2.9 g of a gas at `95^(@)C` occupied the samme volume as 0.184 g of dhydrogen at `17^(@)C`, at the same pressure, what is the molar mass of the gas?

Explanation

Let the molar mass of the unknown gas =M `g*mol^(-1)`
Number of moles for 2.9 g of the gas`=(2.9)/(M)` mol
Number of moles for 0.184g of `H_(2)` gas `=(0.184)/(2)=0.092` mol
As both the gases have same volume at the same pressure and specified temperature, they will have the same value of PV.
to calculate PV, we apply the equation PV=nRT.
For unknown gas, `PV=(2.9)/(M)R(273+95)`
and for `H_(2)` gas, `PV=0.092R(273+17)`
Therefore, `(2.9R)/(M)xx368=0.092Rxx290" "therefore M=40`
Hence, the molar mass of the unknown gas`=40g*mol^(-1)`.