At `25^(@)C` and at a certain pressure, 3.7 g of a gas occupies the same volume as the volume occupied by 0.184 of `H_(2)` gas at `17^(@)C` and the same pressure. Calculate molar mass of the gas.

Explanation

For `H_(2)` gas: `T=(273+17)=290K&n=(0.184)/(2)=0.092mol`
`therefore PV=nRT=0.092molxx0.0821L*atm*mol^(-1)*K^(-1)xx290K`
`=2.19L" "atm`
In case of the unknown gas: `T=(273+25)K=298K,`
`n=(3.7)/(M)mol" "`[M=molar mass of the unknown gas]
`therefore PV=nRT=(3.7)/(M)molxx0.0821L*atm*mol^(-1)*K^(-1)xx298K`
`=((90.52)/(M))L*atm`
Since the values of P and V are the same for both gases, the values of PV will also be the same for both gases.
`therefore (90.52)/(M)=2.19 or, M=41.33`
Therefore, the molar mass of the other gas=`41.33g*mol^(-1)`