 Hello. Hello, good afternoon. Can you start? Sorry, I am late. I got some work to finish in the office. That's why I got not late. Are you there? Please type in your name, all of you. Okay. So, today we are going to solve some questions, numerical questions. There are a few questions which are, you know, subjective here today, but they are not advanced level. They are men's level only. The only thing is the options are not given into that. Okay, so a few questions I have, you know, opted like that also without options. Okay. So, the first question you see, solve this one. The formula weight of an acid is 82, 100 centimeter cube of a solution of this acid containing 39 gram of the acid per liter were completely neutralized by 95 centimeter cube of aqueous NaOH containing 40 gram of NaOH per liter. What is the basicity of the acid? Anyone, you want to get some hints? Just equate number of equivalents. Number of equivalents you equate, round it off. What is the closest value? Round it off. What is the closest value you are getting? Okay. Vastavy is getting 2, Simeon is getting 2. 2 is the correct answer. 2 is the correct answer. Anyone else? If you get 2, resist? Okay. Now I'll just do this. What do you have to do? The equivalence of acid is equals to the number of equivalents of base. Equivalence, you know, what is the equivalence? Number of equivalents will be normality into volume. For acid is equals to normality into volume of base. Okay. So, normality of acid is what? You see the formula weight is given. Formula weight means what? We can consider this as molecular weight. Right? So, number of moles. So, normality will be what? It is number of equivalent. So, for this, you see acid. How do we find out the number of equivalents? Okay. So, first of all, you see mass of acid is given and it's molecular mass we can assume as formula weight. So, number of moles of acid will be what? 39 by 82. It is the number of moles of acid. Right? And from this, we can write down molarity. What is molarity? It is the number of moles which is 39 by 82 divided by volume in litre. And that will be what? That will be 100 into 1000. And from this, we can find out normality. Right? So, normality will be what? Molarity into n factor. Molarity into n factor. Now, this n factor only for acid we have to find out because we have to find out the base city of acid, which is nothing but this n factor. Now, you see the normality will be it will be 39 divided by 82 into n factor. This will be multiplied by 1000 divided by 100 into the volume of the acid will be 100. This is the number of equivalence. And for base, it will be what? This gram of NaOH. Right? And for NaOH, you see, since it is base, so its normality or number of equivalence will be what? Number of equivalence will be molecular mass divided by its acidity, which is 1 here. Right? So, number of equivalence will be 40 divided by 40 divided by 40, which is the number of moles into its base city, which is 1. This becomes the number of equivalence actually. We are not using this formula here. Number of moles into n factor. Right? And when you solve this, you will get n factor. This is what you have done. It will be close to 2. So, we will round it off as 2. You see, in this question, we do not have the options. Options will be in integer only. So, closest option you have to choose. Right? And when you solve this, you will get around 1.99, which is approximately 2 you have to choose. Base city will be 2. 10 to 95. Yes. Volume also will have 10 to 95. Correct. 10 to 95. This is the answer. So, number of equivalence we can find out by various different different methods. Okay? n1 into v1, then mass by equivalent mass. Right? All those factors we have. Different mass, we can find out molar mass divided by n factor. There are many different things. Okay? That is important also. Okay. Next question we will discuss. All of you understood this? Just you have to equate the number of equivalence of acid and base. Next question you see. Solve this one. A mixture of ethane and ethene. One should be ethene. Right? You see, there is a mistake here. A mixture of it is ethane and the second one is ethene. A mixture of ethane and ethene occupies 40 liter at one atmospheric. Temperature is 400 Kelvin. The mixture reacts completely with 130 gram of O2 to produce CO2 and H2O. Assume ideal gas behavior. Calculate the moles fraction of this. Okay. Easy one try, you will get the answer. Write down the equation guys, first of all. The equation of ethane and ethene you write down. Combustion of ethane and ethene write down. Balance the reaction and then you will get the input. 2 by 3 and 1 by 3. Yeah, I think it's right. Yeah, it's right. Close enough. Okay. You see, first of all what we have to do. We will write down the reaction of this with oxygen. So the reaction of ethane will be C2H6 plus O2. It gives CO2 and H2O. For ethane it is C2H4 plus O2. This also gives CO2 and H2O. And then we have to balance the reaction. Then only we will get the mole ratio. Okay. So when you balance this, we will have 2 here and 6 hydrogen so we will have 3H2O. So 4 and 3, 7. So I will write down here 7 by 2. Similarly, this will be 2. This will be 2. 4 plus 2, 6. So we will have here 3. Okay. Now, you see the number of moles of oxygen required total here will be what? If it is suppose x mole, right, then this requires 7x by 2 moles of oxygen. If it is y moles, it requires 3y moles of oxygen here. The total number of moles of oxygen is what? 7x by 2 plus 3y is equals to, this is the total number of moles of oxygen that has been used in this. It is equals to the moles we can find out from here which is 130 divided by the molecular mass of oxygen is 32. So this is equation 1. Now, if I can frame one more equation in x and y, then we can solve those equations, we will get x and y and that will be our answer. Number of moles on this and this, if you get, you can find out mole fraction. Correct? Now, the other relation of what, how can you find out? You see the one more information that they have given, it assumes ideal gas behavior. So what is that? Pv is equals to nRT, right? Pressure it is given, one atmospheric. Volume also given, it is, what is the volume? 40 litre. And we do not know, we will find out n and since it is an ATM, so the R value we use 0.0821 into temperature is 400 Kelvin. Now, when you solve this, you will get number of moles of mixture, that is ethane and ethene. And number of moles when you solve this, I won't solve this, the calculation I won't do, but the number of moles you will get here, it will be approximately 2.19 something, okay? So now out of these number of moles, your total number of moles is this, right? Which is nothing but the number of moles of C2H6 and C2H4. So we can write the, another equation is x plus y is equals to 1.219, this is the equation 2. Little bit of calculation is there, okay? You need to solve these two equations, and then you will get x and y, right? The value of x you will get here will be approximately 0.08 and y also you can find out that will be 1.219 minus 0.08. The mole fraction of ethane, x, I'll write down mole fraction only, mole fraction of ethane will be 0.66 and that of ethene will be approximately 0.34. This is the answer for this question. There is a bit of calculation in this, okay? Whenever calculation is there, you must write down the value up to 3 decimal number, right? And when the question is like this, when there is no option, right? Then better you don't round off, okay? Because this kind of question, when they ask in advance, okay? You have to write down the answer till 3 fractional digit, right? If you round it off too much, then that will give you wrong answer maybe because of rounding off, right? So in subjective question, you don't round off too much. Finally, if you are getting some answer, the final value, there you can round it off if it is required, okay? Or if you want, but you don't round it off in between anywhere, okay? Or suppose if option is given, if the difference in option, there are very large difference in options, then you can round it off approximately, okay? Understood all of you? Next question, can you move on? Okay, now you see the another question we are going to discuss here, which is similar question, right? But a little bit difference is there, but the concept is same actually. Hope you can do this question all of you because you see it is similar question only. Make sure in which the mole ratio of H2O2 is given, reaction is given. The total pressure in the container is given, temperature is given. Determine the final pressure of this, final pressure you have to find out, 80% water. Okay, one small hint. The reaction is given, mole ratio you have, right? So see there are two conditions here, one when 20 degrees Celsius and one 20 degrees Celsius. So what you do, you just write down the ideal gas equation at 20 degrees Celsius, right? Pressure is given, volume you assume, temperature is given, right? Number of moles also it is given, right? When at initial point when the reaction is about and when the reaction starts, for that it is given, 80% yield, right? So when this much reaction takes place, then what is the total number of moles after the reaction done? 80% yield of water is done, then what is the number of moles? That is one case and initial number of moles is what? That is 2 plus 1, 3, or if I take 2n and then n3n, right? So for both the cases at 20 degrees Celsius and 110 degrees Celsius, you write down the ideal gas equation which is PV is equals to nRT and take the ratio. For one case pressure is given, another case you have to find out, you will get the equation there. Try to do this, done or trying. Let me know if you want me to solve this. The pressure is 0.56, no Kushal, check your calculation I think. Okay, try, try. 0.78, viscist is getting 0.78, 0.21, Sundariya no 0.21 is not correct. 0.78 is correct, viscist. Check your calculation all of you. Kushal you also check. 0.56 is close but not that close also. 0.78, viscist write answer. I am doing this you see. I think, hello Povic, how are you? I think you have done some calculation missing. First of all you see the reaction is what? The reaction is this which is given, H2 gas and this gives 2H2 vapor, gas. Now since 2 is to 1 ratio it is given, so I assume the number of moles here, that is the proportionality constant here or 2 is to 1 ratio if you have to maintain. Suppose I am taking this n, so it will be 2n, I will just change my colour. The number of moles for H2 will be 2n moles and for O2 will be n then. So when the reaction is about to start at t is equals to 0, there is no water forms and hence the total number of moles here it will be 3n. But when the reaction starts we know 80% of water form. So for this also 80% reacts and this also 80% reacts. So in that case what happens? The moles of H2 left here it is what I will just do the calculation here. The moles of H2 left will be 2n minus 80% of 2n correct? So when you solve this you will get 160, 1.60 that will be 0.4n okay? And similarly for O2 will be what? n minus 80% of n that will be 0.2n. So the number of moles of H2 left here is 0.4n, 0.2n and the number of moles of H2O formed that will be 1.6n okay? Here if you see the total number of moles will be or add all these you will get 2.2n. Now what we do at these two condition we will apply pv is equals to nrt and then we will take the ratio. So when we take the ratio the pressure initially is 0.8 atmospheric right? That will be 0.8 into v volume is same. In the second condition pressure we have to find out volume will not change but initially the number of moles is 3a, 3a into r and temperature here it is 293, 20 degree Celsius divided by it is 2.2a into r into 120 degree Celsius which is 393. So when you solve this pressure you will find out and that will be approximately 0.787 atmospheric. Anyone has doubt in this how did I write here? 1.6n anyone has doubt in this let me know please otherwise I will move on because here only you will have some doubt otherwise it is very straightforward pv is equals to nrt we use at two different condition we will substitute the value and take the ratio. Anyone has doubt in this how did we get 1.6n tell me yes or no next question see today we are not solving one questions from one specific chapter okay? So it is like almost six seven or eight chapters we are covering today 0.62 gram of nitrate of heavy metal on heating with 18 to constant weight 0.466 gram of is oxide calculate the equivalent weight of the metal from the data okay? 26.5 equivalent mass no I don't think it's correct others tell me the answer 4 week is getting 26.5 check your calculation moving it's not right then okay what you do you just equate the number of equivalents of nitrate and the number of equivalents of oxides means number of equivalents of metal nitrate is equals to the number of equivalents of metal oxide you are trying to do this okay I'll give you one small hint number of equivalent of metal nitrate is equals to number of equivalent of metal oxide because you know equal equivalents react see number for number of moles you must have the reaction without reaction you cannot you know you cannot equate number of moles because it depends on the mole ratio understood the reaction is not given even you do not know which metal we are using here right that's why number of moles you cannot take last class the reaction was given last sorry last example the reaction was given right so here we whatever the metal we have but this is true whatever the reaction we have this is always true right now the number of equivalent of metal oxide will be what it's mass divided by its equivalent mass and that should be equals to again mass divided by equivalent mass now the mass of nitrate is given which is 0.622 so mass of nitrate is 0.622 divided by equivalent mass of nitrate will be what whatever the metal we have for that the equivalent mass of that metal right plus the mass of NO3 that will be 62 this will be equals to mass of metal oxide is 0.466 divided by the equivalent mass of metal which is E plus for oxygen it is 8 equation you have to solve you will get E the equivalent mass of the metal and that will be equals to when you solve this you'll get 120 anyone is getting 120 yeah Sanjana is getting it's 120.4 exactly right so 121 is close enough it's right understood all of you understood this next question you see do this one atomic structure a bulb emits light off wavelength this the bulb is rated as 150 watt and 8% of the energy is emitted as light how many photons are emitted by the bulb per second 2.7 into the power 19 4W is getting 2.7 into the power 19 and that is correct so what we can write here if there are n number of photons right n photons energy emitted e is equals to we can write n h nu okay and that will be equals to e is equals to n h c by lambda and this energy will be equals to 8% of 150 so 150 into 8 by 100 according to the question now h value we know c we know n is equals to n into 6.6 into 10 to the power minus 34 into 3 into 10 to the power 8 divided by 4500 angstrom we have into 10 to the power minus 10 in all these questions the unit is very important what unit you are taking okay usually students makes mistake over here in unit and all right you must take care of that this will be equals to 150 into 8 by 100 that will be 12 right right inside we have 12 when you solve this and you will get 2.7 2 into 10 to the power 19 photons these many photons can emit that energy yeah right or if you have got this next question do this one and a an electron beam can undergo diffraction by crystal through what potential a beam of electrons be accelerated so that its wavelength becomes equals to 1.54 100 volt how 100 world is not correct measures 64 64 is close enough 64 volt is correct see you understand the question first and electron beam can undergo diffraction by crystal through what potential a beam of electrons be accelerated so that its wavelength is this means we have to establish the relation of potential difference okay the relation of potential difference and wavelength lambda this is what we have to do potential difference and lambda lambda we know when a charge e or electron e passes across a potential difference v then its kinetic energy will be what e into v charge into potential and kinetic energy can also be defined by half m v square this v is potential difference and this v is the velocity of mass m right and we also know if a mass m mass m moving with the velocity v it must have a wavelength associated with it and that will be equals to lambda is equals to h by m v what is this law lambda is equals to h by m v right so now when v and lambda if I equate from here v is equals to h by m lambda and this v we can substitute here the velocity so this will be what half into m v square is nothing but h square by m square lambda square is equals to e into v where this e is nothing but the electron right and is nothing but the mass of the electron now you can solve this and you will get the potential difference in this we know all the values at is what 6.6 into 10 to the power minus 34 e value is what 1.6 into 10 to the power minus 19 this potential difference we have to find out the mass of an electron is what 9.1 into 10 to the power minus 31 right and lambda is also given that is 1.54 into 10 to the power minus 10 unit you must take care of in these kind of questions as I use when you solve this you'll get potential v is equals to approximately 63.3 volt which is nothing but you can write as 64.4 but suppose if the option given here like this a is 64 volt v is 63.3 volt c is 64.3 volt like this these options are very close right in that in this kind of cases just try to write down the exact value don't approximate don't round it off okay if 63.3 is not given or if the options are like this 64 and then you know 92 81 and 50 then we can go with 64 depending on the option right anyways so it's the easy one that's the relation of volt and velocity that is it next question you see the correct order of acetic strength options are also given yeah it's D is correct left to right if you go sorry left to right if you go acidity increases right so we can write acidity increases from left to right in a pre-ordered right one more thing is important here if we have oxides of same element oxides of same element for example if I give you this suppose S2O6 2 minus S2O4 2 minus and SO3 2 minus in this case if you have to assign the acidity then what we write for the same elements as percentage of season character increases increases acidity increases right for these molecules you see this one is most acidic and this one is most acidic this is the increasing order of acidity right this is one important thing we can have oxides of phosphorus also if you have oxides of same element percentage oxygen character increases acidity increases how do you assign percentage oxygen character you see in all the molecules we have two sulfur two sulfur and two sulfur but number of oxygen is more so when number of oxygen is more per oxygen per sulfur atom it means it has more percentage oxygen character okay next question tell me this one allowing statements of chemical reactivity of alkali metals and halogens are given which of these statements gives the correct picture read out carefully and then answer is it C some of you are getting B some of you are getting C okay what is option B alkali metals the reactivity increases but the halogens and how it decreases with increase in atomic number with increased atomic number reactivity decreases okay C is what the reactivity increase in alkali metals but increase in halogens with increased atomic number down the group okay you see basically all the options are based on the reactivity of halogens and alkali metals right you see first of all we will discuss about size left to right if you go left to right size decreases up to bottom size increases size has a role on reactivity the other factor is what the other factor is ionization energy and for halogens it is bond dissociation energy or bond energy like for alkali metals you see as they have low ionization energy their reactivity is more reactivity is more they can easily lose sorry more they can easily lose one electron and forms unipostive ion right they are reactive okay reactivity is more forms unipositives unipositive ion okay and as we go down the group for group 1 I am talking about okay alkali metals as we go down the group down the group size increases right so ionization energy even decreases down the group reactivity increases so for alkali metals what we can say its reactivity increases as we go down the group and its reactivity is more reactivity increases but in halogens decreases reactivity decreases in alkali metals so this is not right P is correct I think okay for halogens you see for halogens the consideration is what left to right we are going size decreases right size decreases so halogens are reactive due to their low bond dissociation energy right halogens are reactive due to low bond dissociation energy low bond dissociation energy another one is what they have high electron affinity high electron affinity can easily accept one electron high electron affinity and they have high enthalpy of hydration also right and what happens as we go down the group in halogens right the reactivity decreases with increasing atomic number so down the group if you go reactivity decreases so these are the points you should know to solve this question it depends on E0 value also but you know this E0 value is generally defined for metals okay not for gases the first first group we can define but halogens generally we don't consider E0 value because it is mainly for it is electrode potential right it is mainly for electrode which is usually made up of some metals some rod right so apart from that we have ionization energy because you see ionization energy and E0 value or are interrelated if you see both have relation so if you are talking about ionization energy there is no need to consider E0 values right but from that also you can do ionization energy is low for like alkali metals but for halogens we consider bond dissociation energy right halogens will have low bond dissociation energy and as we go down the group its bond dissociation energy increases right if you combine these two factors we the right one option is alkali metals are reactivity increases but in halogen decreases with increased atomic number down the option B is correct so some of to some of this the reactivity of halogen decreases as we go down the group reactivity of alkali metals increases as we go down the group there are other factors also right but the net effect is this only halogens down the group reactivity decreases alkali metals down the group reactivity increases this is important you must remember next question you see do this one consider the chemical species and point out the correct statement given below one thing is missing here I'll just do the correction NO3 minus NO2 plus we do not have this NO2 minus here right so this this and we have here one more compound which is NO2 minus so the consider the chemical species NO3 minus NO2 plus and NO2 minus point out the correct statement done again 10th one is D the hybridized rate of N in NO3 minus and NO2 minus is the same okay what is the hybridization of nitrogen in all these ions what is the hybridization the state of nitrogen nitrogen in NO3 minus and then NO2 plus and NO2 minus tell me the hybridization is state respectively in all these for this it will be six into three plus five plus one divided by eight so we'll get what three bond pair and zero lone pair hybridization will be sp2 six into two plus five minus one divided by eight so two bond pair and zero lone pair hybridization will be sp and for here it is five plus six into two plus one divided by eight so we'll have 18 no two bond pair and one lone pair sp2 so the hybrid is state of NO3 minus and NO2 minus is same right so this option is correct is there any option correct hybridized rate of N in NO2 plus sp so first one is wrong and in all the species is same wrong shape of NO2 plus and NO2 minus has been what is the shape of NO2 plus linear right sp hybridization it means geometry shape will be same because there's no lone pair so this one is linear first one is what trigonal planar tp and the third one is bent not bent it is geometry will be trigonal planar and shape will be what bent shape V shape so option four is right in this four is correct next one we see what is true about NF5 the molecule does not exist and why is that so the reason answer is correct reason absence of d orbital right nitrogen does not have d orbital can't form can't show extended octet right so not molecule does not exist tell me this one these are P block questions this kind of questions you must go through any book if you have you just go through all the questions and read NCRT is it D BF3 is planar but NF3 is pyramidal yeah it's correct BF3 has trigonal planar geometry and NF3 has pyramidal geometry with one lone pair here but BF3 is like this and since this planar so the net dipole moment will be what mu in this redundant of this two and this side so mu net is equals to zero here is non-polar and here we have mu net value does not equals to zero right it is polar okay next question this one you do now this is states of matter done what do you want to also calculate the partial pressure of helium gas in the cylinder to find out the total pressure and partial pressure okay the total pressure is 0.4 it's getting 0.48 and 0.24 partial pressure yeah it's close enough right I think 0.4926 yeah 0.4926 the exact answer total pressure 0.4926 is the exact answer okay partial pressure is 0.2466 correct all of you have done do I need to solve this just because to nrt you have to use and n will be the number of moles of helium plus number of moles of oxygen plus number of moles of n2 I will write down the equation right calculate the total pressure it's a direct question very basic question we have that is pv is equals to n total rt the pressure volume will be 10 and n total for helium gas will be that will be 0.4 divided by 4 plus 1.6 divided by 16 plus 1.4 divided by 40 this is the number of moles r will be 0.0821 and the temperature here is 316 calculation is also easy 300 Kelvin right so total pressure you'll get here is 0.4926 atmospheric and we have to find out the partial pressure of helium okay the partial pressure will be what mole fraction into total pressure so everything you have you can find out the partial pressure okay next question you see question number 14 for the reaction this calculate the mole fraction of n2 or 5 decomposed at constant volume and temperature initial pressure is given and press it anytime okay 0.37 again check your calculation kushal it's not close enough 0.375 no no it's not right check your calculation 1.66 no not 0.66 I think the first attempt was right only few you know calculation mistake or some you have done round off or something okay okay I'll just do this let me know whether you have done the same thing or not correct answer is 0.40 it is given the 0.375 is close but it's not right because the option is not given you don't write 0.4 it's not correct my answer 0.37 so the reaction is what you see n2o5 gas 2 NO2 plus half of O2 okay now when initially decomposed at constant temperature and pressure correct so suppose if it is one initially like one moles there's no decomposition so 0 and 0 and at this case the pressure is found to be what the pressure is 600 mm of hg when there is no decomposition but when it decomposes it goes 1-x this will be 2x and this will be 0.5x or half of x so the total number of moles here is what this will be 1 plus 1.5x but in this case the pressure is found to be 960 right so we can say these number of moles corresponds to what 960 mm hg of pressure right so one mole corresponds to 960 divided by 1.5x is the pressure and we know the pressure corresponds to one mole is what it is 6600 this should be equals to 600 now when you solve this you'll get 1.5x into 600 is equals to 360 so 900x is equals to 360 x will be 0.4 see exact 0.4 I am getting here yeah 0.4 is correct all of you have done the same thing exactly 0.4 I am getting here you see understood can you move on pressure is correspond is you know proportional to the number of more that's what I have done these number of moles is equal into 960 so one mole is equivalent to what 960 by this and one mole is equivalent to 600 that is what I have done personally and I doubt what should we tell me so under I have done the same thing right Ramchiran you there in terms of pressure of you have done partial pressure see why I haven't done partial pressure and all because you have to find out what you have to find out what fraction of n2o5 if you take pressure here suppose I will take here like you see n2o5 gives n2o2.5 Ramchiran you have done this you have done this okay you solved using ideal gas that is also right okay so initial pressure is suppose p0 right so it is 0 0 so p0 is 600 now when the pressure is p0 minus p2p and half of p right total pressure will be brought here p0 plus p0 plus 1.5p it is p0 plus 1.5p is equals to 960 and p0 value you know already 600 p will be brought 360 divided by 1.5 which will be 3600 by 15 will be what 240 this is the pressure we have now corresponding to this the mole fraction will be what this is the pressure you are getting 4 week tell me right so what is the pressure of n2o5 here that will be p0 minus p which is nothing but p0 is 600 minus p is what 240 which is 360 is the pressure of n2o5 and that will be equals to the mole fraction into total pressure what is the total pressure 900 x will be what you see the same value I am getting you got it 4 week right so pressure is this piece what the pressure which has been decomposed right because of the reaction so the pressure of n2o5 you have to find out first which will be p0 minus p which is 360 now this is the partial pressure of n2o5 and that will be equals to what mole fraction of n2o5 into total pressure total piece 960 or the total piece 960 yeah it's 960 then we are not getting so it will less than 0.4 okay that is why you are getting this x is equals to 0.37 or something oh I have also taken here 900 oh 960 I have taken it's fine only you know I have taken 960 960 minus 600 is 360 by this so you got this here also I have taken 960 this one is correct what is wrong here our initial pressure is 600 okay and then when it reacts p0 minus p2p and half p and this plus this plus this is equals to 960 correct so 2p and p1p plus this so that will be 1.5p equals to 960 p0 is 600 so it is yeah it's 250 mmhd it's correct so this pressure is here and 600 minus 240 is 360 the partial pressure of partial pressure of n2o5 that will be what mole fraction into total pressure and total pressure is 960 only here I have taken 960 only that's why 960 minus 600 is 360 here it is correct you haven't done any approximation also 1.5 into which step I forgot to do here you are saying just a second this one gives 960 by this 600 oh I forgot to multiply this right no no when you multiply this and 960 minus 600 you will get this directly I have written directly I have written 600 into 1 is 600 and 960 minus 600 is 360 I have written it directly 1 step that's not right and whatever number of moles we get this divided by I didn't see we have done any mistake over here what is this value 36 by 96 3 by 8 right that is what you are getting okay 3 by 8 we should not have this much of difference actually the mole fraction of n2o5 decompose at constant volume and decompose okay this is the mole fraction yeah 0.6 by 1.6 is 3 by 8 only okay let it be I'll check this I think otherwise the answer will be close to this the answer will be close to this but we should not have this much of difference 0.40 and 0.37 but here also I can see we did not make any mistake partial pressure is close to mole fraction total pressure 960 is very straight and simple here here you are saying which is here see this oh yeah right so this will be you are saying the mole fraction of mole fraction of n2o5 we have to calculate right so okay so here we are getting 0.37 so that will be 1 minus x that is 0.6 divided by 1.1 1 plus 1.5 this is what you are saying okay yeah mole fraction okay what is this value tell me this value tell me what is this value so this will be 0.6 divided by 1 plus yeah then it's right correct we have to find out mole fraction correct mole fraction we have to find out so 0.6 divided by 1 plus 0.6 right 1 plus 0.6 that will be 3 by yeah it's fine answer will be this 3 by not this one this is the number of moles simply x is the number of moles got it no the both method you will get the same answer that's what I was thinking this much of difference we should not have understood 4 week you understood this pressure method see one thing you understand what where where we got confused here in this question you listen to me very very carefully here you should this reaction is a type where delta ng is greater than 0 what I thought you see what what mistake I made you see I took this x 0.4 then I thought what this is the mole of n2o5 reacts right so this divided by the total number of moles which is one only you see 1 plus 0 plus 0 1 so this the total number of moles initially it is one so what I what I thought that at any point of time this will be some will be one only but that is not the case that is only possible that is only possible when delta ng is equals to 0 in that case only it's possible but here the delta ng is one that's why you see the total number of moles at this point it is more than one we are getting here got it so when it is more than one then mole fraction this will be what this divided by this plus this got it did you understand this right suppose the reaction is given where delta ng is equals to 0 so in that case whatever x you are getting that will be the mole fraction mole fraction that reacts okay but since delta ng greater than equal to 0 that is why this kind of confusion was there anyways good question okay this is the good one not that difficult but it's good to understand the concept okay next question you see calculate the molecular mass of a gas if it's specific heat at constant pressure is this at constant volume is this 166.287 how you know this formula this formula I think probably you do not know that specific heat into molar mass is equals to Cv this formula you know right where this Cv this formula I write down this Cv is molar heat this is molar heat constant constant molar heat constant at constant volume at this right this specific heat is the specific heat constant specific heat constant since this is at constant volume so this would also be at constant volume so molar heat constant molar heat at constant volume is equals to molar mass into a specific heat at constant volume molar heat at constant volume is is equals to the specific heat at constant volume into molar mass. Now you see this CV, the molar heat of constant, this value is nothing but three by two into R. This is molar heat of constant. And how do we know this? This is true for, actually it is there in physics. Okay, these things are there in physics. Degree of freedom based on degree of freedom we'll get this formula of CV, right? CV is equals to three by two R. This formula we get according to degree of freedom, which is not there in chemistry levels, but it is there in physics. Here you see what, here you see, this formula is for monoatomic gas. Now the question is why I am using the formula of monoatomic gas, okay? And for that, how do we know that this gas is monoatomic or diatomic? For that we calculate the gamma value, which is CP by CV. CP value is what? At constant pressure, 0.125 divided by 0.075. When you solve this, you will approximately get the gamma value is 1.66, okay? Which is there for monoatomic gas. The value of gamma 1.66 is possible for monoatomic gas. Now for monoatomic gas, CV, which is the molar heat, molar heat, right? R value we know in calorie will substitute here, that will be three by two, and the value of R is 1.99 calorie. Approximately we can take two, so this is three. Now this three will substitute here, so three will be equals to the molar mass, suppose M into a specific heat at constant volume. The volume is given, which is 0.075. When you solve this, you'll get M is equals to three, divided by 0.075, which is approximately equals to 40 gram for more, right? This question or these concepts rarely they ask in chemistry paper, because it is there in physics and all the degree of freedom, because we don't discuss this degree of freedom. We simply write CVs equals to three by two, or this is only possible for monoatomic gas. Got it? You understood this? Yeah, that's what, 40 gram per mole is the answer for this. This is one different type of question we have. Maybe you will like, mostly the chances are you will not get this question in chemistry, but maybe in physics you'll get this. Okay, so must remember this formula, specific heat into molar mass is equals to CV, where CV is the molar heat at constant volume. Got it? One more question we'll discuss and then we'll take a break. Solve this, the last question before the break. No, Povic, you cannot take RS 8.314 in the CPCB problem. You always take RS 1.99 or two. An open vessel temperature is this, is heated until three by fifth of its air at an open expel. Assuming that the volume of vessel remains constant, fine. These are not options again. 487 or 477, sign here. 87 or 77? Yeah, 750 Kelvin is right. So 750 minus 273, 477, yeah, first one is correct. Don't make this silly mistake, okay? You will lose your marks in the exam. You won't realize that. So whenever you are doing some calculation, do it properly. So for the first one, what we do, see since the vessel is open, right? Open vessel, whenever you have open vessel, then volume and pressure is constant. For open vessel, the volume and pressure is constant, right? And then what we can write? N1 T1, number of moles into temperature is equals to N2 T2. So if N1 is one, temperature is 300 Kelvin. N2 is what? Three by fifth of it expelled out, okay? So what is left over there? One minus three by five into T2. So when you solve this, T2 will be 750 Kelvin. First part, we are done. Second one, the air escaped out of it, if vessel is heated to 900 Kelvin. So when the vessel is heated, so number of moles will change, okay? Again, what we'll use? We'll use the same equation. N1 T1 is equals to N2 T2. N1 is one, T1 is 300, N2 we have to find out, and T2 is what? 900 Kelvin. N2 is what? One by three. So one by three moles is present in the vessel. So what is left out? What is, no, expelled out, left out is this. What is expelled out? Expelled out is one minus one by three, which is two by three moles. Two third moles of the gas expelled out. The temperature at which half of the air escapes. Again, N1, N2, you know now, right? So N1 T1 is equals to N2 T2 again. Temperature we have to find out. So one into 300 is equals to half moles escapes, then half moles left also into T2. So T2 will be 600 Kelvin. So this is the answer for the third one. Okay, easy question, this one, okay? Okay, so we'll take a break now. Right, we'll start after 15 minutes, it's 3.35. So we'll start around 3.50. Okay, and then we'll see the further questions. Fine, okay, take a break now. Okay, guys, can we start now? You're there, can we start? Tell me. Yeah, let me, yeah, I'll be, I'll plug your code. Anyways, okay, fine. Yeah, now you can see. Right, okay, next question you see, one mole of a nitrogen gas at 0.8 atmospheric takes 38 seconds to diffuse through a pinhole based on diffusion-iffusion concept. Okay, what is the answer? One mole of nitrogen gas at 0.8 atmospheric, pressure is also changing, right? 0.8 to 1.6. Use the formula of diffusion-iffusion when difference in pressure is there. Done, anyone? Okay, see, when we have difference in pressure, you have to find out molecular mass, no, it's not 1.5.75. This is a formula of rate of diffusion when pressure changes there. That is R1 by R2 is equals to P1 by P2 into under root M2 by M1. This is the relation we have when change in pressure is there. Okay, the correct answer is 252, right? Now you see the question that one mole of nitrogen gas at 0.8 atmospheric takes 38 seconds to diffuse through a pinhole, whereas one mole of unknown compound of xenon with fluorine at 1.6 atmospheric takes 57 seconds, right? So the rate of, I'll write down here, the rate of nitrogen is what? N2, it is one mole divided by the time, which is 38 seconds, and the rate of another compound that is xenon compound, Xc only, I'll write, that will be one divided by 57 seconds. So this if I substitute here, one by 38 into 57 by one is equals to the pressure of nitrogen is 0.8, right? So we'll have 0.8 divided by 1.6 root over molecular mass of the another compound we have to find out, xenon compound divided by, for nitrogen it is 28. When you solve this you'll get M is equals to 252 gram, this is the answer, direct formula based. Next one, solve this similar concept, tell me the answer, 8 is to 1, okay? So if I write down here, the rate of helium divided by the rate of CH4, this will be equals to molar mixture 4 is to 1, continue this due to hole in the vessel, okay? Now the molar ratio is 4 is to 1, right? It means helium is four mole, then methane is one mole, CH4, right? So partial pressure ratio is what? For helium, partial pressure is what? It is the mole fraction into total pressure, right? And it is directly proportional to the number of moles we have, right? So four moles is equals to, we can write, four gram of helium, sorry, 16 gram of helium, just a second, we have, we have to find out what? We have to find out the mixing of this, okay, fine. I'll write down this first. So we'll have the pressure of HE divided by the pressure of CH4 into under root of, under root of what we write? Molar mass of CH4 divided by molar mass of HE, right? Molar ratio is given, so the ratio of partial pressure will be 16 is to 4, right? So 16 divided by 4 into molar mass of CH4 is 16 divided by 4, we'll do the word of it, right? And when you solve this, you'll get eight is to one, right? This is the answer. Next question is that question, like the pressure related questions are this. Otherwise, like most of the time, we'll get rate of diffusion if you're in, when pressure is constant. So when pressure is constant, you just remove this part, okay, then we'll have directly this one. Okay, accordingly, we can find it out. Okay, next question you see, 19. This is thermodynamics. Six moles of an ideal gas expand isothermally and reversibly. From a volume of one liter to a volume of 10 liter, what is the maximum work done? 34458, 34458, what is the unit for it? It's joule or kilojoule, 34.458, kilojoule, okay. Yeah, it's right. So the value you have taken is 8.314. So the formula of work done in reversible isothermal process minus 2.303 nRT log of B1 by B2 or P2 by P1, right? Since volume is given, so I have taken the expression of volume, okay. Now, all these value you know, right? So we can substitute. When R value you substitute as 0.0821, right, liter ATM mole Kelvin, then the work done will be 340.3 liter ATM, right? This is the answer we have. When R value you substitute 8.314, joule per mole Kelvin, then work done will be 3408.4 kilojoule, 464.8 joule, which is nothing but 34.46 kilojoule, okay. You will get positive. This minus sign we have since I have taken B1 by B2 here, right? Just you solve this expression. Formalize B2 by B1, you'll write, then this minus becomes plus. If you, this you write B2 by B1, suppose, the same thing, if you write B2 by B1, then this minus becomes plus, that is it. Work done by the system is negative, right? No, see, we are talking about magnitude, right? We're talking about magnitude only. What is the work done? Yeah, if you, if you consider work done by the system, then negative sign you have to substitute, you have to put, okay? But the magnitude of work done is what? It's 340 only, approximately this. I'm giving you the magnitude of this, mod of W, right? So this is what, this is subjective question. Option is not given. So it's better that we write mod of W here. We're talking about magnitude. Yeah, it's fine. Expansions to work done by the system should be negative. But this negative, positive, whatever you want, you can write. According to this, B1, B2 will change, okay? So in later ATM, the answer will be this. You can also put the answer into ERG. ERG, and you can also put answer into calorie, okay? Those are the different, different units we have. Okay, depending on the unit of R, you will get different, different values, okay? This is the only formula here. This is important, okay, for example W. Next question you see. This one you tell me. Based on first law of thermodynamics. 14898.7, yeah, it's correct. Right, right. So it's basically based on first law of thermodynamics, and the system you have, right? It does 101.3 Zool of work, right? So work done will be what? Negative of 101.3 Zool, right? And it accepts energy from surroundings, which is equals to 15,000 Zool. So 15,000 Zool energy it accepts, and it expands what? Only 101.3, that's why it is negative. So what is the energy left here? This is the change in internal positive value you get. 4898. Next question. One mole of liquid water at its boiling point vaporizes against a constant external pressure of one atmospheric at the same temperature, assuming ideal behavior and initial volume of water vapors as zero, calculate the work done by the system. 29.84 liter ATM. Negative or positive? Varsity. Answer is close to 30, 30.62. The answer given is 30.62. Negative, that would be negative. Because work done is what? Work done is nothing but minus P delta V. Pressure is given, one atmospheric, right? We have to find out change in volume. And how do we find out change in volume? With the help of what? TV is equals to NRT. Because initial volume is zero. Right, there is no volume mentioned here. So initial volume is zero. So V will be what? One into V. Pressure into volume is equals to one mole that is 0.0821. And temperature is what? Volume point. So what is the temperature? 100 degrees Celsius or 373, right? So when you solve this, you'll get volume as 30.62 liter. Okay. The minus one pressure and change in volume will be 30.62 minus initial volume is zero. So answer will be minus 30.62 liter ATF. Minus 30. Work done is always minus P into delta V, always, right? You always write down negative sign. If you find an answer if you're getting negative, it means work done by the system. If it is positive, then work done on the system, okay? Next question. Question number 22. One mole of H2 and two moles of I2 and three moles of HI are injected in one liter flask. What will be the concentration of H2, I2 and HI at equilibrium? KC is not given. KC, we have to find out. This reaction, KC, we have to find out. I think you did not get the, oh, KC is not given. Let me read out the question. First, one mole of H2, I2 and three moles of HI injected in three moles. What will be the concentration? Achha, concentration, we have to find out, okay? So KC must be given, okay? Let me check. This question number 22. Okay, KC is given actually, yeah. KC for the reaction is, it is 45.9, it's given, yeah, so. Do it fast, we'll solve one more after this. No, it's not right, 0.5, 1 and 4.5, it's not right. The calculation is there, right? So I'll just write down the expression, you just check your expression is right or wrong. H2, I2 gives HI, that is 2 HI. So it is given in the question that we have one mole of H2, two moles of I2 and three moles of HI. When the reaction takes place, it forms one minus X, two minus X and since one gives two, so X gives two X. So number of moles of HI will be three plus two X. Now, KC will be what? Concentration of HI square, concentration of H2, concentration of I2 at equilibrium, and that will be three plus two X square, one minus X, two minus X, two minus X, and two minus X. KC value is given, you have to find out X from this and when you solve, you'll get X is equals to 0.684. When you get X, you'll find out one minus X, two minus X and three plus two X. This is it we have done. Okay, next question you see, which is also based on the same concept actually. The last one for today, this one also the same thing we have, but one thing I wanted to tell you here, first we write down the reaction. BR2 gives two H, BR, okay. So we have 0.6 and 0.2 moles of H2 and BR2 respectively, 0.6, 0.2, it is zero and when the reaction takes place, it is 0.6 minus X, 0.2 minus X and here we'll have two X. So KC will be, I'll write down directly it will be two X square divided by 0.6 minus X and 0.2 minus X, okay. Now KC value is given, just you have to calculate X. Now, since in this expression what happens, you'll get quadratic of X, right? In the previous one also you'll get quadratic of X, but the value of X that you get here, which is here important to understand, will be two value we'll have here, one will be 0.2 and other will be 0.6. These two value of X you'll get here. Now in these two, you cannot consider 0.6 as the value of X here. Why is it not possible? This value is not possible, that's the point you have to understand. 0.6 is not possible here as the value of X and why is it so? Because you see, 0.6 is the number of moles of H2 present initially and one moles of H2 reacts with one moles of Br2, right? So in this case, H2 cannot react completely here, right? Limiting reagent is Br2, right? Yeah, right, that's what, that's what. So limiting reagent is Br2. So product concentration will get according to Br2, right? And since Br2 concentration is 0.2, so we cannot have value of X more than 0.2. Yeah, right, right. So that's why this 0.6 is not possible. Now we have 0.2, you substitute here, you'll get equilibrium concentration, okay? So the question was same, the last question was also the same thing, but this is again one point you have to understand, okay? They can also ask you the question like this, which of this value of, which of this equilibrium concentration of H2 is not possible? Like this also, they can frame the question. Then what do you have to do? You have to take this value over there, right? The value of X, which is not possible, according to that you have to find out the concentration of H2 or anything like that, okay? But this kind of understanding you must have. Understood, if you understood, tell me. Okay, so this is it for today's class. Or tell me who will have exam on 9th of April, Jay Min, any one of you? Any one of you has exam on 9th of April? Visits Sanjana, okay. Okay, so who all has exam because we are not going to have any class between these two, between today and 9th of April? So I wish you all the luck, all of you. Give your best, okay? This is the last three, four days exam. Remember one thing, who will be gone eight, right? Okay, so all of you, okay? And this is the last few days left. So remember, you cannot improve much into physics and maths in these days, okay? But chemistry you can improve. At least in organic part, you must go through once, okay? Previous year questions, NCR to you must revise, you never know that one single line that you are studying now, maybe that particular question they'll ask directly into the exam, okay? So all the best for your exam. Keep revising in organic. Like in what you must revise optical isomerism. That is very important, okay? So in organic part, you must revise these days, okay? So all the best for the exam, okay? See you soon, bye-bye.