 Hi, good morning. All of you please type in your name. Hello Aadmesh, Vishisht. Aadmesh, how is your score? Suresh, Vaishnavi. Okay, so last class we have, you know, discussed a few questions of hydrocarbon, right? So, in the same chapter, okay, you got 96.6. Okay. Okay, work hard for the coming exam in April. Okay, you have enough time now. You can do much better than this. Okay. So, don't lose hope. See, whatever score you get in 96, 97%, 95%ile, from there, if you can improve at least, you know, 40, 50 marks, you'll be having a good score in JMAs. Okay, so at least you'll get into good NITs. Okay. So, that you can, 40, 50 marks, you can easily improve in this coming, in this, you know, the time that you have now. Right? So, the point is, like, you know, the portion where you have, where you have, you know, where you could not score marks in this particular exam, that portion you have to identify and start working on that. Okay. And again, for inorganic, I would suggest, read NCRT properly every day. Okay. Make some plan and read NCRT line by line every day. Okay. And solve some questions. Okay. But they probably ask, most of them they ask directly from NCRT. Okay, so keep that in mind. Anyway, so last class we have discussed few questions of hydrocarbon. Few more we'll discuss today and then we'll move on to states of matter. Okay. So, you see this question, what is the answer for question number 42? Tell me, Aritya is getting B. Why it is B, Aritya? Okay, yeah, that's correct. You see, option B is correct here, because the electron with the drawing nature of this NO2. Okay. So, since we have positive charge of carbocation, right, so stability of carbocation in the ring is directly proportional to minus I group, or we can say minus R also, and inversely proportional to, it is plus I and plus R, and inversely proportional to minus I and minus R, right, electron with the drawing group. So, all these NO2 shows electron with drawing nature, so that will decrease the stability of this positive charge, right. And here we do not have any electron with drawing group present, so comparatively B is the most stable compound if we have, right. Suppose if I write down one more example here, if I write down this molecule, right, here we have positive charge, and here it is suppose C triple bond CH, and all other things are same, HE. Then what is the answer? Suppose this is E, most stable one, right, and most stable compound will be like the least energy, right. Tell me the answer now. Okay. What is the answer? The C is SP hybridized more electron with drawing tendency. It is still B, right, because it is again, if you compare option B and E, right, option B and E, this carbon is also SP hybridized, right. This is again, electron pulling, right. So, again the answer will be B only. Any electron releasing group if you add, like suppose in this case, if I add here CS3, right. All other things are same. And if I add here CS3, then the most stable one will be this one having least energy. So, that is how you can do, okay, so concept, your concept would write, okay. Next question you see, which one of the following the most reactive towards ring nitration, ring nitration. Do you know the formula of this xylene, mesotylene, okay, so I will tell you the formula here, right. Xylene or mesotylene is 135 trimethylbenzene. Remember this, okay, 1,3,5 trimethylbenzene. And xylene is 1,3 dimethylbenzene. 1,3 dimethylbenzene. Now you tell me. But you remember these formula. Yes, right. Why it is A or why it is D? Tell me the reason. Why B, Rithvik, N2 has nucleophilic tendency. Aditya, first of all, it is ring nitration. So, for nitration you must remember, we use HNO2 and H2SF4. And you know, electrophilo, there is NO2 plus. This N2 plus attacks on the ring. If you remember. See the correct answer for this question is option B. Because it has 3 methyl group, right, 3 electron releasing group. And when N2 attacks, you see you will have a positive charge on the ring. 2 attacks here. And you will have a positive charge on the ring, right, because of pi bond attacks on this NO2. So, this positive charge will get stabilized by the electron releasing group, right. More number of electron releasing group, more will be the stability of this compound, right. So, we are following the most reactive towards nitration. Since the formation of carbocation is, depends on the stability of this carbocation, this reactivity. And this carbocation in presence of the electron releasing group is most stable. That's why option B, since it contains 3 electron releasing group, is the most reactive towards nitration. Correct? Option B is right. Yeah, most of you have got the right logic here. So, that's the correct answer option B. So, next question we are moving on, this one. Again, you see this is a nitration reaction. Where N2 attached, that's the question. Why it is B? Sondarya, Sanjana, Suresh, why it is A? Rithvika, Nathmesh? Yes. CF3 is an electron withdrawing group, okay. CF3 is electron withdrawing group. And it is deactivating group also, deactivating group. We can also say it is a meta directing. No, actually, CF3, you must have studied about activation, deactivating group, deactivating group, which groups are ortho-paradirecting and which groups are meta directing group. Okay, so CF3 is a meta directing group. This information you should have actually, okay. It deactivates the ring. It is a meta directing, deactivating group. So, any electrophile in presence of this CF3 group will attack at the meta position of the ring. Okay, that's why we get here option B as the answer. See what happens, any deactivating group or activating group if you have, that will affect the ortho-paraparaparation maximum always, right. So, if you have activating group present, okay, that activates the ortho-paraparation means the electron density at ortho-paraparation will be more. And hence, the molecule will, the electrophile will attack at ortho-paraparation, okay. But when you have deactivating group, that will, you know, decreases the electron density at ortho-paraparation, right. So comparatively, the meta position becomes a more electron rich. That's why the electrophile attacks on the meta position and we call it as meta directing group. Understood? So, this thing you must keep in mind, CF3 is deactivating ortho, sorry, meta directing group. Next question. Just a second, not this one. This is, we'll do one question and not look at it here. I'll give you a question here, not this one, but you solve this question first. One question you see this, then we'll move on to states of matter. The percentage of 1-chloro-lumethyl-propane, 2-methyl-propane, obtained in the chlorination of, chlorination of isobutane is, see, if there is a compound at meta position, then the reaction will not take place. If both meta position is occupied, if one is occupied, then the molecule will attach to the other position. If meta position is occupied, both meta position, then the reaction will not take place. Let me answer this question. Is it, how it is B? The question is getting D. Have you solved this kind of question before? Yeah, only two isomers are there, right from the level. This kind of question, have you solved before? Yeah, this kind of question actually nowadays they are asking also. You see, how do we solve this kind of question first? Okay, so first of all you see this table that I drew here. This one is for primary hydrogen. Okay, this one is for primary this column. This one is secondary and this is tertiary, the last one. And here it is the relative rate of hydrogen atom. Actually it is per hydrogen atom and this is for chlorine or Cl2. This value will be for chlorine. Similarly here also we have the relative rate hydrogen atom and this one is for bromine Br2. To solve this kind of question, these relative rate you have to memorize. Correct. So for chlorine, if the hydrogen is primary, the value is one for one hydrogen atom. For secondary it is 3.8 and for tertiary it is 5. All these are relative rate per hydrogen atom for chlorine. Similarly for bromine it is 182 and 1600. So first of all to solve these questions you have to have these values, 1, 3.8 and 5 for chlorine, 182 and 1600. Now you see the molecule that you have is isobutane. Isobutane is this CS3, CS3, CS3 and one hydrogen here. So in this molecule, how many one degree hydrogen can you tell me? How many one degree hydrogen we have? How many secondary hydrogen we have? How many tertiary hydrogen we have? Can you tell me? Degree hydrogen? Is it 9? Correct. 3, 3 and 3. One degree hydrogen are 9s on the AC. 3, 3 and 3. So there are 9 one degree hydrogen. Do we have any secondary hydrogen? No. 3 degree hydrogen we have only one, correct. So now when you try to do the chlorination of this molecule so what are the two products we get here? One product will be CH3, CH3, CH2, Cl and other product will be CH3, CH3, CH4. this is the only two products possible right so you see this molecule forms by the reaction of 1 degree hydrogen and this is formed by the reaction of tertiary 3 degree hydrogen so for this molecule if you have to find out you know the portion or the composition of this molecule that will be the number of number of 1 degree hydrogen into its reactivity reactivity of what reactivity of 1 degree hydrogen since it is formed by 1 degree hydrogen right it is the composition of this one so how many 1 degree hydrogen we have 1 degree hydrogen we have 9 so that will be 9 into reactivity is what for 1 degree hydrogen for chlorine reactivity is 1 so 9 into 1 is equals to 1 similarly you have to find out the composition of this also so number of 3 degree hydrogen is what 1 and its reactivity is 5 for chlorine so it is 5 okay the two molecules we get a and b suppose this one is a and this one is b right a and b so total will be what a a plus b right so percentage of a will be what a divided by a plus b into 100 that will be equals to percentage of a similarly we can find out percentage of b also or hundred minus a so that will be what a is 9 9 by 14 into 100 so it will be you see it is 450 by 7 so 450 by 7 will be nearly 64 percent you will get is it clear understood so nothing you have to do just you have to find you have to keep this data in mind one three point eight and five one eighty two at sixteen hundred mostly maximum time they'll ask for chlorine only okay so at least one three point eight and five you must keep in mind would you solve one similar question on this yeah yeah you need to keep that in mind because they will not give you these rates one three point eight and five so this you have to memorize with respect to the atoms I too they don't ask usually because the reactivity of iodine is very you know less we need some oxidizing agent for iodination right for iodination we need some oxidizing agent what are those oxidizing agent HNO3n iodic acid HIO4 so generally iodination is very difficult you know is a bit difficult procedure so we don't do iodination so mostly they'll ask chlorine only but I'm giving you one more here that is bromine you can find out B also the question was this only one chloro-2 methyl propane so one chloro-2 methyl propane is this only you see they haven't asked about this yes right they asked about it only you see the questions any doubt you have do you want to solve one similar question on this just for practice okay one more question so question you see you need to find out the percent is yield off so you need to find out the percent is yield off monochlorinated product in the chlorination of of butane tell me the answer see actually chlorinations on the real chlorination of you know alkene what kind of mechanism it shows can you tell me it is free radical mechanism yes and free radical mechanism best order is what 3 degree maximum then 2 degree and then 1 degree yeah right so what is the stability order free radical maximum for 3 degree then 2 degree then 1 degree that's why the activity of tertiary is maximum here answer this question you don't have school later today is it 33.33 can you give me the exact value visit assignment you have finished class to read engine right what time you have finished class engine okay see here is this CH3 CH2 CH2 CH3 so number of primary hydrogen is 6 and secondary hydrogen is 4 there is no tertiary hydrogen here 28.3 that's correct so I mean that's the exact answer to you get okay so you get basically you get here see the answer of this is there are two answers of this okay both you have to write one will not give you the answer marks right because the right to understand the question yield off monochlorinated product so first of all we have to find out that how many monochlorinated product we get you understand this so how many monochlorinated product we get into this monochlorinated product we get here so there are two monochlorinated product we get so we have to find out the percentage competition of both not just one right so one will be supposed x so the will be 100 minus x that's the thing right so why it is to because we have two different position one is this either the coding will attach on this carbon or on this carbon but this two carbon are having the equivalent position and these two carbon is again the same position we have right so there are two possibility we have so one will get when the one degree hydrogen takes part in the reaction then the product will be this CH2C and the other product we get when secondary hydrogen takes part in the reaction CH2CH3 so now the composition of this if you have to find out we again use the same data that we have done in the earlier question so since it is it formed by one degree hydrogen so reactivity of one degree into number of one degree hydrogen that will be for one degree it is 1 and number of one degree hydrogen is 6 that is equals to 6 and for this one number reactivity of secondary hydrogen is 3.8 into 4 that gives you 14.3 so the percentage of this will be 6 divided by 6 plus 14.3 into 100 and this comes around it is 15.2 right sorry it is 15.2 it is 15.2 6 plus 15.2 so when you solve this you will get 28.3 percent right it is 28.3 percent and then the 100 minus this will be the composition of this product right so that is how we solve this kind of question okay generally they ask you the chlorination only they would they won't ask you you know the bromination of the molecule right so if you remember for chlorine you will you can do the question but in the bromine one is not that tough also 182 and 1600 okay so that's the two things you have to keep in mind then you can solve this kind of question is it clear understood haven't asked this kind this type of question in this JEMN exam okay try to you know figure out this thing also that in one particular chapter what kind of question they did not ask okay so now we are moving on to the next chapter that is states of matter solve this one one of you are getting C option C is right since all of you are getting so I'm not going to solve this question if you have any doubt you can ask me then I'll solve this otherwise I'll move on okay I'll move on then option C is correct what about this question the first one is this it's the pressure at constant volume all of you are getting A I think this should be PV only PV versus V it should be PV versus E okay option A is correct here this clear this obviously P1 and 1 by V right so B is 3 B is 3 we have only one option A is correct right this is the graph of pressure and volume so A is 2 this is the graph of PV versus V next question you see all of you are getting D atmosphere is getting C okay see equal weights we have so far number of moles of ethane will be what suppose a mass we are assuming M divided by molecular mass of this which is 30 similarly the number of moles of hydrogen will be M divided by 2 so mole fraction of hydrogen that will be equals to M divided by 2 M by 30 plus M by 2 so this you have to solve and you will get 15 plus 1 16 and then 15 so we'll get 15 by 16 here right and we know the partial pressure of hydrogen will be what mole fraction of hydrogen into total pressure that will be partial pressure of hydrogen really is what 15 by 16 into PT so that 15 by 16 the fraction of total pressure we have right that's why the answer is option D okay next question what happened tell me the answer D is the correct answer option D is right you see how do we do this what is the volume of suppose the volume of mixture I assume volume of mixture is 100 right so what we can write the total mass of the mixture is equals to what the mass of N2O4 plus the mass of NO2 mass of the mixture will be what the density of the mixture into volume which is 100 and that will be equals to the density or vapor density of N2O4 into its volume plus the density of NO2 into its volume right this we have to find out what all it out here that you have to see volume of N2O4 NO2 we can find it out vapor density or density also we can find out since we know the molecular mass of these two elements molecules right so you see here the volume of N2O4 if I ask you what is this can I write this 2 by 3 into 100 ratio is given since the 200 by 3 similarly volume of NO2 is 100 by 3 now since molecular mass of N2O4 is given is we know the density of N2O4 will be what molecular mass that is 96 92 sorry 92 divided by 2 which is 46 and the density of NO2 is 46 is the molar mass divided by 2 is 23 all these value will substitute here and we will find out density of the mixture so density of the mixture will be density of N2O4 46 into 200 by 3 plus 23 into 100 by 3 this divided by 100 so 100 gets cancelled so it is 92 by 3 plus 23 by 3 100 and 15 by 3 that will be 38.3 you'll get okay one more thing you can observe here density of the mixture we can also write directly that the density of NO2 plus density of N2O4 divided by divided by the total NO2 and N2O4 that you have is 3 here in this case if this NO2 and N2O4 is mixed in A is to B ratio then we should write here A plus B if the ratio of N2O4 and NO2 is A is to B by volume in this kind of question they always give the ratio in volume only otherwise it is very difficult to form you should have then some other details this you must keep in mind right and we know this also NO2 and N2O4 molecular mass we know so we know the density also that we can directly use you right so option D is right next question you see Vaishnavi, Saimahir, Atmesh is getting A, Purvi is also getting A, Aditya is getting B A is the right answer Aditya cross check your calculation see the question is the option is 81.1 centimeter from NS3 did you get it from NS3 or HCL Aditya this is HCL and this is NS3 all of you have got answer A, A is the correct one and this is NS3 suppose I assume from a distance from this HCL end will get the white fume over here right this should be obviously 200 minus A because it is given 200 centimeter so we'll directly write here rate of diffusion of HCL R of HCL divided by R of NS3 that will be equals to the molar mass of NS3 divided by molar mass of HCL root over of it and rate is equals to what distance level divided by time so for HCL it is A by T for NS3 it is 200 minus A by T is equals to molar mass of NS3 is 17 for HCL it is 36.5 this is what you have to solve and you will get A because T and T gets answered A value will be 81.1 centimeter from this HCL side hence option A is correct next question you see 46 which is the sixth question a mixture of NS3 and N2H4 is placed in a sealed container okay this is sealed container at 300 Kelvin a total pressure is 0.5 atm the container is heated to 200 Kelvin the both substance decomposes completely according to the equation according to the equation after decomposition the total pressure at 1200 Kelvin is found to be 4.5 1200 Kelvin is found to be 4.5 atmospheric what is a percentage of N2H4 gas in the original volume in the original mixture assuming ideal behavior is it C what about others tell me the answer okay you see for this question option D is correct option D is right you see here first of all we have to find out the percentage of N2H4 in the original mixture right so in the original mixture we do not know the number of moles of N2H4 and NS3 so what we assume N1 is the number of moles of N2H4 initially and N2 is the number of moles of NS3 initially all these data are in the initial condition according to the question what we have to find out we have to find out N1 by N1 plus N2 into this is what we have to find out so either we'll get N1 and N2 then we'll find this or directly if you can get the value of this expression N1 by N1 plus N2 then also we can get the answer okay now this is the question we have now coming back to this question both the molecule dissociates according to this reaction you see from the first reaction what we can say that two moles of NH3 gives four moles of product total I'm talking about and in this four moles we have three moles of H2 and one moles of N2 but two gives four that is what is required here and one moles of one mole of N2H4 gives three moles of product right so since two moles gives four so what is the number of moles we have assumed here and for NS3 it is N2 so N2 moles gives four and two sorry N2 moles gives two and two two gives four one gives two and two gives two and two one gives three so N1 gives three and three right this is the final number of moles we have no this is the number of moles we have for N2 what we say and sorry two moles of NS3 and two no this is N1 sorry this is N1 three N1 right this is the number of moles of product that we get from the first reaction and the second reaction correct so now we can apply since the molecule follows ideal gas behavior so we can apply PV is equals to NRT so in initial condition what we can write you see pressure into volume is equals to NRT correct so pressure is given 0.5 atmospheric so 0.5 into V is equals to N1 plus N2 into RT a total number of most initially is N1 plus N2 only when it dissociates the pressure becomes 4.5 right it is given in the question 4.5 into volume is V the number of moles will be 3N1 plus 2N2 RT now when you take the ratio of these two okay I'll take this is one and this is two so to simplify the you know things I'll take 2 by 1 so it is 3N1 plus 2N2 divided by N1 plus N2 is equals to is equals to what should we write okay we have different temperature no we have 300 and 400 so it is T1 and T2 I should write so this by this is equals to 4 by 2 by 1 so it is 9 by 4 only you need to solve this now you see what we can do next here since N1 plus N2 we have here and we have to find out N1 by N1 plus N2 so I'll write this 3N1 as 2N1 plus N2 plus N1 divided by N1 plus N2 is equals to 9 by 4 when I divide this here so you get 2 plus N1 divided by N1 plus N2 is equals to 9 by 4 from this N1 divided by N1 plus N2 is equals to 1 by 4 which is nothing but 25 percent that's my option D is correct so like this you can solve your calculation will be a bit easier or you can find out N1 plus N2 solve the equation you'll get N1 plus N2 and then you'll again find out N1 by N1 plus N2 any doubt you have in this one question number 7 done Shreyas is getting A so on the way I was getting C tell me the answer of this Orvic may it is getting A atmosphere also getting A right you see option A is correct here in this question now you see how do we solve this question first of all in the question it is given at STP this gram of O2 diffused through a porous partition in this second what volume of CO2 diffuse in the same time under the same condition so what we can write the rate of diffusion of O2 divided by the rate of diffusion of CO2 is equals to the molecular mass of CO2 divided by the molecular mass of O2 root over of it which is nothing but 44 by 32 root under right and rate also we can write what the volume of O2 diffused divided by the time which is given here this divided by volume of CO2 diffused divided by the time and this will be equals to 44 by 32 root under so we have to find out since time is given excuse me time is given which is 1200 right and this is same for both the gases right so we can easily excuse me eliminate the time here correct volume of CO2 we have to find out so if you know the volume of O2 we can find out this right so mass of O2 is given mass of O2 is given which is 0.48 gram so we can find out the volume here because it is at STP right so we know at STP it is given at STP at STP 32 gram of oxygen which is 1 mole 32 gram of oxygen is equals to 22400 ml volume 1 mole is equals to 22.4 liter so these many ml so what is the mass we have here 0.48 gram so volume will be 22 4000 into 0.48 divided by 32 so when you solve this you will get the volume as 336 ml this volume will substitute here 336 divided by volume of CO2 is equals to 44 by 32 root over of it so volume of CO2 when you solve this you will get 286.5 ml okay see one more thing I would like to tell you here you see the volume of CO2 that you are calculating if you are getting any if you are getting any doubt into this you see it is 336 into this by this is what 8 by 11 root under so obviously this value is what it is less than one right first of all this value will be less than one so the volume of CO2 should be less than 336 correct so you cannot get this as an answer option D right and this is less than one but it is not that much small also so that you can get this 113.2 correct so these two options you can eliminate easily you will have doubt in A and B and for that you can you know calculate this and then you can understand understood so like this you can eliminate few options also and avoid a little bit of calculation here understood next question you see all of you are getting B okay yeah option B is correct can I move on or you do you want me to solve this B is right so all of you have got the answer so I'll move on right see the next question yeah I see this dry hydrogen chloride and NH3 right so it depends on the diffusion of gas so which the gas which has the lower molecular mass will diffuse faster right and the ammonium chloride ring forms near that particular gas into that particular gas so since the molecular mass of HCl HCl is more than to that of NH3 right so hence the diffusion of this will be higher diffusion rate is more and hence this ring forms near this NH3 gas end of this NH3 what to write so answer will be option C diffusion rate of this is more so the ring forms near the hydro chloride gas okay so this one is hydrogen chloride and this three will diffuse more to the ring forms near the end of HCl yeah right right so diffusion of NH3 is more so NH3 will travel more distance so ring forms near HCl and that's what it's correct option B is right 10th one A is correct level you tell me 11th one all of you are getting A A is also correct here if you have doubt you tell me I'll just do this otherwise I'll move on let me know do I need to solve this 11th one see you guys E is the kinetic energy and U is the rms velocity the U rms we know or V rms if I write it is equals to root over of 3 pv by m we also know this 3 this pv is equals to kt for n is equals to 1 for one molecule where k is the Boltzmann constant this Boltzmann constant is nothing but r by na r is the gas constant here right kinetic energy formula which is nothing but e is equals to 3 by 2 kt pv or kt right 3 by 2 kt so kinetic energy or kt value 5 by 0 from here that will be 2e divided by 3 right so this is the value of kt or pv we can write this and this I am going to substitute it here so V rms is equals to 2e root over of it hence option A is correct option A is correct understood 12 and 13 why it is B did engine how it is D yes kinetic energy depends only on temperature correct and it does not depend on any other factors right so when the temperature is same the kinetic energy of the molecules gases molecules are same same as that of hydrogen molecule got it engine question number 13 that is also B option B is right kinetic energy is directly proportional to the root over of temperature by molecular mass this is what you have to use you'll get the temperature for SO2 because molecular mass you already know okay 13th one B is correct next one yeah it is same as the previous question right so again option C is correct kinetic energy is directly proportional to temperature right so kinetic energy at 40 degrees Celsius means 313 Kelvin divided by kinetic energy at 20 degrees Celsius means 293 Kelvin will be equals to 313 by 293 option C is correct 16th one it's a direct question okay you must have seen this question at many different books one of these are what we have R A M for root minus square A is for average and M is the most problem so root minus square should have the maximum value if you remember this order you can easily say option B is correct over here root minus square will have maximum value and most problems will have minimum option B is correct solve this one in question number 14 just I will check for what happened in 14 question number 14 you see the first one here VRMS of H2 divided by VRMS of O2 that will be equals to temperature of a molecular mass divided by temperature by molecular mass this is for O2 and this is for H2 root over of it right so H2 is at 50 Kelvin so we'll write 50 here divided by temperature of O2 is 800 okay so this will get cancelled by 16 and it's one only there's one only four big you see I'll get it one that's what I have done you see so option C is correct next question you see what about question number 20 you did not solve question number 20 tell me 21st 21 is right compressibility factor is less than 1 so molar volume is also less than 22.4 Sanjana is getting D question number 20 question number 20 have you finished see the value of critical pressure and temperature we know Z at critical condition is equals to what PC VC divided by RDC okay this is the value of Z so PC is equals to what at critical condition it is A by 27 B square VC is equals to 3B and Tc is equals to 8A by 27RB if you substitute these values you are getting 3 and 8 here so the answer you'll get here is 3 by I have solved question number 22 we'll take a break after this done question number 22 do that question of Sanjana 22 you are getting D you know then what is the mass you are getting molecular mass or simply you said since alpha is there so it should be S8 see in this type of question when you have to find out the molecular mass correct so we know one atom the mass of one atom of that particular element so here it is sulfur so one atom of sulfur the mass is 32 gram if you have the value of molecular mass molecular mass if you find out then we can find out the number of atoms present in this sample right that is molecular mass divided by 32 right suppose the molecular mass you are getting X so number of sulfur atom is what number of sulfur atom is equals to 32 sorry X divided by 32 and suppose it is N or Y so molecular formula will be what Sy that is the answer so point is what we have to find out Y right we have to find out this value Y and for that we need to find out the molecular mass right so here you see what all data are given mass of sulfur is given it is 3.2 gram molecular mass is not given I'll assume M temperature is given 450 degree Celsius it is 450 so it is 450 plus 273 is equal to 723 Kelvin pressure is 723 mm of Hg so this divided by 760 atm volume is also given that is 780 mm so 780 divided by 1000 it is liter so we can apply this pv is equals to nrt because S vapor we have here so pv is equals to nrt so you see pressure we have to find out this molecular mass and we can write mass divided by molecular mass so molecular mass is equals to what mrt which is 3.2 r value is 0.0821 temperature is 723 mrt divided by pv pressure is 723 into 760 volume is 780 into 1000 this is what you have to solve I have converted into atm so that we can get the mass in gram yeah that also you can use so molecular mass when you solve this this will get cancelled right and you will get here because this you have to solve whatever calculation you have you have to do this so when you solve this you'll get molecular mass is 256 gram so the value of y will be what it is 256 divided by 32 that will be 8 hence the molecular formula is s8 option d is right understood this right got it all of you okay so we'll take a break now we'll start at 1135 okay okay so we'll resume the class at 1135 take a break okay can we start the class this question you see solve this 23rd 24th tell me the answer for 24th one 23rd one I'll solve what is the answer of 24th tell me first 23rd okay I'll see this 24th one even graph represents the variation of z versus p for three gases a b and c identify the only correct statement okay so there are two possibilities we have here there are three possibilities actually one is z can be greater than one z can be less than one right it is greater than one when a is equals to zero or it is less than one when b is equals to zero we can derive this also from the no van der Waals equation we can derive this so what happens here this a will be equals to zero when there is no interaction and this b will be equals to zero when there is decrease in pressure right so z will be less than one z will be greater than one the condition is this we have and that is less than one the condition is this we have you see the option a for gas a a is equals to zero and this dependence on on p is linear at all pressure okay so for a is equals to zero you see we'll get z greater than one we can derive this also you see the van der Waals equation is this p plus a by v square v minus b is equals to rt when a is equals to zero so we can eliminate this when there is no intermolecular force we can eliminate this with respect to this so we can we get p into v minus b is equals to rt when you solve this you'll get z is greater than one compressibility factor should be greater than one when we are assuming there is no intermolecular attraction that means a is equals to zero so right a option is correct here for gas a z is increasing when small a is equals to zero now we have to find out the incorrect statement for gas b when b is equals to zero its dependence on p is linear at all temperature that is also linear at all temperature so b option is also true for the gas c which is a typical real gas for which neither a or b is equals to zero by knowing the minima and the point of intersection with that is equals to one a and b can be calculated we cannot calculate this a and b because it depends on what it depends on the gas what gas we are using a value and b value depends on the gas that we have right by this graph we cannot calculate this a and b value right so this is one thing so this option i think it is wrong at high pressure the slope is positive for all real that is also true pressure keeps on increasing and slope is always positive and slope is also d option is also true right so c option you see it is a and b we cannot calculate from this graph hence c is wrong the answer will be from c the graph c represents real gas okay you see when pressure is high the real gas it starts behaving as an ideal gas you see at high pressure the slope is positive we are getting it is true option d is it clear understood question number 23rd you see what is the difference in the density of dry air at one atmosphere at 25 degrees Celsius right and moist air so first of all we will find out the density of dry air what is the molecular mass of dry air md is for dry air right that will be since the air contains what mainly oxygen and nitrogen right air contains mainly oxygen and nitrogen and it is given in dry air the composition is given you see here so what we can write you see molecular mass of dry air is equals to the molecular mass of oxygen that is 32 into its percentage composition 24.5 given in the question plus for nitrogen 28 into its percentage composition 75.5 divided by total we are assuming as 100 this when you solve you will get 28.98 molecular mass of the dry air gram per mole so density of dry air will be what pressure into molecular mass divided by RT formula of density we have this only so pressure it is given 1 molecular mass we have calculated 28.98 divided by R value 0.0821 into temperature is 298 25 degree Celsius 298 and this gives you approximately 1.184 density so this is the density of dry air now for this moist air you see relative humidity is given the formula of relative humidity is what RH the formula is the partial pressure of H2O in air divided by the vapor pressure of H2O vapor pressure of H2O partial pressure by vapor pressure relative humidity is given it is 50 percent and vapor pressure is also given 23.7 ton right so when you solve this you will get the partial pressure of H2O and that will be 50 percent relative humidity 0.50 into vapor pressure of H2O is 23.7 it is given divided by 760 so this becomes atmospheric so this value will get 0.0156 atm pressure of H2O is this the percentage of H2O vapor it is a vapor pressure of H2O right the percentage of H2O vapor will be what percentage of H2O that will be the pressure of H2O into 100 divided by the total pressure we have or atmospheric pressure we have 1 which is equals to 1.56 percent now this is the value of H2O vapor present in air apart from oxygen nitrogen right so what we can write the percentage of N2 and O2 will be what that will be 100 minus 0.156 see moist air what happens we have a little bit of dry air if you have we assume that there is no water content there is no moisture into that but when you have moist moist air so some water also converted into vapor so apart from N2 and O2 will have water vapor also so we have calculated water vapor the percentage of water vapor present 100 minus that gives you the percentage of N2 and O2 so that will be 98.44 percent now the molecular mass of wet air will be what that will be equals to that will be equals to 28.98 right 28.98 into 98.44 plus the water vapor we have there molecular mass of that will be 18 into 1.56 this is nothing but the molecular mass of dry air here this into the percentage composition of N2 and O2 because dry air contains what nitrogen in oxygen so mass of dry air into the percentage of that that's why I have taken this this whole divided by 100 again and take it because the mixture is 100 only so when you solve this you'll get 28.81 molecular mass of wet air right now you see similarly like we have calculated the density of dry air we will also calculate here the density of wet air because the molecular mass we have that is the task we have in this question we have to calculate the molecular mass of dry air and wet air after that fine density formula will use okay so you see the density of wet air will be what again P Mw divided by RT that will be equals to when you substitute all the term here 1 into 28.81 divided by 0.0821 into 298 that this gives you 1.177 so the difference here it is what 1.184 minus 1.77177 and this value gives you 0.007 approximately so answer will be option B so I'll explain this thing again the objective is to find out the density of dry air and wet air correct difference we can find out and density formula we know that is P M by RT density formula we already know that is P M by RT where M is the molecular mass if you are calculating the density of dry air molecular mass will be of dry air for wet air this will be for wet air so our objective is to find out what is the molecular mass of dry air and wet air both correct engine what is the molecular mass of dry air and wet air both so for dry air you see we know the composition of N2 and O2 first of all in dry air we have nitrogen and oxygen only we'll assume this is also given in the question so the molecular mass of N2 into its composition plus molecular mass of O2 into its composition we'll give you the molecular mass of dry air is 28.98 right this value will substitute here you'll get the density of dry air now for moist air the term relative humidity is given and this formula you should know for relative humidity the partial pressure of water in air divided by the vapor pressure of H2O this is the formula of relative humidity right so vapor pressure of H2O also it is given in the question so we'll find out partial pressure of H2O why we need partial pressure of H2O because this is the pressure so when we find out the H2O vapor in you know that will be what partial pressure into total 100 we have divided by the atmospheric pressure which is one atmospheric right so the you know the percentage composition of H2O in vapor into its pressure that is atmospheric pressure is equals to the total amount we are assuming 100 into its partial pressure correct so we'll find out find out the percentage of H2O vapor in air that is this so now in the air apart from nitrogen in oxygen we'll have what we'll have water also water vapor also and that composition we have so 100 minus this gives you the percentage of nitrogen correct now this is the percentage of nitrogen into the molecular mass of dry air which is nothing but 28.98 plus the molecular mass of H2O into its composition by 100 this gives you the molecular mass of what molecular mass of wet air now again we'll use this formula and we'll find out the density of wet air and we get the difference did you understand this so this is a different you know type of question we have generally they don't ask this relative humidity question and all but this formula you must remember okay because this time I'm trying to you know we have already done a set of questions in the previous problem solving session now since you have exam we don't know what question they are going to ask so I'm trying to cover up all those different different types of question that is possible in one particular chapter right so this is a different types of question you must remember these two formula relative humidity and all okay just a second next question you see tell me fast C is correct next one you tell me 26th temperature at which the vapor pressure of a liquid is equal to external pressure is called boiling temperature A is correct at high altitude atmosphere pressure is high wrong so A is correct R is false option C is right right all of you have done this next question you see this one 27th and 28th 27th see the water vapor density is given right 0.0006 gram per centimeter cube when you multiply this is volume occupied by water molecule in one liter right so mass of one liter of steam if you take right so one liter of steam you have here so this into 1000 ml nothing but one liter is equals to 0.6 gram of water centimeter cube 0.6 gram of water we have this is the mass of of one liter steam which is nothing but the mass of the liquid water correct so mass of one liter steam is this density is given 1 gram per centimeter cube this see the amount of steam form that will be equals to the amount of water or liquid water vaporize correct so mass of the steam is equals to the mass of the water correct so mass of the water is also 0.6 gram volume is the density is given so volume is what volume of water is equals to 0.6 into 1 that is 0.6 gram per centimeter cube the same volume is occupied so answer will be option C right 28th one will balance the charge here simply so you see the metal oxide is 0.980 and in this we have m2 plus and m3 plus also so suppose that 3 plus we have to find out so we'll assume this as x so this will be 0.98 minus x so total charge will be what here it is 3x and here it is 0.98 minus x into 2 this plus this plus the charge on oxygen is minus 2 is equals to 0 because of oxide as a whole is neutral this is neutral so total charge and my total charge on this metal is this plus charge on oxygen is minus 2 is equals to 0 so when you solve this x you'll get 0.04 so percentage will be what 0.04 divided by 0.98 into 100 which is nothing but 4.08 percent is last two questions solve these two 28 29th one is based on which concept yeah it's bragg's law so what is the formula of bragg's law that you apply simply all data are given you will find out the wavelength the formula we have here is n lambda is equals to 2d sine theta n is given first order diffraction one set of crystal phase at an angle of 11.8 so theta is 11.8 degree these values are not given in the question it should be given d is what 0.281 nanometer so 0.281 into 10 to the power minus 9 meter all these values put and you'll get lambda is equals to 0.1149 nanometer this is the answer only substitution and calculation you have there yeah c you can use calculator to get sine value sine 11.8 that's not an issue if they ask this question in the exam they'll give you the value of sine theta don't worry with that they'll give you this value in this exam don't worry with that done last question for today question number 30 solve this how many units cell are present in a cube shaped ideal crystal of nscl no b is not correct see the mass of one unit cell if i ask you the mass of one unit cell is equals to the volume of one unit cell into its density density we know volume is a cube where a is the edge length density is what z into m divided by a q into n a avocado number a q a q gets cancelled mass of unit one unit cell is z m by n a nscl it is given so nscl we know it forms fcc lattice so z value is 4 for that mass we also have molecular mass of nscl 58.5 n a is 6.022 10 to the power 23 this will be equals to when you solve this 38.87 into 10 to the power minus 23 gram right so now you see this is the mass of one unit cell right so 38.87 into 10 to the power minus 23 gram is equivalent to one unit cell we have to find out the number of unit cell in one gram so in one gram the number of unit cell will be what one divided by 38.87 into 10 to the power minus 23 so when you solve this the answer you'll get is 2.57 into 10 to the power 20 so it's a did you understand this correct got it it is nscl so fcc so z must be 4 so we'll wind up the class here only okay next class we'll solve some other topic right so the class schedule and that will be probably on friday i will let you know correct thank you all ncrt you must read for inorganic chemistry i'm telling you again and again okay try to finish ncrt at least two times in this two months okay so thank you all bye