Argon makes up 0.93% by volume 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.

Well, well, well, let's dive into some chemistry with a splash of humor, shall we?

To calculate the solubility of Argon in water, we can use Henry's law. Now, Henry's law tells us that the solubility of a gas is directly proportional to its partial pressure.

So, by taking the partial pressure of Argon in air (which is 0.93% of atmospheric pressure) and multiplying it by Henry's law constant, we can determine the solubility. Get ready for some bubbles of knowledge!

First, let's convert the partial pressure of Argon to atm:

Partial pressure of Argon = 0.93% of 1.0 atm = 0.0093 atm

Now, we can plug this value into Henry's law equation:

Solubility = (Henry's Law Constant) x (Partial Pressure of Argon)

Solubility = (1.5 x 10^-3 mol/L*atm) x (0.0093 atm)

Calculating this gives us:

Solubility = 1.395 x 10^-5 mol/L

Looks like Argon is a bit shy when it comes to getting cozy with water molecules, with a solubility of approximately 1.395 x 10^-5 mol/L.

To calculate the solubility (in mol/L) of argon in water at 20°C and 1.0 atm, we can use Henry's law.

Henry's law states that the concentration (in mol/L) of a gas dissolved in a liquid is directly proportional to the partial pressure of the gas above the liquid. The proportionality constant is called Henry's law constant.

The equation we can use to calculate the solubility of argon in water is:
c = k * P

where:
c = concentration of argon in water (in mol/L)
k = Henry's law constant for argon under given conditions (1.5 x 10^-3 mol/L*atm)
P = partial pressure of argon above the water (1.0 atm)

Using this equation, we can substitute the given values to find the solubility of argon in water at 20°C and 1.0 atm.

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.