Colligative Properties Worksheet

Colligative Properties Worksheet The amount of the elevation of the boiling point of a solution is given by Tb = ikbm where kb is the molal boiling point elevation constant of the solvent and m is the molal concentration of the solute. The amount of the depression of freezing point is given by Tf = -ikfm where kf is the molal freezing point depression constant of the solvent and m is the molal concentration of the solute. When working with ionic solutes in water and other polar solvents, one must be careful to take into consideration the fact that the ion concentration is higher than the concentration of the solute. For example, table salt, NaCl dissolves in water to form ions NaCl(s) -> Na+(aq) + Cl-(aq). Each mole of NaCl makes two moles of ions, so if we have a 1 molal solution of NaCl in water, the ion concentration is 2 molal. Remember, nonelectrolyte solutes do not produce any ions. 1. The normal freezing point of water is 0.00 ˚C and it's freezing point depression constant is 1.86 ˚C/m. If we make up a 1.00 molal solution of Na2SO4, what is the freezing point of the mixture. 2. For a 0.222 m aqueous solution of sucrose, C12H22O11, What is the boiling point? kb = 0.512˚C/m 3. What is the normal boiling point of a 0.122 m aqueous solution of KI? 4. Automotive antifreeze consists of ethylene glycol, C2H6O2 , a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass percent solution of ethylene glycol in water. 5. Calculate the freezing point of a solution containing 0.600 kg of CHCl 3 and 42.0 g of eucalyptol (C10H18O), a fragrant substance found in the leaves of eucalyptus tree. Kf for CHCl3 is 4.68 ˚C/molal and normal freezing point is -63.5°C. 6. List the following aqueous solutions in order of their expected freezing points (from lowest to highest): 0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050 m HC2H3O2; 0.10 m C12H22O11 7. A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g in 40.0 g CCl4. The normal boiling point of the resultant solution increased by 0.357˚C. Calculate the molar mass of the solute. Kb= 5.02°C/molal 8. Camphor, C10H16O, melts at 179.8˚C, it has particularly large freezing point depression constant, K f = 40.˚C/molal. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7˚C. What is the approximate molar mass of the solute? Tf= 179.8° C and Kf= 40. °C/molal 9. A compound called pyrene has the empirical formula C8H5 . When 4.04 g of pyrene is dissolved in 10.00 g of benzene, the boiling point of the solution is 85.1°C. Calculate the molar mass of pyrene and determine its molecular formula. The molal boiling-point constant for benzene is 2.53°C/m. Its normal boiling point is 80.1°C 10. A mass of CaCl2 , when dissolved in 100.00 g of water, gives an expected freezing point of -5.0°C. What mass of glucose would give the same result? Answers 1) f =kf•m•i=(1.86˚C/m)(1.00m)(3)= -5.58˚C 2) Bp=100.000˚C+[(0.222m)(.513°C/m)(1)] = 100.114˚C 3) Kb=100.000˚C+[(0.122m)(.513°C/m)(2)] = 100.125°C 4) to convert to m assume 1.00kg of solution boiling point = 102.72 °C freezing point = -9.87 °C 6) NaCl, HCl, CaCl2, C12H22O11, HC2H3O2 7) 8) 110 g/mol (same method as # 7) 9) Use formula in #7. MF=204.4g/mol. EM=101 so MF= C16H10 10) Use formula in # 7 but solve for grams: CaCl2: Sugar: