Aim How does a solute affect the boiling

Aim: How does a solute affect the boiling point and the freezing point of a solvent? H. W. Study pp. 504 -505 (up to sample exercise 11. 8) pp. 506 -507 (up to sample exercise 11. 10) pp. 512 -513 (up to sample exercise 11. 13) Ans. ques. p. 522 # 59, 61, 69 p. 523 # 74, 87 p. 524 # 101

I Colligative Properties – properties that depend upon the total number of solute particles in solution, but not upon what kind of particles they are. e. g. vapor pressure depression boiling point elevation freezing point depression osmotic pressure II Boiling point elevation The presence of a nonvolatile solute raises the b. p. of the solvent by an amount that is proportional to the molal concentration of dissolved solute particles. ΔTb = Kbm where ΔTb = the elevation of the b. p. = t′b – tb Kb = molal b. p. constant m = molal concentration of the solution

The value of Kb is unique for each solvent. For water, Kb = 0. 52 0 C/m Problem: The cooling system of an automobile is filled with a solution formed by mixing equal volumes of water (d= 1. 00 g/m. L), and ethylene glycol, C 2 H 6 O 2, a nonvolatile, nonelectrolyte (d=1. 12 g/m. L). Determine the boiling point of the solution. Ans: Assume a total volume of 2. 00 L VH 2 O = 1. 00 L mass. H 2 O = 1. 00 kg Vglycol = 1. 00 L massglycol = 1120 g moles C 2 H 6 O 2 = 1120 g = 18. 1 62. 02 g ΔTb = Kbm m = 18. 1 mol 1. 00 kg kg ΔTb = (0. 52 0 C/m)(18. 1 m) = 9. 390 C b. p. = 100. 000 C + 9. 390 C = 109. 390 C

Problem: Compare the boiling point of a 2. 00 m solution of glucose, C 6 H 12 O 6, with the boiling point of a 1. 50 m solution of Na. Cl. (The solvent in both cases is water. ) Ans: For glucose, ΔTb = Kbm = (0. 52 0 C/m)(2. 00 m) = 1. 040 C b. p. = 100. 00 0 C + 1. 040 C = 101. 040 C For Na. Cl, Na. Cl(aq) → Na+(aq) + Cl-(aq) ΔTb = Kbm where i = moles of particles in solution moles of solute dissolved in this case, i = 2 ΔTb = (2)(0. 52 0 C/m)(1. 50 m) = 1. 56 0 C b. p. = 100. 00 0 C + 1. 56 0 C = 101. 560 C NOTE: i is called the van’t Hoff factor

III Freezing point depression The presence of a nonvolatile solute lowers the freezing point of the solvent by an amount that is proportional to the molal concentration of dissolved solute particles. ΔTf = Kfm where ΔTf = the depression of the freezing point Kf = molal f. p. constant m = molal concentration of the solution For water, Kf = 1. 86 0 C/m Problem: Which of the following solutions will have the lowest freezing point? Calculate its value. 0. 100 m C 2 H 6 O 2 0. 0500 m Na. Cl 0. 0500 m Ca. Cl 2

Ans: Ca. Cl 2(aq) → Ca 2+(aq) + 2 Cl-(aq) 3(. 0500 m) =. 1500 m >. 1000 m for Na. Cl and. 1 m for C 2 H 6 O 2 ΔTf = i. Kfm = (3)(1. 860 C/m)(. 0500 m) =. 2790 c f. p. = 0. 0000 C -. 2790 C = -. 2790 C Problem: Determine the freezing point of an aqueous solution of glycerin (C 3 H 8 O 3) that is 1. 24 molar and has a density of 1. 26 g/m. L. Ans: Assume 1 L of solution. mol glycerin = 1. 24 mol mass glycerin = (1. 24 mol)(92. 09 g/mol) mass solution = (1. 26 g/m. L)(1000 m. L) = 1260 g mass H 2 O = 1260 g – 114 g = 1146 g = 1. 146 kg m = 1. 24 mol = 1. 08 m C 3 H 8 O 3 1. 146 kg ΔTf = Kfm = (1. 860 C/m)(1. 08 m) = 2. 01 0 C f. p. = -2. 010 C

Solve the following problems: Zumdahl p. 522 # 60 and 62