Determine the solubility of n2 in water exposed to air at 25°c if the atmospheric pressure is 1.2 atm.

assume that the mole fraction of nitrogen is 0.78 in air and the henry's law constant for nitrogen in water at this temperature is 6.1 × 10-4 m/atm.

Solubility of N₂ in water exposed to air at 25°C is 6.09x10⁻⁴ mol/L

Explanation:

Henry's law → C = K . Partial pressure of gas

where C is solubility

and K is Henry's law constant for nitrogen in water

First of all, let's find out the partial pressure for N₂.

We have the mole fraction so

Mole fraction N₂ = Partial pressure N₂ / Total pressure

0.78 = Partial pressure N₂ / 1.28 atm

0.78 . 1.28 atm = Partial pressure N₂ → 0.9984 atm

C = 6.1x10⁻⁴ M /atm . 0.9984 atm

C = 6.09x10⁻⁴ mol/L

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2