Mol fraction
If I add 0.0262 mol of propanoic acid to 80ml of water (and 100cm^3 of kerosene but that isnt important for this calculation), titrate 5 ml of the water-acid mix and calculate that I have 0.02304 mol of acid, do I multiply the moles of acid by 16 before calculating the mol fraction so its equivalent to 80ml or not?
So right now I've got:
0.02304 moles of acid + (80/18) moles of water = 4.6304
Then mol fraction of acid = 0.02304/4.6304 = 0.00516
So right now I've got:
0.02304 moles of acid + (80/18) moles of water = 4.6304
Then mol fraction of acid = 0.02304/4.6304 = 0.00516
If you take 80/18 as a number of moles of water, then yes, you are interested in the number of moles in the original mixture, so you multiply by 80/5=16. You can also calculate molar fraction in a 5 mL sample, dividing 0.02304 by (5/16+0.02304)
Note: molar fraction requires TOTAL number of moles, not just number of moles of solvent.
Note 2: whole procedure is slightly wrong, as 5 mL of the mixture is not exactly 1/16 of the original - you can't be sure volume of the solution didn't change from 80 mL after dissolving the acid. It changed for sure, so the sample taken was no longer exactly 5/80.
Note: molar fraction requires TOTAL number of moles, not just number of moles of solvent.
Note 2: whole procedure is slightly wrong, as 5 mL of the mixture is not exactly 1/16 of the original - you can't be sure volume of the solution didn't change from 80 mL after dissolving the acid. It changed for sure, so the sample taken was no longer exactly 5/80.
Original post by Borek
If you take 80/18 as a number of moles of water, then yes, you are interested in the number of moles in the original mixture, so you multiply by 80/5=16. You can also calculate molar fraction in a 5 mL sample, dividing 0.02304 by (5/16+0.02304)
Note: molar fraction requires TOTAL number of moles, not just number of moles of solvent.
Note 2: whole procedure is slightly wrong, as 5 mL of the mixture is not exactly 1/16 of the original - you can't be sure volume of the solution didn't change from 80 mL after dissolving the acid. It changed for sure, so the sample taken was no longer exactly 5/80.
Note: molar fraction requires TOTAL number of moles, not just number of moles of solvent.
Note 2: whole procedure is slightly wrong, as 5 mL of the mixture is not exactly 1/16 of the original - you can't be sure volume of the solution didn't change from 80 mL after dissolving the acid. It changed for sure, so the sample taken was no longer exactly 5/80.
Thanks!
@Note: Yeah sorry, in my original calculations I used all of it and not just the solvent
@Note 2: Are you referring to the fact that it should actually be 82ml and not 80ml?
Original post by Chris-69
@Note 2: Are you referring to the fact that it should actually be 82ml and not 80ml?
You started with 80 mL of water, you added some propanoic acid - not knowing the solution density it is impossible to predict exact volume, but 82 mL is definitely better than 80 mL.
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