KCET Physical Chemistry Top 50 MCQs | Final Battle Part 3

Added: 2026-05-05

[00:00:00]Hello champions, welcome to the channel.

[00:00:01]And in today's video, we are going to do the complete revision of physical chemistry with top 50 questions from physical chemistry. So, please just watch this complete video because I'm going to cover all the important questions, all the important concepts, so you can easily revise. So, let's get started with our question number one.

[00:00:18]So, the first question here is delta G in the reaction Ag2O gives 2 Ag + 1/2 O2 at a certain temperature is minus 10 kJ per mole. Pick the correct statement here. Now, since this is a negative delta G, that means the reaction will happen automatically, means simultaneous sorry, the reaction will happen spontaneously and Ag2O will get converted to Ag and O2. So, the correct answer here option number A, that is Ag2O decomposes to Ag and O2.

[00:00:49]Next, we have from the given data, find the value of Cr3+ + 3 electrons gives you chromium. So, here I have a equation like this. You have Cr+3, okay? Then, it gets converted to Cr+2.

[00:01:03]And this one gets converted to Cr. And if you see here, the number of electrons involved is one and the value of this one is.424. And the value for Cr2+ to Cr is minus 0.9.

[00:01:16]And this value from Cr+3 to Cr is around uh that is what we have to find out. And here, the number of electrons involved is three. Here, the number of electrons involved is two. Now, what you have to do, let this one be 31, okay? So, 3e1 is equal to here, the number of electrons is one plus three to plus two, so one into this much. So, that is minus.424 plus two into uh 0.9, that is minus 1.8, right? Now, if I add these two, how much I will get?

[00:01:48]So, I have I'll just add here. So, I will have 428 + 4 12. Two. So, I'll have 2.224 is equal to e1, 3e1. Now, if I divide this one, you will get minus.37 are 21 112. Again, [snorts] we will have 34s, so 0.741, which is option number A here.

[00:02:10]Next question, for the reaction 2NO2 gives you N2O2 + O2, the rate of expression of dNO2 by dt is equal to k into NO2 to the power of n, where k value is given. If the rate of O2 is formation is this much, then the molar concentration of NO2 is Now, here if you see the question, okay, just let me check if I am able to record it or not.

[00:02:32]Okay. Now, let's move to the question here.

[00:02:36]Just give me one minute. Yes. So, now we have uh this reaction, we have minus 1/2 of dNO2 by dt is equal to dO2 by dt.

[00:02:52]Right?

[00:02:53]And we also have dNO2 by dt is equal to k into NO2 to the power of n, okay? Now, we have to find the concentration of n NO2. Before that, I have to find the rate of disappearance of NO2. Now, this is equal to I can multiply this two here and dO2 by dt is 1.5 into 10 to the power of minus four. So, that gives me here this one two into 1.5 into 10 to the power of minus four. That gives me three into 10 to the power of minus four here.

[00:03:28]Right? Okay.

[00:03:30]k into Now, if you see the units of k here, mole inverse liter second inverse, that indicates that here it is a second order reaction. So, I'll just put the value of k here, which is 1.5 into uh one second. Sorry, the k value here I'll just erase this one.

[00:03:48]So, what is the value of k here? Uh 3 into 10 to the power of minus three, right? This is 3 into 10 to the power of minus three into NO2 to the power of n. Now, this one got cancelled, so we have 10 to the power of minus one is equal to this is second order reaction, so I will have like this. So, I can write NO2 is equal to root over of this one or that is equal to 10 to the power of minus 1/2, okay?

[00:04:16]So, that you can write it as like 10 to the power of minus one plus 0.5, right?

[00:04:22]That will give me 10 to the power of minus one into 10 to the power of 0.5, okay? Now, which log value is around 10 to the power like 0.5 something like that, you can calculate. So, if you see here, uh log three value is somewhere around point 475, okay? So, it is.47 is very close, so this one will be three and into 10 to the power of minus one. So, somewhere around.3 should be my answer.

[00:04:48]So, I can see option number D here.

[00:04:50]Next question, one mole of an ideal gas expands at constant temperature of 300 K from initial volume 10 L to final volume 20 L. The work done in the expanding of gases. So, work done, we have this formula minus 2.303.

[00:05:05]One mole is there. R value is uh 8.314.

[00:05:10]R T is how much? 300 K, okay? And log of V2 by V1, 20 by 10, so I will get log two. Now, you can see 2.303 into 8.314, it will give me 19.1 into 300 into.301.

[00:05:26]So, that will give me 3 1s are 3, 3 1s are 27, 3 1s are 3 plus four uh 275, so I will have this one 5730 into.301, okay? So, I just take three here, so I'll get three zeros are zero, three threes are nine, three sevens are 21, 15, 16, 17, 1719 something. And here, you have uh So, how much uh I can say um Just a minute. So, that is here we have negative number and there is 1.3 I'm sub calculating, so there will be one decimal place here. So, that gives me option number B is the nearest answer.

[00:06:08]Next question number five here, 9.65 C of electric current is passed through a fused anhydrous magnesium chloride. The magnesium metal obtained is completely converted into Grignard reagent. The number of moles of Grignard reagent obtained is So, first of all, we have to calculate how many moles of magnesium got deposited, that much amount of Grignard reagent will also be formed, okay? So, if I want to calculate here, we have this formula.

[00:06:35]I can write Mg See, Mg2+ plus two electrons gives me Mg. So, I can say that two Faraday gives me one mole, okay? Or two into 96500 C gives me one mole. Then, 9.65 C will be how much or how many moles it will give me? So, that will be one divided by two into 96500 into 965 into 10 to the power of minus two. This this got cancelled, it is 10 to the power of two, this goes up. 1/2 is.5 into 10 to the power of minus four. Now, that if I just arrange five into 10 to the power of minus five, which is option number C here. Next, a solution has an osmotic pressure of this much at 300 K, its concentration would be This is very simple. We have this formula, pi is equal to c into r into t, where c is molarity.

[00:07:31]Uh so, we want to find the concentration, so that will be pi divided by r into t is equal to concentration. So, pi is.0821 uh point sorry, this one. Pi is.821 atm.

[00:07:45]Right? And we have R value is.0821 L atm mole inverse Kelvin inverse. And the temperature is 300 K, right? Kelvin inverse Kelvin inverse get cancelled, atm atm get cancelled. And when it goes up, we will get mole liter inverse, which is the unit of molarity. Now, if you see this one point uh.821 by three zeros and so, I will get this one as 10, right? 10 and 300. 10 by 300, so one zero I can cancel here. So, again I will have 1/30. So, 1/3 if you see is.333 and then again one zero divided by we have 10, so I'll get.03 something, which is option number C here. Next, for a first order reaction A to P, T half is 10 days. The time required for 1/4 conversion of A is This is again very simple question. So, we already have T half, so what is the value of k here?

[00:08:44].693 by T half, okay? Now, the question is asking here time, so we will have 2.303 by k. k is.693 into 10, right? 2.303 by k log of Okay, so you can see in the values are given in terms of ln, so we don't have to uh multiply here 2.303.

[00:09:06]So, what we can take here is um One second, we will take just 1/k. So, it is like 1/k. Uh and if you see that 1/4, the time required for 1/4 conversion, okay? So, that means you have 1/4 has already converted, then remaining will be 3/4. So, 3/4 if I write, it will become 4/3 here, right?

[00:09:28]1/3/4 that becomes 4/3. Now, if I just arrange this one, this will be 0.693 by 10, so 10 goes up here. ln of uh two square minus ln of three. This can be written as like this. Now, this gives me 10 to the power of.693 into two times of ln of two. ln of two is.693, so.693 into two. 2 threes are six, two nines are 18, two sixs are 12, plus 1 13. So, that gives me 1.386 minus ln of 1 3 is 1.1. So, I will have 1.1.

[00:10:07]So, that gives me 0.286, right? So, I will have 10 divided by 0.693 into 0.286.

[00:10:17]So, if you see 2 3's are 6, 2 9's are 18, 2 6's are 12, plus 1 13. Okay, this is different. So, I can write this as 2.8, okay? And 0.69 will be somewhere around equal to 0.7. So, I will get 7 4's are 28. So, somewhere answer will be near to 1 4. So, that is the option number B here. Next, an organic compound C and H percentage is in the ratio of 6:1 and C and O percentage is in the ratio of 3:4. The compound is Now, see, CH percentage is given 6:1 and CO percentage is given 3:4. Now, you can see the carbon is common in between them. So, you have to make the carbon constant here. So, we'll multiply this with two. So, it will become uh 6 and 8. So, if I multiply two here.

[00:11:00]So, now if I see the ratio of C, H, and O, it is 6, 1, and 8, okay? So, what is the total mass here? Uh 15 here. So, if I see the percentage of carbon, it will be how much? 6 by 15 into 100.

[00:11:17]Right? 6 is the mass like carbon and total is 15. So, 6 by 15 into 100. So, 3 2's are 6, 3 5's are 15. Then I have 5 1's are 5 20. So, carbon percentage is 40 here. Now, if I want to calculate the percentage of hydrogen, it will be 1 by 15 into 100, right? So, again here I will have 5 3's are 15, 5 20's are. So, that will give me 6's are 18, 6.66.

[00:11:43]Next, if I have here the percentage of oxygen, it will be 8 by 15 into 100, okay? So, that will give me 5 3's are 15, 5 2's are 20. Then um 160 divided by 3, okay? So, that will be equal to 3 5's are 15, 153.

[00:12:02]Again, so, 53.33 will continue here. So, if I just want to write down these elements now, C, H, and O. Carbon percentage is how much? 40. Hydrogen is 6.66.

[00:12:16]Oxygen is 53.33.

[00:12:19]So, this divided by 12 if I do, 12 3's are 36, 4. So, I'll get 3.33, okay? Now, this one 6.66 divided by 1 you are now converted into number of moles. So, just divide by 12, divide by 1, and this you divide by 16. Now, if you see 16 3's are is 48, okay? So, that will give me again 3.33 here. So, what is the simple ratio?

[00:12:40]1 to 1. So, I will get CH2O, which can be written as HCO or HCHO, okay? That is option number A here.

[00:12:50]Next, the correct order of increasing H3O+ concentration in the aqueous solution. So, the stronger the acid, more will be the H3O+ concentration. So, first of all, if you see H2SO4 should have the highest concentration because it is a strong acid. So, in which option we have H2SO4?

[00:13:06]Here also I have H2SO4. Here this is wrong and this is wrong. Now, after that I need to have H2S and here also I have H2S. After that, the NaCl is a strong salt. So, its pH will be around neutral pH. So, then we have NaCl. And after that, lastly we will have NaNO2. Okay, this order you should Next, we have calculate the extent of dissociation if the equilibrium pressure P for PCl5 gives PCl3 + Cl2 is numerically equal to three times of KP, okay? So, PCl5 is there.

[00:13:40]It is decomposing to give you PCl3 plus Cl2, okay?

[00:13:47]Now, what we can do? Let's assume that this is one, this is zero, this is zero.

[00:13:52]Then we will have 1 minus alpha, this is alpha, and this is alpha. So, what is the n total here? n total is 1 plus alpha. Right? Now, what is the mole fraction of PCl5?

[00:14:05]That will be equal to 1 minus alpha by 1 plus alpha, okay? Now, what is the mole fraction of PCl3? Sorry, mole fraction of PCl3 is alpha by 1 plus alpha. Mole fraction of Cl2 is alpha by 1 plus alpha, okay?

[00:14:21]Now, it is saying that the equilibrium pressure is P. So, what is the partial pressure of PCl5? Partial pressure of PCl5 will be the mole fraction of PCl5 into the total pressure. So, 1 minus alpha by 1 plus alpha into P. Then partial pressure of PCl3 will be alpha by 1 plus alpha into P. Next, partial pressure of Cl2 will be again alpha by 1 plus alpha into P, okay? Now, we have to find that uh what is the value of KP here? Now, if you see KP formula will be uh PCl3 into Cl2 divided by PCl5, right? Now, what I can write here for uh what is given the value? Uh calculate the extent of dissociation if the equilibrium pressure P is numerically numerically equal to three times of KP.

[00:15:18]So, we will substitute that later. This P that we are using is actually three times of KP. According to the question, it is given. Now, let us substitute the value. What is the value of PCl3? I have alpha by 1 plus alpha. Uh and you can see this both the terms are same. So, it is alpha by 1 plus alpha square into P square divided by PCl5 is how much? 1 minus alpha by 1 plus alpha into P. Right? Now, what will happen? We will get alpha square and 1 plus alpha square.

[00:15:51]Then this goes up, I will have 1 plus alpha. And I have here 1 minus alpha, okay? Now, this one will have P square and we have P in the denominator. Now, what will happen? 1 plus alpha, 1 plus alpha will get cancelled. And in the remaining part we will have 1 plus alpha and 1 minus alpha. A plus B A minus B is equal to A square minus B square. So, I have alpha square by 1 minus alpha square and here I have P, okay? P square will be P is equal to KP. Now, I can substitute this P as 3 KP, right? So, what I will do here?

[00:16:28]So, I'll erase the above part so that I can use this space. Now, what is the question saying that the uh equilibrium pressure is actually equal to three times of KP, right? So, whatever P we are having in the equation, that I can substitute as 3 KP.

[00:16:44]And from both the sides now KP can get cancelled. So, if you have this one, that is alpha square by 1 minus alpha square uh 3 KP is equal to KP.

[00:16:58]Okay, I'll just see once that I have got the correct values here. P square and here it will be P. Yeah. So, now KP KP will get cancelled. This I have one here. So, now you can multiply 3 alpha square is equal to 1 minus alpha square.

[00:17:14]So, this side it comes here. So, that gives me 4 alpha square is equal to uh 1 or alpha square is equal to 1 by 4 or alpha is equal to 1 by 2 or 0.5, which is option number A here. Next, at 40°C, the total vapor pressure of methyl alcohol and ethyl alcohol is given by this equation. The vapor pressure of pure ethyl alcohol and pure methyl alcohol at this temperature are respectively, okay? So, first of all, let's assume that uh I have a mole fraction of methyl alcohol, okay? I will write here uh 1 second.

[00:17:53]Let XA be methyl alcohol.

[00:17:57]Okay? And then XB will be ethyl alcohol, okay? So, this will be equal to 1 minus XA. Now, what I can write? P total is equal to P naught of uh A into XA plus P naught of B into 1 minus XA, okay? According to Raoult's law. Now, what is the formula here? P naught of XA. Then I will have here P naught of B minus P naught B XA. Now, if I take common here uh from both of this XA, so, that will give me P naught A minus P naught B into XA plus P naught B. And that gives me P total, okay? Now, according to my equation, this P naught B value is 138.

[00:18:50]Okay? And we have P naught A minus P naught B is equal to 120. Right? So, that means this is 138. So, P naught A is equal to 138 plus 120. That is equal to 258, okay? Now, the question has asked for See, the question is slightly tricky. It is asking first of pure ethyl alcohol and then of methyl alcohol, right? So, pure ethyl alcohol is my PB. PB value is 138 and PA is 258. So, it is 138 and 258. You can see similar option. You may get confused and pick the wrong option here. Next question, we have the equilibrium constant for the given reaction. Now, you can see that how this reaction is actually obtained.

[00:19:39]Uh COO Uh so, this is actually CO like cobalt, okay? Now, if I reverse this reaction and add to the first reaction. So, if I reverse this reaction, I will have CO plus CO2 gives me COO plus CO. Right?

[00:19:56]Now, if I compare this reaction and this reaction, COO COO will get cancelled.

[00:20:01]Cobalt Cobalt will get cancelled and the remaining thing that we have is CO2 plus H2 gives CO plus H2O.

[00:20:11]Right? I think that is what is remaining. I'll just check.

[00:20:16]I have here um Yes. So, I have CO So, all these things are cancelled. So, I have CO2 plus H2 and on this side I have uh CO and H2O.

[00:20:30]Right? So, what I've done I have reversed this reaction and added with this one. So, when you reverse this reaction the equilibrium constant become 1 by 490 and when you multi when you add this one you have to multiply with the equilibrium constant. So, now we have to solve this 67 by 490. So, if you see seven um 7 * 7 = 49, right? So, I'll have 70 and 7 * 9 = 63. So, 67 by 7 will be somewhere around 63. I have 40 then 7 * 5 = 30. So, 9.5 here, right? So, 9.5 divided by 7 if I do uh like I can write it as uh 95 into 10 to the power of this is minus one and this one is there. So, minus two divided by seven. So, 7 * 1 = 7.

[00:21:22]Uh again, I have two So, 7 * 3 = 21. Four will remain. Uh seven five 7 * 7 = 49, right? Again, we have two decimal places here. So, that will give me 0. Okay, one second.

[00:21:38]I will have just I'll calculate like this 67 divided by 490. Okay, I'll just divide this one directly because I got confused here with the calculations. So, let me keep it simple. Here I have 40 and 1 * 490. Okay? And you can see that there's only one option where you're starting with one. So, it has to be one and here also we got 13 something. So, it is option number A here. You can complete this calculation. Next, for the reaction 3A plus B gives 2C plus D the equilibrium constants of A, B, and C are given.

[00:22:11]What is the initial concentration of A?

[00:22:13]Okay. So, if you see the reaction is like 3A plus B uh gives 2C plus D. Okay? Now, let us assume that the concentration of A is somewhere A and this is B. Okay? And so, A minus 3X will react here and here it will be B minus three moles of A reacts with one moles of B. So, B minus X. Here 2X will form and here X will form. Now, the question is saying that A, B, A, B, and C. So, this 2X value is equal to 0.008.

[00:22:49]So, X value will be 0.004.

[00:22:52]Okay? Now, we have the equilibrium concentration of A is 0.03. So, that means A minus three into 0.004 is equal to 0.03.

[00:23:05]So, if I solve here it will be 3.012.

[00:23:10]That is equal to 0.03.

[00:23:13]Now, if I add this one I will have 0.03 plus 0.012.

[00:23:19]So, that will be 0.042 which is option number C here.

[00:23:25]Next, which of the following is a correct statement? When mixture is more volatile there is a positive deviation from Raoult's law. This is a correct statement because when volatility is increasing then the vapors are evaporating and creates a lot of vapor pressure. When mixture is more volatile there is negative deviation. No. Ethanol and water form uh ideal solution. No, they form negative deviation. CH3Cl and water form ideal solution. No. So, option number A here.

[00:23:52]Which of the following represents the enthalpy of formation of water? So, when the formation of there means it should always form one mole and in the natural state. So, H2 exist in the form of liquid state. So, if you see here H2 gas is correct, half O2 gas is correct, but here H2 gas is given. So, this is incorrect. H2 gas, half O2 gas, H2O liquid. Yes.

[00:24:12]This is the correct formula. Here, what is the problem we have? Enthalpy of formation is given positive value.

[00:24:17]Enthalpy of formation is always negative. And here what problem we have?

[00:24:22]It is not for one mole. Okay? Now, the most probable radius for finding the electron in the helium plus atom. So, RN formula is 52.9 into n squared by Z. Okay?

[00:24:37]Now, if you see here um hydrogen helium for uh the n value will be one, but this Z value will be two. So, if I divide here I'll get 26.5 which is option number uh C. Right? 2 * 2 = 4 takes so, 26.5 will be the answer. Next, we have to match here the formulas. So, lambda m is kappa by C. So, A has to three. Only one option. So, I can go with B directly. G star is your uh uh what G star is cell constant, right?

[00:25:07]So, B is equal to one. Kappa is equal to rho into L by A.

[00:25:12]Uh sorry, kappa is equal to this one and R is equal to rho into L by A. Okay?

[00:25:18]Next, we have which of the following statements uh state the law of multiple proportions? So, law of multiple proportion is used when we have two different elements combine to give you more than two products, right? So, carbon forms two oxides CO2 and CO and the masses of oxygen combines with the mass of carbon are in the ratio of 2:1.

[00:25:35]This is law of multiple proportion. Next question is from thermodynamics where the uh salt um the pair of compounds soluble in water. Now, remember whenever the hydration enthalpy is more negative then is if you add hydration and lattice enthalpy if you're getting a negative number in that case the uh the solid will be soluble. Now, if you see first option this is more negative. So, this one will be soluble.

[00:26:01]Now, if you see second option lattice enthalpy is more. If I add these two I'll get a positive number. So, this is not soluble. Here you can see this is more negative. So, this will be soluble.

[00:26:09]Here this is positive. So, this will be not soluble. So, we have just P and R which is option number D here.

[00:26:16]Next, activation energies of two reactions. This is very important question. R EA and EA dash. Okay, two two reactions are there. For first reaction activation energy is EA. For second reaction it is EA dash. And EA is greater than EA dash. Okay?

[00:26:31]As temperature increases from T1 to T2 which is correct?

[00:26:35]Now, see whether temperature you're increasing definitely when you're increase the temperature the rate will increase. But when activation energy of EA is already higher suppose its rate constant is K1, its rate constant is K2 then obviously K2 will be greater than K1. And at higher temperature also K1 dash will be greater than K2 dash. So, I have this relation in option number B here. Next, 4 g of magnesium is burnt in 2 g of O2 gas. What is the amount of MgO formed? So, if I have this reaction 2 Mg plus O2 gives 2 MgO.

[00:27:10]Right? So, now this is 24 48 g and this is 32 g. So, we have 48 g of Mg reacts with 32 g of O2. Now, 1 g of magnesium will react with 32 by 48. Then we have 4 g of magnesium that will react with 32 by 48 into 4.

[00:27:34]Right? So, we have here then again I can cancel 4 * 3 = 4 * 8 = So, you will get 2.

[00:27:43]six something, right? But oxygen mass is already 2 g. It is less than we require 2.6 g, but we have only 2 g. So, oxygen is the limiting reagent. Now, the amount of MgO formed will depend on this oxygen. So, I can say that 32 g of oxygen will form here 40 * 2 80 g of magnesium oxide. So, one will form 80 by 32 and we have 2 g here. So, 2 g will form 2 * 1 = 2 * 16 = 16 * 5 = So, 5 g which is option number D here. Next, 0.1 of M acetic acid is titrated again 0.1 M of NaOH solution. Difference in pH um between 1.4 and 3.4 stages of the neutralization of the acid. Okay. So, now what happens here is if you see this reaction CH3COOH plus NaOH will react to give you CH3COONa plus H2O.

[00:28:44]Right?

[00:28:45]Now, if you see when 1 by 4 reaction has happened. Okay? So, that means uh see uh how much is remaining then? At suppose you have here 0.1 M of NaOH is there.

[00:29:01]Okay? And 0.1 M of acetic acid is there.

[00:29:05]Now, when 1.4 of the acid has been consumed. Okay, 1.4 of the acid has been consumed then at that stage like I will write at 1 by 4. At that stage the acid remaining will be 1 by 4 has been consumed. So, 3 by 4 or 0.75 will be there. And how much has consumed? 0.25 has been consumed. So, that much the salt would have been formed. So, what is the pH value when at 1 by 4 stage? It is pKa, right? Plus log of salt by acid. So, 25 by 75. So, that is 1 by 3. Okay? Now, when you come to 3 by 4 stage. That means 3 by 4 has reacted. Remaining will be 2.5. Now, if 0.25 so if 0.75 has reacted, formed here will be 0.75. So, what is the pH at this case?

[00:29:55]pKa plus log of salt by acid. So, 75 by 2 so log 3. Now, we can take the difference between these two. If you take the difference between these two, you will get pKa plus log of 3 to the power of minus 1 minus [clears throat] Sorry.

[00:30:17]pKa plus log of 3.

[00:30:20]Right? Now, pKa pKa will get cancelled here.

[00:30:24]Then, I have this one here minus So, pKa What's the formula of pH? pH is equal to pKa. So, this is correct, right? So, minus one of log 3 and here also minus log 3. So, that gives me minus 2 log 3. Okay. So, here actually small error is there. This is minus 2 log 3. So, option number B here.

[00:30:47]Next question. Changes in the system from initial to final state were made by different matter in different manner then H plus H value remains same, but Q changes because so because H is a state function and Q is a path function.

[00:31:02]Next at 25° C for combustion of 1 mole of benzene, this is the value of delta H. Calculate the heat of reaction at constant volume. That means it is asking you to calculate delta U here. Now, you can see delta H is equal to delta U plus delta n g R into T. Okay? Now, what is the value of delta H? So, I'll just write down the formula for delta U then. It is delta H minus delta n g R T. Okay? Now, what will be the value of delta H? It is 780980 minus delta n g means gaseous products minus gaseous reactants. So, that is 6 minus 15 by 2. So, that will give me minus 1.5. Right? So, minus 1.5 into R value is 2 and the temperature here is 25 so 298 K.

[00:31:58]Okay. So, this one will be giving me three. Three into this one so 3 * 24 3 * 27 28 29 3 * 6 7 8. Okay? Minus minus I can Okay, here one more thing the enthalpy of combustion is always negative. Okay? So, this one has to be negative. Now, minus minus is positive.

[00:32:15]So, minus 780980 and this side I have 894. So, 10 minus 4 6 17 minus 9 is 8 here.

[00:32:28]Then zero Again, I have zeros 8 and 7. Right? So, that will have we have here three decimal places also, I think. No, okay.

[00:32:40]780086 we will get. Now, to convert it into kilocalorie here one decimal will come.

[00:32:45]So, option number B here.

[00:32:47]Next at 550 K KC for this is given. At equilibrium we have this relation. What is the value of Z in mole inverse? Okay.

[00:32:55]So, if you see KC value will be concentration of Z divided by concentration of X into concentration of Y. Now, here we have 10 to the power of 4.

[00:33:07]Z concentration is already there. Now, you have X is equal to 1 by 2 Y is equal to 1 by 2 Z. So, what is the value of X in terms of Z?

[00:33:20]One second, we have to find the Okay. So, what is the value of X in terms of Z? It is 1 by 2 Z. Right? X is equal to 1 by 2 Z. So, 1 by 2 or I can write Z by 2.

[00:33:36]Right? Next, what is the value of Y with respect to Z? It is just Z because two two will get cancelled. So, I can write this one and this one will get cancelled. So, I will have 10 to the power of 4 is equal to 2 by Z or Z will be equal to 2 into 10 to the power of minus 4. This was a simple question. Option number A here.

[00:33:57]Next, two moles of an ideal gas is expanded isothermally and reversibly at 300 K from 1 L to 10 L. The enthalpy change in the reaction. Now, whenever isothermal process is there, delta U will be equal to zero and delta H is also a function of temperature. This is also zero. So, option number D here.

[00:34:14]Next, to which orbit does the electron in the hydrogen atom jumps on absorbing 12.1 eV of energy? Okay.

[00:34:22]So, we have hydrogen orbitals here.

[00:34:24]Right? Now, the hydrogen in its ground state, okay, has an electron has an energy of 13.6 eV. Now, you have supplied 12.1 eV. Right? So, that gives me how much?

[00:34:36]6 minus 1 5 1.5 Okay, I'll just check the calculation.

[00:34:43]Yeah. So, which energy is having that like which orbital is having that much energy? So, minus 13.6.

[00:34:50]Okay? By n squared is equal to 1.5. That is what we have to calculate.

[00:34:57]Okay.

[00:34:58]So, n squared will be equal to 13.6 by 1.5.

[00:35:05]Right? Or I can write 136 by 15. Now, if you see 15 * 9 9 * 5 is 45 is almost around 9. Right? So, that means n should be equal to 3 here, which is option number B.

[00:35:18]Next, consider the following which is not possible. This is a favorite question from KCET. So, we have to check which is not possible. Right? So, if you see n n is equal to 3 so l can have zero m can have zero. Here, if n is equal to 2 n can l cannot be 2. It has to be zero and one. So, two is not possible. Here, these are all possible. This is if l value is zero, then you cannot have minus one here. So, this is also not possible. Here, n is 3 and l is 2, then you cannot have plus 3. It is minus 2 to plus 2. So, this is also not possible.

[00:35:48]So, two four and five Is there any option? Yes, option number B here.

[00:35:53]Next, van't Hoff factors of X, Y, and Z are given. Which is correct?

[00:35:57]Okay. So, van't Hoff factor boiling point. So, if the van't Hoff factor will be more, then boiling point will be more. Right? So, X is 2.8, Y is 1.8, and Z is 3.5. So, that means Z should be highest, then you should have X, but here it is Y given. So, this is incorrect.

[00:36:15]Freezing point. Now, remember if the I value is more, then depression will be more and freezing point will be least here. So, Z is having the highest value.

[00:36:23]So, it should have the least freezing point, then X, and then Y. So, this is the correct one, which is given here.

[00:36:29]Next, the average osmotic pressure of human blood is 7.8 bar at 27° C. What is the concentration of an aqueous solution that could be used in the bloodstream?

[00:36:40]So, we can just calculate here pi is equal to C into R into T. So, if I have 7.8 bar, C concentration is there. R value will be 0.08 2 1 and temperature is 300. Right? 27 plus 273. So, that gives me 3 * 1 3 3 * 2 6 3 * 24 3 0 Okay. And Okay, sorry. I have two more decimal places also. So, I 1 2 3 4. So, 24.6 something I'll get. So, 78 7.8 divided by 24.6.

[00:37:19]So, that is equal to C here. So, 78 divided by 246.

[00:37:26]Right? So, if you see, we have How can I calculate?

[00:37:32]I'll just check this calculation once again. Okay? 3 * 1 3 3 * 2 6 3 * 24 3 * 0 plus 2. And I have [snorts] 300. So, I'll add two. So, 4 1 2 3 4. Yes, 24.6 and 7.8.

[00:37:45]Okay. Now, one more important thing here is I'll tell. Okay. Now, you see this is actually what? This is NaCl. Right? NaCl has also I value. We should not forget that. Here, I value should also be there and that is two. Right? So, this two will also be divided here. So, I will get 3.9 divided by 24.6 or I can divide 39 by 246.

[00:38:08]Right? So, now you know you have to just take one here 246. So, you can see only one option is there starting from one.

[00:38:15]So, that is the answer.

[00:38:16]Next, nitrogen and hydrogen in the molar ratio of 1:3 was allowed to equilibrate.

[00:38:21]If 50% of each of the reactants at total pressure P then KP for N2 and 3H2 gives 2NH3 is Okay. So, we have this reaction. N2 plus 3H2 gives 2NH3. This is the reaction.

[00:38:36]Okay? Now, 1:3 ratio. So, let this be one. Let this be three. Now, we have 1 minus X. This is 3 minus 3X because 1 mole reacts with 3 moles. So, X moles will react with 3X moles. And this one formed will be 2X.

[00:38:50]Now, um Okay.

[00:38:54]Just a minute. If 50% of each of them has reacted Okay. So, 50% means you will have X will be 0.5. So, this mole will be 0.5. 3 minus 0. This will be 1.5. And this one will be 1. Right? 2 into 0.5 will be 1 here. Now, what I can calculate here is n total.

[00:39:17]n total will be three here.

[00:39:19]So, what is the mole fraction of N2?

[00:39:22]Mole fraction of N2 will be 0.5. That is moles of N2.

[00:39:27]Right? By three.

[00:39:30]Yes. Now, mole fraction of H2 will also be 1.5 by three.

[00:39:37]And mole fraction of NH3 will be 1 by three. Okay?

[00:39:43]Now, what we have to calculate the partial pressure. So, partial pressure of N2 will be uh uh 0.5 by 3 into P. Now, what is the partial pressure of H2? It is 0.5 into P because this I can cancel. And partial pressure of NH3, it is equal to P by 3.

[00:40:02]So, now let me just write down this in terms of fraction, okay? So, this one will be 0.5 means 5 by 30 or I can write 1 by 6 P, right? This one is 1 by 2 P.

[00:40:14]I'm just writing it so that calculation will be uh easier and this one is 1 by 3 into P, right? Now, I can uh write down here. So, here I will have 1 by 3 P square divided by um N2 is one time, so I have just 1 by 6 P. And then I have here 1 by 2 P cube. Right? So, three and four we have, so I'll just arrange this one. This is 1 by 3 P square.

[00:40:43]And uh we have Sorry. 1 by 3 into 1 by 3 here. And this one will go up, so I will have 6 by P.

[00:40:52]And here I will have 8 by P cube. Right? So, here P square P square that will get cancelled and I will have just P here. And if you see 3 1's are 3 2's are you can cancel here and I will have 16 by 3 P square, which is there in option number A here, okay?

[00:41:13]Next, we will move to question number 32 here. That is H2O is given uh delta S1 and delta S2, predict which is correct.

[00:41:21]Now, when you are doing in the uh gaseous state, okay? In that case, the entropy is definitely much more higher, so delta S1 will be greater than delta S2 because we are going from liquid to gas here.

[00:41:34]Next, assuming 100% ionization, the increasing order of freezing point is So, just now we discussed if the uh I value is very higher than the freezing point will be less. So, here the I value is five, here it is three. And we have just And here we have two, right? So, since the I value here is maximum, the freezing point will be minimum. So, this one will be minimum then this one, then this one. So, option number A here.

[00:42:00]Next, value of the equilibrium constant for the reaction here. So, if you see Cu2+ to Cu is undergoing reduction, Sn is undergoing oxidation. So, what is the E0 of the cell? It is cathode minus anode. So, copper minus tin. So, copper is 0.34 and this one is minus 0.15, right?

[00:42:20]Uh uh So, that will give me 14 minus five is nine.

[00:42:24]This one will give me 0.19. I think I'm doing correct. Yes. Now, what I have to calculate delta G? That is equal to minus n f into E0 of the cell. No, sorry. I have to calculate the equilibrium constant. So, we know that E0 of the cell is equal to 0.059 by n log of K. So, this one will be 0.19, right? Uh this one will be somewhere 0.03.

[00:42:52]Yeah, I'll take it to 0.06, so I'll get 19 by 3 log of K.

[00:42:57]Okay? Now, 19 by 3 if you have 3 6's are 18, so 6.3 you will get here. That is equal to log of K. Now, K will be equal to 10 to the power of 6.3. That gives me 10 to the power of 6 into 10 to the power of 0.3, okay? Now, if you uh know that log 2 value is 0.3, so this one will be two and this one will be 10 to the power of 6. That gives me in option number C here.

[00:43:21]Next, we have the uh calculation of kinetics part, so rate order we have to find out. So, if you see uh first I will keep this constant, okay? And I'm here what I'm doing, I'm doubling this one. So, with respect to this one, this is becoming four times.

[00:43:36]So, with respect to NO2 minus, it is a second order reaction and I can find that only one option is like that. So, option number C is the correct answer.

[00:43:45]Next, under the same conditions, the initial concentration 1.386 moles per decimeter cube becomes half in 40 seconds and 20 seconds for zeroth order.

[00:43:55]The ratio of K1 to K0 is So, what you will understand, see? What is the value of K1 means for first order reaction, it is 0.693 by T half. T half for first order given is 40. Now, K0 will be uh A0 by 2K. A0 value is 1.386 by 2 into T half is 20. That is also 40 here. Now, if I divide here, sorry. Now, if I divide um one second, I'll just select the pen here. So, if I divide K1 by K0, okay? If I divide this K1 by K0, it will be equal to uh how much? 1.386 divided by 0.693.

[00:44:36]So, this one will be 2 3's are 6, 2 9's are 18, so you will get a value of two.

[00:44:42]So, option number D here.

[00:44:44]Next, 10.79 g of silver deposited at cathode from AgNO3 solution, same electricity is passed through copper sulfate using the copper sulfate electrodes, weight of copper deposited.

[00:44:54]Here, we can use Faraday's second law of electrolysis. So, mass of silver deposited that is 10.79 by mass of copper deposited is equal to the equivalent mass of silver, so 108 divided by uh equivalent mass of copper that is 63.5 into 2. So, now if I just multiply this one and cross multiply here, so 10.79 I can take it as 10.8. I have 63.5 divided by 2 into 108. So, this one will be I will have 32 3's are 6, 2 1's are 2 uh 7's are 14.

[00:45:33]Yeah, 3.31.75 and there's one divided by 10 also, so that will give me 3.175, which will be equal to 3.2 here. Next, during the charging of lead storage battery, the reaction occurred at So, anode gets oxidized here. So, what will happen? Pb from uh zero gets converted to Pb in plus two. So, answer will be B here, plus four. Let's see the next question. The oxidizing power of chlorine in aqueous solution using the data is given. The energy of half chlorine plus Cl minus will be. So, you can just add here half of dissociation enthalpy first of all, okay? So, half of 240 is 120. Then you have electron gain enthalpy, which is minus 349 and hydration enthalpy minus 381. Now, if you solve this, you will get uh minus 610, which is option number C here.

[00:46:21]Next, we have 12 L of H2 and 11.2 L of Cl2 are mixed and exploded. The composition by volume is a mixture of So, if you see the reaction, you have H2 plus Cl2 gives you 2 HCl, right? So, if you have taken 12 L and here 11.2 L, so this one will completely react, but 0.8 L of HCl will form. Now, 1 mole of chlorine is giving you 2 moles of HCl, so 22.4 moles or double of HCl will form. So, that is in option number C here. Next, 1 L of hydrocarbon at 127°C and 1 atm pressure weighs 2.8 g. It contains 10.5 g of carbon per gram of hydrogen. Then the molecular formula is Okay. So, now this is a bit lengthy question. So, first of all, understand what we have to do. We have to find the empirical formula, okay? C and H I have been given. Uh C is 10.5 g and H is 1 g, right? That is the relation between carbon and hydrogen. So, if I just now divide it and calculate its uh uh number of moles by 12 here and by one here. This one will definitely give me one, but if I want to divide here, so I'll take 105 divided by 120. So, here there will be one zero added. Uh 900 9 2's are 8 9 will be too much, so I'll take 8 here. So, 8 0's are 0, 8 2's are 16. Um 8 2's are 16, then I have uh 8 0's are 0, 8 2's are 8 1's are 8 plus 1 9, right? So, 0 15 minus 6 is 9, so that's all I think I have. Yes. Now, again if I take one more decimal, so [snorts] again you can take 0.87. So, this one will give me 0.87 and this one will give me one. Now, to convert it into whole number, you have to find a multiple to so that you will get here.

[00:48:04]So, I can multiply this one with seven and this one also with seven. So, if I multiply both with seven, what I will get? Sorry. I can multiply this one with eight, this one with eight so that I can get a whole number. So, see 8 7's are 56, five. 8 8's are 64. 64 plus five is 71.

[00:48:23]So, 7.1 I will get here and I will get here eight. So, C8 Sorry. C7H8 is the empirical formula. Now, what is the empirical formula mass here?

[00:48:34]Empirical formula 12 7's are 84. 84 plus eight, this one will be 92. Empirical formula mass is 92. Now, we can find the uh number of moles here. PV by RT.

[00:48:47]Right? So, what is the pressure given?

[00:48:49]Pressure is 1 atm.

[00:48:51]Um I'll write here 1 atm, okay? Volume is 1 L. uh 1 L.

[00:48:57]R value is 0.0821 L atm mole inverse Kelvin inverse. And temperature is how much? 127. So, 127 plus 273.

[00:49:14]That will give me 400.

[00:49:17]Right? So, I have to solve this.

[00:49:19]Okay. So, this is a lengthy calculation.

[00:49:22]You can see 400 I can write it as 10 to the power of minus two so that I don't have to worry about that and I have 0.08 if I just consider 4 8's are 32 4 So, I will have 0.32.

[00:49:35]So, now we have this one. I can write 100 by 32. So, if I just divide 100 by 32, so I will get here three times. 3 * 2 = 6 96 four again. So, 40. So, 3.1 8, okay? So, that will again be three point Sorry, 2 * 2 = 4, 2 * 3 = 6. So, 3.12 can be taken.

[00:50:00]So, I will have 100 divided by 32 into 10 to the power of minus two, right? So, I'll get 0.0312, okay? So, this is the number of moles here. Now, you know number of moles is equal to given mass by molar mass. Now, in the given question, we have 2.8 g as the given mass. So, 2.8 divided by 0.03, okay? 0.0312.

[00:50:29]So, if I divide this, you can calculate.

[00:50:32]You will get around 9 * 3 is 27 something. So, 92 you will get here. And if you see the empirical formula mass is also 92 and the molecular mass is also 92. So, the empirical formula will be C7H8.

[00:50:45]Next, we have to multiply this one.

[00:50:49]Sorry, match the following. Energy of the ground state of helium will be And we have a 13.6 or minus 2.8 into 10 to the power of minus 18 Z squared by N squared. So, ground state will be >> [clears throat] >> here we have Z squared minus 2.18 into 10 to the power of minus 18 Z squared, right?

[00:51:11]Hydrogen helium is two squared, so four.

[00:51:14]So, 4 * 8 = 32 7 4 * 2 = 8. So, 8.72.

[00:51:18]So, A has to be three. Which option is there?

[00:51:22]A3 Okay.

[00:51:26]We can also check in terms because we have option for four. So, let's check if I'm getting in terms of 13.6 into four.

[00:51:33]4 * 6 = 24 12 13 14. 4 * 1 = 4 + 1 5.

[00:51:37]Yes, we are getting minus 54.4. So, A has to be four. So, there are two options. I can eliminate these two, right? Now, the potential energy this value is very is a constant value that is minus 27.2 EV potential energy. So, B has to be two. So, this is wrong. This has to be the correct option.

[00:51:56]Next, N2 and NH2 in 1 is to 3 molar ratio at equilibrium, the total pressure is 10 atm. Mole fraction of NH3 is 0.6.

[00:52:04]The equilibrium constant KP for the reaction is okay.

[00:52:09]Now, see N2 is there.

[00:52:12]Plus 3 H2 gives 2 NH3, okay? Now, let's this be one uh three this is zero. Now, here it will be 1 minus X. This will be 3 minus 3 X and this one will be 2 X.

[00:52:26]Right? Now, the Okay, you we have been also given with the mole fraction of NH3. So, what is the N total here? You will have a minus one 3 plus one four, right? And minus four plus two. So, four minus two X.

[00:52:43]Minus four plus two, yeah. So, this is N total. Then, what is the mole fraction of NH3? 2 X divided by 2 into 2 minus X. I can just take common here.

[00:52:54]So, two two will get cancelled. That is equal to 0.6.

[00:52:58]Now, X is equal to I can multiply here 1.2 plus Sorry, minus 0.6 X. So, this side goes if it comes this side, you will have 1.6 X is equal to 1.2. So, X is equal to 12 by 16. Or, you will have 4 * 3 = 12.

[00:53:17]4 3 by 4. Or, that is 0.75, okay? Now, if this is 0.75, then we can calculate all these values first. So, 1 minus 0.75 will be 0.25.

[00:53:30]3 into this one will be 0.75, I think, because I can take 3 into 1 minus 0.75 and that will be 0.25 and 0.25 into 3 will be again 0.75.

[00:53:44]Okay, next 4 minus 2 X. So, this one is 2 * 5 = 10 1.5. So, 4 minus 1.5 will be now four minus 1.5 will be 2.5.

[00:53:58]Right? I'll just check here. If I multiply 2 * 5 = 10 15, yeah. 4 minus 1.5 I will get 2.5. So, I have got all the values here. Now, the total pressure is given. So, we have to calculate the mole fraction of each of them, right?

[00:54:12]Now, if I see mole fraction of N2, okay?

[00:54:14]That will be 0.25 divided [snorts] by 2.5, right? So, 100 by 10 that will be around like 100 here and that will be 10 to the power of minus one. Now, mole fraction of H2 will be how much? Mole fraction of H2 will be H2 value 0.75, right? So, 0.75 divided by 2.5. So, how can I write this? 75 by 100 into 25 by 10.

[00:54:44]25 * 1 = 25 * = This one this one will get cancelled. So, I will get here 0.3 mole fraction of H2. Now, what is the mole fraction of NH3? That we have already given. That is 0.6. Now, what is the pressures for each of them? So, you have to just multiply with 10 for all the values. Now, let's see the pressure partial pressures of each of this. So, P of N2 will be one and P of H2 will be 0.3 into 10. So, that is three because the total pressure is 10 atm, right? So, you have to multiply this pressure with the mole fraction. And P of NH3 will be equal to six here.

[00:55:20]So, now what we have? We have this formula for KP. Now, the question is asking the KP for the dissociation of NH3. Means this reaction will be reversed. So, you have N2 and 3 H2 on the right hand side and 2 NH3 on the product side. Sorry, on the reactant side. So, N2 pressure is one.

[00:55:36]H2 is three. So, three three times. So, I will have like this.

[00:55:41]Okay?

[00:55:44]Divided by NH3 will be two times.

[00:55:47]Right? So, 3 * 1 = 3 * 2 = 3 * 1 = 3 * 2 = So, I will have 3 by 4. That is equal to 0.75. So, option number B here, okay?

[00:55:58]Now, we will move to question number 44 here. 5.85 g of NaCl is dissolved in 1 L of pure water. The number of ions in 1 ml of the solution is So, if you see first of all molarity if I calculate Oh, sorry. So, molarity if I calculate, it is 5.85 divided by 58.5 into 1 L, right? So, that will give me 100 and will have 10. So, that will be 0.1 or 10 to the power of minus one. So, I can say that 1,000 ml okay? has 0.1 mol of NaCl. Or, the number of ions will be two because each has two moles. Then, in 1 ml it will be how much? 0.2 into 10 to the power of minus three.

[00:56:45]Right? Now, number of This is moles.

[00:56:47]Number of moles I'll multiply 6 into 10 to the power of 23. So, that will be 1.2 into 10 to the power of 20, okay? Which will be option number C here.

[00:56:59]Next, according to collision theory, increase in temperature increases the rate of the reaction because the molecules have more energy than the threshold energy. This is also explained in your Boltzmann curve graph. Which among the following equation represents the Arrhenius equation? It is K equal to A by E A R T.

[00:57:16]Next, 18 g of glucose is present in 1,000 g of the aqueous solution of glucose. It is said to be So, if you see 18 g by 180 into 1,000 by 1,000, right? So, this is 0.1 molar.

[00:57:29]Option number D here. Which of the following reactions correspond to the enthalpy of combustion at 25° C? So, for enthalpy of combustion, the reactant should always be one mole, okay?

[00:57:41]And here we have 2 O2 CO2 plus 2 H2O.

[00:57:45]Okay, here what I have CH4 2 O2 is given liquid. So, this is wrong. Here we have H2O given as a gas. That is wrong. Here we have CH4 plus 2 O2 gives CO2 gas plus 2 H2O. This is given gas here. So, this is also incorrect.

[00:58:03]So, A has to be the correct answer, okay? Now, let's see next question. A 200 g of cricket ball is thrown with a speed of 3 into 10 to the power of 3 cm per second. Its de Broglie wavelength is So, if you see lambda is equal to 6.6 into 10 to the power of minus 34, okay?

[00:58:19]Joule second. Joule kg meter squared second minus two and here I have second, okay? Joule second. And then you have M by V. M is 200 g. So, I will write it as 200 into 10 to the power of minus 3 kg.

[00:58:38]Okay? Then, we have 3 into 10 to the power of 3 cm. Centimeter if I convert it into meter, it will be 10 to the power of minus two, right? 1 meter equal to 100 cm. Then, we have a meter second inverse. Now, this second goes up, it it will become S2 and this is minus two. This all three will get cancelled. Meter will get convert to meter here. And kg kg will get cancelled.

[00:59:07]So, now we have here 6 3 into 2 is 6. 6 divide 6.6 divided by 6 will be 1.1 here, okay? Now, these two zeros are left. And these two these three zeros are left. And this one is left, okay?

[00:59:21]So, I will have 10 to the power of minus So, you can see this one is 3 minus 2 is plus one. This is two. So, I have total zeros will get cancelled in the denominator. So, I will have 1.1 into 10 to the power of minus 34 m. To convert it into centimeter, multiply with 100. So, minus 32 will be the answer here, option number A.

[00:59:42]Next, a mixture of ethyl alcohol and propyl alcohol has a vapor pressure of 290 mm at 300 K. Vapor pressure of propyl alcohol is 200. If the mole fraction of ethyl alcohol is 0.6, its vapor pressure at the same temperature.

[00:59:55]Okay. So, total vapor pressure is 290.

[00:59:59]Now, vapor pressure of propyl alcohol is 200. Okay. Ethyl alcohol mole fraction is 0.6. That means propyl alcohol will be 0.4. Plus 0.6 into the P value. So, 80 here. Okay, 480. So, that will be 290 plus equal to 80 plus point six P. Now, this one will be 210 divided by 0.6 is equal to P here.

[01:00:27]So, that will 2100 divided by six. So, three ones are we have 3700 and two we have 350. Right?

[01:00:35]So, that is option number B here.

[01:00:38]Next, the region in the electromagnetic spectrum where the Balmer series appears is your visible region.

[01:00:44]Next, the reduction of potential hydrogen half cell will be negative. So, we have this formula E is equal to 0 minus 0.0591 by N, right? Log of H2 pressure back P of H2 divided by H+.

[01:01:02]Now, if I want this to be negative, then this value should be more than one. If it is more than one, then the log value will come out to be positive and one minus and positive will give you negative. So, this value should be more than one. Now, let's check here. Here, it will be one by two. So, that will be 0.5 less than one. So, this is not there. Again, one by one. So, this one will become completely zero here.

[01:01:24]Uh So, one by log zero will get next here. You have two by one. Yes. So, if I have two by one, that means I'll get a positive number here. So, option number C here. Okay.

[01:01:36]Uh next, we'll have according to Bronsted-Lowry concept, mark the option in which the conjugate pair is correct correctly matched. Okay.

[01:01:45]Sorry. Yeah.

[01:01:46]Uh if you see here, uh if you see here, conjugate acid if you want to calculate. Okay, conjugate acid means that component should behave as an base. Suppose you want to find the conjugate acid of X, then this X should behave as a base. Base means accepting of proton. So, if I want to calculate the conjugate acid of this one, then it should behave as a base and base is acceptance of proton. So, here I should get H2CO3. So, this is wrong. Now, if you see H HPO4 conjugate acid I want to find out. Okay. Conjugate acid of HPO4 2 minus.

[01:02:20]In that case, what will happen?

[01:02:22]Conjugate acid of one species. That means that should behave as a base. It should accept a proton. So, when this accepts a proton, you will get H2PO4 minus. Now, conjugate base of this one means it should behave as an acid. It should release one H+. So, you'll get PO4 3 minus. So, B is the one which is correctly matched. Then, which of the statement is false here? Okay. The correct order is BaCl2, KCl, and CH3COOH.

[01:02:48]Osmotic pressure given is pi by MRT.

[01:02:52]Raoult's law, vapor pressure of component is proportional to mole fraction. Two sucrose solutions of same molality of different solvents have the same freezing point depression. No, when the solvent is different, freezing point depression will also be different. So, that is why option number D here. So, students, these were the top 50 questions from physical chemistry and I hope you have revised all these questions and stay tuned for the next video.