A) \[56\,c{{m}^{3}}\]
B) \[112\,c{{m}^{3}}\]
C) \[224\,c{{m}^{3}}\]
D) \[448\text{ }c{{m}^{3}}\]
Correct Answer: A
Solution :
m of \[{{O}_{2}}=\frac{E}{96500}\times C\times t\] \[=\frac{8}{96500}\times 5\times 193=0.08\,g\] \[\because \] \[32g\,\,{{O}_{2}}\] contains volume at\[NTP=22400\text{ }c{{m}^{3}}\] \[\therefore \] \[0.08\text{ }g\]\[{{O}_{2}}\] contains volume at NTP \[=\frac{22400\times 0.08}{32}=56\,c{{m}^{3}}\]
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