A) \[11.2\,d{{m}^{3}}\]
B) \[5.6\,d{{m}^{3}}\]
C) \[22.4\text{ }d{{m}^{3}}\]
D) \[1.0\text{ }d{{m}^{3}}\]
Correct Answer: B
Solution :
Weight of \[{{O}_{2}}=\frac{E}{F}\times Q\] \[=\frac{8}{1F}\times F=8g\] Volume of\[8\,g\,{{\text{O}}_{2}}\]on NTP\[=22.4\times \frac{8}{32}d{{m}^{3}}\] \[=5.6\text{ }d{{m}^{3}}\]
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