Question

You determine that at room temperature (25 oC) the Henry’s Law constant of O2gas is 1.3 * 10-3 mol/(L*atm). If you have a 51.4 L of water in your fish tank, and the air pressure is 0.813 atm in the room, how many grams of O2 gas could be dissolved?

Answer 1

As per Henry;s law constant(H), H= C/P

C= concentration of species in the aqueous phase, in this case oxygen in water and P= partial pressure of oxygen inthe gas phase, since air contains 21% O2 and 79%N2, partial pressure of O2= 0.21*0.813 atm =0.171 atm

concentration = moles/L= moles of oxygen/ 51.4

hence 1.3*10^-3 = moles of oxygen/51.4*0.171

moles of oxygen = 51.4*0.171*1.3*10^-3 =0.0114 moles

moles= mass/molar mass, mass of oxygen= 0.0114*32 gm =0.3648 gm