 Hi, Hula there. Hi, Amogha. Please let me know who Hula there online. Good afternoon. Who Hula there? Apnoon. Sir, there's another Siddhant Shirkay. Yeah, he's there. Oh, that's so cool, man. What? There's another Siddhant S. My name is Siddhant Shrikant. Yeah, yeah, yeah, there is one more guy, Siddhant Shirkay, yeah. Yeah, it sounds cool, like, yeah. He's from another school, actually. Anyways, anyways. So, yeah, who? Yeah, Naman, right? Yeah, who are there on Skype? Okay, so if you're there on Skype or YouTube, anywhere, you know, you can join this session, okay? There's no any issue. Okay, last class, Naman, you were there in last class? Yes, sir. Okay, so what have we discussed last class? Yeah. Sir, we finished kinetics and we started solutions. Oh, yeah, solutions. What have we done in solutions? Yeah. Hello, Aditya. Yeah. Okay, so solution, what we have done in solution? Modality. Concentration term. Concentration term. Have I given you some formula into that? Some concentration term? Yes, sir. Like, weight by volume percent, weight by weight percent? Yes, sir. Some modality formula in terms of, okay, okay. So, we'll continue. Since some of you are, you know, not there in the last class, I'll just give you a quick revision of that and then we'll continue, okay? So, like we were discussing about the concentration term last class because again, the calculation of all these concentration thing is important in this chapter, okay? So, we'll keep on revising things also. If you do the questions related to concentration, so you are basically revising your mole concentration part or a mole concept part, right? So, which is also required as your J exam is, you know, in January, right? So, revision is also done. So, basically, we have discussed few of the concentration term and I've given you some relation into that. So, I'll just write down the formula that we have, okay? I didn't share the screen. Wait, wait, wait a second. Oh, you are not there on YouTube, right? Fine now? Okay. Okay, so the first concentration term and the formula that I have given you in the last class that is equals to percentage weight by volume, right? If you have percentage weight by volume, that is equals to what we have? Percentage weight by weight density of the solution, right? Into density. This is what I have given you in the last class probably, right? One more formula I have given you that is the relation of molality and molality, right? Molality and molarity, okay? So, molality is equals to 1000 into molarity divided by 1000 into density minus molarity into molar mass of solute, molar mass of solute. So, basically, when you know density, when you know molarity, you can calculate molality. This is the density of solution, okay? One more formula of, you know, molality we have. Molality, the basic formula we have is this and that is equals to number of moles of solute, right? Divided by mass of solvent. If this mass I'm taking in gram, then we have to multiply 1000 over here, right? This is the formula we have that we have already done in the previous classes, right? One more expression we have, we can write down here. If you divide by total number of moles into the numerator and denominator, then what happens, okay? So, let me write this as number of moles of solute divided by NT, the total number of moles divided by mass of solvent, what we can write? It is a number of moles of solvent into molar mass of, molar mass of solvent, right? Into this 1000 will be as it is and since I divide here by NT, so here also we have to divide by NT, correct? So, the formula becomes what? Number of moles of solute by total number of moles is the mole fraction of solute, X is the mole fraction. So, mole fraction of solute into 1000 divided by mole fraction of solvent into molar mass of solvent. This is another formula we have, correct? This formula probably I have given in the last class. Yes or no? Yes, sir. Okay. Now, one question you just solve onto this, okay? One question all of you write down and the question is an aqueous solution contains chemical kinetics, have you solved chemical kinetics? All question. Yes, sir. Okay, so you are able to solve, right? There's no difficulty. What about AMOG? AMOG, did you solve? Yes. Aditya, did you solve chemical kinetics? Do you have any difficulty or do you got the concept properly? Okay, fine. An aqueous solution contains 28% weight by weight ratio it is, weight by weight thing it is given, right? 28 weight by weight percent of KOH solution density of the solution is given, density of the solution is given and that is 1.25 gram per ml. Okay, you have to find out percent is weight by volume, first thing you have to find out molarity, third you have to find out mol fraction of KOH, so I will write down X of KOH, okay? And the last one you have to find out molality, do it and tell me the answer. Siddhan, are you there? Yes, sir, yes, sir. Did you watch my video, last video did you watch? Why? Why sorry? Why so? You do that, okay, chemical kinetics is important, okay? Yeah, you get some time and do watch it. Okay, we have done this question in the last class, right? Okay, fine, fine, fine. So those who are not there just, okay, we'll do it quickly then, okay? You see, since weight by volume we have to find out, okay? So I have given you the formula, right? So weight by volume is what? Percentage weight by volume, percentage weight by volume, from this formula we can, what we can write down? Percentage weight by weight into density, everything is given, right? So 28 into density is 1.25, okay? Another way we can write down, 28 into 1.25 is nothing but 5 by 4, okay? So that gives you 35, right? 35 weight by volume percent, okay? So what is the meaning of this we have? 35 gram in 100 ml of solution, that is what the definition of weight by volume, okay? 35 gram in 100 ml of solution. So percentage weight by volume, if you write, it is 35 gram in 100 ml, that is the meaning of this data that we have, right? Now in this one thing you see, if I multiply this by 10, so 10 into percentage weight by volume is equals to what we can write? 350 gram present in 1000 ml of solution, right? Everywhere, see, 35 gram present in 100, so 350 present in 1000, the ratio we have to maintain, right? Okay, so now the next thing is what? 10 into percentage volume is nothing but 350 gram present in 1000 ml of solution. Now this is present in 1000 ml of solution, right? So it is weight by volume and this if