A sample of oxygen is collected over water and has a volume of 2.23 mL. The pressure of the collected 'wet' gas is 61.78 kPa and the temperature is 20°C. What will the volume of the 'dry' oxygen be in mL if the pressure is changed to 94.91 kPa, assuming the temperature remains constant?

• A. 2.23 mL
• B. 3.56 mL
• C. 4.12 mL
• D. 1.45 mL

Answer: (D)

Verified step by step guidance

1. Understand that the problem involves a gas collected over water, which means the gas is 'wet' and contains water vapor. To find the volume of 'dry' oxygen, we need to account for the water vapor pressure.
2. Use Dalton's Law of Partial Pressures to find the pressure of the 'dry' oxygen. Dalton's Law states that the total pressure of a gas mixture is the sum of the partial pressures of each component. Therefore, subtract the water vapor pressure at 20°C from the total pressure of the 'wet' gas to find the pressure of the 'dry' oxygen.
3. Look up the water vapor pressure at 20°C, which is approximately 2.34 kPa. Subtract this value from the total pressure of the 'wet' gas (61.78 kPa) to find the pressure of the 'dry' oxygen.
4. Apply Boyle's Law, which states that for a given amount of gas at constant temperature, the pressure and volume are inversely proportional: \( P_1V_1 = P_2V_2 \). Use the initial pressure and volume of the 'dry' oxygen and the final pressure to solve for the final volume.
5. Substitute the known values into Boyle's Law equation: \( V_2 = \frac{P_1V_1}{P_2} \), where \( P_1 \) is the pressure of the 'dry' oxygen, \( V_1 \) is the initial volume, and \( P_2 \) is the final pressure. Calculate \( V_2 \) to find the volume of the 'dry' oxygen at the new pressure.