Question

The Henry’s law constant for He gas in H2O at 30℃ is 3.7 × 10−4 M/atm; that for N2 at 30℃ is 6.0 × 10−4 M/atm. If a gaseous He-N2 mixture that has a He mole fraction of 0.30 is placed over the water

the concentration of dissolved He will be greater than that of dissolved N2.
the concentration ratio of dissolved He to dissolved N2 would be 0.62.
the concentration ratio of dissolved He to dissolved N2 will be 0.26.
the concentration ratio of dissolved He to dissolved N2 will be 0.18.
the concentration ratio of dissolved He to dissolved N2 will be 0.49.

Answer 1

the concentration ratio of dissolved He to dissolved N2 will be 0.49.

Step-by-step explanation:

Henry’s Law:

• Pi = Ci×K

∴ Pi: partial pressure gas

∴ Ci: Concentration

∴ K: Henry’s law constant

⇒ He: KHe = 3.7 E-4 M/atm

⇒ N2: KN2 = 6.0 E-4 M/atm

mix He-N2:

∴ XHe = 0.30

⇒ XN2 = 1 - XHe = 0.70

Raoult:

• Pi = Xi×Pi*

∴ Pi*: vapor pressure to T=30°C

Antoine:

⇒ P*He (30°C) ≅ 1.64 atm

⇒ P*N2 (30°C) ≅ 2.72 atm

⇒ PHe = (0.30)(1.64 atm) = 0.492 atm

⇒ PN2 = (0.70)(2.72 atm) = 1.904 atm

Henry’s Law:

He:

∴ PHe = (CHe)(3.7 E-4 M/atm) = 0.492 atm

⇒ CHe = 1329.73 M

N2:

∴ PN2 = (CN2)(6.0 E-4 M/atm) = 1.904 atm

⇒ CN2 = 3173.333 M

concentration ratio:

CHe/CN2 = 1329.73 M / 3173.333 M = 0.42 ≅ 0.49

⇒ 0.42 atm is close to 0.49 atm, the difference could be in the value of the vapor pressure at the given temperature.