Answer to Question #52339 in Inorganic Chemistry for Candidate

Question #52339
Dry air is a mixture, by volume, of 78% nitrogen (N2), 21% Oxygen (O2) and Argon (Ar 40 g). a) Compute the gram molecular weight of dry air b) Assuming water vapor saturation the new mixture being at 20 deg.Celius contains 2.4% water vapor by volume. Compute the new gram molecular weight of the mixture c) In this last case compute the percentage of Oxygen, by weight, of the mixture
a) The gram molecular weight of dry air equels its molecular weight. Lets find molecular weight of each gas in air:
M(N2) = (78% x 28)/100% = 21.84 g/mol
M(O2) = (21% x 32)/100% = 6.72 g/mol
M(Ar) = (1% x 40)/100% = 0.4 g/mol
Total molecular weight of air is 21.84 + 6.72 + 0.4 = 28.96 g/mol

b) Water is lighter than other gases in air so water will replace them. Now gases will occupy only 100%-2.4%=97.6% of volume.
M(H2O) = (2.4% x 18)/100% = 0.43 g/m
M(N2,O2,Ar) = (97.6% x 28.96)/100% = 28.265 g/mol
Total molecular weight of the mixture is 28.265 + 0.43 = 28.7 g/mol

c) Percentage of Oxygen by weight in last mixture is:
1) molecular weight of Oxygen in new mixture is M(O2) = (6.72 x 97.6%)/100% = 6.56 g/mol
2) percentage of Oxygen is 6.56 x 100%/32 = 20.5%

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