2.9 g of a gas at `90^(@)C` occupie the same volume as 0.184 g of `H_(2)` at `17^(@)C` at the same pressure. What is the molar mass of the gas ?

Explanation

Let the molar mass be M
`thereforen(g)=(2.9)/(M)`
`n(H_(2))=(0.184)/(2)=0.092`
volume of `H_(2)=(0.092xxRxx290)/(P)=(26.68R)/(P)`
volume of `(2.9)/(M)` mol of gas `=(2.9xxRxx368)/(MP)=(1067.2)/(MP)`
Now `(1067.2R)/(MP)=(26.68R)/(P)` or `M=40` g `mol^(-1)`