Question

6) Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at -14.5 °C? The freezing point for pure water is 0.0°C and Kf is equal to 1.86 °C/m.

Answer 1

Given solute = KNO3 Solvent = Water Mass of Solvent = 275 g or ml 0.275 kg Mass of Solute = ? Molar mass of solute = 101.103 g/mol Freezing point dipression constant Kf = 1.86 °C Kg/mol Van't hoff factor for KNO3 = 2.00 Depression in freezing point ATf=Kf xm xi ATf=To - T501 = 14.50 °C Note: 0-(-14.5) = 14.5 deg where To = freezing point of water and Tsol = freezing point of solution SO ATf= To - Tsol = Kf xm xi Molality m = ATf/ (Kfxi) 14.500 °C (1.86 °C Kg mol-1 x 3.8978 mol. Kg-1 2.00) Molality = Mass of solute in g Molar mass in g/mol x 1 Mass of solvent in Kg Mass of solute = (Molality in mol kg-1) X (Molar mass in g mol-1) x (Mass of solvent in kg) = 3.8978 mol kg-1 101.10 g mol-1 x 0.275 kg = 108.4 g Mass of KNO3 = 108 g