The atmospheric pressure in a lab is calculated as 1.3 atm. If oxygen gas contributes 62% of this atmospheric pressure, determine its mass (in g) dissolved at room temperature in 25 L of water. The Henry's Law Constant for oxygen in water at this temperature is 5.3 × 10^–5 M/atm.

• A. 1.4 × 10^–3 g
• B. 6.9 × 10^–5 g
• C. 0.055 g
• D. 0.034 g

Answer: (D)

Verified step by step guidance

1. Calculate the partial pressure of oxygen gas using the given percentage: \( P_{O_2} = 0.62 \times 1.3 \text{ atm} \).
2. Apply Henry's Law to find the concentration of dissolved oxygen: \( C = k_H \times P_{O_2} \), where \( k_H = 5.3 \times 10^{-5} \text{ M/atm} \).
3. Substitute the values into Henry's Law equation to find the molarity of dissolved oxygen: \( C = 5.3 \times 10^{-5} \text{ M/atm} \times P_{O_2} \).
4. Calculate the number of moles of oxygen in 25 L of water using the molarity: \( ext{moles of } O_2 = C \times 25 \text{ L} \).
5. Convert the moles of oxygen to grams using the molar mass of oxygen (32.00 g/mol): \( ext{mass of } O_2 = ext{moles of } O_2 \times 32.00 \text{ g/mol} \).