1. Calculations Involving Colligative Properties

2. Objectives When you complete this presentation, you will be able to o calculate the molality of a solution o calculate the mole fraction of the components of a solution o calculate the freezing point depression of a solution using the molality or mole fraction of the components of the solution o calculate the boiling point elevation of a solution using the molality or mole fraction of the components of the solution

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 sodium chloride dissolved in 2.500 kg of water. m NaCl = 87.66 g m H2O = 2.500 kg M NaCl = 58.44 g/mol m = n NaCl /m H2O n NaCl = m NaCl / M NaCl =87.66 g/58.44 g/mol n NaCl = 1.500 mol =1.500 mol/2.500 kg m = 0.600 mol/kg First, we write down our known values. This value come from the sum of the atomic mass of Na (22.99 g/mol) and the atomic mass of Cl (35.45 g/mol) Next, we write down the equation for molality. We have a value for m H2O, but we need to calculate n NaCl. We know m NaCl and M NaCl, so we can calculate n NaCl. Now, we can complete our calculation for molality.

8. Molality How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution? m solution = 0.060 m m H2O = 500.0 g = 0.5000 kg M KI = 166.0 g/mol m = n KI /m H2O =n KI = m ∙ m H2O = 0.030 mol(0.060)(0.5000) mol m KI = n KI x M KI =(0.030)(166.0) g=5.0 g Example 2

9. 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 X to represent the mole fraction. X solute = n solute n solute + n solvent

10. Mole Fraction Example 3 Ethylene glycol, 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 ethylene glycol (EG) and 4.00 mol of water? n EG = 1.25 mol n H2O = 4.00 mol X EG = n EG n EG + n H2O = 1.25 mol 1.25 mol + 4.00 mol =0.238 X H2O = n H2O n EG + n H2O = 4.00 mol 1.25 mol + 4.00 mol =0.762

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

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

13. Colligative Calculations ∆T f = K f x m o 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

14. Colligative Calculations ∆T b = K b x m o 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

15. Example 4 What is the freezing point depression of a benzene (C 6 H 6, BZ) solution containing 400 g of benzene and 200 g of the compound acetone (C 3 H 6 O, AC). K f for benzene is 5.12°C/m. m BZ = 400 g = 0.400 kg m AC = 200 g M AC = 58.0 g/mol K f = 5.12°C/m n AC = m AC M AC = 200 g 58.0 g/mol 3.45 mol m = n AC m BZ = 3.45 mol 0.400 kg =8.62 m Colligative Calculations n AC = ∆T f = K f x m= (5.12°C/m)(8.62 m) = 44.1°C

16. NaCl produces 2 mols of particles for each mol of salt added; m = 2(1.50 m) = 3.00 m Example 5 What is the boiling point of a 1.50 m NaCl solution? m = 1.50 m K b = 0.512°C/m T b = 100.0°C Colligative Calculations ∆T b = K b x m= (0.512°C/m)(3.00 m) = 1.54°C T = T b + ∆T b = 100.0°C + 1.54°C = 101.5°C

17. 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.

18. 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 Summary

19. ∆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, and 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, and m is the molal concentration of the solution. Summary