A) \[0.16\text{ }g\]
B) \[0.8\text{ }g\]
C) \[0.08g\]
D) \[1.6g\]
Correct Answer: C
Solution :
At STP \[22400c{{m}^{3}}\]of \[C{{H}_{4}}=12+4=16gm\,(C{{H}_{4}})\] At STP \[112c{{m}^{3}}\] of \[C{{H}_{4}}=\frac{16\times 112}{22400}=0.08gm\]
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