Calculate % of 'free volume' available in 1 mol gaseous water at 1.0 atm and `100^(@)C` .Density of liquid water at `100^(@)C` is 0.958 `g//mol`.

Explanation

We know that PV = nRT
Volume occupied by 1 mol gaseous water at `100^(@)C` and 1 atm pressure `= ( 1 xx 0.0821 xx 373 )/( 1) = 30.62 ` litre
volume of 1 mol liquid water `= ( " mass")/( " density ") = ( 18) /( 0.958) = 1.88 xx 10^(-2) ` litre
% of volume occupied by liquid water `= ( 1. 88 xx 10^(-2))/( 30.62 ) xx 100 = 0.0613 %`
% of free volume = 100 - 0.0613 = 99.93%