Asked by Anonymous
What volume ( L ) of the solute methanol should be mixed with 3.41 L of the solvent ethanol to make a solution that is 11.0 m methanol?
Am I supposed to use the molality eq'n?
Am I supposed to use the molality eq'n?
Answers
Answered by
DrBob222
yes.
m = mols solute/kg solvent
You will need the densities to convert grams to volume and volume to grams.
m = mols solute/kg solvent
You will need the densities to convert grams to volume and volume to grams.
Answered by
Anonymous
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?
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?
Answered by
DrBob222
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.
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.
Answered by
Anonymous
Hmm okay, I followed that and is the answer that you got 119.795 L?
Answered by
DrBob222
Hang on a second. I've messed up in a couple of places. Let me get it straight an I will post back here.
Answered by
DrBob222
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.