Molality chemistry confusion help please?

Question

What are the mole fraction (X) and molality (m) of ethanol (C2H5OH) in an aqueous solution that is 35.0% ethanol by volume? The density of water is 1.00 g/mL and that of ethanol is 0.789 g/mL.

Mole fraction=____

Molality=______

I don't know how to go about finding mole fraction, but I found the molality to be 17.14, which ended up being wrong.

I found it by dividing 6 mols of C2H5OH over .35 kg of H2)....

Answer 1

mole fraction is moles ethanol per mole solution

ie.. moles ethanol / (moles ethanol + moles water)

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assume 100 ml for simplicity.  since etoh = 35.0%... you therefore have....

35.0 ml Etoh

65.0 ml H2O

molar mass etoh = 46.1 g/mole

molar mass H2O = 18.0 g/mole

moles Etoh = 35.0 ml x (0.789 g/ml) x (1 mole / 46.1g) = 0.599 moles

moles H2O = 65.0 ml x (1.00 g/ml) x (1 moles / 18.0g) = 3.61 moles

mole fraction etoh = 0.599 / (0.599+3.61) = 0.142

mole fraction water = 1-mole fraction etoh = 0.858

molality = moles solute / kg solvent = moles etoh / kg water

kg water = 65.0 ml h2O x (1 g / ml) x (1kg / 1000 g) = 0.065 kg

so molality = 0.599 moles / 0.065 kg = 9.22 moles / kg = 9.22 molal