At `100^(@)C` and `1 atm`, if the density of the 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 L` of steam at this temperature is

Options

A. `6 cc`
B. `60 cc`
C. `0.6 cc`
D. `0.06 cc`

Explanation

For water vapours, `P=0.0006 g c c^(-1)`
`0.0006=(Mass)/(Volume)=(Mass)/(1000)`
Mass `=1000xx0.0006=0.6 g`
Density of liquid water `=1 g c c^(-1)`
Volume occupied by water `=(Mass)/(Density)=(0.6)/(1)=0.6 cc`