# chemistry

An aqueous solution is prepared by diluting 3.30 mL acetone,(d= 0.789g/mL) with water to a final volume of 75.0 mL. The density of the solution is 0.993 g/mL. What is the molarity, molality and mole fraction of acetone in this solution?

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1. molality = moles/kg solvent. Ac2O is acetone.

First the moles Ac2O.
grams = 3.30 mL x 0.789 g/mL = 2.604 g Ac2O.
moles = g/molar mass = 2.604/58.08 = 0.045 g (approximate--you should do it more carefully).

Next for solvent (H2O)
grams soln = 75.0mL x 0.993 g/mL = about 74 grams
We had 2.6 g Ac2O in that ; therefore, g H2O = 74-2.6 = about 71 g or 0.071 kg
Then molality = moles/kg solvent.
I assume you can convert from molality to M and mole fraction.
M = moles/L soln
mole fraction = moles Ac2O/total moles.

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