Air contains 23% oxygen and 77% nitrogen by w | vedclass

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(B) Assume we have $100 \, g$ of air mixture.Volume occupied by $23 \, g$ of $O_2 = \frac{22.4}{32} 	imes 23 = 16.1 \, L$ (since $32 \, g$ of $O_2$ o

Vedclass · Accepted Answer

$20.8$

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(B) Assume we have $100 \, g$ of air mixture.Volume occupied by $23 \, g$ of $O_2 = \frac{22.4}{32} 	imes 23 = 16.1 \, L$ (since $32 \, g$ of $O_2$ occupies $22.4 \, L$ at $STP$).Volume occupied by $77 \, g$ of $N_2 = \frac{22.4}{28} 	imes 77 = 61.6 \, L$ (since $28 \, g$ of $N_2$ occupies $22.4 \, L$ at $STP$).Total volume of the mixture $= 16.1 + 61.6 = 77.7 \, L$.Percentage by volume of $O_2 = \frac{16.1}{77.7} 	imes 100 \approx 20.72 \% \approx 20.8 \% $.

Air contains $23\%$ oxygen and $77\%$ nitrogen by weight. What is the percentage by volume of $O_2$?

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