1. Calculations Involving Colligative Properties Freezing Point Depression and Boiling Point Elevation Calculations

2. Objectives When you complete this presentation, you will be able to calculate the freezing point depression or boiling point elevation of a solution when you know its molality calculate the molality of a solution when you know its freezing point depression or boiling point elevation

3. Introduction We now understand colligative properties. To use this knowledge, we need to be able to predict these colligative properties. Freezing Point Depression Boiling Point Elevation

4. Introduction We also need to use a different kind of concentration determination. Instead of molarity, we will use molality, m mole fraction, X

5. Molality Molarity - we measure the number of mols of solute in the volume of the solution. M solution = n solute /V solution Molality - we measure the number of mols of solute in the mass of solvent. m solution = n solute /m solvent

6. Molality m solution = n solute /m solvent The mass of the solvent is measured in kilograms, kg. 1 mole of solute in 1,000 g of solvent gives a 1 m solution.

7. Molality Example 1: Find the molality of 87.66 g of NaCl dissolved in 2.500 kg of H 2 O. m NaCl = 87.66 g m H2O = 2.500 kg M NaCl = 58.45 g/mol n NaCl = m NaCl / M NaCl m = n NaCl /m H2O First, we write down the known values. This value comes from the sum of the atomic masses of Na, 22.99 g/mol, and Cl, 35.45 g/mol. To calculate molality, m, we need to know the number of mols of NaCl. We only know the mass of NaCl. Therefore, we calculate the number of mols of NaCl. We substitute in known values … = (87.66 g)/(58.44 g/mol) … and do the calculation. = 1.500 mol Now, we can calculate the molality, m of the solution. We put in the known values … = (1.500 mol)/(2.500 kg) … and do the calculation. = 0.600 mol/kg

8. Molality Example Problems: 1.Find the molality of 100.0 g of NaOH in 1.500 kg of H 2 O. 2.Find the molality of 32.00 g of KCl in 0.5000 kg of H 2 O. 3.Find the molality of 142.5 g of CuCl 2 in 2.000 kg of H 2 O. 4.Find the molality of 180.0 g of H 2 O in 1.500 kg of ethyl alcohol. 1.667 m 0.8585 m 1.667 m 6.667 m

9. Molality Example 2: How many grams of potassium iodide, KI, M = 166.0 g/mol, must be dissolved in 500.0 g of water to produce a 0.060 molal solution? m soln = 0.060 mol/kg m H2O = 0.500 kg M KI = 166.0 g/mol m soln = n KI /m H2O m KI = M KI n KI First, we write down the known values. To calculate m KI, the mass of KI, we need to know the number of mols of KI, n KI. We only know the molality of the solution, m soln. … we substitute in known values … n KI = (0.060 mol/kg)(0.500 kg) … and do the calculation. = 0.030 mol Now, we can calculate m KI, the mass of KI needed to prepare the solution. We put in the known values … = (166.0 g/mol)(0.030 mol) … and do the calculation. = 5.0 g We can rearrange the equation to solve for n KI … n KI = m soln m H20

10. Molality Example Problems: 1.Find the mass of NaCl (58.5 g/mol) needed to prepare a 0.600 m solution with a mass of 2.50 kg of water. 2.Find the mass of NaOH (40.0 g/mol) needed to prepare a 1.20 m solutionwith a mass of 0.250 kg of water. 3.Find the mass of CrF 2 (90.0 g/mol) needed to prepare a 1.50 m solution with a mass of 0.400 kg of water. 4.Find the mass of KNO 3 (101 g/mol) needed to prepare a 1.01 m solution with a mass of 0.100 kg of water. 87.8 g NaCl 12.0 g NaOH 54.0 g CrF 2 10.2 g KNO 3

11. Mole Fraction Mole Fraction is the ratio of number of mols of the solute to the total number of mols of the solute plus the solvent. We use the symbol Χ (the Greek letter chi) to represent the mole fraction. Χ solute = n solute n solute + n solvent

12. Mole Fraction Example 3 Ethylene glycol (EG), C 2 H 6 O 2, is added to automobile cooling systems to protect against cold weather. What is the mole fraction of each component in a solution containing 1.25 mols of EG and 4.00 mols of water? n EG = 1.25 mol n H2O = 4.00 mol Χ EG = Χ H2O = n EG n EG + n H2O n EG n EG + n H2O 1.25 mol (1.25 + 4.00) mol = = 0.238 4.00 mol (1.25 + 4.00) mol = = 0.762

13. Mole Fraction Example Problems: 1.Find the mole fraction of 1.00 mol of NaCl in 55.6 mols of water. 2.Find the mole fraction of 5.00 mol of BaCl 2 in 10.0 mols of water. 3.Find the mole fraction of 0.240 mol of C 6 H 12 O 6 in 4.76 mols of ethanol. 4.Find the mole fraction of 0.0320 mol of KNO 3 in 10.2 mols of water. Χ = 0.177 Χ = 0.333 Χ = 0.0480 Χ = 0.00319

14. Colligative Calculations The magnitudes of freezing point depression (∆T f ) and boiling point elevation (∆T b ) are directly proportional to the molal concentration of the solute, if the solute is molecular and not ionic.

15. Colligative Calculations The magnitudes of freezing point depression (∆T f ) and boiling point elevation (∆T b ) are directly proportional to the molal concentration of all solute ions in the solution, if the solute is ionic.

16. Colligative Calculations ∆T f = −K f x m where ∆T f is the freezing point depression of the solution. K f is the molal freezing point constant for the solvent. m is the molal concentration of the solution.

17. Colligative Calculations ∆T b = K b x m where ∆T b is the boiling point elevation of the solution. K b is the molal boiling point constant for the solvent. m is the molal concentration of the solution.

18. Colligative Calculations Example 4 What is the freezing point depression, ∆T f, of a benzene (C 6 H 6, 78.1 g/mol, Bz) solution containing 400 g of Bz and 200 g of the compound acetone (C 3 H 6 O, 58.0 g/mol, Ac)? K f for Bz is 5.12°C/m. m Bz = 400 g = 0.400 kgm Ac = 200 g M Ac = 58.0 g/molK f = 5.12°C/m n Ac = m soln = m Ac M Ac n Ac m Bz 200 g 58.0 g/mol = = 3.45 mol 3.45 mol 0.400 kg = = 8.62 m ∆T f = −K f × m = (5.12°C/m)(8.62 m)= −44.1°C This is a non-ionic compound The number of particles in solution is equal to the number of mols of the compound.

19. Colligative Calculations Example 5 What is the freezing point depression, ∆T f, of a sodium chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of NaCl in 1.20 kg water? K f for water is 1.86°C/m. m H2O = 1.20 kgm NaCl = 87.6 g M NaCl = 58.4 g/molK f = 1.86°C/m n NaCl = m soln = m NaCl M NaCl n NaCl m H2O 87.6 g 58.4 g/mol = = 1.50 mol NaCl 3.00 mol 1.20 kg = = 2.50 m ∆T f = −K f × m = (1.86°C/m)(2.50 m)= −4.65°C This is an ionic compound The number of particles in solution is equal to the number of ions of the compound. For each mol of NaCl in solution, there is 1 mol of Na + and 1 mol of Cl − ions. We need to multiply n by 2. = 3.00 mol

20. Colligative Calculations Example Problems: Find the freezing point depression of: 1.58. 44 g of NaCl ( M = 58.44 g/mol) in 1.00 kg of H 2 O. (K f = 1.86°C/m) 2.228 g of octane, C 8 H 18 ( M = 114 g/mol), in 0.500 kg of benzene. (K f = 5.12°C/m) 3.33.5 g of NaC 2 H 3 OH ( M = 67.0 g/mol) in 1.25 kg of acetic acid. (K f = 3.90°C/m) 4.217 g of C 5 H 12 ( M = 72.2 g/mol) in 0.600 kg of cyclohexane. (K f = 20.2°C/m) ∆T f = −3.72°C ∆T f = −20.5°C ∆T f = −3.12°C ∆T f = −101°C

21. Colligative Calculations Example 6 What is the boiling point, T, of a benzene (Bz) solution containing 0.600 kg of Bz and 200 g of the compound phenol (C 6 H 5 OH, 94.1 g/mol, Ph)? K b for Bz is 2.53°C/m. Normal boiling point for Bz is T b = 80.1°C. m Bz = 0.600 kgm Ph = 200 gT b = 80.1°C M Ph = 94.1 g/molK f = 5.12°C/m n Ph = m soln = m Ph M Ph n Ph m Bz 200 g 94.1 g/mol = = 2.13 mol 2.13mol 0.600kg = = 3.54m ∆T b = K b × m = (2.53°C/m)(3.54 m)= 8.96°C T = T b + ∆T b = 80.1°C + 8.96°C= 89.1°C

22. Colligative Calculations Example 7 What is the boiling point, T, of a sodium chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of NaCl in 0.800 kg of acetic acid (AcOH)? K b for AcOH is 3.07°C/m. The boiling point of AcOH is T b = 118.5°C. m H2O = 1.20 kgm NaCl = 87.6 gT b = 118.5°C M NaCl = 58.4 g/molK b = 3.07°C/m n NaCl = m soln = m NaCl M NaCl n NaCl m AcOH 87.6 g 58.4 g/mol = = 1.50 mol NaCl 3.00 mol 0.800 kg = = 3.75 m ∆T b = K b × m = (3.07°C/m)(3.75m)= 11.5°C = 3.00 mol particles T = T b + ∆T b = 118.5°C + 11.5°C= 130.0°C

23. Colligative Calculations Example Problems: Find the boiling point elevation of: 1.58. 44 g of NaCl ( M = 58.44 g/mol) in 1.00 kg of H 2 O. (K b = 0.512°C/m) 2.228 g of octane, C 8 H 18 ( M = 114 g/mol), in 0.500 kg of benzene. (K b = 2.53°C/m) 3.33.5 g of NaC 2 H 3 OH ( M = 67.0 g/mol) in 1.25 kg of acetic acid. (K b = 3.07°C/m) 4.217 g of C 5 H 12 ( M = 72.2 g/mol) in 0.600 kg of cyclohexane. (K b = 2.79°C/m) ∆T b = 1.02°C ∆T b = 10.1°C ∆T b = 2.46°C ∆T b = 14.0°C

24. Summary Molality - we measure the number of mols of solute in the mass of solvent. m solution = n solute /m solvent The mass of the solvent is measured in kilograms, kg. Mole Fraction is the ratio of number of mols of the solute to the total number of mols of the solute plus the solvent. We use the symbol X to represent the mole fraction. X solute = (n solute )/(n solute + n solvent )

25. Summary ∆T f = K f x m ∆T f is the freezing point depression of the solution K f is the molal freezing point constant for the solute m is the molal concentration of the solution. ∆T b = K b x m ∆T b is the boiling point elevation of the solution K b is the molal boiling point constant for the solute m is the molal concentration of the solution.