Answer to Question #144053 in General Chemistry for Ten

Question #144053

The density of dry air at 0.9866 bar and 27°C is 1.146 g/dm³. Calculate the percentage nitrogen by mole.

Expert's answer

PV = nRT

0.9866 bar to atm = 0.974atm

27°C = 300K

n= PV/RT = 0.974 ×1 / (0.0821×300) = 0.040mole

Percentage by mole = 78/100 × 0.040 = 0.0312

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