One litre of water is placed in a closed room $$2 \times 10^4$$ liter capacity containing dry air at $$27^{\circ}\mathrm{C}$$. If aqueous tension at $$27^{\circ}\mathrm{C}$$ is $$23\mathrm{mm}$$ of $$\mathrm{Hg}$$ and density of water is $$1\mathrm{g/mL}$$, then volume of water that will evaporate is

At equilibrium, water vapour in the room will have partial pressure equal to the aqueous tension. $$P = \frac{23}{760}\,\mathrm{atm}$$ Room volume is $$V = 2 \times 10^4\,\mathrm{L}$$ Temperature is $$T = 300\,\mathrm{K}$$ Using the ideal gas equation for water vapour, $$n = \frac{PV}{RT}$$ $$n = \frac{\left(\frac{23}{760}\right)\left(2 \times 10^4\right)}{(0.0821)(300)}$$ $$n \approx 24.56$$ Mass of water evaporated is $$m = 24.56 \times 18 \approx 442.08\,\mathrm{g}$$ Since density of water is $$1\,\mathrm{g/mL}$$, the evaporated volume is $$442.08\,\mathrm{mL}$$ This matches option $$442.8\mathrm{mL}$$ most closely.