A) 0.32 g
B) 0.64 g
C) 0.16 g
D) 0.96 g
Correct Answer: C
Solution :
22400 ml is the volume of \[{{O}_{2}}\]at N.T.P =32gm of \[{{O}_{2}}\] 1ml is the volume of \[{{O}_{2}}\] at NTP = \[\frac{32}{22400}\] 112 ml is the volume of \[{{O}_{2}}\] at NTP = \[\frac{32}{22400}\times 112\] \[=0.16gm\] of \[{{O}_{2}}\]
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