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