Question

17% 02/27/18 1 Question ( point) a See page 154 Most rum sold in liquor stores is about 30.0-40.0% ethanol, C2H5OH, by volume. However, overproof rum, which is used to prepare flaming drinks might contain anywhere between 60.0 95.0% alcohol by volume. A t 200°C the densities of ethanol and water are 0789 g/mL and 0.998 g/mL, respectively. 1st attempt 63 Part 1 (0.3 point) Feedback hl See Periodic Table See Hint Calculate the number of moles of ethanol in a 100.0 mL sample of an overproof rum that is a 1.0% ethanol by volume. Assume that the remainder is water mol ethanol Part 2 (0.3 point) O Feedback Calculate the number of moles of water in a 100 mL sample of an overproof rum that is 810% ethanol by volume. Assume that the remainder is water mol water 0106 > VIEW SOLUTION

Answer 1

Part 1

81% v/v ethanol meaning that 100 ml soltuion contain 81 ml ethanol

density of ethanol = 0.789 gm/ml

mass = volume X density

mass of ethanol = 81 X 0.789 = 63.909 gm

molar mass of ethanol = 46.07 g/mol then 63.909 gm of ethanol = 63.909/46.07 = 1.387 mole

100 ml overproof rum contain 1.387 mole of ethanol

part 2

81% v/v ethanol meaning that 100 ml soltuion contain 81 ml ethanol and 19 ml water

density of water = 0.998 gm/ml

mass = volume X density

mass of water = 19 X 0.998 = 18.962 gm

molar mass of water = 18.01528 g/mol then 18.962 gm of water = 18.962/18.01528 = 1.053 mole

100 ml overproof rum contain 1.053 mole of water