The solubility of O₂ in water is 0.590 g/L at an oxygen pressure of around 14.5 atm.What is the Henry's law constant for O₂ (in units of mol/L·atm)? A)4.07 $\times$ 10^-² B)1.27 $\times$ 10^-³ C)7.86 $\times$ 10² D)2.67 $\times$ 10^-¹ E)None of the above are within 5% of the correct answer.

Question

Quizplus · Accepted Answer

Henry's law constant (k) is given by the equation: k = C/P Where C is the solubility of the gas in the solvent and P is the partial pressure of the gas above the solvent. From the given data, C = 0.590 g/L P = 14.5 atm To convert the solubility from g/L to mol/L, we need to divide by the molar mass of O₂, which is 32 g/mol. So, C = 0.590 g/L ÷ 32 g/mol = 0.0184 mol/L Therefore, k = 0.0184 mol/L ÷ 14.5 atm = 1.27 × 10^−3 mol/L·atm This matches option B.

The Solubility of O2 in Water Is 0 ×\times× 10-2 B)1 ×\times×

The Solubility of O₂ in Water Is 0 $\times$ 10^-²
B)1 $\times$