posted by Forrest .
Mass of unknown solution=25.671
Mass ethanol =8.243
Assuming that prior experiments have shown the single-step extraction removes only 72.0 % of the ethanol, the mass of ethanol in your original unknown is 11.45
Mass percent ethanol =44.6 %
Given a density of 0.789 g/mL, what would have been the volume of the ethanol in the original unknown?
I suppose all of those numbers are in grams.
I agree with 11.45 g EtOH in the unknown.
I agree with 44.6% EtOH in the original soln.
So mass = volume x density
You have mass EtOH and density EtOH, solve for volume EtOH.
Round to 3 s.f. I get approximately 14 mL