Answer:
\[PM=dRT\]
Gaseous oxide \[2\text{
}\times M=dRT~~~~~~~~~~~~~~~~~~~...\left( \text{i} \right)\]
Nitrogen \[5\text{
}x\text{ }28\text{ }=dRT~~~~~~~~~~~~~~~~~~~~~~~~...\left( \text{ii} \right)\]
From Eqns.
(i) and (ii)
\[2\times M=5\times
28\]
\[M=70g\,mo{{l}^{-1}}\]
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