Calculate the solubility of O2 (in mol/L) in water at 25 degrees C at a partial

Question

Calculate the solubility of O2 (in mol/L) in water at 25 degrees C at a partial pressure of O2 equal to 120 torr. The Henry's Law constant for O2 is 7.8x10^2 atm L/mol at 25 degrees C. A)1.1x10^-2 B)6.5 C)0.15 D)1.1x10^-5 E)none of these are correct Calculate the solubility of O2 (in mol/L) in water at 25 degrees C at a partial pressure of O2 equal to 120 torr. The Henry's Law constant for O2 is 7.8x10^2 atm L/mol at 25 degrees C. A)1.1x10^-2 B)6.5 C)0.15 D)1.1x10^-5 E)none of these are correct A)1.1x10^-2 B)6.5 C)0.15 D)1.1x10^-5 E)none of these are correct

Explanation / Answer

7.8x10^2 atm L/mol

H = P°/M

P° = 120 torr = 120/760 = 0.1578 atm

M = P°/H = (0.1578 )/(7.8*10^2) = 0.0002023076 mol / L

M = 2.03*10^-4 M

choose E, none of these

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.