A) \[{{H}_{2}}\]
B) He
C) \[C{{H}_{4}}\]
D) \[S{{O}_{2}}\]
Correct Answer: C
Solution :
\[{{v}_{rms}}\propto \frac{1}{\sqrt{M}}\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,\,\,\,\,\frac{{{\nu }_{1}}}{{{\nu }_{2}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}\] \[\therefore \,\,\,\,\,\,\frac{1}{\sqrt{2}}=\sqrt{\frac{{{M}_{2}}}{32}}\,\,\,\,\,\,\,\Rightarrow \,\,\,\,{{M}_{2}}=16\] Hence the gas is \[C{{H}_{4}}\].
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