Vapour density of a gas, relative to air is 1.528. what is the mass of 2L of the gas at `27^(@)C` temperature and 750 mm Hg pressure? [vapour density of air, relative to hydrogen=14.4]

Explanation

Vapour density of a gas`=("mass of certain volume of a gas")/("mass of same volume "H_(2)" gas ")`
[at certain temperature and pressure]
`therefore`Vapour density of the gas `=("mass of V "cm^(3)" of gas")/("mass of V "cm^(3)" of air")xx("mass of V "cm^(3)" of air")/("mass of V "cm^(3)" off "H_(2)" gas")`
[at the same temperature and pressure]
`=1.528xx14.4=22`
`therefore`Molecular mass of the gas=`2xx`vapour density=44
Again, volume of 1 gram-mole of any gas at STP=22.4 L
So, mass of 22.4L of given gas at STP=44g
Now, let the volume of the gas be V L at STP. i.e.,
`P_(1)=750mm,P_(2)=760mm,V_(1)=2L,V_(2)=VL,`
`T_(1)=27^(@)C=(27+273)=300K,T_(2)=273K`
`therefore (760xx2)/(300)=(760xxV)/(273) or, V=(750xx2xx273)/(300xx760)=1.796L`
Now, mass of 22.4 L of the gas at STP=44g.
`therefore`Mass of 1.796g of the gas at STP`=(44xx1.796)/(22.4)=3.528g`
Hence, the mass of the gas =3.528g.