Ah, so we're talking about the science of bubbly beverages! Let's pop this question open, shall we?
First, we can calculate the amount of ethanol produced by converting the volume percent to moles. Since the total volume is 825 mL, and we want 12% of it to be ethanol, we have:
825 mL * 0.12 = 99 mL ethanol
Next, we need to convert this volume of ethanol to grams. We know the density of ethanol is 0.79 g/cm3, so:
99 mL * 0.79 g/cm3 = 78.21 g ethanol
Now, let's determine the moles of ethanol using its molar mass. The molar mass of ethanol (C2H5OH) is:
2*(12.01 g/mol) + 6*(1.01 g/mol) + 1*(16.00 g/mol) = 46.07 g/mol
So, the moles of ethanol produced are:
78.21 g / 46.07 g/mol = 1.70 mol ethanol
According to the balanced equation, for every 1 mole of ethanol produced, 2 moles of CO2 are also produced. Therefore, we have:
2 moles CO2 / 1.70 mol ethanol = 1.18 mol CO2
Now, let's calculate the partial pressure of CO2 using Henry's law. Given that the Henry's law constant for CO2 is 32 L·atm/mol, we can use the equation P = kC, where C is the concentration of CO2 in mol/L.
Here, our concentration is:
1.18 mol CO2 / (0.746 L - 0.099 L) = 1.82 mol/L
Plugging it into Henry's law equation gives:
P = (32 L·atm/mol) * (1.82 mol/L) = 58.24 atm
So, the partial pressure of CO2 in the gas phase is approximately 58.24 atm.
Lastly, we need to find the solubility of CO2 in the wine at 25°C. Since Henry's law relates the concentration of the gas to its partial pressure, we can rearrange the equation to find the concentration:
C = P / k
C = 58.24 atm / (32 L·atm/mol) = 1.82 mol/L
So, the solubility of CO2 in the wine at 25°C is approximately 1.82 mol/L.
Cheers to the science behind sparkling wine!