What volume is occupied by $100.0$ grams of oxygen gas $(O_{2})$ at STP?

140 L

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35 L

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280 L

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70 L

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Solution

The correct option is D $70 L$
$n u m b e r o f m o l e s = \frac{w e i g h t}{m o l e c u l a r w e i g h t}$
$100.0 g O_{2}$ at STP $= \frac{100}{32} = 3.125 m o l e s$
$1 m o l e$ gas at STP occupies volume $= 22.4 L$
$3.125 m o l e s = 22.4 \times 3.125 = 70 L$

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