Posted by Colleen on .
At 298 K, the Henry\'s law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm? At 298 K, what is the solubility of oxygen in water exposed to air at 0.896 atm? If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 K, how much oxygen will be released from 4.80 L of water in an unsealed container?
C = k*P
k = 0.00130
PO2 = 0.21(1 atm)
Solve for C = (O2) at 1 atm total P.
C = 0.00130*(0.896)(0.21) = (O2) M
Solve for C = (O2) M.
I would calculate mols O2 in 4.80 L H2O at 1 atm and 0.896 atm. Subtract top find mols O2 released if the pressure changed from 1.00 to 0.896 atm. The problem doesn't specify units