Argon makes up 0.93% by volumee of air. Calculate its solubility (mol/L) in water at 20C and 1.0 atm. Henry's law constant for Argon under these conditions is 1.5 x 10^-3 mol/L*atm.

To calculate the solubility of Argon in water, we can use Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Henry's law can be represented by the equation:

C = k * P

Where:
C is the molar concentration (solubility) of the gas in the liquid (in mol/L),
k is Henry's law constant (in mol/L*atm), and
P is the partial pressure of the gas (in atm).

In this case, we want to calculate the solubility of Argon (Ar) in water at a temperature of 20°C (293K) and a pressure of 1.0 atm. Given that the Henry's law constant for Argon under these conditions is 1.5 x 10^-3 mol/L*atm, we can substitute the values into the equation:

C = k * P
C = (1.5 x 10^-3 mol/L*atm) * (1.0 atm)
C = 1.5 x 10^-3 mol/L

Therefore, the solubility of Argon in water at 20°C and 1.0 atm is 1.5 x 10^-3 mol/L.