Solution:
At STP, 1 mole of an ideal gas occupies $22.4 \text{ L}$. The number of moles in $4.5 \text{ kg}$ of water is $n = \frac{4500 \text{ g}}{18 \text{ g/mol}} = 250 \text{ moles}$. The volume is $V = 250 \times 22.4 \text{ L} = 5600 \text{ L} = 5.6 \text{ m}^3$.
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