What is the freezing point of a 4.00 molality Calcium Chloride?

Chemistry question on how to find molality, freezing point and boiling point.?

• 275g Sodium chloride is dissolved in 1.50 kg water a)find molality b)find new freezing point c)find new boiling point

• Answer:

  *** a *** 275 g NaCl x (1 mole / 58.45g) = 4.70 moles NaCl molality = moles solute / kg solvent = 4.70 moles / 1.50 kg = 3.13 m *** b *** dTf = Kf x m x i dTf = change in FP = pure FP - depressed FP Kf = cryoscopic constant for the solvent = 1.86 C/m for water i = van't hoff factor = # ions the solute dissociates into in solution since 1 NaCl ----> 1 Na+ + 1 Cl-... NaCl dissociates into "2" ions so i = 2 so dTf = 1.86 C/m x 3.13 m x 2 = 11.6 C since dTf = pure FP - depressed FP depressed FP = pure FP - dTf = 0 C - 11.6 C = -11.6C *** c *** dTb = Kb x m x i dTb = change in BP = elevated BP - pure BP Kb = ebullioscopic constant for the solvent = 0.512 C/m for water i = van't hoff factor = # ions the solute dissociates into in solution again, i = 2 dTb = 0.512 C/m x 3.13 m x 2 = 3.21 C since dTb = elevated BP - pure BP elevated BP = dTb + pure BP = 3.21 C + 100 C = 103.21C