Consider a carbonated drink in a bottle at \(37^{\circ} \mathrm{C}\) and \(130 \mathrm{kPa}\). Assuming the gas space above the liquid consists of a saturated mixture of \(\mathrm{CO}_{2}\) and water vapor and treating the drink as water, determine \((a)\) the mole fraction of the water vapor in the \(\mathrm{CO}_{2}\) gas and \((b)\) the mass of dissolved \(\mathrm{CO}_{2}\) in a 200-ml drink.

Assuming the gas space above the liquid consists of a saturated mixture of CO₂ and water vapor, we can find the partial pressures of each component using Dalton's law of partial pressures, which states that the total pressure is equal to the sum of the partial pressures of each component: \(P_{total} = P_{CO_2} + P_{H_2O}\) The total pressure in the bottle is given as 130 kPa. We can find the partial pressure of water vapor, \(P_{H_2O}\), by using the saturation pressure, \(p_g\), at the given temperature: \(p_g (37^{\circ}C) = 6.54 \mathrm{kPa}\) (from steam tables) This means that the partial pressure of the CO₂ gas can be determined as follows: \(P_{CO_2} = P_{total} - P_{H_2O} = 130\,\mathrm{kPa} - 6.54\, \mathrm{kPa} = 123.46\,\mathrm{kPa}\)

Now that we have the partial pressures of both CO₂ and water vapor, we can calculate the mole fraction of water vapor, which is given by: \(x_{H_2O} = \frac{P_{H_2O}}{P_{total}}\) Substituting the values: \(x_{H_2O} = \frac{6.54 \, \mathrm{kPa}}{130\, \mathrm{kPa}} = 0.0503\) So, the mole fraction of water vapor in the CO₂ gas is 0.0503.

To determine the mass of dissolved CO₂, we will use Henry's law, which states that the concentration of a gas in a liquid is proportional to the partial pressure of the gas above the liquid: \(C = k_H P_{CO_2}\) Here, \(C\) is the concentration of dissolved CO₂ in the drink, \(k_H\) is Henry's law constant, and \(P_{CO_2}\) is the partial pressure of CO₂ determined in Step 1. Given the temperature of 37°C and considering the drink as water, we find the Henry's law constant for CO₂ from reference tables: \(k_H (37^{\circ}C) = 2.1 \times 10^{-2} \, \mathrm{mol/kg \cdot kPa}\) Now we can calculate the concentration of dissolved CO₂: \(C = (2.1 \times 10^{-2}\,\mathrm{mol/kg \cdot kPa})(123.46\,\mathrm{kPa}) = 2.5897\, \mathrm{mol/kg}\) To find the mass of dissolved CO₂ in a 200-ml drink, we need to multiply the concentration by the mass of the drink (assuming a density of 1 kg/L for water at room temperature): \(\mathrm{Mass\,of\,CO_2} = (2.5897\, \mathrm{mol/kg}) * (200\, \mathrm{ml}) * \frac{1\, \mathrm{kg}}{1000\, \mathrm{ml}} * \frac{44.01\, \mathrm{g}}{1\, \mathrm{mol}}\) \(\mathrm{Mass\,of\,CO_2} = 2.30\, \mathrm{g}\) So, the mass of dissolved CO₂ in a 200-ml drink is 2.30 g.