Law of Chemical Combination | Class 11 | Chemistry | NEET 2026 & 2027 | Diksha Ma'am

0:20 Hi everyone, welcome to Vdant Neat English. I'm Vika Koshal, your chemistry master teacher. So everybody today is going to be the first lecture of some basic concepts of chemistry. Right where we are going to discuss about law of chemical combinations, right? So basically we have different different laws and just not the definition see by just looking at the definitions it will be like zero right because we are not in ninth or 10th class right usually we used to ignore this topic right but why at this point looking at what has happened in me 2025 right in chemistry paper I would not say it was very difficult or something like that it was

1:07 obviously lengy but The questions were from the syabus only right and some questions were from those topics which we usually ignore right so that's why in neat 2026 you are not supposed to ignore anything very simple topic to very difficult topic you have to do everything for your neat 2026 right so for that that's why we have a proper like uh lecture for detailed lecture for law of chemical combinations right okay so in the last lecture we discussed about unit and conversions and I'll also show you there were some students who did their homework you can see uh Tishwar Sharma he did his

1:54 homework future he did his homework isobia he did she did his she or he did the homework right so these are the only three students who did their homework rest of the students I can clearly see are not serious regarding having homeworks, right? So students, we have started our chapter, right? My idea is in every three to four sessions, we are going to complete one chapter, right? So that will be the flow of class 11 and class 12 lectures I'm taking at 8:00 p.m. Right? So you can simultaneously for droppers you can go for class 11 class 12 lectures simultaneously. And for if I talk about only the students of

2:40 class 11 right you have 6 PM slot daily for you but for that you also need to practice questions okay so for that I really want to provide you with the DPSPs but you know even I need to see efforts from your side right guys if there is efforts from only my side then I am not going to write the exam I am not doing this for myself right I want you to be disciplined I want you to be consistent right for that if you do your homework on daily basis you write it down in the comment section and I'll get to know that these are the students who are serious and if I see that let's say if I see that in every single session more than 10 comments are there right not only one student that means uh different different students are

3:28 commenting answering not comments answer homeworks if you're writing right in the end of the chapter Every chapter I'll provide with a set of uh DPPS right for every single topic right that you will practice and we can also have a doubt session to answer the those DPPS but for that you also need to put your efforts okay yes everybody in need thinking of drop one year uh Ashish I would want you to ask your mom I think she would be really really good at it so you can ask her right? She obviously has given birth to you. So, uh yeah, that's that would be great. You can have a conversation with your sister, with your mother. That would be

4:13 great, right? Okay. Uh yes, finally. Thank you very much, Vicksha ma'am, for this video. Finally on regarding class 11 since I have problem. Okay. Awesome. Awesome. Good evening. Good evening. Good evening Rosie and uh Yes. Now let us see will you do your homework from next class or not? Right. Okay. What was the homework ma'am? Homework was very simple homework I was had given you. S unit in context of wavelength that will be per meter. Right. In context of your energy that will be jewel. So answer you can easily know per meter and jewel. Right? Next is SI unit of volume that is me cube. So answer will be C. Right? So that's how you can

4:59 easily answer right. This whole thing about units and conversion I have already given you in the last lecture. Okay. Yes. Now now let us start with the law of chemical combination. That is the first topic of your class 11. Right? Law of chemical combinations. In chemistry for chemical reactions, there are certain laws that are obeyed. Right? We have one 2 3 4 5 six laws in total. Right? Most of you have learned law of conservation, law of constant, law of multiple proportion, wave loss and ever getross. Right? Then law of reciprocal is also very very important. reciprocal

5:44 law or law of reciprocal proportion whatever you call it that is also very very important right so these are the six laws that we are going to study one by one firstly the flow of the lecture will be firstly I'll give you definition I'll explain you that you'll find it very easy trust me right till here you will find it easy maybe you will find this last one difficult right but when we will solve the questions might be you get stuck in these two the questions of constant proportion, multiple proportion, right? Because they are sometimes confusing. But once you know that how to solve this, you will find that also very easy. Okay? Yes, students, shall I start with the questions? Everybody, shall I start with the

6:32 questions or not? Okay. So, first is law of conservation of mass, right? So, conservation of mass means the mass has to be conserved, right? That means the particles the reactants the mass of the reactant should be equal to the mass of the products. Right? So we can say that in any physical or chemical change mass can neither be created and it can nor be destroyed. Right? That means total mass of reactant will be equals to total mass of the product. Right? And if you know that if I'm saying this thing and we also talk about balancing of equations, balancing of equations, you know what

7:19 balancing of equation is based on this thing only that the atoms see how the mass will be conserved. If I say the mass of reactant is equals to mass of products, it simply means that can I just simply write that it means that uh number of atom or not number of atoms atoms on reactant side the number the type is equals to atoms on atoms on atoms on product side. Okay, atoms on product side. Let's take an example. For example, you have H2 plus O2 that give rise to H2.

8:07 Right? Now, we always tell you that in chemistry you have to balance the equation, right? Stoometry, limiting reagent, all the concepts that you're going to see, right? in this chapter and throughout physical chemistry that will be based on the balancing of the reaction that you have to do. So similarly here also we will balance the reaction. How we will multiply this by half and now this will be the balanced reaction. Okay. Now can you see one thing why we get the balancing? We did the balancing so that number of atoms on reactant and product side should be same. Right? So balancing is based on law of conservation of mass only that's why we balance atoms right now ma'am if

8:53 number of atoms and types of atom on reactant in product side is same can you see what is happening here there is rearrangement of atoms right whenever a chemical reaction takes place there is rearrangement of atoms Right? It's not like some new atom will be generated that is also generated. These are those are nuclear reactions and nuclear reactions do not follow law of conservation of mass. Okay. We will also see what are the drawbacks of law of conservation of mass. What are the things that don't follow law of conservation of mass but generally law of conservation of mass is followed.

9:41 Right. Okay. Yes. Uh Sakshi maybe after three four lectures right after three four lectures of some basic concepts of chemistry. Okay. Now if I say that this is CH4 this is 22 this is CO2 and 2 H2. Right? Now we have balanced this equation by multiplying O2 with 2 by multiplying H2O with 2. Right? And can you see there is rejangement of the atoms. Here you have CH4 and here you have O2 and O2. Now here you have C. This is your C and two oxygen are linked here that is CO2. And

10:26 you have H2O and that of two molecules of H2O that is you have one oxygen two hydrogen from here one H2O. Right? Then you have two o two hydrogen one oxygen from here that is H2. That means there is rearrangement of atoms because of law of conservation of mass. Right? So if anyone asks you law of conservation of mass the application is rearrangement. The application is balancing of uh any chemical equation is based on law of conservation of mass. Right? Now example is you have mercuric oxide, you have mercury and you have oxygen, right? So mercuric oxide is mg2

11:13 right? Okay. any mercury oxide will be MGO right because this is two positive this is two negative it will cancel out it will be MGO right so this is giving rise to mg mercury I'm writing magnesium oxide mercuric oxide mercuric tell me mercuric this also I have told you in the basics of chemistry where I taught you where is US o where is IC so mercury has two oxidation states state + 2 and + 1. Lower oxidation state is O US. Higher oxidation state is IC. Always in any case, ferrus, feric, mercuric, mercurus, right? Anything you have higher oxidation state will have IC,

12:01 right? So that means mercuric is plus. Mercuric is plus2. So you can say Hg2 positive. This is O2 negative. That will give rise to a compound that is HGO. And here you have Hg and here you have O2 that will be half. Okay, this is the balancing of the equation. If you see that ma'am uh how come we can say that there is law of conservation of mass applied here. If you see this here, this is 100 g. Right? Here the reactants are the products are going to be 92.6 plus this which is nothing but 100 g only. So the mass of reactant is equals to mass of the product. That means it follows law of conservation of mass. Right?

12:48 Okay. Now there is exception also in this case. Right? So the exception as you can see students that is your nuclear reactions. In the nuclear reactions right large amount of heat is produced atomic bombs right in all these cases very large amount of heat is produced. That means that mass that was in that reaction right that is being it's not like in normal reactions energy is not produced but energy is very minimal we can also ignore it but in case of nuclear reactions the energy is so massive right you can see these attacks nuclear attacks that has happened in wars right the attack is so

13:33 fatal that it is now even we study those things right even now People living in those areas are suffering from diseases, right? Can you see the amount of energy that must be released, right? So that amount of energy we can't ignore. That means a large mass has been converted to energy. In this case, mass is not conserved. Mass is getting converted to another form that is energy. Right? And the answer of this thing is given by E= to MC² that is the Einstein equation that definitely do not follow law of conservation of energy right you will study this more in physics right so

14:19 Einstein gave us E= to MC² right so he actually gave you the conversion that you can convert energy to mass and mass to energy so that was the uh opposition of the law of conservation of mass. Okay. Thank you so much. Uh Si, thank you so much. Okay. Yes. Till now any problem students. Now I know that definitions and all is clear to you. Let's try to do some questions based on that. See law of conservation of mass questions will be very easy right? So they will not be asked to you but still I'm just doing that. Law of definite proportion, multiple proportion then a ros gix reciprocal these questions you will find like more interesting right

15:05 this first law is very basic law right so if you see this here 10 g of ca3 on heating gave 4.4 4 g of CO2. Right? First you have to see the question. They are saying that 10 g of you have 10 g of Ca CO3 that is giving you 4.4 g of CO2 right and 5.6 g of 5.6 g of Cao. Show that this observation is in accordance with law of conservation of that's not a big deal right what you can do directly here see beta very simple such a into question CO2 plus Cao that's

15:52 a very that's why I told you initially we will do very basic questions you have to be very patient when you are in the uh lecture you have to keep the patience while you are in the class right so here this is CA2 is 10 This is CO2 is nothing but 4.4. This is 5.6. Can you see the mass of product and the mass of reactant are nothing but same? So that is definitely in accordance with law of conservation of mass. Okay students got it. Now next question here. Next question that is next question is 6.3 g of Na CO3 are

16:38 added to CH3. Now beta if you see Na H3 is added to CH3 CO first thing is ma'am we don't know the reaction. Now if you remember in your 10th class see I'm assuming that you are also there are students who have just come out of class 10th stepped out of class 10th and right now they are in class 11th and I want you to relate it to your class 10th. So students if you remember this is your biccarbonate HO3 biccarbonate. Whenever bicarbonate you used to study the reaction with HCl. Whenever bicarbonate reacts with any acid it has the formation of CO2 H2O and the salt of caroxilic acid that is

17:28 CH3 CO N. So that is your reaction. First thing is that you should know that what is going what we are going to form. Even if you have biccarbonate or simple carbonate, you will always have salt plus carbon dioxide plus water as the products. Okay. Now if you see you need to balance this reaction. Okay. So after balancing this reaction right I think it is already balanced. 1 N 1 N C C 1 2 3 It is already balanced. Okay. So you need to balance this equation. It is already balanced. Now 6.3 g of NH3 okay

18:13 are added to 15 g of CH3. The residue is fine to have 18 g. Residue see this will be the gas. Okay. Gas residue means that is left in the beaker. Right? That is your residue. Carbon dioxide is not a residue. Carbon dioxide is eliminated. It is like eliminated. That means it is a gas. It is not there in the beaker. Right? So that's why residue when we talk about we are talking about only CH3, COA and water that is 18 g. Now they are asking you the mass of CO2 released. So mass of CO2 we can take it as X. We know that these two will be equal to each other. Why? because of law

19:00 of uh mass law of conservation of mass right that is conserved here. So x is equ= to 6.3 + 15 - of 18. Right? That is going to be the answer. Let's see who is going to be tell me the answer here. Right? Okay. 15 16 17 18 minus 3 it is going to be 3.3 g. Okay. Yes. Everybody 3.3 g. Very good. Uh very good swarm. Very good. H good evening. Good evening Santos. Good evening. Okay. Got it.

19:46 Yes, we have more laws. Yes, we have more laws. We are going to talk about that. Don't worry. There is no hurry. Right. Okay. Okay. Yes. Now, let's see the next question here. Right. Now this question see this is not a direct question here you have to use a little bit of brain right this was the very simple question like ninth 10th you have studied these kind of questions right so that's why I have started with the basic level now we will not advance this is not at all advanced level this is little bit high right so that you can increase the level right so this much of hydrocarbon they are saying that there is some

20:33 hydrocarbon Let's say C X H Y on complete combustion. Combustion means there is some oxygen here right with oxygen gave CO2 definitely whenever a hydrocarbon is under combustion it only gives CO2 and H2 right now he's saying that we know that what is the we know that what is the mass of hydrocarbon right we know that what is the mass of this uh ox no not oxygen CO2 to what is the mass of water? This is given to us. Okay, all these things are given to us in the question. You can see I just uh I have just written what is given in the question. Okay, now they are saying show that

21:20 these reserves are in agreement with the law of conservation of mass. This could have been a very very easy question, right? If you also know what is the mass of the oxygen ma'am how do we find out ma'am not a big deal ma'am add this add this you will get the value but they are saying show that these are in accordance right so to show that these are in accordance we can't add them we can't simply say that yes they are in accordance you have to tell that it is in accordance now what to do right so now let us think that what we are going to do here, right? Simple thing

22:06 if you talk about this is your hydrocarbon, right? You need to find out how much carbon is there, how much hydrogen is there. If you're able to find out in this percentage, in this gram, how much carbon you have in this gram, how much hydrogen you have. If the addition of these two things are equal to 66 then can you say that this is in accordance? Can you say that this is in accordance? But for that ma'am how to find out this thing? See you might feel like ma'am these are very simple things but the base of unitary method you are need aspirants and for you the biggest problem is ma'am we don't know how to

22:51 apply unitary method. You get confused in these things you just want to jump to the advanced level directly. That's not how things work. You have to do basic questions to understand advanced questions. Okay. So now if you see this here I'll apply unitary method. Okay. What I'll do I'll just divide it into two parts. Right here I will say that in 44 g of CO2 ma'am how you have 44 g always always remember when we are applying unitary method we are comparing the standard thing what is the standard thing standard thing is the

23:40 molar mass right so 44 g of carbon dioxide has how much oxygen no How much carbon? 12 g of carbon. Right? 1 g of CO2 will have 12x 44 of carbon and 1.32 will have 12x 44 into 1.32 g. Okay. Here what you will say? You will say that 18 g of water will have how much gram of hydrogen guys? How much gram of hydrogen? How much gram of hydrogen? It has 2 g of hydrogen.

24:26 Right? Two hydrogen's are there. Right? 1 g will have 2 by8 and 2.7 g will have 2 by8 ult* 2.7. Now you have to tell me what is the answer. Is the answer equal to let me just calculate it for you 12 / 44.2 * 1.32 that is coming out to be.36. Perfect. What about that? 2 / 18 1 by 9 ultiplied by 2.7 that is coming out to be.3. It is coming out to be.3.

25:12 Now you add these two things. This is this is what 36. This is.3 right? When you add these two things 36 plus.3 that is nothing but 66. So is it in accordance with this thing? Yes. Now we can say it is in accordance with law of conservation of mass. Is that clear? Right? Okay. You can call me Sharma. Yes, I'll call you Sharma. That would be great. Easy. Right. So is that clear students? Is that clear what we are discussing here? Why these things are important? What if in neat exam they ask you simple questions not simple

25:57 questions from law of chemical combination you can answer everything if you have understood these kind of question your expansion right brain expansion that this question also I can solve these kind of question also I can solve that would be clear okay right now let us talk about the next question here this is a simpler one this I know that you can do easily. How much mass of silver nitrate will react with 5.8 g of sodium chloride? Silver nitrate. Silver is Ag nitrate is NO3. Ag positive NO3 negative is nothing but N Ag NOO3. This reacts with sodium chloride. Okay.

26:44 Sodium chloride is NaCl. Right? Produce this much of silver chloride. Silver chloride it is producing. Then it is producing silver nitrate, silver chloride and silver nitrate. I think it should be sodium nitrate. Okay, it should be sodium nitrate. It should be sodium nitrate. Typing mistake, right? It should be sodium nitrate. That is Na N. Right now you have these things. Now what you're going to do? Just put in the values here. How much of this of silver chloride? That means this is 14.35. This is 8.5. NaCCl is 5.85. And they are saying that law of

27:30 conservation of mass is followed. Let's say this is x. Now you have to calculate the value of x that is 14.35 8.5 - 5.85 85 and answer this and you will have your answer here. Right everybody clear. Okay, not a big deal. Right. Simple. Simple. Let's do answer will be 17. Very good. Very good. Very good. That's great. Now we will move towards the second law. Okay students, let's move towards the second law and let me see how many target you have achieved. How many likes you have given. Let's make it to 50 plus likes. Okay.

28:17 Okay. Now, next is law of definite proportion. Right. Definite proportion. We are saying I am drinking water. You are drinking water. Right? Someone else is drinking water. Right? Some Canadian is drinking water. Some American is drinking water. Some African is drinking water. Some Indian is drinking water. Water is saying everyone is drinking H2 only. Right? So someone is saying that ma'am H2 ma'am no ma'am Amani's wife ma'am is a different H2. That is the different thing. No, I'm saying when you are talking about normal

29:06 water, right? H2 only we are drinking, there may be more nutrients, there may be less minerals, there will be more minerals, there will be more impurities, less impurities. But the formula of water is nothing but H2 only. No matter Amani is drinking it, me drinking, you drinking, any other person drinking, the formula water will always be H2 only. Right? The impurities may be more impurities maybe less minerals maybe more minerals maybe less we are all drinking water right I really I'm so happy that right now at least water is free right it's not completely free right but uh at least we don't have to pay for water in our houses right though

29:52 we pay water bill but still it is minimal right so if you say sources of water it's ma'am river it's ma'am lake mountains ma'am oceans ma'am groundwater from a well anywhere you pick out water water is always going to be H2O always and always it is going to be H2O right so that is called law of definite proportion law of constant proportion law of fixed proportion that And the proportion in which the atoms are there is going to be fixed. Right? Now which proportion they are talking about they

30:39 are talking about same proportion by mass. Right? Let's discuss this thing here. You have you have H2O. Okay. This is H2O. Now I am saying proportion by atoms proportion by mass. He is saying specially mentioning that by mass I am saying I'll tell you why I am again stressing on by mass. Okay. So let's see you have you have H2 H2 hydrogen and oxygen by atoms is 21 right

31:24 by masses H2O is hydrogen and oxygen ratio is this is hydrogen is 2 oxygen 1 is 16 so it will be 1 ratio 8 okay so always and Always water is going to have 1 8 for H ratio oxygen and for water it for atoms it is 21 but let's say I am talking about D2O heavy water right we also call dutarium as hydrogen so butium if you talk about hydrogen here hydrogen ratio oxygen it is 2 1 if you take it as a isotope of hydrogen but here it will change here. It will change B2O hydrogen

32:12 ratio oxygen is going to be this is going to be 46. Understood? This is going to be 46 because if you see uh here you will have one neutron one proton that is mass is going to be 2 for one dutium. So 2 into 2 is 4 ratio 16 that is 1 4. So that's why I'm specially mentioning that we have to focus on the isotope also right. So if I say that ma'am here is ma'am I'll write dutarium only ma'am I'll write dutaterium only here right then you can say all the dutaterium ratio oxygen is going to be 1 ratio 4 D2

32:59 is always going to be 1 4 by mass and by atom it is 2 1 now by atoms H2O and D2 have the same ratio okay so there might be other elements also which will have the same ratio. Okay. So that's why to emit this confusion we don't apply by atoms rather we talk about biomass because by mass there will be more clarity right so that's why we talk about by mass same proportion of uh elements by mass clear everybody why we are talking about uh okay yes everybody same water only

33:44 great okay that is clear by mass not by atoms though by atoms also it is same but we are specifying on by mass to end up the confusion here right okay now let's talk about some questions on law of definite proportion one thing you should know it is also called as constant proportion it is also called as fixed proportion fixed proportion right okay Now let's talk about the question here right see law of definite and multiple proportion are the way of solving the question will be quite similar only so you have to pay attention here right okay so you have to pay full attention

34:33 right beta okay cool so let's see this question firstly they are saying that this much of copper oxide copper oxide cup oxide cup means CO2 2 positive this is O2 negative so that is CO this much of cubric oxide reduced by heating in a current of hydrogen and the weight of copper copper remained was this now if I see this experiment here in cupric oxide you have 1.375 and copper that remained was this that means 1.098 098. Okay. If I talk about the second experiment here, if I talk about the second

35:21 experiment here, that is in another experiment this much of copper was dissolved. Okay. Now we are talking about copper. How much of copper was dissolved guys? that is 1.179 was dissolved in nitric acid and the resulting copper nitrate converted into cupriic oxide by ignition. The weight of cupri oxide formed is this. Okay. Now the weight of cupri oxide is 1 bit. Just take out what is important for you. You don't have to write everything. Right? See what is important here. You had cupriic oxide right? you had copper. Similarly, here you had copper, right?

36:07 This is your hubric oxide. Okay. Now they are saying that you have to show that this is in accordance with law of definite proportion. Right? That's why I'm focusing less on definition and more on questions. Okay? So expansion of brain how thing is that if I say that this and this oxide has the same mass who can see that it has the same mass but you have to prove it with these with this data. How do you prove it? How we will prove it? We will simply say that if the percentage of oxygen is the same

36:54 in both the cases then that means you see if I fix the copper let's say I'm fixing the copper I am saying that for 100 g of copper how much oxygen you require right I'm fixing one thing then I want to find the next thing you can fix anything here Right? When you fix the copper, you fix copper oxide with there are three four methods to prove this. One thing is you can simply say that you can fix the amount of copper you can say for 100 g of copper how much oxygen is required. If in both the cases same amount of oxygen is required for 100 g of copper that means we are on the right track right. Or you can also

37:42 calculate the percentage of oxygen. If percentage of oxygen is same then also we can say that definitely we are calculating the same thing only. It's just the way of finding the answer is different. Right? So I would want to find the oxygen firstly if you see oxygen will be oxygen here will be equals to uh 1.375 - 1.098. Here oxygen will be 1.4. 476 - 1.179 right percentage of oxygen would be a better idea right because you know you have to fix one thing in your head otherwise you will end up getting confused in so many things right so - 1.098

38:29 that is 2777 now let's do this 1.4 4 beta in exam you will not get any calculator there okay I'm just showing you so that we can do things fastly 297 okay subtraction simple it is now if you find the percentage of oxygen percentage of oxygen will be 277 divided by how much you have initially 1.75 multiplied by 100 right and here percentage of oxygen will be they can also also ask you what is the percentage of oxygen in this. That's how to calculate the percentage of oxygen. 297 divided by 1.476

39:17 * 100. And when you will solve this, the answer is going to be right when you will solve this, the answer is going to be 20% in both the cases. Okay? I think it should be 20% only. Let me see if it is so right. Uh yes it is going to be yes it is coming out to be 20% only that means it is in accordance with the law of definite proportion right uh these kind of question can come in

40:02 your board exam can come. We can't say that I can't assure you that these kind of question will not come because actually it's not that of a difficult question. Okay? These kind of questions can come in your exam. Right? Because in competitive exam they will ask you the same question in the format of MCQ. Right? Okay. So you will can definitely see these kind of questions in your exams. Right. Next is same kind of question we have. Copper oxide was prepared. Now what we are saying copper oxide was prepared by three different methods. In case so you have three different methods here. Okay. So I'll just draw the line of three different methods here. In the first method uh metal gave this much of oxide. Metal is

40:51 copper. Copper is how much? 1.75. Copper oxide is how much? 2.19. Okay. In the second, copper is 1.14 and oxide is 1.43. Okay. In the third experiment, same thing you can do on your own also. Co copper is this much and copper oxide is copper oxide is 1.83. Okay. Now they are again asking you tell that this illustrate the law of constant proportion. So again what you have to do here simple is you will find the value of oxygen here that is 2.19 - 1.75.

41:40 Here oxygen is 1.43 - 1.14. Here oxygen is 1.83 - 1.45. Subtract this you will get the answer for oxygen here. Right? Then you have to find the percentage of oxygen. So I'll just do that for you. Uh -1.75. It is coming out to be.44. I am calculating the percentage of oxygen here. This is44 divided by the full copper is 1. This so you can see the answer is going to be divided by 2.19 again it is coming out to be 20% only. Right? So in all the other cases also what you will see it will come out to be 20%.

42:26 And 20% only you can also calculate this yourself. Okay understood m that's why I have picked out three four same type of questions so you can understand metal oxide right calculate oxygen then calculate the percentage of oxygen. Percentage of oxygen is mass of oxygen divided by total mass multiplied by 100. Same thing you will do here. Same thing you will do here and you will get the percentage as 20%. Is that clear to everybody? Is that clear students? Okay. Right everyone that is clear. Shall we move to the third one that is

43:11 law of multiple proportions? Law of multiple proportions. Third law that is law of multiple proportion. Law of multiple proportions states that you have two atoms right they can combine in different different proportions to form entirely different different products. Now this is water. This is hydrogen peroxide. You can drink water, you can't drink hydrogen peroxide, you will die. Right? So just one X, just one oxygen extra and entire properties are changed. That is multiple proportion. That means when two elements

43:57 combine to form two or more compounds the masses of now very important thing they are saying the masses of one of the element that combines with fixed mass of other element always be a simple whole number ratio like ma'am what you just said I have no idea right let's understand this thing so here you have water here you have H2O Now he is saying that two elements combine to form two or more than two compounds. Here I'm talking about two compounds, right? Masses of one of the element that combines with the fixed mass of other. So out of this one element has to be

44:44 fixed mass. So I'm taking this as fixed. This one is your fixed mass because in both the cases H2 is two right so I have fixed hydrogen here right so we have fixed hydrogen here right now he's saying that fixed masses of other always bear a simple whole number ratio right so if this is fixed here then the other one oxygen ratio oxygen here is going to be nothing but this is 16 32 that is 1 ratio 2 that is a simple whole number ratio simple whole number

45:32 ratio. Okay. So when you fix the masses of one of the element that combines with fixed mass of other this is fixed mass of so this element and this element will be a simple whole number ratio what you have to do fix the mass of one thing then other thing will have a simple whole number ratio I'll tell you this with the help of one more example let's say you have CO and CO2 okay you have CO and CO2 let's talk about each of them one by one, right? Okay. So, what we can do is I can write here as well CO and CO2. So, should I erase this thing here? Yes, CO

46:19 and CO2. Right? Let's write down CO and CO2. Simple thing is you have CO. This is CO2. Which mass I can fix? I can fix carbon here. Fix carbon. It is fixed. Now oxygen ratio oxygen 2 is nothing but 16 ratio 32 that is 1 ratio 2. That is the simple whole number ratio. This is what they have illustrated with the help of diagrams. You have carbon and oxygen. It forms two kind of compounds. One is CO and the other is CO2. Right? Okay. Understood? Now what he is saying that if you take the fixed mass

47:06 carbon is fixed here then this oxygen and the other oxygen from here they will bear a simple whole number ratio. Can you see this ratio this is nothing but one ratio too. Okay. Clear beta from the diagrams also. Is it clear that what I am talking about? Right. Okay. from the diagram also is it clear that what I'm talking about now if you see N2 N2 N23 let us see what we are going to do here right let's say you have N2 let's say you have NO2 you have N2 you have N2O3

47:53 you have N25 okay let's take an example of all these things Okay. So this is nitrogen. This is oxygen and this is the ratio. This is nitrogen. This is oxygen. You need to fix one mass. Which mass can you fix? Which mass? Which mass? You can easily fix nitrogen or oxygen. Say nitrogen is here also. N2. Here also N2 here also N2. Here it is one. Oxygen is 2 1 3 5. So oxygen is very difficult to fix. But nitrogen you can fix by multiplying this with 2. Now here

48:38 nitrogen will be 14 into 22 28. 14 into 22 28 28 28 and oxygen is going to be 32 into 2 that is 64. Here it is 16. Oxygen into 3 48 right? Oxygen into 5 is I uh huh 80. Yeah. Okay. So you have this thing. So can you see all these are in the simple whole number ratio 64 ratio 16 ratio 48 ratio 80. If you uh take the simple hole, this will

49:28 be 1. This will be 4 4 16 this is three. No, this is Wait a minute. Huh? So this is four. This is five. So is it a simple whole number ratio? Is it a simple whole number ratio or not. Is it a simple whole number ratio guys? Yes. No. So what you have here you have some oxide you fixed this mass. One mass you have to fix. When you fix one mass then you take the ratio of other masses right

50:13 of which element you are left with then what you will see that will be now I have fixed nitrogen here. So the ratio of oxygen in all the cases is coming out to be in the simple whole number ratio. This is the law of multiple proportion. Okay. This is the law of multiple proportion. Got it? Yes. Now let us see some questions based on this. So the question is on analysis it was found that black oxide of copper and red oxide of copper contains 80 89%. Okay. So what you have here is you have here black oxide of copper let's say you have no idea about that

51:02 what is black okay black oxide this is red ox I'm assuming that you have no idea you will say ma'am we actually have no idea right so black oxide of copper and red oxide of copper contain 80% of copper Black oxide is copper is 80%. So if I take copper oxide as 100 g then copper will be 80 g and oxide will be 20 g. Okay. Here if I take copper oxide it is let's say 100 g right here copper is going to be 100 minus

51:48 8 copper is going to be 89 right and oxide is going to be 100 minus 89 that is nothing but 11 g. Okay is that clear? What is copper? What is what we have g now if you see this here we have black oxide red oxide now you have to tell that this is multiple proportion see if I want to tell that this is in accordance with definite proportion I will say that percentage of oxygen should be same in this case I will say that the percentage of oxygen has to be different can you see that here percentage of oxy

52:34 oxygen in the same amount 100 g the percentage of oxygen is different. Can't you see that if in the same mass the percent if you want to calculate the percent also you can calculate but it is visible here that it is visible here that with the same amount if you take this also as 100 this also has 100 the answer of oxygen is coming out to be different that means here the oxide is also different both are different different oxides if you want to calculate the percentage of oxygen how you will calculate it. It will be 20 by 100 * 100 that is 20%. Here it is percentage of oxygen is 11

53:23 divided by 100 * 100 that is 11%. So if the percentage of oxygen is different that is definitely law of multiple proportion. Okay understood. You just have to see for a constant value of CO if the oxygen value is coming out to be different which is true. And now if you see copper oxide black oxide is Cu and red oxide is Cu O. Wait a minute. Cu2. Yes. Okay. Black oxide is this. Red oxide is yes. Okay. Yes, everybody

54:09 understood. Let's see the next question here. That is hydrogen and oxygen are known to form two compounds. Two compounds they can form. Okay. So let's see this here. But I'm just going through these questions. Okay. This is hydrogen and oxygen. Two compounds. Hydrogen content in one of these compound is 5.93 while in other it is hydrogen and oxygen other it is 11.2 2 percentage data is in accordance with multiple proportion. Is it a big deal? Everybody tell me is it a big deal? Total we have taken whatever is the I don't know what is this oxide but I am taking total HO whatever is

54:55 this this is 100 total is 100 so oxygen will be 100 minus 5.93 here oxygen will be 100 minus this is 11.2 2. Now this is 11.2. So you can easily see that for 100 g the answer of oxygen is different. Right? So it is definitely not this is definitely the percentage of oxygen is not same. That means this is in accordance with law of multiple proportion. Okay everybody? Yes Nandu. Okay guys, I think now you are find you must be finding it easy because we have done a lot of questions based on this

55:41 multiple proportion will have different percentage of the metal if you fix the full compound right let's say the compound you fix it that you're talking about 100 g then the percentage of any of the element should be same in definite proportion but in the multiple proportion the percentage of any of the element has to be different right okay yes everybody good now next law is Glokix's law okay next law is going to be Glokix's law now what is Glokix's law and evocu's law are comparatively compared to each other a little bit so we'll talk about them gay

56:29 states that when gases combine are produced in chemical reaction or they are produced and we are talking about at constant pressure and temperature at same temperature and pressure. What happens they do so in simple whole number simple ratio by volume provided that all gases are at same temperature pressure. Simple law is that if this is the reaction then the ratio of 1 ratio half ratio 1 this is nothing but 2 ratio 1 ratio 2 which is a simple whole number ratio. So if the gases are produced or the gases are combining in a chemical reaction, they will do so in a simple whole number ratio. Right? This is what we say in

57:16 Gusk's law. Provided that the temperature and the pressure is constant here. Ratio by volume of hydrogen oxygen and water is 1 this and this. That is nothing but that is nothing but 2 1 2 that is simple whole number ratio. Okay everybody? Yes everyone got it? Now one volume of nitrogen another example is habber's process all of these are gases. So here also the ratio is 1 ratio 3 ratio 2 1 3 and 2 that is also in the simple whole number ratio right now this 1 this 3 this two

58:03 this represents the volume only right if you're thinking about volume this also represents the mole this we will study in stoometry right this we will study in stochometry in this chapter only why we are saying that this is the volume right because of stochometry and stoometry we will study in some basic concepts of chemistry only one volume of nitrogen combined with three volume of hydrogen to form two volume of ammonia gas so the ratio will be same right stoometry means you can tell anything you can also say one molecule 3 molecule 2 molecule you can also say 1 volume 3 volume 2 volume you can also say 1 mole 3 mole 2 mole anything you can say one particle Three particle, two particle, anything. Right? So that we will study mole concept in

58:52 stoometry. Right? Now if you see this here, one volume of hydrogen, this one volume react with one volume of chlorine to form two volume of HCl. Can you see this here? How in the beaker how it is? This one molecule reacted with this one molecule to form two volumes of HCl. That means this one and this one. Right? So that's why what you will see here in total you have 1 2 3 4 5 here 1 2 3 4 5 and here you have 1 2 3 4 5 6 7 8 9 10 because 1 and 1 they are giving you two volumes of HCl that's why we write it as

59:41 H2 plus CL2 give rise to 2 of HCl okay understood everybody understood everyone or not what is a lok law that you can see from the diagram what is happening inside the beaker. Okay, what is exactly happening inside the beaker? Right students, now let us do this question here that says when 10 ml of propane gas is combusted completely. Propane is nothing but propane. This you will study in organic chemistry. This is nothing but this is the formula. So this is C3 32678. So it is nothing but C3 H8. Okay. So you can directly write it as C3 H8 that is combusted means oxygen. So you

1:00:28 can write it oxygen here. Right? This will give you carbon dioxide and water always in combustion this happens. Carbon is three you will write three here. Oxygen hydrogen is eight. So you can write four there. Okay. uh now oxygen also we want to make it six six and 10 we want to make it 10 so it will be five I think now it's done h now the this is complete balance reaction in chemistry you need to have balance equations okay so 10 ml of propane this is 10 ml

1:01:14 right completed volume of CO2 gas obtained. CO2 gas. See this is also gas. This is also gas. Right? If one mole is giving you three mole that means one volume is giving you three volume then 10 will give you 3 into 10 that is 30 ml. Okay. Then 10 will give you because what is GI's law? Gay Lusk's law say that the ratio is going to be same. That is going to be the simple whole number ratio. So if this is one, this is giving you three, right? That is simple whole number ratio. So ratio is 1 ratio 3, right? So if this is 10, this will be

1:02:00 30. So answer will be C is the correct answer here. Right? Okay. Next is Evogedru's law. Next we have is Evogedru's law. uh in Avagadro's law. This is also

1:06:55 working. Is it now working? We are deaf. Mic is muted. I'm so sorry. Right. I think my mic is not charged and I'm not able to find my other mic. Right. So I'll try to complete the session right I think it will be just five 10 minutes more it can work right so let's see to complete the session by then right so what I was saying that if you have if this is H2 this is CL2 right if I say that volume is double then number of molecules will be doubled if the volume is tripled then the number of

1:07:42 molecules will also be tripled If I say this is 3, this will also be 3 N. Okay. So that means if the volume is same, the number of molecules is same. If the volume is doubled, the molecules is doubled. If the volume is tripled, the molecules is triple. Now it doesn't matter that the uh that the gases should be same or different. It doesn't matter. Right? That is your Evogros law. Okay? Understood? Now guys, we have the last law. We have the last law here that is if the same weight of one element combine separately with two other elements but this definition you will find it very weird. Right? I'll just tell you directly here with the example.

1:08:30 Right? You have two elements. Let's say you have carbon and sulfur. two elements combine with the third element. Third element where the mass of that third element is fixed. Now the mass of this third element is fixed here. Okay? The mass of this third element is fixed here. Now this thing is fixed here. This thing is fixed. Right? If you see the ratio of carbon and sulfur here will be carbon is nothing but it is 12 sulfur is 32 right that is going to be the ratio that

1:09:17 is going to be the ratio but and simultaneously he is saying that these two elements that that have uh made a compound with the third element if you combine them together they can also make element. Right? Now the ratio here the ratio will be 12 ratio 32 into 2 that is 64. Okay. Here the ratio of carbon ratio sulfur will be 12 into 32 ratio 64. Understood? Now he is saying that these two ratios if you fix carbon and carbon here I have fixed this thing here then the ratio of sulfur and sulfur in these two cases

1:10:05 will be in the simple whole number ratio. Got it or not? Everybody got it or not? Now see this equation. See this definition. If the same weight of one element, same weight of one element, this is the same element, this is the same weight of one element, combine separately with two other elements, it is combining with two elements separately, right? Then the weight of those two elements that combine with each other are related to the weights that combine with the first element. They're saying then the weight of those two elements that combine with each other, right? So when they are combining with each other,

1:10:52 this weight ratio are related to the weights that combine with the first element is related to when they combine with the first element. Now in this the ratio is 12 32. In this the ratio is 12 64. Right? The ratio of these two things is going to be the simple whole number ratio that is law of reciprocal uh uh law of reciprocal proportion this you can say right so this law is not there in the NCERTT so that's why this becomes very very important okay students is that clear to everybody or not now let us do some questions based on that right now first is among the following pairs of compounds the one

1:11:38 that illustrates The law of multiple proportion is see in multiple proportion you need to have two elements that are same but combining in different proportions. So here the elements are different. Here also elements are different. In this case the elements are same. In this also elements are different. Here elements are same but the combining ratio is different. Right? So answer is going to be C. Next is which of the following pair shows law of multiple proportion? Same thing. Elements are different different different here the elements are same that's why they are combining but they are combining in a different proportion so that's why the answer is A is the correct answer next is there are two common oxides of sulfur

1:12:26 one contains 50% oxygen by weight other contains 60%. the weight of sulfur which combines with 1 g of oxygen are in the fixed ratio. Now beta we are saying that you have two oxides two oxides of sulfur usually we know that it is SO2 and SO3 right if you want to equate this here in one contains 50% of oxygen one contains by weight 50% oxygen means uh by weight it is containing 50% oxygen if you see they are talking about SO2 because here if you see this weight is 32 32 here this is also 32 so it is 50%. Right? So we

1:13:14 can say this is 50% of oxygen. In this case it is 60% because this is 16 into 3 that is 48 and this is 32. So that is nothing but 60% of oxygen. Now they are saying the weight of sulfur which combines with 1 g of oxygen is fixed is in the ratio. See in both the cases you have to tell in 1 g oxygen how how much amount you have the weight of sulfur they are asking you in which combines with 1 g of oxygen right. So oxygen is fixed as 1 g. Okay this also you take this also you take 1 g. So simple thing is 32 g of

1:14:02 oxygen gives 32 g of sulfur. 1 g of oxygen will give 32 by 32 that is 1 g of sulfur. Here it is 32 g of n 48 g of oxygen give 32 g of sulfur. 1 g will give you 32x 48. What is 32x 48? That is 1 and this is 16. This is 2 4. This is 8 1 2 2 4 6 2x 3. Right? So we can say that this is 1 g. Here you require 1 g of sulfur.

1:14:50 Here it is 2x3 g of sulfur. So the ratio is 1 2x3 that is nothing but 3 ratio 2. So answer is going to be D is the correct answer here. Okay students D is the correct answer. Got it? Now how we answer it? Understood how we are going to answer it. This is 1 g. They are saying in both the cases how much percentage of sulfur is there? We are fixing oxygen here. You have to fix. You can say that 32 g of oxygen gives this much of sulfur. This much of oxygen gives this much of sulfur. Then calculate the sulfur in both the cases. Then take the ratio. You will have your answer here. So answer is going to be very good. Manos craft. The answer is going to be a D is the correct answer

1:15:36 here. Right? Okay students. Next is CO and CO2. C2O3 follows which thing? CO you have CO2 beta this so you can easily tell me elements are same the proportion is different elements are same the proportion is different so we can say this belongs to law of multiple proportion that is B okay next question is two samples of lead oxides were separately reduced to metallic lead by heating uh in the current of this okay two samples of lead oxide the weight of lead from one oxide oxide was half of the weight of lead from other oxide. This is an example. See that means the

1:16:23 weight of lead percentage of lead is different. I told you that if you have uh two different compounds then the percentage of any metal or oxide is going to be different there. So that means they are talking about very good they are talking about law of multiple proportions here. Chemical equations are balanced based on which thing beta chemical equations are balanced based on law of reciprocal conservation of mass, conservation of energy or law of definite proportions. They are balanced based on the law of conservation of mass meaning conservation of mass. Now yes h we talk about mass conservation of mass. So answer will be B is the correct

1:17:08 answer. We are talking about mass that's why we talk about rearrangement right. So answer will be B is the correct answer here. Very good. Who proposed equal volume of all gases at same temperature and given pressure should contain equal number of molecules. Who proposed volumes and if you have equal volume it will have equal molecules. Who proposed this last fifth law that we did that was evoggate rose that is C is going to be the correct answer there. Okay students now students we have your homework take the screenshot now so many questions I have solved in front of you. Now do some hard work yourself as well. Put in some

1:17:54 efforts. Okay. So this is the question that you have to solve. First question. Okay. Then you have the second question here. PDF will be provided to you. You can take the screenshot of the second question here as well. Right? Okay. Yes. Everybody take the screenshot of the second question as well here. Right. Okay. You have to tell that which is the law that it is stating, right? And write down all your answers in the comment section. This is our telegram channel. you can subscribe where all the uh live sessions every update regarding meat is given on the telegram channel and I want you guys to write it down in the comment section the answers right I am waiting for your answers in the comment section

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