced chemical reaction. Okay. So yeah this is the very simple thing that as I've told you that it involves or it deals with the calculation of what mole mass volume right either it can be of what reactant and it can be of what product. Okay.

[00:52:19]Okay.

[00:52:22]So this is this is what we have to go through and and the stochometric uses all the laws of what chemical combination interesting point and that is one of the important point which we used to you know explain whenever you are going through stoometry calculation and all you have to know the balanced chemical reaction Whenever we talk about reaction probably the reaction may not given to you right probably the reaction may not given to you. Suppose if I say the decomposition of calcium carbonate so the reaction may not given to you and so you need to know that how to write the balanced chemical reaction for the same okay related let's let's go through some of the thing here let's say suppose I'm having a reaction Uh I'm I'm for example I'm taking a reaction let's say 2 H2 plus O2 gives us what 2 H2 correct 2 H2 correct now now if we if we talk about different parameter have a look here what is thetric value Say two mole say two mole yeah one mole yeah two mole to whatever value we are having that stochometric value in general represents what mole number of molec volume What we can what we can talk about this reaction? Can we say like this that two mole of H2 you know react with react with one mole of O2 to give you know 2 mole of H2 I mean right I mean definitely in terms of Correct. At the same time, at the same time, can we talk about mass number?

[00:55:05]Look H2 one one mole of H2 becomes what? Two, right? 2 g. So 2.

[00:55:17]So in terms of mass 32 G, 36 G.

[00:55:28]Can we say like this that 4 g of H2 right ree equip right 32 g of O2 to give what? 36 g of H2.

[00:55:56]But the simple thing is that you need to know exactly.

[00:56:02]Okay. And rel 22.4. So 2 into 22.4 4 liter of H2 reacts with you know 22.4 liter of O2.

[00:56:23]So this is also how we can pronounce.

[00:56:25]Okay. So yes it talks about mass, it talks about volume, it talks about number of molec.

[00:56:33]So interesting during this calculation during stochometric calculation during stochometric calculation like reactant.

[00:56:57]So focus on what?

[00:57:00]Focus on reactance.

[00:57:03]Okay, focus on reactant. Let's say let's say A + B SA and we are getting C. So these are the reactant, right? And whenever question is asked.

[00:57:25]Do it amount.

[00:57:30]If if amount of both the reactant or more than one reactant is given Then among the two what do we have to do?

[00:58:05]Find out the limiting reason.

[00:58:14]Find out the limiting reason. Okay.

[00:58:17]Because because the amount of because because the amount of product forms depends on the depends on the limiting reaction.

[00:58:48]You should know this. This is very much important for you. Sometime suppose if I say that 2 g of A and 4 g of B is reacting to give C. Right? So what is the amount of C form? So if such type of questions are given to you that means mass or amount or any in terms of mole that means if amount of both the reactant or more than one reactant is given to you then our job is to find out the limiting reaction. Okay. Our job is to find out the limiting reason among the two because limiting reaction defines the amount of product form.

[00:59:26]One if there is there is what mass of only or the amount of only one substance is given or one sorry reactant is given substance in terms of reactant is given then what you just directly compare with that so you don't have to find out the you know limiting reason there limiting reasoning come Okay. So limiting reaction basically what it it it talk about or it limits about the amount of product form. So usually as per the balance reaction as per the balance reaction we have to look after the balanc equation balancing if I give the amount of A and B both then yes you have to find out the limiting reactant.

[01:00:38]So this is what you have to do of a react with B to give C. So what is the amount of C for amount?

[01:00:53]So no problem with that. You just directly relate A with B sorry A with C as the compare. So that's why it is much more important to look after the balanced chemical equation mass volume number of molecule. always have to look after the balanced chemical reaction first. Okay, let's go through an very simple example. Let's go through in very simple example what I'm talking about. Okay, have a look here. Suppose example, let's say we having this reaction 2 H2 + O2 gives us what? 2 H2, right? 2 H2. Now, as per the balanced chemical equation, we know this that it has what? 4 g. This is as per the balanced chemical reaction. So this is what 32 g and we know it is what 36 g correct 36 g. Now let's have a look here. probably if if 10 g H2 and let's say uh let's say let's say 30 g 30 g O2 sorry 30 g of O2 reacts to form water right so the question is what is the mass of water formed let's say so such type of questions are asked to you what will we do right now if you look at the question carefully then but the mass of both the reactants are given to you so one is what 10 g of H2 and another one is what 30 sorry 30 g of what O2 30 g of what O2 is given to us. Now whenever such type of questions are given to you as I've already told you that you have to find out what the limiting reasoning okay very simple if you have understood the thing or understood what exactly the question try to you know ask you that is very simple that is very simple As per the balanced chemical reaction, as per the balanced chemical reaction, if hydrogen is 4 g, then it requires 32 g oxygen to give what? 36 g water. This is as per the balance ratio. So what is the mass of hydrogen given to you? That is 10 g. So if you look at this value that means if you are having if you're having 4 g to here 32 so if you simply calculate in your mind that if I'm having if I'm having a hydrogen x amount then it is what it required what so 8 isn't it 8 * right so if hydrogen is x amount then O2 required is what 8x As per the reaction 80 as per the amount of hydrogen provided to us it doesn't seems that our oxygen required amount of oxygen is less current 4 Hydrogen 32 g oxygen 4 g hydrogen 32. So 10 obviously 10 oxygen definitely According to the reaction, according to the balance, chemical reaction 4 g H2 requires 32 g O2. So therefore hydrogen amount 10 g. So 10 g H2 requires what? So 32 into 10 by what? Four. Now so definitely it is what?

[01:05:53]Definitely it is what? 80 g. So 80 g. So do we have 80 g? So we saying no we don't have 80 g. That means what? But but the given to supplied amount of O2 is 30 g. So O2 is limiting reagent.

[01:06:24]O2 is what? Limiting reagent. Okay. So if O2 is limiting reagent then what? The amount of product form depends on what?

[01:06:32]depends on what O2 the amount of O2 got the point.

[01:06:43]So according to reaction according to reaction directly oxygen limiting reason 326 direct that is what? So 32 g O2 gives what? 36 g H2. Therefore 30 g O2 gives to 36 into 30 by what? 32 g.

[01:07:20]So if you solve this if you solve this solve 36 into 32 sorry yes no 30 by 32 right. So this is it.

[01:07:43]So 216 218 right?

[01:07:48]So 28 29 270 by 8 3 * 24 then 4 0 8 14 now if I'm not wrong I'm true right.

[01:08:08]So 35 g H2.

[01:08:19]Let's have a look. So 32 gives 36, right? Yes. So 30 gives this 36 into 30 by 32. Okay, that is true. Next calculation.

[01:08:35]H [clears throat] 16 29 28 27 3 again 24 and then what 6 0 7 56 4 Now five now to ultimately 33.75 g 30 to 36 obviously I don't 36 32 32 36 33.75.

[01:09:51]Okay.

[01:09:53]So yes, Pavitra. Hi. Hi Pavitra.

[01:10:11]Okay.

[01:10:14]Okay.

[01:10:27]Okay.

[01:10:31]Highable. Hi.

[01:10:44]Try to solve this. The amount of water produced by the combustion of 32 g of methane.

[01:10:51]So at first you need to know the reaction, right? Methane if it undergo combustion it requires 2 mole of oxygen which will give us what? 1 mole of CO2 plus 2 mole of H2O. That's it.

[01:11:04]Say amount of water in gram. Now, okay, let's solve it.

[01:11:21]Am 16 as per the reaction 64 mass 36.

[01:11:44]The amount of water produced by the combustion of 32 g of methane.

[01:11:51]water. So 16 right to amount of only CH4 is given to us to 16 gives what? 36 to 32 this that mean it is double. So it will become what to double to 16 gives what? 36 g to that is what 32 g. So 32 gives what? 36 into 32 by what? 16 36 into 2 * ultimately 72 option C is the correct answer. Okay.

[01:12:43]So as per the reaction 16 36 simple what is the number of moles of oxygen gas produced during the electrolytic decomposition of 180 of that's decomposition decomposition 2 H2O right 2 H2 P O2 P So as per the as per the g value you will say sir 36 you will say 32 36 32 that means what 32 means what ultimate value this is What? 2.

[01:14:11]So 2 is giving what? 1 to 180. So 180 by 36.

[01:14:18]So 94. Huh? 94 36 9 this it will what?

[01:14:24]Five times.

[01:14:29]So 2.

[01:14:36]So 2 definitely 2.5 yes 2.5 2.5 everyone understood look amount 36 balance reaction 2 oxygen 32 H2 will decompose to give what O2 So if you calculate mole here right we have calculated mole here that's it.5 that's it.

[01:15:32]Okay. So, ah, you people have done mistake. Huh? Or did I done mistake? No, that is true.

[01:15:48]So 180 g of water means what? Five M.

[01:15:52]So 2 is to 1 ratio.

[01:16:00]Okay.

[01:16:03]Yes.

[01:16:06]So uh so if you look after the summary to we have gone through what mole concept.

[01:16:13]Okay.

[01:16:15]So next class okay then right concept what is mole? How mole is calculated and all. Okay. So you people have to go through this things huh mo. So just for your practice okay I am just providing some of the question as in homework to calculate the number of moles in each of the following 11 g of CO2 3.01 into 10^ 22 moleculate number of moles and 1.12 L of CO2.

[01:16:51]Okay, homework. How many moles of methan are required to produce 22 g carbon dioxide after combustion? So these are some homeworks for you people. Okay, definitely right.

[01:17:21]participate. Then definitely you people have to you know download one application that is called pil. Okay. So yes, thank you everyone. See you in the next class. Okay. Bye-bye. Ajima. Good evening. Good evening. Okay. So see you people in the next class. Okay. Thank you.