A) 1.43 g
B) 2.24 g
C) 11.2 g
D) 22.4 g
Correct Answer: A
Solution :
\[22.4l\ {{O}_{2}}\] at S.T.P. \[=32gm\] of \[{{O}_{2}}\] \[1l\]\[{{O}_{2}}\] at S.T.P. \[=\frac{32}{22.4}=1.43gm\] of \[{{O}_{2}}\]
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