Step 1: Plan. Freezing point depression is a colligative property given by \(\Delta T_f = K_f \, m\). We need the molality of ethylene glycol in water.
Step 2: Convert mass to moles. Molar mass of ethylene glycol \((HOCH_2CH_2OH)\) is \(62\ \text{g mol}^{-1}\). Moles \(= 45 \div 62 = 0.726\ \text{mol}\).
Step 3: Molality. Solvent mass \(= 600\ \text{g} = 0.6\ \text{kg}\), so \(m = 0.726 \div 0.6 = 1.21\ \text{mol kg}^{-1}\).
Step 4: Final value. Multiply by \(K_f\): \(\Delta T_f = 1.86 \times 1.21 \approx 2.25\ \text{K}\). \[ \boxed{\Delta T_f \approx 2.25\ \text{K}} \] The water thus freezes about \(2.25\ \text{K}\) lower than \(273.15\ \text{K}\).