Question

Air is a mixture of gases that is about 78.0% N₂ by volume. When air is at standard pressure and 25.0 ºC, 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? [Sg = k Pg]

A) 4.88 x 10⁻⁴ M/L-atm
B) 6.26 x 10⁻⁴ M/ L-atm
C) 4.88 x 10⁻⁴ M/ L-mmHg
D) 6.26 x 10⁻⁴ M/L-mmHg

Answer 1

The value of Henry's law constant for N₂ under the given conditions is approximately 6.26 × 10⁻⁴ M/L-atm.

Step-by-step explanation:

The value of Henry's law constant for N₂ under the given conditions can be calculated using the formula: Cg = k * Pg, where Cg is the solubility of N₂ in water, k is the Henry's law constant, and Pg is the partial pressure of N₂. In this case, the solubility of N₂ is given as 4.88×10⁻⁴ M and the partial pressure of N₂ can be calculated using the mole fraction of N₂ in air (78.0%) and the standard pressure (1 atm). So, the Henry's law constant can be calculated as:

k = Cg / Pg

Substituting the given values, we get:

k = (4.88×10⁻⁴ M) / [0.78 * 1 atm]

k ≈ 6.26 × 10⁻⁴ M/L-atm