Most rum sold in liquor stores is about 30-40% ethanol, C_2H_5OH, by volume. How
ID: 505363 • Letter: M
Question
Most rum sold in liquor stores is about 30-40% ethanol, C_2H_5OH, by volume. However, overproof rum, which is used to prepare flaming drinks, might contain anywhere between 60-95% alcohol by volume. Calculate the number of moles of ethanol and moles of water in a 100.0 mL sample of an overproof rum that is 82% ethanol by volume. Assume that the remainder is water. Which is the solvent, and which is the solute? At 20.0 degree C, the densities of ethanol and water are 0.789 g/mL and 0.998 g/mL, respectively. ethanol waterExplanation / Answer
Precentage of alcohol by volume in the proof run we have divide it by 2 the percentage of ethanol by volume
82 / 2 = 41 %
volume of pure ethanol = 100 mL * 41 %
= 100 * 41 / 100 = 41 mL
the density of ethanol is 0.789 g/mL and molar mass of ethanol (CH3CH2OH) is 46.068 g/mL
we know
Density of ethanol = mass of ethanol / volume of ethanol
Mass of ethanol = Density of ethanol * volume of ethanol
Number of moles = mass of ethanol / molar mass of ethanol
Number of moles = Density of ethanol * volume of ethanol / molar mass of ethanol
Number of moles = 0.789 g/mL * 41 mL / 46.068 g/mL
Number of moles = 32.35 g / 46.068 g/mol
Number of moles = 0.70 moles
OR
Precentage of alcohol by volume 82 %
volume of pure ethanol = 100 mL * 82 %
= 100 * 82 / 100 = 82 mL
the density of ethanol is 0.789 g/mL and molar mass of ethanol (CH3CH2OH) is 46.068 g/mL
we know
Density of ethanol = mass of ethanol / volume of ethanol
Mass of ethanol = Density of ethanol * volume of ethanol
Number of moles = mass of ethanol / molar mass of ethanol
Number of moles = Density of ethanol * volume of ethanol / molar mass of ethanol
Number of moles = 0.789 g/mL * 82 mL / 46.068 g/mL
Number of moles = 64.70 g / 46.068 g/mol
Number of moles = 1.40 moles
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