Question

Item 3 Constants | Periodic Table In wine making, aqueous glucose (Cs H1206) from grapes undergoes fermentation to produce liquid ethanol and carbon dioxide gas. A bottle of vintage port wine has a volume of 750 mL and contains 135 mL of ethanol (C2 Hs O). Ethanol has a density of 0.789 g/mL. In 1.5 lb of grapes, there is 27 g of glucose. When the sugar in grapes is fermented ethanol is produced 20 MacBo

Answer 1

A)

volume% ethanol = (volume ethanol / volume of wine) x 100

= (135 / 750) x 100% = 18%

B)

mass of ethanol = density of ethanol x volume of ethanol

mass of ethanol = 0.789 g/mL x 135 mL = 106.515 grams

Molar mass of of ethanol = 46 g /mol

L of wine = 750 mL x (1L / 1000 mL) = 0.750 L

Now since

Molarity = (mass of ethanol / molar mass of ethanol) / Volume of solution in litres

Molarity = (106.515 g / 46 g/mol) /0.750 L = 3.09 molar

D)

The balanced chemical equation would be,

C6H12O6 =====> 2C2H6O + 2CO2

mass of glucose required = (mass of ethanol / molar mass of ethanol) x (mole ratio of glucose over ethanol based on the given chemical equation) x molar mass of of glucose

mass of glucose required = 106. 515 / 46.07 x (1/2) x 180

= 208.08 grams

E.)

Mass of glucose in 3.5 ton of grapes = 3.5 ton x (2000 lbs / 1 ton) x (27 g glucose / 1.5 lbs)

mass of glucose in 3.5 ton of grapes = 1,26000 g

Now,

mass of ethanol = (mass of glucose/ molar mass of glucose) x (mole ratio of ethanol over glucose) x molar mass of ethanol

mass of ethanol = (126000 grams / 180) x (2 /1) x 46

= 64400 g

And,

volume of ethanol = mass/density = 64400 / 0.789 g /mL = 81622.37 mL

Now,

In 750 mL of wine, ethanol present = 135 mL

In 81622.37 mL of ethanol, wine = (750 / 135) x 81622.37

= 453457.61 mL

750 mL wine = 1 bottle

453457.61 mL wine = (1 / 750) x 453457.61 bottles

= 604 bottles approx