A) \[293K\]
B) \[298K\]
C) \[301K\]
D) \[304K\]
Correct Answer: A
Solution :
We know that \[PV=nRT\] or \[PV=\frac{w}{m}RT\] or \[m=\frac{w}{v}\frac{RT}{P}\] or \[M=d\frac{RT}{P}\] \[d=1.964g/d{{m}^{3}}=1.964\times {{10}^{-3\text{ }}}g/cc.\] \[P=76cm=1\text{ }atm\] \[R=0.0812L\,\,atm\,{{K}^{-1}}mo{{l}^{-1}}\] \[=8.21cc\,atm\,{{K}^{-1}}\,mo{{l}^{-1}}\] \[T=273K\] \[m=\frac{1.964\times {{10}^{-3}}\times 82.1\times 273}{1}=44.\] The molecular weight of \[C{{O}_{2}}\] is 44. So, the gas is \[C{{O}_{2}}\].You need to login to perform this action.
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