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 \[M=d\frac{RT}{p}\] \[d=1.964\,\,g/d{{m}^{3}}=1.964\times {{10}^{-3}}\,\,g/cc\] \[p=76\,\,cm=1\,\,atm\] \[R=0.0821\,\,L\,\,atm\,\,{{K}^{-1}}mo{{l}^{-1}}\] \[=82.1\,\,cc\,\,atm\,\,{{K}^{-1}}mo{{l}^{-1}}\] \[T=273\,\,K\] \[M=\frac{1.964\times {{10}^{-3}}\times 82.1\times 273}{1}=44\] The molecular weight of\[C{{O}_{2}}\]is 44. So, the gas is\[C{{O}_{2}}\].
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