Question

At 100^o C and 1 atm, if the density of liquid water is 1.0 g/cm³ and that of water vapour is 0.0006 g/m³ , then the volume occupied by water molecules in 1 litre of steam at that temperature is

• A. 6 cm³
• B. 60 cm³
• C. 0.6 cm³
• D. 0.06 cm³

Answer 1

The correct option is C 0.6 cm³

Volume of steam = 1 lit = ${10}^{3}c{m}^3$
$\because$ m = d.V
$\therefore$ Mass of ${10}^{3}c{m}^3$ steam = density x Volume =$\frac{0.0006gm}{c{m}^3}$ x ${10}^{3}c{m}^3$ = 0.6gm
Actual volume occupied by $H_{2}O$ molecules is equal to volume of water of same mass
$\therefore$ Actual volume of $H_{2}O$ molecules in 0.6gm Steam= mass of steam/density of water= 0.6 gm /1 gm/$c{m}^3\Rightarrow 0.6c{m}^3$