An aqueous solution of an electrolyte AB ₂ (molar mass 200 g mol ⁻¹ ) has a molarity of 3 M and a density of 1.2 g mL ⁻¹ . The electrolyte dissociates to the extent of 50 % in the solution. The freezing point of the solution in Kelvin is ________. (Nearest integer) [Given: K_f for water = 1.8 K kg mol ⁻¹ , Freezing point of pure water = 273 K ]

Step-by-step solution

Consider 1 L ( 1000 mL ) of the solution. Moles of solute ( AB ₂ ) = 3 mol Mass of solute = 3 200 = 600 g Mass of solution = Volume Density = 1000 1.2 = 1200 g Mass of solvent = 1200 - 600 = 600 g = 0.6 kg Molality ( m ) = Moles of solute Mass of solvent in kg = 3 0.6 = 5 m The electrolyte AB ₂ dissociates into 3 ions (e.g., 1 A ²⁺ and 2 B ^- ). van't Hoff factor i = 1 + (n - 1) Given n = 3 and = 0.5 : i = 1 + (3 - 1) 0.5 = 1 + 1 = 2 Depression in freezing point ( T_f ) = i K_f m = 2 1.8 5 = 18 K Freezing point of