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 `0.00006 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

For water vapours, `P = 0.0006 g c c^(-1)`
`:. 0.0006 = ("Mass")/("Volume") = ("Mass")/(1000)`
`"Mass" = 1000 xx 0.0006 = 0.6 g`
The density of liquid water is `1 g c c^(-1)`
So, the volume occupied by water is
`("Mass")/("Density"0 = (0.6)/(1) = 0.6 cc`