 Hi everyone in last lecture I introduced you to a law which is called dilution law and this video is just a continuation of that so there are two particular cases which we need to know in case of dilution law. So it may happen that I am mixing two solutions and I want to know the resultant molarity of the solution so I am talking about those two cases pertaining to mixture of two solutions. So when we mix two solutions there can be case one in which there is only and basically in these types of questions we will be seeing mainly in case when two acids are mixed or two bases are mixed or an acid or a base are mixed. So case one include acid solution plus acid solution or base basic solution plus basic solution. Okay so two different like same substance we are mixing together and I want to know the molarity of the resultant solution that time what we need to do is the resultant molarity will be equal to m1v1 that is the molarity of the first solution into the volume of the first solution plus m2v2 upon v1 plus v2. This is the formula that we are going to follow in this particular case one when both of them are acids or both of them are our base. Now in case two what can happen in case two we are taking an acid plus base plus base. Now here let's say if it is the acid or the base both are the gram equivalent if the gram equivalent of the acid is equal to the gram equivalent of base then it is complete neutralization okay. So there will be no H plus or OH minus left in the solution whose molarity I can find out everything will be consumed and it will be forming water and salt but if the gram equivalent of the acid and the base is not equal then there is a way to find out the molarity and what is that the resultant molarity will be equal to m1v1 minus m2v2 upon v1 plus v2. So which one should be our m1v1? m1v1 should be the one which is more in quantity m1v1 it can be either acid also or base also how do we choose that m1v1 whoever who's ever millimoles or moles are higher that will be m1v1 that will be m1v1 and also that means what that particular solution suppose acid is more acid concentration is more than that of base so all of the base will take up the acid that is there and neutralize and what will be left in the solution acid will be left so whatever molarity we are finding out that is the molarity of what basically of the acid that is left with us. Now if this m1v1 and m2v2 is same that means gram equivalent of acid is equal to the gram equivalent of base we can easily see that the concentration is getting zero. Now if the m1v1 of acid is greater than that of m1v1 of base then definitely the resultant solution will be acidic and if m1v1 of base is greater than that of m so this one will be m1v1 another one will be m2v2 just for it will be confusing take both as one so if m1v1 of base let's say is greater than m2v2 of acid then the resultant solution will be basic why base will be left in the solution so here h plus concentration we will be finding out what concentration m are and here we are finding out basically the OH minus concentration so these are the two cases of dilution that we are seeing here. Now we will be doing questions pertaining to that let's say we will do this question only we will do yes 9 points so this is this is a question from J.E.Advance 2012 so 29.2% weight by weight HCl stock solution has a density of okay before starting this question we need to know something very very important that is there are certain relations between the concentration term let's go and write down the relation between the concentration then terms then we will come out come on to this question so write down relations between concentration terms first relation is percentage weight by volume is equal to percentage weight by weight into density this is the first formula second formula is molarity the relation between molarity and percentage weight by weight so molarity is equal to 10 into percentage weight by weight into density upon molecular mass of solute molecular mass of solute so this is the relation between molarity and percentage weight upon weight now just now we saw that percentage weight upon weight into density is nothing but percentage weight upon volume so molarity is equal to 10 into percentage weight upon volume upon molecular mass of solute both are correct the next relation is between molarity and molarity so small m is denoted with small m denotes molarity so molarity is equal to 1000 into molarity upon 1000 into density minus molarity into molecular mass of solute this is the relation between molality and molarity okay coming on to the next relation very important relation between molality and mole fraction so molarity is equal to mole fraction of solute this is what this is chi into 1000 divided by mole fraction of solvent into molecular mass of solvent now suppose only mole fraction of solute is given you can easily find out by replacing this by 1 minus mole fraction of solute this also you can replace right so these are all the few important relations between the concentration terms we should know before starting with with the numericals because this will be given okay one convergence can be given and with this formulas it will get much more easier now quickly we will just see an example for this and then we will go forward with that question j advance question now for example let's say we will do with this equation only the last one find mole fraction find mole fraction of solute in one molal aqueous solution one molal aqueous solution so what should we do one molal aqueous solution is given and they are telling us to find out mole fraction just we will put it in the formula so 1 is equal to chi solute into 1000 divided by 1 minus chi solute into as it is an aqueous solution so definitely the molecular mass of solvent solvent is here h2o so it will be 18 now let's solve this how much it will be so it will be around 18 minus 18 chi s is equal to chi s into 1000 so 18 chi s plus 1000 chi s is equal to 18 so chi of solute will be equal to 18 divided by 1018 okay so this will be the mole fraction of one molal aqueous solution this is the correct answer now let's let's move on into the next question that question j advance question so they are telling that we have a hcl stock solution now what do we mean by stock solution it is nothing but for example when we have guests at home have you seen your parent your mother is making a stock solution of orange juice it is very concentrated okay it is very concentrated and while serving what she does she takes this a little amount of this stock solution and mix it with water and gives it to the guest so this is our stock solution the concentrated solution that is already made that is our stock solution and from here little of the amount let's say 10 ml i am taking and mixing it with 100 ml and serving to our guests okay so this is our stock solution and from the stock i am making a diluted version okay so this is called as stock solution now so they have told that 29.2 percent weight upon weight hcl stock solution has a density of 1.25 gram per ml okay first things first we have to find out the molarity of this stock solution the stock solution has some molarity right but they have given the percentage weight by weight now see why have i given you certain equations you can see the relation between molarity and percentage weight by weight from here so we will apply that formula here molarity is equal to 10 into percentage weight by weight that is 29.2 into density is given as 1.25 divided by molecular mass of the solute that means hcl hcl molecular mass is 36.5 now when you multiply this you will get it as 365 upon 36.5 that will be around 10 molar so the stock solution that is given to us has a molarity of 10 molar now what do i have to do the volume of stock solution now they are asking how much amount of this stock solution how much amount of that orange juice i should take out in order to make 200 in order to make 200 ml solution of 0.4 molar okay so now i will apply m1 v1 is equal to m2 v2 because moles will be same only right the moles of the solute will be same in both of them now m1 is our 10 molar into v1 this v1 only we need to find out how much volume i should take in order to make this particular solution with solution 200 ml of which what concentration 0.4 molar so here we will put 0.4 into 200 so v1 will be how much 0.4 into 200 divided by 10 that will give you a value of 8 8 ml so that means i should take up 8 ml of this stock solution and i have to make it up to 200 i have to make it up to 200 ml in order to make the concentration as 0.4 molar i hope you have understood this particular problem right coming on to the next question that is a compound h2x now this also has come in j advance 2014 a compound h2x with molar weight of 80 grams is dissolved in a solvent having density of 0.4 gram per ml okay assuming no change in volume now see this is a very crucial statement that is no change in volume means there will be no change in concentration as well because moles will anyway not change okay so that means they are asking what they have given molarity and they are simply asking molality so which formula should we use m is equal to 1000 into molarity upon 1000 into d minus molarity into molecular mass of solute using this formula we will be easily able to calculate the molarity directly so put the values 1000 into 3.2 divided by 1000 into 0.4 minus 3.2 into 80 so that will be around 3200 divided by 400 minus 32 into 80 is 256 so we will get around 3200 upon 144 so that is around 22.22 molar so that's what they have asked so it is not that easy see in j advance these kinds of questions have come in which involves direct formula waste questions okay so I hope you have understood the application of this particular formulas and the questions now coming on to the next question this is very easy a 5 molar solution of H2SO4 is diluted from 1 liter to a volume of 10 liters so that means m1v1 is equal to m2v2 okay m2v2 now what we are doing 5 into 1 that is equal to m2 the new molarity I need to find out it has been diluted to 10 liters now so m2 will be equal to 5 into 1 upon 10 that is 0.5 now this is our molarity what they're asking normality of the solution now we all know normality one formula we have seen that is equal to molarity into n factor and what is the n factor of H2SO4 that is nothing but 2 when nothing is given we should take the highest one so normality will be equal to molarity into 2 so that is 0.5 into 2 it will be 1 normal this is in molar this is the easiest question that is there now now we did certain case one and case two mixture of solutions right so these kinds of questions will be generally seen in acid base reactions or titrations more to be more precise so how the question will come is let's say they will give how much how much volume of I'm just making a question right away how much volume of 0.2 molar H2SO4 is required to completely neutralize to completely neutralize base let's say 0.1 molar 8 not 8 cl NaOH this is the question so what we will do here is we know that for complete neutralization the gram equivalent the gram equivalent of acid should be equal to the gram equivalent of base so the gram equivalent of acid is equal to the gram equivalent of base okay so that time what will happen so we can say that MA into VA into n factor of A should be equal to MB into VB into n factor of B where A is our acid and B is our base so from here what is the molarity of the acid given 0.2 VA I need to calculate because volume how much volume of acid I should require that only they are asking so volume I have to find out what will be the n factor of this particular acid again it will be 2 just now we we saw it and molarity of the base is given as 0.1 and okay here volume I didn't mention neutralize let's say 500 ml of 0.1 molar NaOH so here I will put 500 and what is the n factor of NaOH it is nothing but 1 so from here VA will be equal to 0.1 into 500 into 1 divided by 0.2 into 2 so it will be 50 upon 0.4 okay so it will be 500 upon 4 this will be the volume that will be required this is in ml so this much amount of volume of the acid will be required in order to neutralize completely 500 ml of 0.1 molar NaOH this is one type of question from those two cases and another question will be direct that you have been given let's say 200 ml of 0.2 molar NaOH and this has been mixed with 400 or let's say 300 let's make it 300 300 ml of 0.1 molar HCl okay so now what is the resultant molarity this is a question so what you will do first is first you will find out milli moles how will you find milli moles molarity into volume in ml that will give you the answer in milli moles so what will be it will be for NaOH it will be 0.2 into 200 okay so what will be the answer to this 40 40 milli moles and if we find out this this is for NaOH and if I find out the same thing for HCl what it will be 0.1 into 300 milli moles that is 30 milli moles so which one is in excess now definitely the base is in excess so resultant molarity will be 40 minus 30 divided by 500 by 500 because v1 plus v2 so it will be 10 upon 500 that is nothing but one upon 50 one upon 50 molar will be our solution strength after we mix this two which solution OH minus strength basically who is left in excess NaOH and the nature of the solution will be definitely basic okay I hope you have understood this entire dilution law coming on to our next topic and the last topic of the chapter is volumetric volumetric strength of H2O2 volumetric strength of H2O2 now generally when we see a bottle of hydrogen peroxide or inside the like on the top of the bottle on the label there is certain things written like 10 v 20 v 30 v so definitely we can understand it has something to do with the concentration but it is a different kind of way of expressing the concentration of H2O2 so this is another type of concentration only right let's try to understand this H2O2 like basically the concentration of H2O2 is measured in terms of how much oxygen it is liberating okay so when we heat it we get 2 H2O plus O2 now from this equation I can see that 1 mole of O2 is given by is given by 2 moles of H2O2 2 moles of H2O2 so I can say 20 1 mole of O2 is equivalent to 22.4 liters of O2 so 22.4 liters of O2 is given by basically 2 moles of H2O2 now 1 liter 1 liter of O2 is given by how much moles of H2O2 2 upon 22.4 moles of H2O2 now if I say liters of O2 now this x why did I take x suppose I have first claimed that this is x volume H2O2 okay x volume H2O2 that means v means volume strength the concentration of H2O2 so x liters of O2 is given by 2 upon 22.4 into x moles of H2O2 okay so when we cancel it it will become 11 x upon 11.2 moles of H2O2 so when I say x volume it means that 1 mole 1 mole of H2O2 is liberating x ml x liters x liters of O2 that is the meaning of this x volume so now let's try to find out the molarity molarity will be equal to number of moles of H2O2 upon volume of H2O2 now I just now told you x volume means 1 liter H2O2 giving x liters of O2 okay so x liters of O2 is given by how much moles this many moles so we will place the value x upon 11.2 divided by how many how many liters is there 1 liter of H2O2 so it will be x upon 11.2 so ultimately the volume strength is given by the formula 11.2 into molarity this is the formula that we are getting now let's analyze it a little more you can see if the molarity is 1 of H2O2 then x is equal to 11.2 volume right so I can have a number of variations number of relations rather I can say 11.2 volume of H2O2 is equivalent to is equivalent to 1 molar H2O2 that is also this is very very important if you know this relations you don't need to put your so much of thoughts into this calculation that I have done so far this is just for your understanding more in-depth understanding but if you want to solve questions this part whatever I'm giving now that will be enough so 11.2 volume of H2O2 is equivalent to 1 molar of 1 molar H2O2 that is equivalent to 2 normal H2O2 why because the n factor of H2O2 is 2 that is also equivalent to 3.4 weight upon volume percent of H2O2 that is also equivalent to 3.4 grams of H2O2 in 100 ml of the solution that is also equivalent to 3.34 grams of H2O2 in 1000 ml of solution these are all the relations that you need to remember and using all these relations all the numericals will be done another important thing that is the question is asked is volume of O2 liberated by the decomposition of decomposition of H2O2 they will ask so what is the formula for that physical quantity that means the amount of volume of H2O2 they have given physical quantity of H2O2 in ml millilitre into volume strength of H2O2 so for example if they say they have given 10 ml of 25 volume H2O2 and they are asking how much of O2 will be liberated by this particular substance of H2O2 so it will be nothing but 10 into 25 that will be 250 ml. 250 ml of O2 will be liberated when 10 ml of 25 volume H2O2 is decomposed simple and what are the other type of question that can come from this relation let's say they are asking find percentage weight by weight of 5.6 volume H2O2 density they have given us 1 gram per ml now from the relations above where are the relations see here percentage weight by volume is equal to percentage weight by weight into density 11.2 volume H2O2 is equivalent to is equivalent to how much how much weight by volume percent 3 point weight by volume percent so 5.6 volume H2O2 will be equivalent to 3.4 divided by 11.2 into 5.6 this will get cancelled that is equal to 1.7 weight by volume percent now from the relations we have seen that percentage weight by weight is equal to percentage weight by volume upon density and as because here density is 1 this will be same percentage weight by weight in is will be equal to percentage weight by volume so this is only equal to 1.7 weight by weight percentage so these kinds of relation questions will come okay so you can just use unitary method to solve these kinds of questions and when they ask about volume of O2 liberated you can just simply use this formula to find out the amount of O2 that is liberated so with this we have come to the end of this chapter I hope you have understood all the concepts I will be uploading the assignments soon so see you in the next chapter then thank you so much have a great day