A) \[{{C}_{2}}{{H}_{4}}\]
B) \[{{C}_{3}}{{H}_{6}}\]
C) \[{{C}_{6}}{{H}_{12}}\]
D) \[{{C}_{4}}{{H}_{8}}\]
Correct Answer: A
Solution :
[a] : Empirical formula weight \[=12+2\times 1=14\] 22.4 L of \[{{N}_{2}}\equiv 28g\]\[\Rightarrow \]1 L of \[{{N}_{2}}=\frac{28}{22.4}g\] Now, 1 L of organic gas \[=\frac{28}{22.4}g\] \[\Rightarrow \]22.4 L of organic gas = 28 g \[\therefore \]Molecular weight of gas = 28 g (Empirical weight)\[_{n}\] = Molecular weight\[n=\frac{28}{14}=2\] Molecular formula = (Empirical formula)\[_{n}\] \[={{(C{{H}_{2}})}_{2}}={{C}_{2}}{{H}_{4}}\]
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