A box contains 0.90 g of liquid water in equilibrium with water vapour at 27°C. The equilibrium vapour pressure of water at 27°C 32.0 Torr. When the volume of the box is increased, some of the liquid water evaporates to maintain the equilibrium pressure. If all the liquid water evaporates, then the volume of the box must be ______ litre. [nearest integer]
(Given: R = 0.082 L atm K$$^{-1}$$ mol$$^{-1}$$)
(Ignore the volume of the liquid water and assume water vapours behave as an ideal gas.)

We need to find the volume at which all liquid water evaporates while maintaining equilibrium vapour pressure. The mass of water is 0.90 g and its molar mass is 18 g/mol, so $$n = \frac{0.90}{18} = 0.05 \text{ mol}$$. The vapour pressure is 32.0 Torr, which is $$P = 32.0 \text{ Torr} = \frac{32.0}{760} \text{ atm} = 0.04211 \text{ atm}$$.

When all liquid water evaporates, all 0.05 mol exists as vapour at the equilibrium pressure. By the ideal gas law, $$PV = nRT$$, so $$V = \frac{nRT}{P} = \frac{0.05 \times 0.082 \times 300}{32/760}$$, which gives $$= \frac{0.05 \times 0.082 \times 300}{0.04211}$$ and then $$= \frac{1.23}{0.04211} = 29.2 \approx 29 \text{ litres}$$.

The answer is 29 litres.