At 100$^\circ$C and 1 atm. if the density of liquid water is 1.0 g $cm^{-3}$ and that of water vapour is 0.0006 g $cm^{-3}$, then the volume occupied by water molecules in 1 litre of steam at that temperature
1L = 1000 mL = 1000 $cm^3$ Mass = Density $\times$ volume = (0.0006 g $cm^{-3}) \times$ (1000 $cm^3$) = 0.6 g 18 g of water = 18 $cm^3$ $\therefore$ 0.6 g of water = 0.6 $cm^3$ $\therefore$ 0.6 g of water = 0.6 $cm^3$ Actual volume occupied by molecules = 0.6 $cm^3$