To solve this problem, we need to find the volume occupied by the molecules in 4.5 kg of water if the intermolecular forces vanish. This is equivalent to finding the volume of the water in its gaseous state at STP (Standard Temperature and Pressure).$\\$First, we need to determine the number of moles of water in 4.5 kg.$\\$The molar mass of water (H$_2$O) is approximately 18 g/mol.$\\$Convert the mass of water from kg to g:$\\ 4.5$ kg$= 4500$ g$\\$Calculate the number of moles n of water:$\newline n = \frac{4500 \ g}{18 \ g/mol} \\ n = 250$ moles$\\$At STP, 1 mole of an ideal gas occupies 22.4 L.$\\$Therefore, the volume occupied by 250 moles of water vapor at STP is:$\\ V = 250 \times 22.4$ L$\\ V = 5600$ L$\\$Convert the volume from liters to cubic meters (1 m$^3 = 1000$ L):$\\ V = \frac{5600 L}{1000} \\ V = 5.6$ m$^3 \\$Thus, the volume occupied by the molecules contained in 4.5 kg of water at STP, if the intermolecular forces vanish, is $5.6$ m$^3. \\$This corresponds to Option 4.