I would use the volume of the bottle minus the volume of the ethanol. We're assuming all of the grape juice was glucose and all of it was converted to ethanol. Actually, since the grape juice was only 12% ethanol, I suppose the rest of it is water. The author of the problem may expect you to use volume as volume of bottle - volume of ethanol - volume of the water that wasn't ethanol. Frankly, I don't know. With so many false assumptions at the start of the problem it's difficult to know what the author expects you to do.
BTW, the second post you put up is not correct. The density is given in g/mL and the volume of ethanol is in mL so the correct value for mass is 85.4 mL x 0.79 g/mL = ? grams.
In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is given below. C6H12O6(aq) 2 C2H5OH(aq) + 2 CO2(g) Fermentation of 712 mL grape juice (density = 1.0 g/cm3) is allowed to take place in a bottle with a total volume of 825 mL until 12% by volume of the liquid is ethanol (C2H5OH). Assuming that the CO2 is insoluble in H2O (actually, a wrong assumption), what would be the pressure of CO2 inside the wine bottle at 30.°C? (The density of ethanol is 0.79 g/cm3.)
Chemistry - DrBob222, Wednesday, November 14, 2012 at 12:18am
You need to find the arrow button and use it with equations. --> or ==> or >>>.
I think the first thing to do is to convert 12% v/v to grams ethanol. That is 12 mL ethanol/100 mL soln. Scale that up to 712 (I guess we assume ALL of the grape juice is glucose which of course is not quite right so we lump that in with the error made in assuming CO2 is not soluble in water).
12 mL ethanol x 712/100 = 85.4 mL ethanol. Convert that to grams using density. mass = volume x density alcohol which is given at the end of the problem.
mols ethanol = grams/molar mass
Use the coefficients in the balanced equation to convert mols ethanol to mols CO2.
Then use PV = nRT to solve for P CO2 at the conditions listed. Post your work if you get stuck.
Chemistry - Zachary, Wednesday, November 14, 2012 at 11:49pm
I found the moles of CO2 and I got 1.47. 85.4 L ethanol * .79 g/cm3 = 67.5g ethanol. 67.5/(12+12+5+16+1)= 1.47 moles ethanol. 1.47 moles C2H5OH *(2 moles C2H5OH/ 2 moles CO2) = 1.47 moles CO2. What do I use for the volume? do I use the volume of the bottle, the volume of the bottle minus the volume of the ethanol, or the volume occupied by the grape juice?
Chemistry - Zachary, Thursday, November 15, 2012 at 12:01am
Sorry, I made a mistake. 85.4 mL = .0854 L. .0854 L * .79 g/cm3 = .0675g. .0675/46(MM of C2H5OH) = .0147 moles ethanol. .0147 moles ethanol = .0147 moles CO2.
Chemistry - Zachary, Thursday, November 15, 2012 at 12:03am
I'm still not sure which volume to use though. would it be 825 mL, 825mL - 85.4 mL, or 712 mL?
5 answers
Okay. I'll try both. 825mL - 712= 113. You have to convert mL to L in order to use the PV = nRT correct? if so,113mL=.113L P(.1130) = 1.47 moles CO2 * .0821 * 303 K
.1130P=36.57
P = 323.6 atm
-OR-
825-85.4 = 739.6 mL = .7396L
.7369P=1.47(.0821)(303)
.7369P=36.57
P=49.6atm
.1130P=36.57
P = 323.6 atm
-OR-
825-85.4 = 739.6 mL = .7396L
.7369P=1.47(.0821)(303)
.7369P=36.57
P=49.6atm
The first one was correct! I just used the incorrect number of sig figs. Thank you so much!
The one that makes the most sense to me is to use the 825-712. If your prof is touchy about the number of significant figures, you want to go over this very carefully because that answer of 323.6 has too many
Thanks for sharing that information. If I had been just a little faster on my end you would have caught the too many s.f. I was typing while you were checking.