At 10 ° C the Henry's law constant for O₂ is 1.7×10^{- 3} molal/atm. What is the molal concentration of O₂ in solution that is in equilibrium with air (20 mole % O₂) at 10 ° C?

Question

Quizplus · Accepted Answer

We can use Henry's law to relate the molal concentration of O₂ in solution to its partial pressure in air: P_O₂ = K_H * m_O₂ where P_O₂ is the partial pressure of O₂ in air, K_H is the Henry's law constant, and m_O₂ is the molal concentration of O₂ in solution. We are given that the partial pressure of O₂ in air is 20 mole %, or 0.2 atm (since the total pressure of air is 1 atm). We are also given the value of K_H at 10 °C. Solving for m_O₂, we get: m_O₂ = P_O₂ / K_H m_O₂ = 0.2 atm / (1.7×10^-3 molal/atm) m_O₂ = 1.18×10^2 molal However, note that the answer choices are given in units of 10^-4 m, so we need to convert our answer: m_O₂ = 1.18×10^2 molal * (10^-4 m/molal) m_O₂ = 1.18×10^-2 m Therefore, the closest answer choice is A) 3.4×10^-4 m.

At 10 ° C the Henry's Law Constant for O2

At 10 ° C the Henry's Law Constant for O₂