Assuming ideal gas behaviour and using density of liquid water at 100°C equal to 0.958g(mL)−, find the percentagevalue of empty space available in one mole of H2O(g) at 760mm and 100oC

no. of mol, n=1  ;T=100+273=373KR=0.0821LatmK−mol−P= 760mm =1 atmV=?. We know that :(i)     PV=nRT ; V=nRTP=1mol×0.0821LatmK−mol−×373K1atmor V=30.62L(ii) Density of H2O(l)=0.958g(mL)−Mass of 1 mole of H2O=(2×1)+16=18g∴ Volume of 1 mole of H2O(l)= Mass  density =18g0.958g(mL)−=18.79mL=18.79mL×1L1000mL=0.01879L∴ Percentage of volume occupied byH2O(g) molecules =0.01879L30.62L×100=0.0614Hence, percentage of empty space available= 100 - 0.0614 = 99.9386 ≈99.94