Question

At ${100}^{\circ }$C and 1 atm, if the density of liquid water is 1.0 g/$c{m}^3$ and that of water vapour is 0.0006 g/$m^3$ , then the volume occupied by water molecules in 1 litre of steam at that temperature is

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

Answer 1

The correct option is C 0.6cm³

Mass of water vapor = density x Volume = 0.0006 g/$c{m}^3$ x 1 litre = 0.0006 g/$c{m}^3$ x ${10}^3$ $c{m}^3$
$\Rightarrow$ Mass of water vapor = 0.6 g
Volume occupied by water molecules of vapor
= volume of same amount of water in liquid phase
= $\frac{m}{densityofwater}=\frac{0.6g}{1gc{m}^{-3}}$ = 0.6 $c{m}^3$