AIIMS 2018 Chemistry Colligative Properties MCQ Question

To determine the mass of ethylene glycol (C₂H₆O₂) needed to lower the freezing point of 2 L of water (H₂O) to -2.4°C, we first calculate the required molality using the freezing point depression formula: ΔTf = Kf × m. Here, ΔTf = 2.4°C and Kf for water is 1.86 K·kg/mol, leading to a molality (m) of approximately 1.29 mol/kg. For 2 kg of water, we need about 2.58 moles of ethylene glycol, which corresponds to a mass of roughly 160 g (using the molar mass of C₂H₆O₂ ≈ 62 g/mol).

Options A (16 kg), C (1.60 kg), and D (16 g) are incorrect as they either overestimate or underestimate the required mass of ethylene glycol based on the calculations, demonstrating a misunderstanding of the freezing point depression concept and the relationship between molality and mass.