Assuming that the solubility of radon in water with 1 atm pressure of the gas over the water at 30 ∘c is 7.27×10−3m, what is the henry's law constant for radon in water at this temperature?

K = 137.55 atm/M.

Explanation:

P = (K)(C)

where P is the partial pressure of the gaseous  solute above the solution (P = 1.0 atm).

k is a constant (Henry’s constant).

C is the concentration of the dissolved gas (C = 7.27 x 10⁻³ M).

∴ K = P/C = (1.0 atm)/(7.27 x 10⁻³ M) = 137.55 atm/M.