2.9 g of a gas at `95^(@)C` 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

We are given that
`m_(g)=2.9g`
`m_(H_(2)) =0.184 g `
`T_(g)=95^(@)C=95+273=368K`
`T_(H_(2))=17^(@)C =273 +17=290K`
We are to calculate the molar mass of the gas,
We know that `pV = nRT = (m)/(M)RT`
For gas, `pV=(m_(g))/(M_(g))R xx 368" " ...(i)`
For `H_(2), pV= (m_(H_(2)))/(M_(H_(2)))xx R xx 290" " ...(ii)`
From (i) and (ii)
`(m_(g))/(M_(g)) xx R xx 368 = (m_(H_(2)))/(M_(H_(2)))xx R xx 290`
`(2.9xx 368)/(M_(g))=(0.184 xx 290)/(2)`
`M_(g)= (2xx2.9xx 368)/(0.184 xx 290)=40` g/mol