Question

Henry's law states that the solubility of a gas is directly proportional to its partial pressure, LaTeX: S_g=k\cdot P_gS g = k ⋅ P g, where Sg is the solubility of the gas, Pg is the partial pressure of the gas, and k is the Henry's law constant for a specific gas at a specified temperature. What is the millimolar solubility of oxygen when the atmospheric pressure is 705 mmHg? Air is composed of 21% oxygen by moles. The Henry's law constant for oxygen is 1.3x10-3 mol/L-atm.
Enter your answer numerically in units of mM to three significant figures.

Answer 1

Answer:

253 mmol

Explanation:

The solubility is calculated by the given equation:

S(g) = k P(g)

where k =  1.3 x 10⁻³ mol/L-atm

and P(g) = 0.21 (705 mmHg/760 mmHg/atm) = 0.195 atm

Notice  the pressure is  converted to atm because  Henry´s constant is in units of atmospheres , and also we multiplied by 0.21  oxygen composition.

S(g) = 1.3 x 10⁻³ mol/L-atm x 1000 mmol/mol x 0.195 atm x  =253 mmol

Here we did multiply by the factor 1000 mmol/mol since the mmol concentration is required in the answer.