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