Calculate the percentage of free volume available in 1 mole gaseous water at 1.0atm and $100^{\large\circ}C$. Density of liquid $H_2O$ at $100^{\large\circ}C$ is 0.958 g/mL.

$(a)\;100\qquad(b)\;99.9386\qquad(c)\;98\qquad(d)\;0.9938$

For gaseous water PV = nRT
Thus Volume occupied by 1 mole gaseous water can be derived as
T = 273 + 100 = 373 K
$V = \large\frac{nRT}{P} = \large\frac{1\times0.0821\times373}{1}$
$\;\;\;\;\;\;\;\;\;\;\; = 30.62 litre$
Also Volume of 1 mole of liquid water = $\large\frac{mass}{density}$
$= \large\frac{18}{0.958}$
=18.79 mL
$=18.79\times10^{-3}$ litre
Thus Volume Percentage occupied by water molecules in gaseous state
$=\large\frac{18.79\times10^{-3}}{30.62}\times100$
= 0.0614
$\therefore \%$ of free volume = 100 - 0.0614
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=99.9386$