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 \[m=\frac{m}{V}\frac{RT}{P}\] Or \[M=d\frac{RT}{P}\] \[d=1.964g/d{{m}^{3}}=1.964\times {{10}^{-3}}g/cc.\] \[P=76cm=1atm\] \[R=0.0812\text{ }L\text{ }atm\text{ }{{K}^{-1}}mo{{l}^{-1}}\] \[=82.1\,cc\,atm\,{{K}^{-1}}mo{{l}^{-1}}\] \[T=273K\] \[m=\frac{1.964\times {{10}^{-3}}\times 82.1\times 273}{1}=44\] The molecular weight of\[C{{O}_{2}}\]is 44. So, the gas is\[C{{O}_{2}}\].
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