At `27^(@)C` and 780mm pressure 1L of a gas weighs 1.125g. Find its vapour density.

Explanation

Let, at `27^(@)C` temperature and 780 mm Hg pressure the volume of 1L of a gas at STP be VL, then
`(760xxV)/(273)=(780xx1)/(273+27)[(P_(1)V)/(T_(1))=(P_(2)V_(2))/(T_(2))]" "thereforeV=0.934L`
`therefore`At STP, mass of 0.934L of the gas=1.125g
`therefore`At STP, mass of 22.4L of the gas`=(1.125)/(0.934)xx22.4=26.98g
`therefore`Molecular mass=26.98 and vapour density
`=(26.98)/(2)=13.49`