Argon makes up 0.93% by volume of air. Calculate its solubility in water a 20 C and 1.0 atm. The henry's law constant for argon under these conditions is 0.0015 mol/L*atm

C = K*p

C = (0.0015 mol/L*atm)*0.0093 = ? M.

To calculate the solubility of argon in water at 20°C and 1.0 atm, we can use Henry's Law. Henry's Law states that the solubility of a gas is directly proportional to its partial pressure. The equation is as follows:

S = k * P

Where:
S = solubility of the gas in water (in mol/L)
k = Henry's Law constant (in mol/L*atm)
P = partial pressure of the gas (in atm)

Given:
Henry's Law constant, k = 0.0015 mol/L*atm
Partial pressure of argon, P = 0.0093 atm (0.93% of air)

Substituting the values into the Henry's Law equation:

S = (0.0015 mol/L*atm) * (0.0093 atm)
S = 0.00001395 mol/L

Therefore, at 20°C and 1.0 atm, the solubility of argon in water is approximately 0.00001395 mol/L.

To calculate the solubility of argon in water at 20°C and 1.0 atm using Henry's Law, we can use the equation:

C = k * P

where:
C is the solubility of argon in water (in mol/L)
k is the Henry's Law constant (0.0015 mol/L*atm)
P is the partial pressure of argon (in atm)

First, we need to determine the partial pressure of argon in air. Given that argon makes up 0.93% by volume of air, we can assume that the partial pressure of argon is equal to its mole fraction (0.0093) multiplied by the total pressure of air (1.0 atm):

P = 0.0093 * 1.0 atm
P = 0.0093 atm

Now, we can substitute the values into the equation:

C = (0.0015 mol/L*atm) * (0.0093 atm)
C = 0.00001395 mol/L

Therefore, the solubility of argon in water at 20°C and 1.0 atm is approximately 0.00001395 mol/L.