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

C

Explanation

`1 L = 1000 mL = 1000 cm^(3)` Mass = Density `xx` volume
`= (0.006 g cm^(-3)) xx (1000 cm^(3))`
`= 0.6 g`
18 g of water `= 18 cm^(3)`
`:.` 0.6 g of water `= 0.6 cm^(3)`
`:.` 0.6 g of water `= 0.6 cm^(3)`
Actual volume occupied by molecules `= 0.6 cm^(3)`