Molarity Problems – Solution and Colligative Properties – Chemistry Class 12

Do subscribe to Ekeeda Channel and press bell icon to get updates about latest engineering HSC and IIT JEE main and advanced videos Hello students in the last lecture we have did a numerical based on how to calculate the percentage by volume of a solution now we are going to do a numerical based on how to calculate the mole fraction of a solute as well as solvent so as on your screen you can see a numerical which is based on how to calculate the mole fraction of a solute and solvent present in the solution let us see the numerical in very deep so the caution is we have to calculate the mole fraction of solute and solvent of a solution that contains 60 gram of urea present in 180 gram of water and we have also the molecular weight of each component that is solute and solvent which is the molecular weight of area is 60 gram per mole and molecule made of h2 which is one it example so the thing that you should consider is we should consider what as solute and what as solvent as you can see there are two values which are present in them that is 60 gram and one it again since we know 180 gram is more than 60 M so therefore Vanita gram will be the solvent which is nothing but water so the weight of solvent which is 180 gram of water and that of the weight of solute is 60 gram of durian as well as we have also given the molecular weight of urea as well as border with denote the monitor weight of the solvent as ma which is nothing but 18 gram per mole and that of solute is 60 gram per moon so before finding the intersection of solute and solvent we should know the number of moles of solute and the number of moles of solvent and how to calculate it let us do if I calculate the number of moles of solvent that is and we have the formula weight of solvent divided by molecular weight of solvent so in this case the w/e is 1 it again and the molecular weight is 18 gram therefore 180 divided by 18 is nothing but and what will be the unit C it is number of moles so therefore the unit will be smooth so the value of the number of moles of water will be 10 and now how can we calculate the number of moles of solute that is urea let us say so the form of EB and B is equal to WB developer and B in this case the WB is 60 and the MB is also 60 therefore but the problem is not completed we have calculated the number of moles of solute and the number of moles of solvent which are 1 and 10 respectively let us do further so to calculate the number of moles of solvent we have the formula X a which is nothing but molecular mole fraction of solvent X a is equal to and a divided by any plus MB and recently we have calculated the value of m8s then and okay it has a value of n a which is N and n B we have bought the value as 1 so the value of this would be 0.909 and what will be the unit as it is mole fraction it is nothing but the ratio of number of moles of solvent divided by the number of moles of solution it will be homogeneous but the work is not done yet we have to calculate also the number of moles that we have calculated the number of moles of solute and we have to calculate now the mole fraction of solute we have the formula the mole fraction of solute is nothing but a ratio of number of moles of solute divided by the number of moles of solution that is an A plus and V and we have got the value of NB as 1 and the value of n a was then and again NB is 1 so therefore the value that will get is 0.090 and what will be the unit again it is unitless so this is how we have calculated a numerical based on mole fraction for solute as well as for Sun so let me give you one more detail that the value or the sum of the mole fraction of solute and more section of solvent is always 1 that is we have got the value of XA s 0.909 and that of mole fraction of solute as zero point zero nine zero and we free calculate the both of it then we’ll get zero point nine nine nine which is approximately one so thank you guys for watching this video I hope you have liked this video and please don’t forget to subscribe here or come and share with your friends too thank you

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