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

Expert's answer

M(N

M(O

M(Ar) = (1% x 40)/100% = 0.4 g/mol

Total molecular weight of air is 21.84 + 6.72 + 0.4 =

M(H

M(N

Total molecular weight of the mixture is 28.265 + 0.43 =

1) molecular weight of Oxygen in new mixture is M(O

2) percentage of Oxygen is 6.56 x 100%/32 =

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