To calculate the grams of oxygen that will dissolve in water, we can use Henry's Law and the given information. Given: Henry's Law constant for oxygen in water (kH) = 1.28 x 10^-3 M/atm Partial pressure of oxygen (P) = 1.32 atm Volume of water (V) = 1.75 L Step 1: Convert the volume of water from liters to moles using the ideal gas law equation, PV = nRT: n = PV / RT Since we are dealing with dissolved gases, we can approximate the gas constant (R) as 0.0821 L.atm/mol.K: n = (1.32 atm) * (1.75 L) / (0.0821 L.atm/mol.K * 298 K) Step 2: Calculate the concentration of dissolved oxygen using Henry's Law: Concentration of oxygen (C) = kH * P Substituting the given values: C = (1.28 x 10^-3 M/atm) * (1.32 atm) Step 3: Calculate the number of moles of oxygen that will dissolve in water: n = C * V Substituting the values: n = (1.28 x 10^-3 M/atm) * (1.32 atm) * (1.75 L) Step 4: Convert moles to grams using the molar mass of oxygen (O2), which is approximately 32 g/mol: Mass of oxygen = n * molar mass of O2 Substituting the values: Mass of oxygen = (1.28 x 10^-3 M/atm) * (1.32 atm) * (1.75 L) * (32 g/mol) Step 5: Calculate the mass of oxygen: Mass of oxygen = 0.074496 g Therefore, approximately 0.074496 grams of oxygen will dissolve in 1.75 L of water when the partial pressure of oxygen is 1.32 atm, based on Henry's Law.
