A) 1.43 g
B) 2.24 g
C) 11.2 g
D) 22.4 g
Correct Answer: A
Solution :
Given, volume of \[{{O}_{2}}=1\,L\] \[\because \]\[22.4\,L\]of \[{{O}_{2}}\]\[STP=32\,g\] \[\therefore \]\[1L\]of \[{{O}_{2}}\]at \[STP=\frac{32}{22.4}g\] \[=1.43\,g\]
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