Question

What will be the % of "Free volume" available in 1 mole of water vapour at 1 atm and 373 K ? Density of liquid water at 373 K is 1 gm/ml approximately-

• A. 100 %
• B. 99.985 %
• C. 99.92 %
• D. 50 %

Answer 1

Answer: (C)

99.92 %

Answer 2

The total volume occupied by the gas

= $\frac{1 \times 0.082 \times 300}{1}$ = 24.6 L volume occupied by 1 mole of water = 18 ml

\ % of free vol = $\frac{(24600 - 18)}{24600}$ × 100 = 99.926.