$112 m l$ of oxygen at STP is subjected to liquefaction. The mass of liquid oxygen obtained is:

0.64

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0.16

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0.32

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0.96

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Solution

The correct option is D $0.16$ g
Moles of $O_{2}$ subjected to liquefaction= $\frac{112}{22400} = 5 \times 10^{- 3}$
This is equal to moles of $O_{2}$ obtained
$∴$ Mass of liquid $O_{2}$ obtained= $5 \times 10^{- 3} \times 32 = 0.16 g$

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112 m l

The mass of 112 ml of N₂ at STP on liquefaction is:

(C_{6} H_{6})

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