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

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:

${\text{C"_6"H"_12"O"_6 → "2C"_2"H"_5"OH" + "2CO}}_{2}$

$\text{Mass of glucose" = 1000 color(red)(cancel(color(black)("g soln"))) × "20.0 g glucose"/(100 color(red)(cancel(color(black)("g soln")))) = "200 g glucose}$

$\text{Moles of glucose" = 200 color(red)(cancel(color(black)("g glucose"))) × (1 color(red)(cancel(color(black)("mol glucose"))))/(180.16 color(red)(cancel(color(black)("g glucose")))) = "1.110 mol glucose}$

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

4(b) Mass percent of alcohol

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

$\text{Total mass" = "(102 + 800) g" = "902 g}$

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

4(c) Volume of alcohol

$\text{Volume of alcohol" = 102 color(red)(cancel(color(black)("g alcohol"))) × "1 mL alcohol"/(0.789 color(red)(cancel(color(black)("g alcohol")))) = "130 mL alcohol}$

4(d) Concentration of alcohol

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

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

$\text{Proof" = 14.1 color(red)(cancel(color(black)("% ABV"))) × "2 proof"/(1 color(red)(cancel(color(black)("% ABV")))) = "28.2 proof}$