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` steam at this temperature is

Options

A. `6 cm^(3)`
B. `60 cm^(3)`
C. `0.6 cm^(3)`
D. `0.06 cm^(3)`

Answer

D

Explanation

Mass of water in `1 L` steam `= 1000 xx 0.00006 g`
`= 0.06 g`
Volume of `0.06 g` water `= (0.06)/(1 g cm^(-3)) = 0.06 cm^(3)`