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

Answer

C

Explanation

Volume of steam = 1L = `10^(3) x cm^(3)`
` :. ` m = d.V
` :. ` mass of `10^(3) cm^(3)` steam = density `xx ` Volume
= `(0.0006g)/(cm^(3)) xx 10^(3) cm^(3) = 0.6 g `
Actual volume occupied by `H_(2)O` molecules is equal to volume of water of same mass
` :. ` Actual volume of `H_(2)O` molecules in 0.6 g steam
= mass of steam /density of water
= 0.6 g/1 `g//cm^(3) rArr 0.6 cm^(3)`