Mass of $112 m l$ of oxygen at NTP on liquefaction would be:

0.32

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0.64 g

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0.96 g

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0.16 g

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Solution

The correct option is D $0.16 g$
22400 ml is the volume of $O_{2}$ at NTP=32g of $O_{2}$

1 ml is the volume of $O_{2}$ = $\frac{32}{22400}$

112 ml is the volume of $O_{2}$ at NTP = $\frac{32}{22400}$ X 112 =0.16g of $O_{2}$

Therefore 0.16g is the answer.

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

112 m l

0.32

112

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