Question

Fermentation of 826 mL grape juice (density is 1.0 ) is allowed to take place in a bottle with a total volume of 885 mL until 19% by volume is ethanol (). Assuming that obeys Henry’s law, calculate the partial pressure of in the gas phase and the solubility of in the wine at 25°C. The Henry’s law constant for is mol/L ⋅ atm at 25°C with Henry’s law in the form , where is the concentration of the gas in mol/L. (The density of ethanol is 0.79 .)

Answer 1

Here is the full and correct question

In the "Methode Champenoise", grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is:

C₆H₁₂O₆(aq)  ----------->    2C₂H₅OH(aq)    +     2CO₂(g)

Fermentation of 826 mL grape juice (density is 1.0 g/cm³) is allowed to take place in a bottle with a total volume of 885 mL until 19% by volume is ethanol(C₂H₅OH).

Assuming that:

CO₂ obeys Henry's Law, calculate the partial pressure of CO₂ in the gas phase and the Solubility of CO₂ in the wine at 25°C

The Henry Law Constant for CO₂ is 3.1 × 10⁻² mol/L. atm at 25°C with Henry's Law in the form C = KP;

where:

C = concentration of the gas in mol/L.

(The solubility of ethanol is 0.79 g/cm³)

Answer:

Partial Pressure of CO₂ in the gas phase is 104.84 atm

Solubility of CO₂ in the wine at 25°C is  3.25004 M

Explanation:

The equation for the reaction is given as:

C₆H₁₂O₆(aq)  ----------->    2C₂H₅OH(aq)    +     2CO₂(g)

Given that:

The total volume = 885 mL

Volume of ethanol is by 19% of the total volume = 0.19 × 885

Density = 0.79

density = $\frac{mass}{volume}$

the mass of ethanol can be calculated as = 0.79 × 0.19 × 885 = 132.84 g

number of moles of ethanol = $\frac{massof ethanol}{molarmass}$

molar mass of ethanol = 46.07

∴

number of moles of ethanol = $\frac{132.84}{46.07}$

= 2.8834 mole

Henry's Law posits that the solubility of a gas is directly proportional to the partial pressure of the gas above the solution.

NOW, Assuming that CO₂ obeys Henry's Law;

then numbers of moles of ethanol = numbers of mole of CO₂

So, molar mass of CO₂ = 44.01

then mass of CO₂ = number of moles of CO₂ ×  molar mass

mass of CO₂ =  2.8834 mole × 44.01

mass of CO₂ = 126.90 g

Since total volume = 885 mL  = 0.885 m

Concentration of CO₂ = $\frac{numbers of moles of CO_2}{Total volume}$

Concentration of CO₂ =  $\frac{2.8834}{0.885}$

Concentration of CO₂ = 3.258 M

C = $K*P$

$3.258 = 3.1*10^{-2}*P$

P = $\frac{3.25}{3.1*10^{-2}}$

P = 104.84 atm

∴ the partial pressure of in the gas phase = 104.84 atm

b)

Solubility  of ethanol in the wine = $K_H*P$

Solubility  of ethanol in the wine = $3.1*10^{-2}*104.84$

Solubility  of ethanol in the wine = 3.25004 M