1. Given data

• Mass of liquid water $= 0.90\,\text{g}$
• Temperature $= 27^\circ \text{C} = 300\,\text{K}$
• Vapour pressure of water at $27^\circ \text{C}$ is $32.0\,\text{Torr}$
• $R = 0.082\,\text{L atm K}^{-1}\text{ mol}^{-1}$
• Molar mass of water $= 18\,\text{g mol}^{-1}$

We are asked for the box volume when all the liquid just evaporates.

At that limiting condition, the whole $0.90\,$g of water is present as vapour, and since liquid is just exhausted, the vapour is still at equilibrium pressure:

$P = 32.0\,\text{Torr}$

2. Calculate moles of water

$n = \frac{0.90}{18} = 0.05\,\text{mol}$

3. Convert pressure into atm

$P = \frac{32.0}{760}\,\text{atm}$

$P \approx 0.0421\,\text{atm}$

4. Use ideal gas equation

$PV = nRT$

So,

$V = \frac{nRT}{P}$

Substitute values:

$V = \frac{(0.05)(0.082)(300)}{32/760}$

First calculate numerator:

$0.05 \times 0.082 \times 300 = 1.23$

Thus,

$V = \frac{1.23}{32/760} = 1.23 \times \frac{760}{32}$

$V = 29.21\,\text{L}$

5. Nearest integer

$V \approx 29\,\text{L}$

6. Comparison with stored answer

Stored correct answer: $29$

Our derived answer is also $29$.