If 1 g of each of the following gases are taken at STP, which of the gases will occupy (a) greatest volume and (b) smallest volume ?
`CO, H_(2)O, CH_(4), NO`

Explanation

From Avogardro.s law, we know that
Bolume of 1 mole of gas = gram molecular mass = 22.4 L at STP.
Volume of occupied by 28 g CO (1 mol CO) = 22.4 L at STP.
(`because` Molar mass of `CO=12+16=28 g mol^(-1)`)
`terefore` Volume occupied by 1 g `CO=(22.4)/(28)` L at STP.
Similarly, Volume occupied by 1 g `H_(2)O=(22.4)/(18)` L at STP. (`because` Molar mass of `H_(2)O = (2xx1)+16)=18 g mol^(-1)`)
Volume occupied by 1 g `CH_(4)=(22.4)/(16)` L at STP. (`because` Molar mass of `CH_(4)=12+(4xx1)=16 g mol^(-1)`)
Volume occupied by 1 g NO `= (22.4)/(30)` L at STP. (`because` Molar mass of NO `= 14+16 = 30 g mol^(-1)`)
Hence, 1 g `CH_(4)` will occupy maximum volume and 1 g of NO will occupy minimum volume at STP.