[Tony]

As far as molality goes  I know that it's (moles solute)/(kg solvent), but how do you get the amount of each if all that is given is the is the molality of the product? Do you just assume the weight of solvent to be 1 kg and suppose that moles of solute is equivalent? This assumption just doesn't seem viable. Basically I know that the molality is .207 mol kg^-1 and I need to find the amount of moles of 1-hexanol and H_2O there is. I know that this may be very elementary or phrased poorly, but I'm fairly new to P-CHEM. Any help is welcome. If clarification is needed please ask!

[mjc123]

If that's all the information you have, you know that you have 0.207 moles of hexanol per kg of water. If you want the total number of moles of each, you can't tell unless you know how much solution you have. Can you quote the exact question?

[Hunter2]

Why can't?
If you have 0,207 mol/kg  means the ratio is 0,207 mol hexanol and 55,5 mol H₂O. Of cause you need the present mass to calculate.

[mjc123]

Of cause you need the present mass to calculate.

Yes, that's exactly my point.

[Tony]

So I'm pulling this out of this paper
http://www.bf.uni-lj.si/fileadmin/groups/2747/Zaposleni/Natasa_Segatin/2004_2._Monatsh._Chem..pdf
and I'm just trying to convert the data so I can insert it into a program. I couldn't find anything regarding the amount of solute. I think I figured out another way I can get the data though.

[orthoformate]

My understanding of Molality is that you have one kg of material, and within that one kg is your 0.207 mol of 1-hexanol.

0.207mol*102.18g/mol=21.15g 1-hexanol

1000g-21.15g=978.85g of water

978.85g*1mol/18g=54.38 mol water

[mjc123]

No, molality is moles solute per kg solvent.
Anyway, looking at the paper, 0.207 mol/kg is the molality of water in hexanol, not the other way round. but that is not the solubility of water in hexanol. It's the solubility at a water activity of 0.1105. Is that really what you wanted?
What data do you need to insert into your program?