Can anyone help out in solving these 4 questions under Question 4?

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1 Answer
Nov 24, 2016

(a) 102 g ethanol; (b) 11.3 % m/m; (c) 130 mL ethanol; (d) 28.2 proof.

Explanation:

4(a) Mass of ethanol

The overall chemical equation for alcoholic fermentation is:

"C"_6"H"_12"O"_6 → "2C"_2"H"_5"OH" + "2CO"_2

"Mass of glucose" = 1000 color(red)(cancel(color(black)("g soln"))) × "20.0 g glucose"/(100 color(red)(cancel(color(black)("g soln")))) = "200 g glucose"

"Moles of glucose" = 200 color(red)(cancel(color(black)("g glucose"))) × (1 color(red)(cancel(color(black)("mol glucose"))))/(180.16 color(red)(cancel(color(black)("g glucose")))) = "1.110 mol glucose"

"Mass of ethanol" = 1.110 color(red)(cancel(color(black)("mol glucose"))) × (2 color(red)(cancel(color(black)("mol ethanol"))))/(1 color(red)(cancel(color(black)("mol glucose")))) × "46.07 g ethanol"/(1 color(red)(cancel(color(black)("mol ethanol")))) = "102 g ethanol"

4(b) Mass percent of alcohol

The solution contains 102 g of ethanol and 800 g of water.

"Total mass" = "(102 + 800) g" = "902 g"

"Percent by mass" = "mass of alcohol"/"total mass" × 100 % = (102 color(red)(cancel(color(black)("g"))))/(902 color(red)(cancel(color(black)("g")))) × 100 % = "11.3 % m/m"

4(c) Volume of alcohol

"Volume of alcohol" = 102 color(red)(cancel(color(black)("g alcohol"))) × "1 mL alcohol"/(0.789 color(red)(cancel(color(black)("g alcohol")))) = "130 mL alcohol"

4(d) Concentration of alcohol

"Volume of solution" = 902 color(red)(cancel(color(black)("g soln"))) × "1 mL soln"/(0.982 color(red)(cancel(color(black)("g soln")))) = "919 mL"

"% ABV" = "volume of alcohol"/"total volume" × 100 % = (130 color(red)(cancel(color(black)("mL"))))/(919 color(red)(cancel(color(black)("mL")))) × 100 % = "14.1 % ABV"

"Proof" = 14.1 color(red)(cancel(color(black)("% ABV"))) × "2 proof"/(1 color(red)(cancel(color(black)("% ABV")))) = "28.2 proof"