In the "Methode Champenoise," grape juice is fermented in a wine bottle to produce sparkling wine. The reaction is $$\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q) \longrightarrow 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q)+2 \mathrm{CO}_{2}(g)$$ Fermentation of \(750 .\) mL grape juice (density $=1.0 \mathrm{g} / \mathrm{cm}^{3} )$ is allowed to take place in a bottle with a total volume of 825 \(\mathrm{mL}\) until 12\(\%\) by volume is ethanol $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\( . Assuming that the \)\mathrm{CO}_{2}$ is insoluble in \(\mathrm{H}_{2} \mathrm{O}\) (actually, a wrong assumption), what would be the pressure of \(\mathrm{CO}_{2}\) inside the wine bottle at \(25^{\circ} \mathrm{C} ?\) (The density of ethanol is $0.79 \mathrm{g} / \mathrm{cm}^{3} . )$

Given that the grape juice is converted into 12% by volume of ethanol after fermentation, we can calculate the volume of ethanol produced as follows: Volume of ethanol = (12 / 100) * Volume of grape juice = (12 / 100) * 750 mL = 90 mL

Now, let's find the amount (in moles) of glucose consumed and ethanol produced using their densities and molar masses: Density of grape juice = 1.0 g/cm³ Density of ethanol = 0.79 g/cm³ Molar mass of glucose (C₆H₁₂O₆) = 180.16 g/mol Molar mass of ethanol (C₂H₅OH) = 46.07 g/mol First, we calculate the mass of glucose consumed: Mass of grape juice = Volume of grape juice * Density of grape juice = 750 mL * 1.0 g/cm³ (since 1 mL = 1 cm³) = 750 g The mass of glucose consumed = mass of grape juice = 750 g. Now, we utilize the stoichiometry of the balanced equation to find the moles of glucose consumed: Moles of glucose consumed = mass of glucose consumed / Molar mass of glucose = 750 g / 180.16 g/mol = 4.164 moles From the balanced equation, we know that 2 moles of ethanol are produced for each mole of glucose consumed. So, moles of ethanol produced: Moles of ethanol produced = 2 * Moles of glucose consumed = 2 * 4.164 moles = 8.328 moles

From the balanced equation, we know that 2 moles of CO₂ are produced for each mole of glucose consumed. So, moles of CO₂ produced: Moles of CO₂ produced = 2 * Moles of glucose consumed = 2 * 4.164 moles = 8.328 moles

Now, we have the moles of CO₂, the temperature, and the volume of the bottle, so we can calculate the pressure using the Ideal Gas Law: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.08206 L atm/mol K), and T is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin: T(K) = 25 + 273.15 = 298.15 K The volume of the bottle (V) = 825 mL = 0.825 L Next, plug the values into the equation and solve for P: P = (nRT) / V P = (8.328 moles * 0.08206 L atm/mol K * 298.15 K) / 0.825 L P ≈ 24.8 atm Thus, the pressure of CO₂ inside the wine bottle would be approximately 24.8 atm.