Q.20 - An Introduction to General, Organic, and Biological Chemistry

Question

In wine making, aqueous glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ 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 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)

Step-by-Step Solution

Answer

a. volume per cent of ethanol in the port wine is $18\%$.

b. $3.09 \mathrm{M}$

c. ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\to 2{\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

d. $208.8 g$

e. $16$ bottles

Step 1: 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.

Step 2: Explanation Part (a)

Given,

The volume of ethanol = $135 ml$ and volume of wine =$750 ml$

Volume per cent  = $\frac{135\times 100}{750}=18\%$

Step 3: Explanation Part (b)

Given,

The volume of ethanol= $135 ml$and The volume of wine =$750 ml$

The density of ethanol = $0.789 \mathrm{g}/\mathrm{mL}$

Mass of ethanol = Density $\times$ Volume

$135 \mathrm{mL}\times 0.789 \mathrm{g}/\mathrm{mL}=106.52\text{ g }$

The molar mass of ethanol ${\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}$ =$(2\times 12)+(6\times 1)+(1\times 16)=46 \mathrm{g}/\text{ mole }$

Number of moles of ethanol =$\frac{mass}{molar mass}=\frac{106.52}{46}=2.32\text{ moles}$

Calculating the molarity =$\frac{no. of moles}{volume}=\frac{2.32\times 1000}{750}=3.09 M$

Step 4: Explanation Part (c)

The chemical reaction is -

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\to 2{\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

Step 5: Explanation Part (d)

we know, ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\to 2{\mathrm{C}}_2{\mathrm{H}}_6\mathrm{O}\left(\mathrm{l}\right)+2{\mathrm{CO}}_2\left(\mathrm{g}\right)$

From the reaction,

One mole of ethanol = $\frac{1}{2}$ mole of glucose

$2.32$ moles of ethanol = $2.32\times \frac{1}{2}$ mole of glucose = $1.16$ mole of glucose

The molar mass of glucose$\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$= $=(6\times 12)+(6\times 1)+(6\times 16)=180 \mathrm{g}/\mathrm{mole}$

Hence the mass of glucose = $1.16\times 180=208.8 \text{g }$

Step 6: Explanation Part (e)

Given,

$26 g$ of glucose from = $1.5 lb$ of grapes

$200 lb$of grapes = $\frac{26}{1.5}\times 200=3466 g$ of glucose

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

Amount of ethanol from $3466 g$ of glucose =$\frac{92}{3466}=1771.5 g$

The volume of ethanol = $\frac{mass}{density}=\frac{1771.5}{0.789}=2245.26 ml$

Volume in one bottle = $135 ml$

Hence the number of bottles = $\frac{2245.6}{135}=16.6\approx 16$