Argon makes up 0.93% by volume of air. calculate its solubility (in mol/l) in water at 20°c and 1.0 atm. the henry's law constant for ar under these conditions is 1.5 × 10−3 mol/l·atm. enter your answer in scientific notation.

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

(1)

where S: is the solubility or concentration of Ar in water, : is Henry's law constant and p: is the pressure of the Ar

Since the argon is 0.93%, we need to multiply the equation (1) by this percent:

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

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