Question

Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 ∘C, the N2 component will dissolve in water with a solubility of 4.88×10−4 M. What is the value of Henry's law constant for N2 under these conditions?

Answer 1

Answer:

The value of Henry's law constant for N2 = 6.26 * 10^-4  M/ atm

Explanation:

Step 1: Data given

Temperature = 25.0 °C

The N2 component will dissolve in water with a solubility of 4.88*10^−4 M

air is at standard pressure = 1.00 atm

Step 2: Calculate Henry's law constant for N2

C=k*Pgas

⇒ with  C = the solubility of a gas at a fixed temperature in a particular solvent

⇒  with k = Henry's law constant

 ⇒  with Pgas = the partial pressure of the gas

k = C/Pgas

Since 78.0 % of the gas is N2

P(N2) = 0.78 atm

k = C/P(N2) = (4.88*10^−4 M)/(0.78 atm)

k = 6.26 * 10^-4  M/ atm

The value of Henry's law constant for N2 = 6.26 * 10^-4  M/ atm