The enthalpy of vaporisation is the energy required to convert one mole of a liquid into its vapour state at constant pressure. It can be calculated from standard enthalpies of formation using: (\Delta_{vap}H^\circ = \Delta_f H^\circ(H_2O, g) - \Delta_f H^\circ(H_2O, l)). This is a direct application of Hess's Law, since enthalpy is a state function. Substituting the given values: (\Delta_{vap}H^\circ = (-241.8) - (-285.8) = -241.8 + 285.8 = +44.0) kJ/mol. The positive sign confirms that vaporisation is an endothermic process — energy must be supplied to overcome intermolecular hydrogen bonds in liquid water. Option -44.0 kJ/mol is the enthalpy of condensation (the reverse process), not vaporisation. Option 527.6 kJ/mol incorrectly sums the absolute values of both enthalpies of formation. Option 22.0 kJ/mol would be the correct answer if only 0.5 mol of water vaporised. This is a standard NCERT thermochemical calculation and is frequently tested in JEE Main. Plausibility check: the known enthalpy of vaporisation of water at 298 K is approximately 44 kJ/mol, consistent with the strong hydrogen bonding in liquid water, confirming both sign and magnitude are correct.