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}](https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D)
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}](https://tex.z-dn.net/?f=%5Cfrac%7Bmassof%20ethanol%7D%7Bmolarmass%7D)
molar mass of ethanol = 46.07
∴
number of moles of ethanol = ![\frac{132.84}{46.07}](https://tex.z-dn.net/?f=%5Cfrac%7B132.84%7D%7B46.07%7D)
= 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}](https://tex.z-dn.net/?f=%5Cfrac%7Bnumbers%20of%20moles%20of%20CO_2%7D%7BTotal%20volume%7D)
Concentration of CO₂ = ![\frac{2.8834}{0.885}](https://tex.z-dn.net/?f=%5Cfrac%7B2.8834%7D%7B0.885%7D)
Concentration of CO₂ = 3.258 M
C = ![K*P](https://tex.z-dn.net/?f=K%2AP)
![3.258 = 3.1*10^{-2}*P](https://tex.z-dn.net/?f=3.258%20%3D%203.1%2A10%5E%7B-2%7D%2AP)
P = ![\frac{3.25}{3.1*10^{-2}}](https://tex.z-dn.net/?f=%5Cfrac%7B3.25%7D%7B3.1%2A10%5E%7B-2%7D%7D)
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](https://tex.z-dn.net/?f=K_H%2AP)
Solubility of ethanol in the wine = ![3.1*10^{-2}*104.84](https://tex.z-dn.net/?f=3.1%2A10%5E%7B-2%7D%2A104.84)
Solubility of ethanol in the wine = 3.25004 M