Odebral se 12,0-litrový vzorek vzduchu o teplotě 25 °C a tlaku 1,00 atm a vysušil se. Po vysušení
byl objem vzorku přesně 11,50 l. Jaké bylo hmotnostní procento vody v původním vzorku vzduchu?


A 12.0-liter sample of air at 25°C and 1.00 atm pressure was collected and dried. After
drying, the volume of the sample was exactly 11.50 L. What was the percentage by mass
of water in the original air sample?
Answer: Under the conditions given, a mole of dry air occupies:
22.4L×(298 K/273 K) = 24.5 L/mol
mol dry air = 11.5L/(24.5 L/mol) = 0.469 mol
The average molar mass of dry air is 29.1 g/mol
The mass of dry air = 0.469 mol × 29.1 g/mol = 13.6 g
mols of H2O = 0.5L/(24.5 L/mol) = 0.0204 mol
mass of H2O = 0.0204 mol ’ 18.0 g/mol = 0.367 g
% H2O by mass = 100 × 0.367 g/(13.6 g + 0.367 g) = 2.62 %