Quark:
Perhaps if we go into more detail for 25362's calculations, it might be helpful. First, some assumptions:
(1) Bif's aqua ammonia is a solution of 5 weight percent of ammonia in water. (He didn't explicitly state that it was weight percent.)
(2) The storage temperature of the aqua ammonia is 20 degree C.
Then, some givens:
(1) Atmospheric pressure is 14.696 psia
(2) mole % = volume % = partial pressure % for gases
(3) Molecular weights: NH3 = 17, H20 = 18, air = 29
Evidently, 25362 found these vapor pressure data for a 5 weight % solution of ammonia in water at 20 deg C:
(1) vapor pressure of NH3 above the solution = 0.735 psia
(2) vapor pressure of H2O above the solution = 0.295 psia
(3) total vapor pressure of the solution = 1.030 psia
Thus, the composition of the vapor above the solution:
NH3 = 100(0.735/14.696) = 5 partial press. % = 5 mol %
H20 = 100(0.295/14.690) = 2 partial press. % = 2 mol %
Air = 100 - 5 - 2 = 93 mol %
Converting the vapor composition from mol % to wt. %:
NH3 = (5 mols)(17 lb/lb-mol) = 85 lbs = 3.0 %
H2O = (2 mols)(18 lb/lb-mol) = 36 lbs = 1.3 %
Air = (93 mols)(29 lb/lb-mol) = 2697 lbs = 95.7 %
Note: 100 total mols with a total mass of 2818 lbs means that the vapor above the solution has a molecular weight of 28.18 lb/lb-mol.
Also note: The vapor pressures that I found in Perry's Chemical Engineers' Handbook are slightly different from those found by 25362, but not enough to be significant.
I hope this helps,
Milton Beychok
(Contact me at www.air-dispersion.com)