A) \[\text{C}{{\text{H}}_{\text{4}}}\]
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 \[M=d\frac{RT}{p}\] \[d=1.964\,g/d{{m}^{3}}=1.964\times {{10}^{-3}}\,g/cc\] \[p=76\text{ }cm=1\text{ }atm\] \[R=0.0821\,\text{L}\,\,\text{atm}\,{{\text{K}}^{-1}}\text{mo}{{\text{l}}^{-1}}\] \[=82.1\,\text{cc atm}\,{{\text{K}}^{-1}}\,\text{mo}{{\text{l}}^{-1}}\] \[T=273\,K\] \[M=\frac{1.964\times {{10}^{-3}}\times 82.1\times 273}{1}=44\] The molecular weight of \[\text{C}{{\text{O}}_{\text{2}}}\] is 44. So, the gas is \[\text{C}{{\text{O}}_{\text{2}}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
