If `3.7 g` of a gas at `25^(@)C` occupies the same volume as `0.814 g` of hydrogen at `17^(@)C` and at the same pressure, then what is the molecular weight of the gas?

Explanation

`PV=nRT`
`P` and `V` are the same.
So, `n_(1)T_(1)=nT_(2)`
or `(w_(1))/(M_(1))T_(1)=(w_(2))/(M_(2))T_(2)`
or `(3.7)/(M_(1))xx298=(0.184)/(2)xx290`
or `M_(1)=41.33 g`