Asked by hershi
The solubility of Ar in water at 321.9 torr is 0.02538 g/L. What is Henry's Law constant (M/atm) for Ar in water?
Answers
Answered by
DrBob222
C = p*k
k = C/p
Convert 321.9 torr to atm. atm = torr/760
Convert 0.02538 g/L to mols/L = M
mols = grams/molar mass
Substitute into equation and solve for k.
k = C/p
Convert 321.9 torr to atm. atm = torr/760
Convert 0.02538 g/L to mols/L = M
mols = grams/molar mass
Substitute into equation and solve for k.
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