Colligative Properties of Solvents 8.HW Colligative Properties.doc Use the Equations given in your notes to solve the Colligative Property Questions. ΔT b m K b, ΔT f m Solvent Formula Freezing Point ( C) Boiling Point ( C) ( C/m) K b ( C/m) Water H 2 O 0.000 00.000.858 0.52 Acetic acid HC 2 H 3 O 2 6.60 8.5 3.59 3.08 Benzene C 6 H 6 5.455 80.2 5.065 2.6 Camphor C 0 H 6 O 79.5... 40... Carbon disulfide CS 2... 46.3... 2.40 Cyclohexane C 6 H 2 6.55 80.74 20.0 2.79 Ethanol C 2 H 5 OH... 78.3....07. Which solvent s freezing point is depressed the most by the addition of a solute? This is determined by the Freezing Point Depression constant,. The substance with the highest value for will be affected the most. This would be Camphor with a constant of 40. 2. Which solvent s freezing point is depressed the least by the addition of a solute? By the same logic as above, the substance with the lowest value for will be affected the least. This is water. Certainly the case could be made that Carbon disulfide and Ethanol are affected the least as they do not have a constant. 3. Which solvent s boiling point is elevated the least by the addition of a solute? Water 4. Which solvent s boiling point is elevated the most by the addition of a solute? Acetic Acid 5. How does relate to K b? > K b (fill in the blank) The freezing point constant is always greater. 6. A solution of 58.5 grams of NaCl in,000 grams of water is made. At what temperature will the solution freeze? Calculate the Moles of Salt in the Solution Molar Mass of NaCl 58.5 (you should know how to find this by now) 58.5g _ NaCl _ mole _ mole This gives the # of moles in the solution. 58.5 _ grams Calculate the molality of the Solution
# _ of _ moles _ of _ solute _ mole molal solution kg _ of _ solution _ kg m molal (from chart above),.858 ΔT f change in freezing temp Write the Equation: ΔT f m ΔT f m ( m)(.858 C/m).858 C (this is the change in the temperature not the temp) T f 0 -.858 -.858 degrees Celcius 7. A solution of 46 grams of NaCl in,000 grams of water is made. At what temperature will the solution freeze? Calculate the Moles of Salt in the Solution Molar Mass of NaCl 58.5 (you should know how to find this by now) 46g _ NaCl _ mole 2.5_ mole This gives the # of moles in the solution. 58.5_ grams Calculate the molality of the Solution # _ of _ moles _ of _ solute 2.5_ mole 2.5 molal solution kg _ of _ solution _ kg m 2.5 molal (from chart above),.858 ΔT f change in freezing temp Write the Equation ΔT f m ΔT f m ( m)(.858 C/m) 4.6 C (this is the change in the temperature not the temp) T f 0 4.6-4.6 degrees Celcius
8. At what temperature will the solution described in number 6 boil? m molal K b (from chart above),.52 ΔT b change in freezing temp Write the Equation: ΔT b m K b ΔT b m K b ( m)(.52 C/m).52 C (this is the change in the temperature not the temp) T b 00 +.52 00.52 degrees Celcius 9. kilogram of a water solution has an unknown amount of salt in it. To find quantity of salt the sample is cooled until it freezes. It is found to freeze at -8ºC. Show all the calculations to find the number of moles in the solution. (You will need to work backwards) m?, # of moles? (from chart above),.858 ΔT f -8ºC Write the Equation ΔT f m Solve the Equation for m: m T f to Determine Molality m T f 8 C.858 m 4.3 molal solution Since this is in one kilogram of solvent it is 4.3 moles of solute. 0. At what temperature will a molal solution of salt in Acetic Acid (Vinegar) boil? m molal K b (from chart above),.3.08 ΔT b change in freezing temp Write the Equation: ΔT b m K b ΔT b m K b ( m)(3.08 C/m) 3.08 C (this is the change in the temperature not the temp) T b 8.5 + 3.08 2.58 degrees Celcius
. At what temperature will a 2.5 molal solution of salt in Acetic Acid (Vinegar) boil? m 2.5 molal K b (from chart above),.3.08 ΔT b change in freezing temp Write the Equation: ΔT b m K b ΔT b m K b (2.5 m)(3.08 C/m) 7.7 C (this is the change in the temperature not the temp) T b 8.5 + 7.7 26.2 degrees Celcius 2. A sample liquid is given to a student in order to find its K b. The student makes various concentration solutions and heats each to measure its boiling point. The solvent itself (with no solute boils at 50 ºC) Using the data below, calculate its K b. Molarity of Boiling Solution Temp (Cº) 58.8 2.5 72 3 76.4 The slope or the graph represents the constant K, Slope Rise Run 76 C 3 molal 58.7 C m
3. A chief ingredient of antifreeze (for your car) is liquid ethylene glycol, C 2 H 4 (OH) 2 (MM 62 g/mole). Assume C 2 H 4 (OH) 2 is added to a car radiator which holds 5 liters of water. A. How many moles should be added to 5 liters of water to lower the freezing point from 0 C to -8 C? m?, # of moles? (from chart above),.858 ΔT f -8ºC Write the Equation ΔT f m Solve the Equation for m: m T f to Determine Molality m T f 8 C.858 m 9.7 molal solution Use the molal Equation to determine total Moles molality moles kg of solvent 9.7 mole kg X 5 kg, so x 4.8 moles of C 2H 4 (OH) 2 B. How many grams is this? 4.8 moles C 2 H 4 (OH) 2 62 g C 2H 4 (OH) 2 mole C 2 H 4 (OH) 2 298 grams C. If C 2 H 4 (OH) 2 has a density of. kg/liter, how many liters is this? 298 g C 2 H 4 (OH) 2 kg,000 g liter. kg.27 liters of C 2H 4 (OH) 2 D. What happens to the volume of water in the radiator as you add the antifreeze? The radiator and cooling system in the car is a fixed volume so when the antifreeze is added an equal volume of water is removed. This acts to increase the molality of the solution. This could be factored in but it would require a second equation be created.