`3.7 gm` of gas at `25^(@)C` occupied the same volume as `0.184 gm` of hydrogen at `17^(@)C ` and at the same pressure. What is the molecular mass of the gas ?

Answer

41.3

Explanation

Volume occupied by 0.184 g of `H_(2)` gas at `17^(@)C` and at pressure p(say) may be calculated as
`pV_(1)=nRT=(w)/(M)RT`
`"or "V_(1)=(w)/(Mp)RT`
`=(0.184xx0.0821xx(273+17))/(2p)=(2.1904)/(p)`
Volume occupied by 3.7 g of gas at `25^(@)C` and at same pressure.
`V_(2)=(w)/(Mp)RT=(3.7xx0.0821xx298)/(Mp)=(90.523)/(Mp)`
Now `V_(1)=V_(2)=(2.1904)/(p)=(90.523)/(Mp)`
`orM=(90.523)/(2.1904)=41.3`