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