Question

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.

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the solubility of argon in water at 25 degrees celcius is 0.0150M. what is the henry's law constant... Argon makes up 0.93% by volumee of air. Calculate its solubility (mol/L) in water at 20C and 1.0 atm... The solubility of argon in water at 25°C is 0.0150 mol/L. What is the Henry's Law constant for argon...