The solubility of argon in water at 25°C is 0.0150 mol/L.What is the Henry's Law constant for argon if the partial pressure of argon in air is 0.00934 atm? A)1.40 × 10^-4 mol/(L ∙ atm) B)0.623 mol/(L ∙ atm) C)1.61 mol/(L ∙ atm) D)4.10 mol/(L ∙ atm)

Question

Quizplus · Accepted Answer

Henry's Law states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. The equation for Henry's Law is:

C = kH * P

where C is the concentration of the gas in the liquid, P is the partial pressure of the gas above the liquid, and kH is the Henry's Law constant.

To find kH for argon, we can rearrange the equation to:

kH = C / P

Substituting the given values, we get:

kH = 0.0150 mol/L / 0.00934 atm

kH = 1.61 mol/(L ∙ atm)

Therefore, the correct answer is C.

The Solubility of Argon in Water at 25°C Is 0