 Hi everyone, in this lecture let's continue with the study of equivalent weight, as you can see I have already mentioned a particular definition here, now let's understand this better, so equivalent weight of a substance is defined as the number of parts by weight, when I say parts by weight I mean gram okay, so grams of a given substance which combines, now see here if this X is the substance that we have and I want to find out the equivalent weight of this substance X, so it is that amount let's say X grams okay, now let's say X grams of this substance is displacing, is displacing 1 gram of H2 that is equivalent to 11.2 liters of H2, it is also that mass which displaces 8 grams of O2 equivalent to 5.6 liters of O2, now this is for you to find out that whether 8 grams of O2 is actually 5.6 liters or not this is 5.6 liters okay, now it is also that amount of grams that is displacing 35.5 grams of Cl2 which is equivalent to 11.2 liters of Cl2 again right, so now this is what actually equivalent weight is, now coming to that part which we studied already that equivalent weight is equal to molecular weight upon n factor, this is definitely true, this is true for acids, bases, salts, now when we are trying to find out the equivalent weight of a particular element or a substance then this particular concept is also required, why because the question will be as such right, now try to understand one relation, we will be finding out three different relations of equivalent weight and different substances, so first thing let's say X gram of O2 is getting displaced by Y grams of a metal okay, this is the statement that is given, so now X grams is displaced by Y grams of the metal, so 8 grams of O2 will get displaced by Y by X into 8, so if I say X grams is getting displaced by Y grams, 1 gram will get displaced by Y by X grams and then 8 grams of O2 will get displaced by how much gram of the metal Y upon X into 8, so from here we reach a definite formula that is formula number one that is our equivalent weight of metal divided by equivalent weight of oxygen is equal to mass of the metal displaced, mass of the metal which is displacing the oxygen upon mass of oxygen displaced okay, so this is a method which is known as oxide method, oxide method, method to do what to find equivalent weight of any metal okay, so this is the first formula that we are getting, so simply if I have written this formula then you would have thought a lot of things but it is mainly coming from unitary method, similarly you can see here this 8 gram of oxygen is displaced by this much amount of metal, so this is our equivalent weight of metal, so EWX is equal to Y upon X into 8, so EWX that is equivalent weight of metal upon equivalent weight of oxygen is equal to mass of the metal upon mass of O2 that is displaced, so this thing and the equation that I have written below you can tally it, it is same only okay, so basic unitary methods we are getting this formula, so this is our oxide method, this is our first method okay, second method is hydride method, now this we will again not use unitary because same method may the formulas are getting derived, so you can do it on your own, I am just giving you the formula this time that is equivalent weight of the metal upon equivalent weight of hydrogen that is equal to the mass of metal which is displacing the hydrogen upon the mass of hydrogen that is getting displaced, this is by hydride method. Next method is our chloride method, chloride method, what is chloride method again one formula we have to write down, equivalent weight of the metal upon equivalent weight of chlorine that chlorine means here equivalent weight of hydrogen means H2 gas okay, here also oxygen means we are talking about O2 here, so here also it should be Cl2 okay, just do the corrections, now that is equal to mass of the metal that is displacing the chlorine upon mass of Cl2 displaced by the metal, so these are the three methods with which we can calculate equivalent weight of any particular metal, okay, now suppose one small example equivalent weight of a metal is 9, 9 grams it is given, so what will be the equivalent weight of its oxide, the equivalent weight of its oxide will be the equivalent weight of the metal plus the equivalent weight of oxygen, why, why is it like that? Because in oxide we only have the metal and the oxygen, so whenever we are given any particular substance that is equivalent weight of AB, let us say that is equal to equivalent weight of A plus equivalent weight of B okay, according to this particular formula only I am finding this out, so if the equivalent weight of the metal is 9 then what will be the equivalent weight of its oxide 9 plus 8 that will be our 17 grams, similarly equivalent weight of its chloride and hydride, both we have to find out, so for chloride it will be equivalent weight of the metal plus 35.5 that will be 44.5 grams and for its hydride it will be 9 plus 1 that is equal to 10 grams, so I hope you have understood this equivalent weight concept, now one more thing that is let us take the ratio of two substances okay, now for example let us say equivalent weight of AB upon equivalent weight of CD that is equal to mass of AB divided by mass of CD, now what is this, this is the case when we need to find the equivalent weight of any substance in a double displacement reaction, double displacement reaction, so for example if the question says that there is an oxide which is getting converted into its chloride okay, so what will be the equivalent weight of our metal okay, metal oxide, let us say certain metal oxide is getting converted to let us say x grams of metal oxide is getting converted to y grams of metal chloride okay, when the question comes like this and they are asking the equivalent weight of metal, this is the one type of question that can, this is one type of question that can come up okay, so at that time we will be using this particular formula here, we will be doing questions based on this but before that one more thing we will learn that is, so I hope this is entirely clear you can see there are only four formulas, one is this oxide method then hydride method then chloride method and then this is our double displacement method when the question type will be somewhat like this okay, x gram of metal oxide getting transformed into x, y grams of metal chloride find out the equivalent weight of metal, relating to this we will practice numerical soup, now coming to an important law that is law of chemical equivalence, for example A plus B is giving me C plus D, let us say I have balanced it and in front of A N1 is there, B N2 is there, C N3 and D N4, so this N1, N2, N3, N4 can be any digit okay, it is not always that N1 has to be equal to N2, N2 has to be equal to N3 and N3 has to be equal to N4, it is nothing like that, N1, N2, N3, N4 may or may not be equal, may not be equal okay and what does this N represent, number of moles, number of moles but what about the number of gram equivalent, now previously we have been taught that gram equivalent is equal to N into N factor right, so if I say, now according to this law the number of gram equivalents of A is equal to the number of gram equivalents of B that is equal to the number of gram equivalents of C and that is equal to the gram equivalents of D, this is law of chemical equivalents that means moles can vary, may or may not be it can be same but the number of gram equivalents of all the reactants as well as products should be always same in any balanced chemical equation, so from here I can write down pertaining to this particular equation that N1 into N factor of A should be equal to N2 into N factor of B that should be equal to N3 into N factor of C that should be equal to N4 into N factor of D, so if we know the gram equivalent of even one reactant or one product we will be easily able to calculate all the N factor of the reactants and the products okay, so this is the significance of our law of chemical equivalents, now coming on to the next part that is now there are certain important pointers as well but before that we will be solving certain questions, coming on to the first question that is yes that type of question that I was telling you, 3 gram of a metal oxide is converted to 5 gram of a metal chloride okay, so this is let's say our x and this is our y okay, now which formula to follow? This is the formula to follow okay, so now this is let's say we are taking as x and this is as y, so here in the numerator first we will write down the formula equivalent weight of AB divided by equivalent weight of, instead of writing AB we can write it as metal oxide MO and equivalent weight of MCL okay, that is equal to mass of metal oxide that we have taken as x divided by mass of metal chloride that we have taken as y, now equivalent weight of the metal oxide will be equal to equivalent weight of metal plus equivalent weight of oxygen that is 8 divided by equivalent weight of metal plus equivalent weight of chlorine that is 35.5 okay that is equal to 3 upon 5, now we just have to solve for let's take the equivalent weight of M as M, so M plus 35.5 that is equal to 3 upon 5, 3 upon 5, so when you calculate it the equivalent weight of the metal will come around 33.25 grams, 33.25 grams right, coming on to the next question, 2 grams of a base whose equivalent weight is equal to 40 reacts with 3 gram of an acid find equivalent weight of the acid, now again I told you the law of chemical equivalence in which the number of gram equivalence will be same, so in this case also the gram equivalence for complete neutralization, one note point we will write that is for complete neutralization, for complete neutralization the number of gram equivalence of acid should be equal to the number of gram equivalence of base, number of gram equivalence of base, so what does this mean that N of acid into N factor of acid should be equal to moles of base into N factor of base, now what is our moles of acid that they have given, let's keep it as moles of acid only or we can change it into weight upon molecular weight of acid into, now N factor of acid we already know that equivalent weight is equal to molecular weight upon N factor, so from there N factor will be equal to molecular weight upon equivalent weight, so weight upon molecular weight of acid into molecular weight of acid upon equivalent weight of acid, so now this 2 gets cancelled, here also same thing will happen, so this is weight of acid, weight of acid upon molecular weight of, this will not be acid right, this will be base, this is moles of base right, so we will be writing base by molecular weight of base into molecular weight of base upon equivalent weight of base, again these 2 gets cancelled giving the final value as weight of acid upon equivalent weight of acid that is equal to weight of base upon equivalent weight of base, so now we will put the values, so weight of acid is given that is 3, equivalent weight of acid we need to find out that is equal to 2 upon 40, so from here equivalent weight of acid we are getting as 40 into 3 divided by 2 that is our 60 grams, I hope you have understood this, on to the next question, see here there is one equation that has been given and what is the question, calculate the N factor or valence, N factor is also known as valency factor, so calculate the N factor for HNO3, TIN and H2O, now SN 12 by itself, now here if you knew redox you would have done this in no time but as you do not know for now we can still do it with the help of mole concept, so you can see here there are 12 HCL and we know that HCL's N factor is how much, 1, so that means N into N factor, what is N for HCL? 12, so N into N factor that is nothing but gram equivalent will be equal to 12 into 1 that is 12 gram equivalent, 80% of the thing is done, so if HCL is 12 gram equivalent then all the substance here present here will be 12 gram equivalent only, now we need to calculate for HNO3, so N into N factor of HNO3 will be equal to 12, what is N for HNO3? 4, 4 into N factor is equal to 12, so N factor specifically for HNO3 is coming out as 3, now for TIN, TIN will be N into N factor that is equal to 12 again, so for TIN the N is equal to 3, so it will be 3 into N factor is equal to 12, so N factor will be equal to 12 upon 3 that is 4, coming to water again N into N factor will be equal to 12 and how many moles of water is present? 8 moles, so 8 into N factor is equal to 12, so N factor will be equal to 12 upon 8 that will be around 6 by 4 that is nothing but 3 by 2, so mole the N factor can also be in fraction at times, so please be careful while we are calculating the valency factor. Now coming on to the next question, read the question and write on your own first, right, so 3 litres of propane and butane mixture on complete combustion give 10 litres of CO2, find the ratio of propane and butane in the mixture, see here that first we will write down the equation combustion reaction of propane and butane, so these are the combustion reactions for propane and butane. Now we can clearly see in case of this propane mixture that 1 litre of propane gives how much amount of CO2, 3 litres of CO2 according to that Avogadro's law sorry Gelusax law it will be volume, in terms of volume, so 1 litre of C3H8 will give 3 litres of CO2. Now X litres of C3H8 will give how much amount of CO2, 3 X litres of CO2, ok so far done. Now coming to the butane reaction we can see 2 litres of C4H10 that is butane gives how many litres of CO2, 8 litres of CO2 according to Gelusax law. Now total mixture was 3 litres, right and X litres have been used as propane, so how much is left, 3 minus X litres of butane will give me how much amount of CO2, 8 into 3 minus X litres of CO2, right. Now ok we did some mistake it is 2 litres is giving 8 litres, so 3 minus X litres will give how much we have to divide it by 2 also now, so we were doing here 8 into 3 minus X divided by 2 litres, why because 2 litres is giving 8 litres mean 1 litre will give 4 litres, right so that means 3 minus X should give 4 into 3 minus X litres of CO2 and the amount of CO2 that is formed at the end is given 10 litres of CO2. So what should we do we should just add these 2 and solve for X, 3 X plus 4 3 Z12 minus 4 X is equal to 10 when we solve this X we will get it as 2. So that means the amount of propane that is present in the mixture is 2 litres, this is propane C3 H8, so how much will be butane, butane will be 3 minus 2 that is 1, so what will be the ratio of propane is to butane it will be 2 is to 1, very important question this is. Now coming on to the next one again a very important type of question. So this question says that we have to find out the ratio of Fe plus 3 and Fe plus 2 in Fe 0.9 S, okay, S, now this Fe 0.9 S is definitely a neutral compound. So I am considering out of 100 sulphide ion, okay, there are 90 iron ion, okay, so out of this 90 iron ion how many there are 2 oxidation states of Fe plus 2 and Fe plus 3. So out of this 90 let us say I am taking Fe plus 2 to be X and Fe plus 3 to be 90 minus X. Now we will just use this neutral oxidation and like the entire charge of the compound as neutral concept, let us say X into plus 2 plus 90 minus X into plus 3 that should be equal to 2 into 100, why 2 into 100 because sulphide is minus 2, okay, minus 2. So these some should be equal that means the total positive charge in the compound should be equal to the total negative charge in the compound. So from that we will solve it, so 2X plus 270 minus 3X is equal to 200, so from here we are getting X is equal to 70, X is equal to 70. So from here what we are getting X is our 70 molecules, 70 ions out of 90 and this which one is 70, Fe2 plus is 70 and Fe3 plus is our 20, 90 minus 70 that is 20 ions. So what will be the ratio of Fe3 plus upon Fe2 plus it will be 20 is to 70 that is 2 by 7 that will be 0.28. So I hope all these important questions are clear to you. Now moving forward there is one law, very important law, dilution law. What does dilution states, what is dilution? Basically when we dilute something on dilution what happens to concentration? The concentration of solute decreases whenever we dilute any particular substance, particular solution the concentration of the solution is decreasing. What will happen to the volume? The volume will increase, the volume will increase. What will happen to the number of moles of solute? It will remain same, number of moles of solute remains same, this point is very important based on this point all the numericals will be done. So I am saying molarity that is concentration is decreasing, volume is increasing. But what is staying constant? Number of moles that means m into V is equal to constant because m into V is ultimately equal to moles, right n was equal to m was equal to n upon V. So m into V will be equal to constant that is number of moles of solute remains constant. Now let us understand this properly, we have a beaker, we have certain solute particles let us say we have 10 moles of solute okay and then we have let us say the molarity of this is 1 and the volume is let us say how much volume should be there 100, 100 ml, okay. 100 ml means how many litres 10 to the power minus 1 litre, 10 to the power minus 1 litre. So that is how there are 10 moles, okay. Now let us say if I am diluting it, okay I have diluted it and now it has become 1000 ml. So now what will be the new molarity for this? What will be the new molarity for this? Definitely the molarity will be decreasing, if I am the volume has been increased the molarity should decrease, so the molarity should it should have decreased, so the molarity has now become initially it was 1 and in 1 let us say there was 100 ml, 100 ml solution definitely when I am increasing the volume the molarity will decrease. So what will happen to the number of moles? Did I change anything in the number of moles? I did not change anything in the number of moles, the moles are as it is let us take it as x only, okay it is not better to write not a digit, we are not calculating anything here let us take x moles, okay when there is 1 molar solution 100 ml solution mean that is x moles, so when I increase the volume and I am increasing the volume that means I am diluting it, that time concentration will decrease but what will happen to the number of solute particles it will remain as it is, okay so that particular concept we will be using, so here we can say that M1 V1 is equal to M2 V2 after dilution this is let us say before dilution this is after dilution. So after dilution and before dilution the number of moles this is moles, moles of the solute will remain same. Now let us quickly do one question based on this, how much water should be added, should be added in 250 ml of semi molar solution, what is semi molar solution now, semi molar solution to make it desimolar, now you know what is desimolar solution, last video I have told you what is desimolar solution, now a desimolar solution means the molarity is 0.1 molar and when I say semi molar, semi molar means the molarity is 0.5 molar, so now what they have told 250 ml of a semi molar solution, let us take this as V1, this as M1, this as M2, so apparently M1 V1 should be equal to M2 V2, so it should be 0.5 into 250 is equal to 0.1 into V2, so from here volume is coming as how much, 0.5 into 250 divided by 0.1, so it will be around 1, 2, 5, 0 ml, now this is the volume of the new solution, so for making initially I had 250 ml, I had 250 ml whose molarity was 0.5, now the solution has become 1, 2, 5, 0 ml, now the molarity is 0.1, so how much of extra water did I add, that should be 1, 2, 5, 0 minus 250 that will be how much, 1,000 ml, 1,000 ml of extra water I have added into the solution to make to dilute this particular solution, okay, now I hope you have understood this so far, now there are two categories of this particular dilution, two cases are there which we are going to continue in the next lecture, thank you so much.