A) \[C{{H}_{4}}\]
B) \[{{C}_{2}}{{H}_{6}}\]
C) \[C{{O}_{2}}\]
D) \[Xe\]
Correct Answer: C
Solution :
We know that \[PV=nRT\] or \[PV=\frac{w}{m}RT\] or \[m=\frac{w}{V}\frac{RT}{P}\] or \[M=d\frac{RT}{P}\] \[d=1.964\,g/d{{m}^{3}}=1.964\times {{10}^{-3}}g/cc\]. \[P=76\,cm=1\]atm \[R=0.0812\,L\]atm \[{{K}^{-1}}mo{{l}^{-1}}\] \[=82.1\,\,cc\]atm \[{{K}^{-1}}mo{{l}^{-1}}\] \[T=273\,\,K\] \[m=\frac{1.964\times {{10}^{-3}}\times 82.1\times 273}{1}=44\]. The molecular weight of \[C{{O}_{2}}\] is 44. So, the gas is \[C{{O}_{2}}\].
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