Question

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?

Answer 1

The Henry's law constant for radon in water at 30 °C is calculated using the given solubility of 7.27×10⁻³M at 1 atm pressure, resulting in a value of 7.27×10⁻³ M/atm.

Step-by-step explanation:

The student has asked for the Henry's law constant for radon in water at 30 °C given the solubility of radon is 7.27×10⁻³M with a pressure of 1 atm of the gas over the water. Henry's law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid. The Henry's law constant (kH) can be calculated using the formula Cg = kHPg, where Cg is the concentration of the gas in solution (molarity), and Pg is the partial pressure of the gas.

Given that the solubility (Cg) is 7.27×10⁻³M and the pressure (Pg) is 1 atm, the Henry's law constant for radon can be calculated as follows:
kH = Cg/Pg = 7.27×10⁻³ M / 1 atm = 7.27×10⁻³ M/atm.

Answer 2

K = 137.55 atm/M.

Step-by-step explanation:

• The relationship between gas pressure and the concentration of dissolved gas is given by Henry’s law:

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.