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

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

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=99.9386$

Hence answer is (b)

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