m = mols solute/kg solvent
You will need the densities to convert grams to volume and volume to grams.
I converted ethanol using it's density, so I got mass=(.7893 kg/L)(3.41 L)=2.691513 g ethanol
then plugged it into m=#mols solute/kg mass solvent to solve for # mols of solute which I got to be #mols=(11 m)(2.691513 g ethanol)=29.6066
Now do I take this and multiply it with the molar mass of methanol, and then use the density relation to solve for volume?
I wouldn't go that way. What you have done so far is very good. I would then determine how much methanol was needed.
molality = moles CH3OH/0.269 kg solvent
moles CH3OH = 11 x 0.269 = ??
Then moles CH3OH x molar mass CH3OH = grams and convert that with density to volume CH3OH.
Hmm okay, I followed that and is the answer that you got 119.795 L?
Hang on a second. I've messed up in a couple of places. Let me get it straight an I will post back here.
I have been using the wrong molar masses (I thought I knew them by memory but I didn't). Also, I used the wrong volume of ethanol. So here goes the correct way.
3.41 L ethanol x 1000 = 3410 mL ethanol.
m = v x d
m = 3410 x 0.789 = 2960 grams or 2.96 kg ethanol.
How much solute do we need?
m = moles solute/kg solvent.
11 = moles solute/2.69
11 x 2.69 = 29.59 mols solute = 29.59 moles methanol.
20.59 moles methanol x molar mass (32) = 946.9 grams methanol.
m = v x d
946.9 = v x 0.792 g/mL
v = 1196 mL or 1.196 L.
You may get slightly different values if you use slightly different molar masses and/or slightly different densities.
Now let's check it to make sure we are right.
1.196 L methanol/3.41 L ethanol.
1.196 L methanol = 1196 mL
1196 mL x 0.792 g/mL = 947 grams methanol.
947/32 = 29.6 moles methanol.
3.41 L ethanol solvent = 3410 grams.
m = v*d
m = 3410 x 0.789 g/mL = 2690 g ethanol or 2.69 kg ethanol.
Then 29.6 mols solute/2.69 kg = 11.00 m.
I'm glad you wrote that Hmmm--it made me go back and check the problem and my numbers. I hope I haven't confused you.