If the osmotic pressure of a 3.32times10-2-M aqueous solution of K2S04 was found

Question

If the osmotic pressure of a 3.32times10-2-M aqueous solution of K2S04 was found to be 2.31 atm at 20DegreeC, what would be the "observed" van't Hoff factor? 2.89 cf Zumdahl "Chemical Principles" 6th cd., secl7.6 "Osmotic Pressure" and secl7.7 "Colligative Properties of Electrolyte Solutions", pp867-873. You are correct. Computer's answer now shown above. Your rececipt is 157-6802 Use the above van't Hoff factor to predict the freezing point of this solution. Note that the corresponding molality would be 3.35times10-2 m, given that the density of the solution is 0.997 g/cm3. Degree C cf Zumdahl "Chemical Principles" 6th cd., sccl7.7 "Colligative Properties of Electrolyte Solutions", p872. Tries 0/5 Use the above van't Hoff factor to predict by how many degrees the boiling point of this solution will be elevated above that for pure water. Degree C

Explanation / Answer

1) Osmotic Pressure = van'tHoff*concentration*RT van't HOff = Osmotic Pressure / (concentration *RT) van't Hoff = 2.31atm /(3.32e-2 M *0.0821 L*atm/mol/K *298K) van't Hoff = 2.84 2) van't Hoff = 2.84 Kf = 1.86 C/m in water delT = van'tHOff* Kf*m = 2.84*(1.86 C/m)*3.35e-2 m = 0.177 C new freezing point: -0.177 C 3) Kb = 0.512 C/m delT = van'tHoff*Kb*m = (2.84)*(0.512 C/m)*(3.35e-2 m) = 0.0487C Tboiling = Tboiling_water + delT = 100 C + 0.0487 C = 100.05C

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