4.) A solution is prepared by mixing 1.00g ethanol (C2H5OH)...

4.) A solution is prepared by mixing 1.00g ethanol (C2H5OH) with 100.0g water to give a final volume of 101 mL. a) calculate the molarity. b) calculate the mass percent. c) Mole fraction of ethanol. d) Molality:

Transcript

00:01 First of all molar mass of c2h5oh so that will be 2 multiplied by atomic mass of carbon plus 5 multiplied by atomic mass of hydrogen atomic mass of oxygen plus atomic mass of hydrogen so this will be equal to 2 multiplied by 12 plus 5 multiplied by 1 plus 16 multiplied multiplied by plus 1 okay so this will be after summing up all these we will get 46 gram per mole okay now first of all we have to calculate for molarity so molarity is the number of moles of dissolved in one liter of solution so that is the number of moles of solute divided by the volume of solution in liter so now putting the values furthermore we know that number of moles is the ratio of mass divided by molar mass right so this will be mass divided by molar mass multiplied by volume in liter okay so now mass we have is 1 .00 gram divided by molar mass is 46 gram per mole and now we have volume that is 101 milliliter to convert it into liter dividing this by thousand milliliter because one liter consists of thousand milliliter okay so now this will be equal to this will be 0 .215 molar okay that is the required molarity moving on to next we have v part mass percent so mass percent that will be equal to mass of the solute divided by mass of solution okay and now since this is percentage so that's why we have to multiply this by hundred now solving this putting the values we will 1 .00 gram the solute is ethanol divided by 1 .00 plus hundred point zero gram of water we have here right so if solution is formed by solute and solvent right when we mix solute into the solvent right so the solute mass is 1 .00 gram solvent mass is 100 so after adding the solute and solvent mass we get the mass of solution and multiplied by 100 now solving this we will get 0 .99 percent percent okay moving on to next c part so that was the required mass percent mole fraction of ethanol okay so let me write it as x so that is the symbol for mole fraction and e for the subscript of ethanol okay e for ethanol this is the subscript here and now that will be equal to number of moles of ethanol divided by number of moles of ethanol plus number of moles of water.

04:25 See mole fraction is the ratio of the number of moles of given component divided by total number of moles of the component present in the solution.

04:33 Here we have ethanol and water in the solution.

04:38 So that's why for ethanol this will be the ratio of number of moles of ethanol divided by the total number of moles that is the number of ethanol and number of moles of water furthermore this can also be elaborated as number of moles is the ratio of mass divided by molar mass as i told you so this will be mass of ethanol divided by molar mass of ethanol divided by mass of ethanol divided by molar mass of ethanol plus mass of water divided by molar mass of water.

05:13 Okay and now this will be equal to so we have 1 .00 gram divided by 46 gram per mole divided by 1 .00 gram divided by 46 gram per mole and this will be plus mass of the water we have is 100 gram divided by we have 18 gram per mole the molar mass for water we all know the sum of atomic mass is right.

05:55 So, now solving this, this will be equal to, this will be 0 .0217 mole divided by 0 .0217 mole plus 5 .55 mole...

Question

4.) A solution is prepared by mixing 1.00g ethanol (C2H5OH) with 100.0g water to give a final volume of 101 mL. a) calculate the molarity. b) calculate the mass percent. c) Mole fraction of ethanol. d) Molality:

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