Hello students let us continue our discussion on mole and equivalent Concept now in this module first we will take an example on percentage purity and then we will discuss what is the percentage yield of a reaction let’s take the example first and impure six grams of NACL is dissolved in water and then treated with excess of silver nitrate the mass of AGCL precipitate is found to be 14 . 35 grams what is the percentage purity of NACL now Aise cases me sabse pehele reaction likh lete hai ho ky raha hai  it is a double displacement reaction NACL plus silver nitrate gives yeild a AGCL plus NANO3 and now we will first develop the strategy here strategy ky honi chahiye hamari Jo bhi agcl precipitate hona hai  to  actual NACL se precipitate Hona hai  so using the mass of the AGCL precipitate will first calculate the mass of  NACL that is present and then using this information we find the percentage purity jitna actual hai  therotically hume  a given a sixth gram we can calculate so we write down the reaction and then we first do the shoso metric analysis one mole NACL it gives yeild one mole AGCL so using the molar mass data about atomic weights hume given hai silver ke liye hai  108, sodium 23, chlorine  35.5 so from here we can calculate the molar mass of any NACL 58.5 and that of AGCL ls 143.5 now from here converting the moles into mass we can write down 58.5 grams of NACL is producing 143.5 grams of AGCL up hum reverse ka answer analysis kr lete hai 143.5 grams of agcl is going to be produced by 58.5 grams of NACL so one gram of AGCL is going to be produced by 58.5 / 143.5 grams of NACL AB hume kitna mil raha hai  14.35 grams of AGCL it so that is going to produce the other that is going to be produced by 58.5 by 143.5 *14 .35 grams of NACL it it comes out as five point eight grams of NACL ye tha actual amount theoretically  kitna given hai given six grams so percent a purity is simply what 5.85/6 *100 that is equal to 97.5 now let us discuss a very important percentage yield what is percentage yeild sometimes the reaction may not get completed according to the initial amount taken , matlab aapne job hi amount liya use aap jo sochte hai banega wo banta hi nahi  What are the  possible reason is the first possible reason  establishment of equilibrium now this chapter you will read later in the portion of chemistry but let me tell you jitna reactant aap lete ho woe k time ke baad step stabilized ho jata hai wo aage badh jata hai And  that is called as equilibrium and therefore a poora  reaction the corresponding ke  product expect krte hai wo aata hi nahi hai   now for the second reason some of the reactants are not able to react to the formation of a protective layer suppose karo ek solid uske upar humne kuch liquid dala   that reacts with the surface and then cover set up not that we meaning so it cannot react and therefore we will not get the actual amount that we are thinking now the third possible case would be when we take gaseous reactants they may simply escape out of the reaction chamber and will not give you the products and the fourth possible reason can be other reactions happening in parallel for the same reactants Jaise abhi tak aapne padha  n2 and h2 can give you ammonia NH3 but they can also combine to give you N2h 4 which is called as hydrazine so you may not get ammonia in total that you’re  thinking now in such cases the efficiency of the process is measured using percentage yield what is percentage yeild it is the product mass that you actually get divided by the product mass that you theoretically predict times hundred now let us take an example for this A on controlled oxidation gives x according to the reaction hume so-so metric  given it they’re saying some of the reactant a oxidizes and only seventy percent of maximum yield is obtained what mass excess is produced if 200 grams of A is taken now what should be a strategy sub se pehle normal  calculate krte hai  Hum  maan  ke chalet hai hundred percent yeild there and let us find out how much mass A that we can produce and then using the mass of X that we have calculated we will calculate the actual mass by using the percentage yeild seventy percent so let us see how we do it right down the So-so  metric  of the reaction to 2a plus 9b gives yeild 2 x + 4 y plus 5 z it but interesting parts hamare liye ky hai  2A and 2X  two mole so we write down  2a is going to produce two mole X so one mole A is going to produce one mole X and the first convert to moles into mass so 128 gram of a will give you 128 grams of X based on the molecular mass data that was given to us so from here we can say 128 gram of a produces 140 grams of x and therefore one gram of a is going to produce 148 x 128 grams of X hundred-person yield Mante hue  weigh 200 grams of is going to produce 148 x 128  times 200 grams of X now since percentage yield is equal to product mass actually / product mass Theoretical  times hundred this equation re arrange kr lete hai  kyu ki mass actual aata hai  a so we can write it down as mass actually equal to percentage yield by hundred times mass actual now putting in the value yeild the key value Hai  seventy percent so 70 by hundred times this much mass and that is going to give us 161 .875 grams so first we calculate the actual amount or rather the theoretical amount and then we calculate the actual amount in that yeild’ thank you