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
To find the volume of ethanol in the original unknown, we can use the formula:
Volume = Mass / Density
First, we need to find the mass of the ethanol in the original unknown. From the information given, we know that the mass of ethanol removed in the single-step extraction is 8.243 g and that 72.0% of the ethanol was removed. Therefore, we can calculate the mass of ethanol in the original unknown as follows:
Mass of ethanol in original unknown = (Mass of ethanol removed) / (Percentage of ethanol removed)
Mass of ethanol in original unknown = 8.243 g / 72.0% = 11.45 g
Now, we can calculate the volume of ethanol using the given density of ethanol, which is 0.789 g/mL:
Volume of ethanol = Mass of ethanol / Density
Volume of ethanol = 11.45 g / 0.789 g/mL
Calculating this gives us the volume of ethanol in the original unknown.