Question

At ${100}^{o}C$ and $1$ atm, if the density of the liquid water is $1.0g$ $c{m}^{-3}$ and that of water vapour is $0.0006g$ $c{m}^{-3}$, then the volume occupied by water molecules in $1$ L of steam at this temperature is:

• A. $6c{m}^3$
• B. $60c{m}^3$
• C. $0.6c{m}^3$
• D. $0.06c{m}^3$

Answer 1

Answer: (C)

$0.6c{m}^3$

For water vapour, Density$=0.0006gc{c}^{-1}$

$\therefore 0.0006=\frac{Mass}{Volume}=\frac{Mass}{1000}$

$Mass=1000\times 0.0006=0.6g$

The density of liquid water is $1gc{c}^{-1}$.
So, the volume occupied by water is $\frac{Mass}{Density}=\frac{0.6}{1}=0.6c{m}^3$.