Asked by charles
Calculate the mass of oxygen (in mg) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.21 and the Henry's law constant for oxygen in water at this temperature to be 1.3 x 10-3 M/atm.
Answers
Answered by
bobpursley
massO2=Hconstant*pressure*volume*molmass
= 1.3E-3*1.13*5L*32 grams=235mg\\check it.
= 1.3E-3*1.13*5L*32 grams=235mg\\check it.
Answered by
Ben
This answer worked better for me
P = P°*X
=> C = P°*X*k = 1.13atm * 0.21 * 1.3*10^-3M/atm = 3.085*10^-4M
3.085*10^-4M * 5L = 1.54*10^-3mol = 1.54mmol
1.54*MM(O2) = 1.54mmol * 32mg/mmol = 49.4mg
P = P°*X
=> C = P°*X*k = 1.13atm * 0.21 * 1.3*10^-3M/atm = 3.085*10^-4M
3.085*10^-4M * 5L = 1.54*10^-3mol = 1.54mmol
1.54*MM(O2) = 1.54mmol * 32mg/mmol = 49.4mg
Answered by
quit I
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Answered by
quit I
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