2. Example Calculations involving Colligative Properties (Solution Properties)

3. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. i.Read question carefully. What are you trying to determine?

4. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. i.Read question carefully. What are you trying to determine? Freezing point ii.Identify the solute and solvent: Solute = CaCl 2, Solvent = H 2 O iii.Identify the appropriate equation: Fp depression: FP soln = FP solv –  T f where:  T f = i k f m

5. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. iii.Identify the appropriate equation: Fp depression: FP soln = FP solv –  T f where:  T f = i k f m iv.Organize Information: FP H2O = 0.0  C (must know) i = 3 (Ca 2+, 2 Cl - ; 3 ions formed)

6. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. iii.Identify the appropriate equation: Fp depression: FP soln = FP solv –  T f where:  T f = i k f m iv.Organize Information: FP H2O = 0.0  C (must know) i = 3 (Ca 2+, 2 Cl - ; 3 ions formed) k f = 1.858  C/m m = ????? (molality not given!!!!!)

7. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. v.Must convert given concentration to molality (m): vi.Given Concentration unit?

8. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. v.Must convert given concentration to molality (m) vi.Given Concentration unit? 32.0% by mass CaCl 2 vii.Use 32% by mass as the starting point 32% CaCl 2 by mass = 32 g CaCl 2 / 100 g soln so input: (1) 32 g CaCl 2 and (2) 100 g soln into the table

9. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Input: (1) 32 g CaCl 2 and (2) 100 g soln into the appropriate cells! CaCl 2 H2OH2OSoln mole gram ml

10. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. input: (1) 32 g CaCl 2 and (2) 100 g soln into the appropriate cells! CaCl 2 H2OH2OSoln mole gram 32 g ml

11. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. input: (1) 32 g CaCl 2 and (2) 100 g soln into the appropriate cells! CaCl 2 H2OH2OSoln mole gram 32 g100 g ml

12. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Now use the definition of molality (m) to find what’s needed: m = mol CaCl 2 / Kg H 2 O We want the following boxes: (1) mol CaCl 2 and (2) Kg H 2 O CaCl 2 H2OH2OSoln mole gram 32 g100 g ml

13. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Now use the definition of molality (m) to find what’s needed: m = mol CaCl 2 / Kg H 2 O We want the following boxes: (1) mol CaCl 2 and (2) Kg H 2 O CaCl 2 H2OH2OSoln mole ??? gram 32 g???100 g ml

14. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Use CaCl 2 molar mass to convert g CaCl 2 to mol CaCl 2 32 g CaCl 2 (1 mol CaCl 2 / 110.99 g CaCl 2 ) = 0.28831 mol CaCl 2 CaCl 2 H2OH2OSoln mole ??? gram 32 g100 g ml

15. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Use CaCl 2 molar mass to convert g CaCl 2 to mol CaCl 2 32 g CaCl 2 (1 mol CaCl 2 / 110.99 g CaCl 2 ) = 0.28831 mol CaCl 2 CaCl 2 H2OH2OSoln mole.28831 mol gram 32 g100 g ml

16. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Now find Kg H 2 O by subtracting the mass of soln and solute 100 g soln – 32 g CaCl 2 = 68 g or 0.068 Kg H 2 O CaCl 2 H2OH2OSoln mole.28831 mol gram 32 g???100 g ml

17. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. Now find Kg H 2 O by subtracting the mass of soln and solute 100 g soln – 32 g CaCl 2 = 68 g or 0.068 Kg H 2 O CaCl 2 H2OH2OSoln mole.28831 mol gram 32 g68 g100 g ml

18. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. We can now calculate molarity (m) m = mol CaCl 2 / Kg H2O =.28831 mol /.068 Kg = 4.2340 m CaCl 2 H2OH2OSoln mole.28831 mol gram 32 g68 g100 g ml

19. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. iii.Identify the appropriate equation: Fp depression: FP soln = FP solv –  T f where:  T f = i k f m iv.Organize Information: FP H2O = 0.0  C (must know) i = 3 (Ca 2+, 2 Cl - ; 3 ions formed) k f = 1.858  C/m m = 4.2399 m viii.Plug in numbers and calculate (Plug and Chug!)

20. Calcium Chloride (CaCl 2, Mol. Mass = 110.99 g/mol, d = 2.15 g/ml) is often used as a de-icing agent for sidewalks. Given that a saturated aqueous solution of CaCl 2 is 32.0% by mass and has a density of 1.56 g/ml, determine the freezing point of this saturated solution. Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. ix.FP H2O = 0.0  C (must know) i = 3 (Ca 2+, 2 Cl - ; 3 ions formed) k f = 1.858  C/m m = 4.2399 m x.  T f = i k f m :  T f = 3 (1.858  C/m ) (4.2399 m) = 23.63  C xi.FP soln = FP H20 –  T f : FP soln = 0.0  C – 23.63  C = -23.6  C (FINISHED!)

21. Hemoglobin (HG) is the oxygen transport protein in humans. An aqueous solution of HG was prepared by dissolving 8.75 g HG into enough water to prepare 250.0 ml of solution (d soln = 1.05 g/ml). The osmotic pressure of this solution was measured at 25.0 C and determined to be 10.0 torr. Determine the molar mass of Hemoglobin (HG). Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. i.Read the question carefully. What are you trying to determine?

22. Hemoglobin (HG) is the oxygen transport protein in humans. An aqueous solution of HG was prepared by dissolving 8.75 g HG into enough water to prepare 250.0 ml of solution (d soln = 1.05 g/ml). The osmotic pressure of this solution was measured at 25.0 C and determined to be 10.0 torr. Determine the molar mass of Hemoglobin (HG). Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. i.Read the question carefully. What are you trying to determine? Molar mass of HG ii.Identify the solute and solvent: Solute = HG, Solvent = H 2 O iii.Identify the appropriate solution property given and equation:

23. Hemoglobin (HG) is the oxygen transport protein in humans. An aqueous solution of HG was prepared by dissolving 8.75 g HG into enough water to prepare 250.0 ml of solution (d soln = 1.05 g/ml). The osmotic pressure of this solution was measured at 25.0 C and determined to be 10.0 torr. Determine the molar mass of Hemoglobin (HG). Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. i.Read the question carefully. What are you trying to determine? Molar mass of HG ii.Identify the solute and solvent: Solute = HG, Solvent = H 2 O iii.Identify the appropriate solution property given and equation: Osmotic Pressure  = i R T M

24. Hemoglobin (HG) is the oxygen transport protein in humans. An aqueous solution of HG was prepared by dissolving 8.75 g HG into enough water to prepare 250.0 ml of solution (d soln = 1.05 g/ml). The osmotic pressure of this solution was measured at 25.0 C and determined to be 10.0 torr. Determine the molar mass of Hemoglobin (HG). Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. iv.Organize information:  = i R T M  = 10 torr (1 atm / 760 torr) = 0.013158 atm (must be atm) i = 1 (proteins are extremely large molecules) R = 0.08206 (L atm/mol K) (gas constant) T = 25.0 C + 273.15 = 298.15 K (Temp must be kelvin) v.Solve for concentration M: M =  / (i R T) M =.013159 / (1 .08205  298.15) = 5.3865 x 10 -4 M

25. Hemoglobin (HG) is the oxygen transport protein in humans. An aqueous solution of HG was prepared by dissolving 8.75 g HG into enough water to prepare 250.0 ml of solution (d soln = 1.05 g/ml). The osmotic pressure of this solution was measured at 25.0 C and determined to be 10.0 torr. Determine the molar mass of Hemoglobin (HG). Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. iv.Use the concentration now to find moles HG: M = mol HG / L soln where M = 5.3865x10 -4 mol / L (calc) and L soln = 0.250 L (given) Solving for mol HG mol HG = M (L soln) = 5.3865 x 10 -4 mol/L (0.25 L) = 1.34466 x 10 -4 mol

26. Hemoglobin (HG) is the oxygen transport protein in humans. An aqueous solution of HG was prepared by dissolving 8.75 g HG into enough water to prepare 250.0 ml of solution (d soln = 1.05 g/ml). The osmotic pressure of this solution was measured at 25.0 C and determined to be 10.0 torr. Determine the molar mass of Hemoglobin (HG). Note that for H 2 O, mol. mass = 18.016 g/mol, k b = 0.512  C/m, k f = 1.858  C/m, d= 0.9979 g/ml. v.Finally use the mol Hg and g HG to find molar mass: Molar mass HG = g HG / mol HG g HG = 8.75 g (given above) mol HG = 1.34466 x 10 -4 mol Mol. Mass HG = 8.75 g / 1.34466 x 10 -4 mol = 65100 g/mol FINISHED!