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.
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
Answered by quit I
0O0
Answered by quit I



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