question_answer

The density of a gas is \[1.964\,\,d{{m}^{-3}}\] at \[273\,\,K\]and\[76\,\,cm\]\[Hg\]. The gas is

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            \[M=d\frac{RT}{p}\] \[d=1.964\,\,g/d{{m}^{3}}=1.964\times {{10}^{-3}}g/cc\] \[p=76\,\,cm=1\,\,atm\] \[R=0.0821\,\,L\,\,atm\,\,{{K}^{-1}}mo{{l}^{-1}}\] \[=82.1\,\,cc\,\,atm\,\,{{K}^{-1}}at{{m}^{-1}}\] \[T=273\,\,K\] \[M=\frac{1.964\times {{10}^{-3}}\times 82.1\times 273}{1}=44\] The molecular weight of \[C{{O}_{2}}\] is\[44\]. So, the gas is\[C{{O}_{2}}\]. \[3bp+1lp=s{{p}^{3}}\]hybridisation. pyramidal (due to presence of one lone pi electron)