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

Explanation

For unknown gas
`V_1=V`
`n_1=2.9/m`
`T_1=95^@+273=368 K`
`P_1V_1=n_1RT_1`
`P_1=(n_1RT_1)/V_1=2.9/m times (R times 368)/V`
For Hydrogen gas
`V_2=V_0`
`n_2=0.184/2`
`T_2=17^@C+273=290K`
`P_2V_2=n_2RT_2`
`P_2=(n_2RT_2)/V_2=0.184/2 times (R times 290)/V`
Since the pressure are same
`2.9/m times (R times 368)/V=0.184/2 times 290`
`m=(2.9 times 368)/(0.184 times145)=40gm//ml`.