Find the freezing point of a solution containing 27 g urea in 1 dm ^3 of the aqueous solution. [ K_f = 1.8 K kg mol ⁻¹ and molar mass of urea = 60 g mol ⁻¹ ]

Step-by-step solution

Number of moles of urea = Mass Molar mass = 27 60 = 0.45 mol Given the volume of the aqueous solution is 1 dm ^3 ( 1 L ). For a dilute aqueous solution, we can approximate the mass of the solvent to be equal to the mass of 1 L of water, which is 1 kg . Molality ( m ) = Moles of solute Mass of solvent in kg = 0.45 1 = 0.45 mol kg ⁻¹ Depression in freezing point is given by: T_f = K_f m T_f = 1.8 0.45 = 0.81 K The freezing point of pure water is 0 ^ C . Therefore, the freezing point of the solution is: T_f = 0 - T_f