 Hello students, today's lecture is about concentration terms and it is one of the most important topic in mole concept, so let's get started. Before starting, we need to know what is the meaning of this term concentration. So we know a solution, right? A solution is made up of a particular solute and a solvent. Now let's talk about what is a solute. Now two cases can be deciphered here. This one, when both the solute and solvent both are in the same phase. Same phase means that either they are liquid or they are both are gases. So that means both the solute and the solvent are in the same phase. So at that time how we will decipher which one is this suppose I have mixed let's say generalize it a plus b I have mixed and I don't know which one is solute, which one is solvent. So case one is when both a and our b, a and b both of this are in same phase. That time how we will decipher which one is solute and which one is solvent. The one which is in lower concentration, lower concentration. Now that's when the term concentration comes into the picture and gets very important. What is the meaning of concentration? That means the amount of that one particular thing that is present. So if I am mixing a plus b suppose I have mixed 4 grams of a and I have mixed 10 grams of b. So 4 grams of a I have mixed with 10 grams of b. So now which one is lower in quantity, substance a is lower in quantity. That means substance a will be our solute in this case. So we cannot always say that okay substance a is 4 grams. So we cannot always say like that or substance b is 10 grams or it is very difficult to express that way. So that's when concentration terms comes into the picture. We will learn that later but let's complete this case one and case two first. So in the case when a and b both are in the same phase then which one will be the solute? The one which is in lower concentration will be the solute will be the solute and whichever is in the higher concentration that will be our solvent. Now let's say we have mixed ethanol and water. So let's say that I have mixed 5 percent like the in the entire solution of ethanol and water 5 percent is ethanol and 95 percent is water. So who is lower in concentration definitely ethanol and this ethanol will be our solute and water will be solvent. If it was the opposite way that water was 5 percent and ethanol was 95 percent then definitely water will have been the solute in that case okay. So it is not that always water will be our solvent it depends on the quantity in which it is taken. So this was our case one in deciphering which one is solute and which one is solvent. Next coming up on to the case two. In case two a plus b whichever we are mixing both the phases are different. So a is in one phase and b is in another phase. So both are in different phases okay. So when both are in different phases then when we mix those two the one which loses its identity which loses its identity in the solution upon mixing in the solution that will be our solute and the one which retains its identity retains the identity that will be our solvent. Now we will understand this with the help of an example small example let's say we have let's say we have taken water anything we can take suppose sugar plus water we have taken sugar plus water. Now upon mixing this two this is our solution that is present this is the solution. Now if somebody has not told us that this is sugar solution we will be just seeing that okay this is water we won't be able to decipher whether it is salt or sugar or any other substance that is mixed in the water. So in order to know it we have to actually taste it if it is sugary then definitely it is sugar solution. Now so we can see here that sugar is we know that it is solid crystals right and water here will be in the liquid phase now who is losing its identity upon going in the solution definitely sugar is losing its identity upon going in the solution that mean sugar is losing its identity and water is retaining its identity that means if we see the solution we can say that okay it's a liquid. So water is retaining its identity and sugar is losing its identity so in this case sugar will be our solute and water will be our solvent. I hope what is solute and what is solvent that is clear blindly we will not say that okay solute is something which we dissolve in water that's a very wrong definition of a solute okay so there are two cases we have to decipher which one is the solute and solvent according to this cases that I have given here. Now so clearly we can understand that this concentration is a way that we denote that how much of solute is present in how much of solvent okay so we are quantifying we are quantifying how much of the solute is present in how much of solvent. Now there are two kinds of concentration terms one is our temperature independent temperature independent concentration term and the second is temperature dependent concentration term what a temperature independent concentration term that means it is not depending on temperature so whichever concentration term will not depend on volume does not depend on volume that will be temperature independent why because we know that is like in the next sessions we will be studying one equation that is ideal gas equation Pv is equal to nRt. Now why am I saying this right now is you just need to know just a vague idea that pressure volume and temperature these three are very good friends among themselves okay so if you tell anything to temperature definitely volume and pressure is going to react okay so this pressure volume and temperature if you are changing any one of these the pressure the pressure volume and temperature whatever is left if you are changing let's say volume then the pressure and temperature is also is also will change okay so if it is not depending on volume then definitely it will not depend on temperature also and those concentration terms will be known as temperature independent concentration terms what are the some examples of temperature independent concentration terms percentage weight upon weight then one more fraction mole fraction then the third one molality molality these three are concentration terms which does not depend on temperature and there are concentration terms which depends on temperature why because they depend on volume so what whatever equation we will write of this concentration term we will be seeing that in the equation itself there is volume involved so first is our percentage volume by volume so this this you can actually see in the term itself that there is volume involved then second is percentage weight upon volume third one is molarity fourth is normality normality so all these are the concentration terms that we are going to study today in details so all these are so this is also important they can ask which one is the temperature independent concentration term and which one is the temperature dependent concentration term so under temperature independent percentage weight by weight mole fraction molality and under temperature dependent percentage volume by volume percentage weight by volume molarity normality now now we will start studying each one of them in details one by one starting off with these three this one percentage weight by weight percentage volume upon volume and percentage weight upon volume this three we will study together so I am marking it as A, A, A then we will study regarding mole fraction which is our B and these three we will be studying together molality molarity and normality now starting off the first one is our percentage weight upon weight okay the formula is weight of solute upon weight of the solution weight of the solution and as because it is a percentage into 100 okay so percentage weight by weight means how much of the solute the weight of the solute that is presented the total weight of the solution so this is the first formula that we have got for percentage weight by weight next is our we will do we will solve numericals based on this but first I want you to note down the formulas properly then we have percentage weight upon volume what is the formula for that weight of solute weight of solute upon volume volume of solution in ml so remember this here the unit weight of solute it is gram you can take it in kg also as because it is a ratio it will get cancelled now when this percentage weight upon volume are fighting the weight of the solute will be in grams and the volume of the solution you take it in ml and of course percentage is there so into 100 so this is our second formula that we have got the third one is percentage volume upon volume that will be nothing but volume of solute upon volume of solution volume of solution in ml now both of this can be either in ml or in liters as it is an issue it doesn't matter again into 100 as because it is a percentage so these are the formulas for this particular concentration terms concentration terms now I hope you have understood these three particular formulas now based on this formulas we have numericals that we need to do so starting off with the first one the weight of urea in 90 grams of solution is 9 grams ok 90 grams of the solution they have given already ok so this is the weight of solution that is given and the weight of urea that is given as 9 grams so this is our weight of solute so now you have to find the percentage weight by weight of urea so what we will do we will just simply put the put it in the formula percentage weight by weight is equal to weight of solute upon weight of solution into 100 so what we will get 9 upon 90 into 100 so the answer will be 10 percent there is 10 percent urea in 90 gram of urea solution ok so when we are adding 9 grams of urea in in a particular solution and the weight of the solution becomes 90 grams then the percentage weight by weight of urea inside that solution will be 10 percent now I hope you have understood this similar kind of questions are gonna come from the next two formulas also for example let's see the weight of urea in 100 ml of solution they can change the question also like this ok that time what will happen 9 upon 100 into 100 so it will be 9 gram per ml ok 9 gram per ml so I hope you have understood this or you can also say it as 9 percent 9 percent means whenever you are given with a 9 percent solution suppose in in the lab also sometimes in the like lab work we will be given questions like make a 12 percent solution or make a 9 percent solution when in the lab we are doing generally in the lab we make this particular solution we work with this particular solution so what we need to do suppose they have asked us to make 9 percent solution of urea let's say of urea only so what we will do we will take 9 grams of urea we will take that in a beaker and we will make it up to 100 ml with water make it up to 100 ml the beaker will be marked right it will be calibrated it will be calibrated beaker let's say this is 100 so once we have taken this solute 9 grams of urea we will just make it up this solution to this 100 mark so that's how we make a 9 percent solution of urea I hope you have understood this moving on moving on to the next particular concentration term for today that is d mole fraction mole fraction now let's understand this mole fraction properly what is mole fraction for example let's say we have a particular beaker we have a beaker inside the beaker we have three substances and I'm generalizing it by taking three a mixture can be made up of more than two substances as well so let's take three a b plus c now inside this solution I have let's say I have put certain amount of a b and c and I have found out the moles the amount of moles that I have put inside the solution let's say I have put n a moles of a n b moles of b and n c moles of c so mole fraction is defined as the ratio of the number of moles of the substance that I am suppose I want to find out the mole fraction of b so here I am talking about ratio of the number of moles of the substance that means b substance to the total number of moles that is present in the solution total number of moles in the solution now when we say a plus b plus c I am talking specifically about a ternary solution ternary solution ternary solution means where we are mixing three different substances if it was a plus b we call it as a binary solution binary solution so here specifically I am talking about a ternary solution so now suppose I need to find out the mole fraction of a b and c separately we denote the mole fraction as this elongated x which is known as chi okay chi now I want to find out the mole fraction of a so chi it is noted by chi a chi a will be equal to number of moles of a upon total moles in the solution so if I take the total moles that is n a plus n b plus n c equal to n total okay I am taking this I am considering this so chi a will be equal to n a upon n total correct similarly I can find out chi b also chi b will be equal to n b upon n total and chi c will be equal to n c upon n total so this is how we can find out mole fraction when I say n total n total like this I can also write it as n a upon n a plus n b plus n c similarly this can also be written as n b upon n a plus n b plus n c similarly this can be written as n c upon n a plus n b plus n c so this is the formula for mole fraction that we have got now one thing we need to know that we have been given this solution so I can say from this solution that chi a plus chi b plus chi c that will be equal to one we can prove it also chi a was how much n a upon n total plus chi b was n b upon n total chi c was n c upon n total so now we can see the denominator is common so we can take it as n a plus n b plus n c upon n total and what is this n a plus n b plus n c n total only so ultimately n total upon n total will give me one and it is true for any solution that means a binary solution a ternary solution any kind of solution how no matter how many substances you mix together so if I generalize it so now I am generalizing the equation so chi a is equal to n a upon n a plus n b plus n c plus dot dot dot n x that means x is the total number of substances that you have mixed together so this is a generalized way of representing representing mole fraction similarly same thing will happen chi a plus chi b plus chi c plus dot dot dot plus chi x that will be equal to one it will always be equal to one so we have to keep in mind these two particular formulas in case of mole fraction now speaking about binary solution generally we will be working with binary solution generally in binary solution chi a will be equal to n a upon n a plus n b and chi b will be equal to n b upon n a plus n b and we know chi a plus chi b is equal to one so suppose in the question if chi a is given chi b I will already know how chi b will be equal to one minus chi a so if any one of the mole fraction is given in a binary solution I will be knowing the other one as well so chi a will be equal to one minus chi b these two things we have to keep it in mind while we are solving questions related to mole fraction now I hope we all the formulas are clear to you so far so this is the these are the three formulas that we have got previously and this is the mole fraction formula that we have got so four concentration terms I hope it is clear now coming to a question let's say the question is 60 grams of urea 60 grams of urea is mixed with is mixed with 90 grams of water I have to find out chi of urea which I am taking as chi a here and chi of water which I am taking as chi b first try it on your own it's very easy you can easily do it okay so 60 gram of urea is mixed with 90 grams of water first things first we need to find out the moles okay moles so n a that means moles of urea will be equal to weight that is given upon molecular weight of a okay so weight of a given as 60 and please note that the molecular weight of urea is again 60 so the moles is our one n a is equal to one now coming to the moles of water n b weight of b upon molecular weight of b so that will be 90 upon 18 that will be how much 5 5 so now we know that n a is equal to number of moles of urea is 1 number of moles of water is 5 now it is easy chi a will be equal to n a upon n a plus n b that is 1 upon 1 plus 5 that is 1 upon 6 so the mole fraction of urea in this particular solution that is made is 1 upon 6 and what about chi b chi b you can find out using this formula also or you can just do 1 minus chi a that is 1 minus 1 upon 6 that is equal to 5 upon 6 I hope it's clear now moving on to the last three concentration terms molarity normality molarity so these three we are gonna do it like in a tabular form comparative form okay now what is molarity molarity is defined as the number of moles per liter volume of solution so molarity is denoted with capital M so our m will be equal to the number of moles upon the volume of the solution in liters now little bit I will expand this formula moles was what moles was equal to weight upon molecular weight and generally in the questions the volume will be given in milliliters so I will just change it also so it will be volume in liters if I have to change it in ml I have to divide it by 1000 so ultimately when we make the reciprocal and everything we will get weight upon molecular weight into 1000 upon volume in ml so one mistake we have done here this will not be in liter this will be in ml if it is given in ml for changing it in for changing it in liters I have to divide it by 1000 the volume in the question will be given in milliliters now I know that molarity is moles upon volume in liters so this entire thing should be volume in liters this entire thing should be the moles moles of the solute okay now so this volume should be definitely in ml and not in liters now the formula that we have obtained is this molarity is equal to weight by molecular weight and what is this weight weight when we say w w is the weight of the solute that is put in the solution and molecular weight is also of that particular solute itself okay and this volume is the volume of the solution so this is all about molarity this is one particular concentration term which is very very important and molarity some few important points we will see but let's complete that table first moving on to normality if you have understood molarity the other two will be very very easy normality is the number of gram equivalents I'm writing it in a full way first because you're listening it for the first time number of gram equivalents per liter volume of solution now the question is what is gram equivalent gram equivalent is very very similar to moles it's it is also a measure of the amount of substance okay now let's talk about a little bit more regarding gram equivalents before we move further so basically if we talk about normality normality is nothing but number of gram equivalents upon volume in liters now if we split this gram equivalent I'm doing it here gram equivalent was nothing but weight upon equivalent weight moles was what moles was weight upon molecular weight so now we got another new term that is our equivalent weight equivalent weight and what is equivalent weight equivalent weight is nothing but molecular weight that is equal to molecular weight upon upon n factor n factor now we need to know again from one of one one one term one new term is coming up so first we know what is gram equivalent gram equivalent is nothing but very similar to moles itself it is also a amount of it is also a like measure of amount of substance where moles is the weight by molecular weight and gram equivalent is weight by equivalent weight now we know that equivalent weight is equal to molecular weight upon n factor so what is the most important thing that we need to know here is what is n factor now coming on to this special thing that is n factor we need to know basically how to find out the n factor properly for each substance now this thing we will be studying more in details in redox but so far three main things we need to find out three main for three main substances we need to know how to find out n factor first is our acids now second bases and the third is our salts now how to find out n factor in case of acids number of H plus ions released in aqueous solution of that acid okay for example if I talk about HCl if it if it goes in the aqueous solution it will dissociate into how many ions H plus and Cl minus so in the aqueous solution what it is giving 1 H plus plus Cl minus Cl minus according to the definition n factor is the number of H plus ions released in aqueous solution so how many H plus ions it is giving 1 so n factor for HCl will be equal to 1 similarly for HNO3 it will be again n factor will be again 1 what about H2SO4 H2SO4 now H2SO4 it can dissociate into 2H plus plus SO4 2 minus or it can also dissociate as H plus plus HSO4 minus so both are possible for H2SO4 n factor can be 1 also 2 also but suppose nothing is mentioned only H2SO4 is given to you and you have to use the n factor of H2SO4 then we will be taking H n factor as 2 only the highest one that is possible that we will be taking but suppose they have given us this equation and you can see that in the aqueous solution it is giving only 1 H plus and that time to definitely we have to take n factor as 1 because they have mentioned it so similar thing will go for this as well H3PO4 so for H3PO4 either it can dissociate as 3H plus plus PO4 3 minus or it can dissociate as 2H plus plus HPO4 2 minus or it can dissociate as H plus plus H2PO4 minus so if any of the equation is mentioned then we will see the equation that how many H plus is released and we will go with that n factor but if nothing is mentioned we will go with 3 itself but the possibilities are 1 2 3 and if nothing is mentioned we will be taking as 3 next H2CO3 n factor possible is 1 and 2 then H2C2O4 n factor possible is again 1 and 2 then H3PO3 okay now this is an exception n factor is equal to high as possible is 1 and 2 only okay so if we go and see the structure of this one hydrogen is directly attached to phosphorus so that is not ionizable okay which one is ionizable which is related to this oxygen okay so this this is based on the structure similarly if we see H H3PO3 H3PO2 sorry H3PO2 it will be what n factor will be only 1 2 is not possible okay 2 is not possible so if I show you the structure here it is like this OH OH OH so all this hydrogen which is related to this which is connected to the oxygen those are ionizable so that's why all the three are getting ionized in H3PO4 but if I show you the structure of H3PO3 then it will be something like this and one H is directly attached to the phosphorus so this H particularly is not ionizable similarly H3PO2 structure is 2H are directly connected to phosphorus and one is ionizable I hope you understood why the n factor varies for H3PO4 H3PO3 and H3PO2 I hope you have noted down all the structures I'm gonna rub it so please note down if you have not okay now coming on to the next one that is our H3BO3 H3BO3 what about H3BO3 H3BO3 also same thing n factor is equal to 1 1 now almost all of the assets we have covered n factor now coming to the basis basis number of OH minus ions released released in the aqueous solution there is no such exception here so basically NaOH may n factor will be 1 then KoH n factor is also 1 CaOH whole twice n factor will be possible n factors will be 1 and 2 if nothing is mentioned again here also we are going to take the highest one that is 2 next Mg OH whole twice here also n factor possible is 1 and 2 next AoH whole thrice here also n factor possible is 1 2 and 3 but if nothing is mentioned we are going to take the highest one so I hope this n factor thing is clear for the assets and basis properly now coming on to the salt part coming on to the salt part so salt I am showing it here itself salts in salts n factor is equal to the total positive charge or not and or the total negative charge that means we are not concerned about the charge here negative or positive we are not concerned whatever it is we are just taking the numeral value for example NaCl what is the total positive charge here plus 1 and for Cl it is minus 1 so that means either of them we can take so n factor will be equal to 1 next example CaCl2 so total positive charge calcium for calcium it is plus 2 and for 1 chlorine it is minus 1 and how many chlorine are there 2 so 2 into minus 1 is minus 2 so n factor here will be 2 now next AoH whole thrice so here AL is plus 3 1 OH is minus 1 so minus 1 into 3 that is our minus 3 so n factor will be equal to only 3 next Ca3 PO4 whole thrice yes so how much is the total positive charge 1 calcium is plus 2 into 3 that will give me how much plus 6 and 1 PO4 is minus 3 so there are 2 of them so minus 3 into 2 that will give me again minus 6 so here the n factor value will be 6 now coming to NaHCO3 this is called an acid salt in case of acid salts also the n factor will be equal to number of replaceable replaceable hydrogen okay so NaHCO3 may how many replaceable hydrogen is there 1 so it will be n factor is equal to 1 so this is how we find out the n factor in cases in case of salts now I am coming back to the equivalent part again so equivalent weight is equal to molecular weight upon n factor okay so that means for example one example small example we can take let's say H2SO4 nothing is mentioned means n factor of H2SO4 is 2 so what will be the equivalent weight of H2SO4 it will be molecular weight of H2SO4 upon 2 and we know that for H2SO4 the molecular weight is 98 so 98 upon 2 that will give me an equivalent weight of 49 so please note that whenever the n factor is greater than 1 definitely the equivalent weight will be lesser than that of molecular weight so for H2SO4 the equivalent weight is our 49 okay now we will see certain relations I hope this part is clear to you that equivalent weight and n factor part we will be practicing questions but before that we need to establish certain relationships between gram equivalent and moles and n factor let's come back to normality then so in normality we can write as weight upon equivalent weight divided by volume in ml upon 1000 so if we expand this particular formula what we will get normality is equal to weight upon equivalent weight into 1000 divided by volume in ml this is the formula that we have got for normality wait this is not the end okay so I can place I can modify a little bit I can in place of this equivalent weight I can place molecular weight upon n factor because just now we saw that equivalent weight EW is equal to molecular weight upon n factor let's try to incorporate the n factor in this equation so it will be molecular weight upon n factor in 2000 divided by volume in ml now if I take this n factor take the reciprocal it will go up so weight upon molecular weight into n factor in 2000 divided by volume in ml now try to understand this entire part was what this part was gram equivalent number of gram equivalent so apparently whatever is left other than this 100 upon volume in ml so this part is also amount of number of gram equivalent this is also number of gram equivalent so if I write this now see here writing it here so I can say gram equivalent is equal to this particular color is not visible properly let's go with this yes so number of gram equivalent is equal to I can say this entire thing weight upon molecular weight into n factor so so now from this equation what is this this is nothing but moles we have seen no here see here this is moles so same thing we can say here also gram equivalent is equal to then moles into n factor very very important moles into n factor now come back to normality again if this is moles now try to imagine it normality is equal to this n factor I am putting it at the last I am bringing this part along with this okay so w upon mw into 1000 divided by volume in ml into n factor now what is this is in this part the same as this part check it so what is this part molarity so from here what we are getting normality is equal to molarity into n factor so see so many equations we have got so from here the first equation is the normal equation that is molarity equation normality equation and one more we will see that is molarity equation don't mix this molality and molarity and normality now just see molarity and normality for now so this is the two equations and from here three more equations we have got the first one being this one equivalent weight is equal to molecular weight upon n factor second equation is this gram equivalent is equal to n into n factor and third is normality is equal to molarity into n factor so please note down all these formulas I will provide you with formula sheet as well but still taking your running notes is very very important okay so far we have understood a lot of things molarity normality then we have come on to equivalent weight and n factor now the last concentration term that is our molality what is molality molality is nothing but number of moles number of moles per kg of solvent per kg of solvent here there is no solution involved okay so number of moles per kg of solvent that means if I try to write the formula now this is denoted as small m okay so small m is equal to number of moles upon the mass of capital M of solvent in kg this is the particular formula of molality now if I again expand it it will become molality is equal to weight upon molecular weight divided by mass of solvent in grams upon 1000 so why we have taken this in grams because we are dividing it with 1000 so if in the question it is given in grams only generally so we have to convert it into kilograms so ultimately what formula we are getting weight upon molecular weight into 1000 upon mass of solvent in grams so this is the formula for molality molality now we can clearly see that here is volume term here is volume term so these two will be temperature dependent but can you see any volume term in this molality equation no so it will be temperature independent okay so I hope all this concentration terms are very very clear to you and all the equations you have noted down so again I am going through once see these three formulas were there percentage weight by weight weight by volume volume by volume these three are important then we had mole fraction these two formulas this is generally if it is binary you take till b only if it is ternary you take till c okay next these two generally if we go for binary this is the one you have to follow we have solved certain questions and then we have come up with three different equations one for molarity one for normality one for molality and if we talk about molarity and normality again we have come up with three more equation that is this particular equivalent weight and then we have come up with gram equivalent and then we have come up with the relation between normality and molarity and we have also learned how to calculate the n factor for different different substances like acids bases and salts redox right now is not not that important so if we know this three that is enough now so with this we have completed all the important concentration terms that are there now let's move on to certain questions quickly find molality of 49 percent weight upon weight H2SO4 now what is 49 percent when I say 49 percent that time I mean that 49 percent means what 49 grams in 100 grams of solution okay so 49 grams of solute in 100 grams of solution then only it will get multiplied with 100 and give me 49 percent weight by weight correct so this is the meaning of this 49 percent weight by weight now you know that we have 49 grams of solute and we have 100 grams of solution so now tell me how much will be solvent in that how much will be solvent solvent will be 100 minus 49 that is 51 grams now you can easily solve it using the formula m is equal to weight upon molecular weight into 1000 divided by volume sorry mass of solvent in grams that is 51 and what was the weight here 49 and here the molecular weight of H2SO4 is 98 so just calculate it and find the answer okay so it will be 49 upon 98 into 1000 upon 51 so this will get cancelled with 2 and this by 500 so the answer will come as 500 upon 51 molal 500 upon 51 molal next question calculate the weight of H2SO4 present in 200 ml of desimolar solution what do you mean by desimolar solution desimolar solution please note this very important desimolar solution means 0.1 molar solution 0.1 molar so when they say desimolar solution actually they have already given you the molarity that is 0.1 molar now they are asking me to calculate weight of H2SO4 present so first I will write the formula molarity is equal to weight upon molecular weight into 1000 divided by volume in ml see in the question the volume is given in ml so this is given that is 200 molecular weight of H2SO4 is 98 molarity is also given that is 0.1 what do you need to calculate weight see how simple it's very easy 0.1 is equal to weight I am taking as w only upon 98 into 1000 divided by 200 okay 200 so this gets cancelled with 5 so ultimately weight will be equal to 0.1 into 98 divided by 0.5 so when we calculate this we will get an answer equivalent to 1.96 grams so to make a solution to make like the weight of H2SO4 that is present in 200 ml of desimolar solution is this much grams okay next question the number of the number of moles of kcl in 1000 ml of 3 molar solution 3 molar molarity is 3 they have given and 1000 ml means the volume is 1 litre so what is the meaning of this molarity number of moles per litre of solution okay per litre of solution now so that means if molarity is equal to 3 that means 3 moles of solute is present in 1 litre of solution and here also in the question they have mentioned it is 1000 ml so definitely the number of moles is of kcl that is present in 1000 ml of 3 molar solution is x is equal to 3 simple now what do we need to find out the value of x plus 27 so it will be 3 plus 27 that is equal to 30 so this will be our final answer next question if the density of methanol is 0.793 kg per litre what is the volume needed for making okay first we will find out so they have told that the volume needed for making 2.5 litres of its 0.25 molar solution first we will find out first first things first we will write down formula weight upon molecular weight is into 1000 upon volume in ml okay this was the question now what they are asking is molarity they have given that is 0.25 and weight I need to find out weight upon molecular weight of methanol is how much ch3 OH so it will be 12 plus 3 plus 16 plus 1 how much is it 16 plus 16 correct that is 32 so 32 into 1000 divided by now 2.5 litres they have already given it in litres so we do not need to we do not need this 1000 basically so we can directly write it as 1 upon 2.5 now weight will be equal to 0.25 into 32 into 2.5 this is the weight okay so your homework will be to calculate this value okay calculate this value so it will definitely come in grams okay grams now after that you convert that into kilograms you convert that into kilograms now next you know that density is equal to mass upon volume and what do we need to find out the volume so if the density of this particular thing is given to us and we know the mass this is the mass so what should we do to find out the volume volume should be equal to mass upon density so whatever let's say this is x kg you have found out so it will be volume will be equal to x upon 0.793 litres okay we are finding out in litres so this is your work this is your homework find out the value for this we will continue this lecture this particular topic in the next lecture thank you everyone see you in the next video