A volume of 6 l H₂O is placed in a closed evacuated room of volume 827 l at the temperature 300 K. T — States of Matter and Gaseous State Chemistry Question
Question
A volume of 6 l H₂O is placed in a closed evacuated room of volume 827 l at the temperature 300 K. The density of liquid water at 300 K is 1.0 g/ml. The vapour pressure of water at 300 K is 22.8 mm Hg. Neglect the change in volume of liquid water by vaporization.
💡 Solution & Explanation
Free volume for vapour $V = 827 - 6 = 821 \text{ L}$. Pressure $P = 22.8 / 760 = 0.03 \text{ atm}$. Moles of vapour $n = PV/RT = (0.03 \times 821) / (0.0821 \times 300) = 1 \text{ mole}$ (B $\rightarrow$ S). Mass of vapour $= 1 \text{ mol} \times 18 \text{ g/mol} = 18 \text{ g}$ (A $\rightarrow$ Q). Initial mass of liquid $= 6 \text{ L} \times 1 \text{ kg/L} = 6 \text{ kg}$. Leftover mass $= 6 \text{ kg} - 0.018 \text{ kg} \approx 6 \text{ kg}$ (C $\rightarrow$ P). One mole of H₂O contains 3 moles of atoms (2 H + 1 O), so total moles of atoms = 3 (D $\rightarrow$ R). Therefore, correct answer is 1-B, 2-D, 3-A, 4-C.