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In wine making, aqueous glucose C6H12O6from 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 135ml of ethanol. Ethanol has a density of 0.789g/ml. In 1.5 lb of grapes, there is 26g of glucose.

a. Calculate the volume percent (v/v) of ethanol in the port wine.

b. What is the molarity (M) of ethanol in the port wine?

c. Write a balanced chemical equation for the fermentation reaction of glucose.

d. How many grams of glucose are required to produce one bottle of port wine?

e. How many bottles of port wine can be produced from 1 ton of grapes. (1 ton = 2000lb)

Short Answer

Expert verified

a. volume per cent of ethanol in the port wine is18%.

b.localid="1654284058658" 3.09M

c.C6H12O62C2H6O(l)+2CO2(g)

d.localid="1654284063701" 208.8g

e.16bottles

Step by step solution

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01

Given Information

Ethanol is an essential liquor that is ethane in which one of the hydrogens is subbed by a hydroxy gathering. It plays a part as a germ-free medication, a polar dissolvable, a neurotoxin, a focal sensory system depressant, and a teratogenic specialist.

02

Explanation Part (a)

Given,

The volume of ethanol = 135mland volume of wine =750ml

Volume per cent =135×100750=18%

03

Explanation Part (b)

Given,

The volume of ethanol= 135mland The volume of wine =750ml

The density of ethanol = 0.789g/mL

Mass of ethanol = Density ×Volume

localid="1654284099213" 135mL×0.789g/mL=106.52g

The molar mass of ethanol C2H6O=localid="1654284106380" (2×12)+(6×1)+(1×16)=46g/mole

Number of moles of ethanol =localid="1654284125120" massmolarmass=106.5246=2.32moles

Calculating the molarity =localid="1654284116533" no.ofmolesvolume=2.32×1000750=3.09M

04

Explanation Part (c)

The chemical reaction is -

C6H12O62C2H6O(l)+2CO2(g)

05

Explanation Part (d)

we know, C6H12O62C2H6O(l)+2CO2(g)

From the reaction,

One mole of ethanol = 12mole of glucose

2.32moles of ethanol = 2.32×12mole of glucose = 1.16mole of glucose

The molar mass of glucoseC6H12O6= localid="1654284149243" =(6×12)+(6×1)+(6×16)=180g/mole

Hence the mass of glucose = localid="1654284163484" role="math" 1.16×180=208.8g

06

Explanation Part (e)

Given,

26gof glucose from = 1.5lbof grapes

200lbof grapes = 261.5×200=3466gof glucose

From the reaction, one mole of glucose = two moles of ethanol

Amount of ethanol from 3466gof glucose =923466=1771.5g

The volume of ethanol = massdensity=1771.50.789=2245.26ml

Volume in one bottle = 135ml

Hence the number of bottles =2245.6135=16.616

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