Question

At 100${}^0$C and 1 atm, if the density of liquid water is $1.0g{cm}^{-3}$ and that of water vapour is $0.0006g{cm}^{-3}$, then the volume occupied by water molecules in one litre of steam at that temperature is :

Answer 1

Density of liquid water $={1gcm}^{-3}$

Density of water vapour $={0.0006gcm}^{-3}$

Mass of $1000ml$ stream $=1000\times 0.0006=0.6g$

$\because$ Volume of liquid water $=\frac{\text{mass of 1000ml stream}}{\text{density of liquid water}}$

$V=\frac{0.6g}{1g.c{m}^{-3}}$

$V=0.6c{m}^3$

$\therefore$ The answer is option C.