Question

At ${100}^{o}C$ and $1$ atm, if the density of the liquid water is $1.0gc{m}^{-3}$ and that of water vapour is $0.0006$ $gc{m}^{-3}$, 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)

$0.6c{m}^3$

The number of moles of water molecules is $n=\frac{PV}{RT}=\frac{1\times 1}{0.0821\times 373}=0.0326mol$.

The mass of water molecules is $0.0326mol\times 18g/mol=0.5877g$.

The volume occupied by water molecules in 1 L of steam is $\frac{0.5877g}{(1-0.0006)gc{m}^{-3}}=0.6c{m}^3$.

Hence, the correct answer is option $\text{C}$.