Question

Henry's law constant for transferring \( O_2 \) from air into water, at room temperature, is 1.3 \( \frac{\text{mmol}}{\text{liter-atm}} \) . Given that the partial pressure of \( O_2 \) in the atmosphere is 0.21 atm, the concentration of dissolved oxygen (mg/liter) in water in equilibrium with the atmosphere at room temperature is \[ \text{(Consider the molecular weight of O}_2 \text{ as 32 g/mol)} \]

• A. 8.7
• B. 0.8
• C. 198.1
• D. 0.2

Hint:

To convert from mmol/liter to mg/liter, multiply by the molar mass of the substance in mg/mol.

Answer 1

The Correct Option is A

We can use Henry's law to calculate the concentration of dissolved oxygen: \[ C = k_H \cdot P \] where \( k_H \) is Henry's law constant, and \( P \) is the partial pressure of \( O_2 \). We are given \( k_H = 1.3 \, \frac{\text{mmol}}{\text{liter-atm}} \) and \( P = 0.21 \, \text{atm} \).
Now, to convert from mmol/liter to mg/liter, we use the molar mass of \( O_2 \), which is 32 g/mol (or 32,000 mg/mol): \[ C = 1.3 \cdot 0.21 \cdot 32 = 8.7 \, \text{mg/liter} \] Final Answer: (A) 8.7